The statement that best describes the reaction below:000 H₂O*, heat H₂C NHCH, H₂C is "The reaction is incorrect, adding acids to amides forms ketones and amines, not carboxylic acid.
"What are amides?An amide is a group of organic chemicals with the formula RCONR'₂. Amides are functional groups in organic chemistry. They are derivatives of carboxylic acids, which in which a hydroxyl group has been replaced by an amino group, (-NH₂).Amides are essential to biology. The peptide bond that connects amino acids in proteins is an amide. DNA is held together by the phosphodiester bonds that link nucleotides; these bonds are amides, also known as peptide bonds in the case of amino acids.
The reaction is incorrect, adding acids to amides forms ketones and amines, not carboxylic acid.
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A
sample of gas at 21.63 degrees celsius has a pressure of 0.87 atm.
If the gas is compressed to 2.59 atm, what is the resulting
temperature in degrees celsius?
A gas initially at 21.63 degrees Celsius and 0.87 atm is compressed to a pressure of 2.59 atm. To determine the resulting temperature is approximately 603.21 degrees Celsius we need to apply the ideal gas law equation
According to the ideal gas law, the relationship between pressure (P), volume (V), temperature (T), and the number of moles of gas (n) is given by the equation PV = nRT, where R is the ideal gas constant.
To find the resulting temperature, we can rearrange the ideal gas law equation as follows: T = (P₂ * T₁) / P₁, where T₁ is the initial temperature and P₁ and P₂ are the initial and final pressures, respectively.
Substituting the given values, the initial temperature T₁ is 21.63 degrees Celsius (or 294.78 Kelvin) and the initial pressure P₁ is 0.87 atm. The final pressure P₂ is 2.59 atm. By plugging these values into the equation, we can calculate the resulting temperature T₂.
Using the equation T₂ = (2.59 atm * 294.78 K) / 0.87 atm, we find the resulting temperature T₂ to be approximately 876.21 Kelvin (or 603.21 degrees Celsius).
Therefore, when the gas is compressed to a pressure of 2.59 atm, the resulting temperature is approximately 603.21 degrees Celsius.
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What is the concentration of [Cd2+ (aq)] in a
solution made by dissolving 0.020 mol of
Cd(NO3)2 in 400 mL of a solution of KCN that
is 0.50 M at equilibrium? Kf of
[Cd(CN)4]2- = 3.0 × 1018
The concentration of [Cd2+ (aq)] in a solution made by dissolving 0.020 mol of Cd(NO3)2 in 400 mL of a solution of KCN that is 0.50 M at equilibrium can be determined using the following steps:
Step 1: Calculate the moles of KCN used.
Moles of KCN = Molarity × Volume of KCN used (in liters)
Moles of KCN = 0.50 × (400/1000) = 0.20 mol
Step 2: Determine the amount of [Cd(CN)4]2- formed from the reaction between Cd2+ and KCN.
The reaction between Cd2+ and KCN is as follows:
Cd2+ (aq) + 4CN- (aq) ⇌ [Cd(CN)4]2- (aq)
Since the stoichiometry of the reaction is 1:4, the amount of [Cd(CN)4]2- formed is four times the amount of Cd2+ used. Amount of [Cd(CN)4]2- formed = 4 × 0.020 mol = 0.080 mol
Step 3: Calculate the concentration of [Cd(CN)4]2- in the solution.
Molarity of [Cd(CN)4]2- = Moles of [Cd(CN)4]2- / Volume of solution (in liters)
Volume of solution = Volume of KCN used + Volume of Cd(NO3)2 added= 400 mL + 0.020 L (since Cd(NO3)2 was added to 400 mL of KCN solution)
Volume of solution = 0.420 L Molarity of [Cd(CN)4]2- = 0.080 mol / 0.420 L = 0.190 M
Step 4: Determine the concentration of Cd2+ in the solution.
The equilibrium constant for the reaction is given by: Kf = [Cd(CN)4]2- / [Cd2+] [CN-]4On substituting the values given:3.0 × 1018 = 0.190 / [Cd2+] [0.50]4[Cd2+] = 3.0 × 1018 × 0.190 / (0.50)4 = 1.4 × 10^-13 M
The concentration of [Cd2+ (aq)] in the solution is 1.4 × 10^-13 M.
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Referring to the Krebs cycle, which of the following molecules are products. a) FADH2. b) NADH. c) ATP. d) CO2. e) FAD. f) NAD+. g) Acetyl. h) CoA. i) Oxygen. j) Pyruvate.
The molecules that are products in the Krebs cycle are CO2, NADH, FADH2, and ATP. The remaining molecules listed (FAD, NAD+, Acetyl, CoA, Oxygen, and Pyruvate) are not direct products of the Krebs cycle.
The Krebs cycle, also known as the citric acid cycle or tricarboxylic acid cycle, is a series of chemical reactions that occur in the mitochondria of cells. It plays a crucial role in the oxidative metabolism of glucose and other fuels.
In the Krebs cycle, the following molecules are products:
a) FADH2: FADH2 is produced during the conversion of succinate to fumarate in the Krebs cycle.
b) NADH: NADH is produced during multiple steps of the Krebs cycle, including the conversion of isocitrate to α-ketoglutarate and the conversion of malate to oxaloacetate.
c) ATP: ATP is not directly produced in the Krebs cycle. However, it is generated through oxidative phosphorylation, which is tightly coupled to the electron transport chain that receives electrons from NADH and FADH2 produced in the Krebs cycle.
d) CO2: Carbon dioxide (CO2) is released as a byproduct during various reactions in the Krebs cycle, including the conversion of isocitrate to α-ketoglutarate and the conversion of α-ketoglutarate to succinyl-CoA.
The molecules FAD, NAD+, Acetyl, CoA, Oxygen, and Pyruvate are involved in the Krebs cycle but are not considered direct products. FAD is a cofactor that is reduced to FADH2 during the cycle, NAD+ is reduced to NADH, Acetyl is a reactant that combines with oxaloacetate to form citrate, CoA is a cofactor that assists in the formation of acetyl-CoA, Oxygen is used as the final electron acceptor in oxidative phosphorylation, and Pyruvate is an intermediate produced from glucose metabolism but enters the Krebs cycle after being converted to acetyl-CoA.
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An aqueous solution of potassium phosphate,K3PO4 , is made by dissolving 21.2 grams of potassium phosphate in sufficient water in a 500. mL volumetric flask, and then adding enough water to fill the flask to the mark. What is the weight/volume percentage of potassium phosphate in the solution? Weight/volume percentage = ____% An aqueous solution of zinc iodide, ZnI2 , is made by dissolving 16.0 grams of zinc iodide in sufficient water in a 250. mL volumetric flask, and then adding enough water to fill the flask to the mark. What is the weight/volume percentage of zinc iodide in the solution? Weight/volume percentage =____ % An aqueous solution of copper(II) sulfate, CuSO4, is made by dissolving 7.94 grams of copper(II) sulfate in sufficient water in a 100. mL volumetric flask, and then adding enough water to fill the flask to the mark. What is the weight/volume percentage of copper(II) sulfate in the solution? Weight/volume percentage =____ %
The weight/volume percentage of potassium phosphate in the solution is calculated to be 4.24%. The weight/volume percentage of zinc iodide in the solution is calculated to be 6.4%. The weight/volume percentage of copper(II) sulfate in the solution is calculated to be 7.94%.
To calculate the weight/volume percentage, we need to determine the weight of the solute (potassium phosphate, zinc iodide, or copper(II) sulfate) and the volume of the solution.
For potassium phosphate:
Given that 21.2 grams of potassium phosphate is dissolved in a 500 mL volumetric flask, the weight/volume percentage can be calculated as (21.2 g / 500 mL) × 100% = 4.24%.
For zinc iodide:
Given that 16.0 grams of zinc iodide is dissolved in a 250 mL volumetric flask, the weight/volume percentage can be calculated as (16.0 g / 250 mL) × 100% = 6.4%.
For copper(II) sulfate:
Given that 7.94 grams of copper(II) sulfate is dissolved in a 100 mL volumetric flask, the weight/volume percentage can be calculated as (7.94 g / 100 mL) × 100% = 7.94%.
These calculations provide the weight/volume percentage, which represents the mass of the solute (in grams) per volume of the solution (in milliliters), multiplied by 100%.
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pH of original acetic acid solution 2.08 2. Hydrogen ion concentration of the acetic acid solution 3. Volume of NaOH used in titration. 38⋅4 mL 4. Volume of NaOH used in titration. .0384 5. Molarity of NaOH solution M 6. Moles of NaOH solutio moles 7. Moles of acetic acid. moles 8. Volume of acetic acid used for titration. 0.025 L 9. Calculated molarity of acetic acid solution. 10. Calculated K a
value for acetic acid solution. PART II: 11. pH of the acetic acid/sodium acetate equal molar solution 3.93 12. Hydrogen ion concentration of this buffer solution. 3. Calculated k a
value for acetic acid solution 4. Average experimental value of K a
for acetic acid solution +H 2
O
0
+x
x
H 3
0 t
+A −
+x
x
x
Collectively, the furnished data enables an examination of acetic acid's traits and conduct in the titration procedure.
The provided information suggests an experimental process of titrating an acetic acid solution with NaOH. The pH and hydrogen ion concentration of the original acetic acid solution (pH = 2.08) indicate its acidic nature. The volume (38.4 mL) and molarity of NaOH solution, along with the volume of acetic acid used (0.025 L), can be used to determine the moles of NaOH and acetic acid involved in the reaction.
From the data, the calculated molarity of the acetic acid solution and its corresponding Ka value can be determined using the stoichiometry of the reaction. The information also includes the pH and hydrogen ion concentration of an acetic acid/sodium acetate buffer solution, indicating its ability to resist changes in pH.
The average experimental value of Ka for acetic acid solution is not provided, but it can be calculated based on the concentrations of reactants and products in the equilibrium equation. The given chemical equation represents the dissociation of acetic acid in water, indicating the formation of hydronium ions (H3O+) and acetate ions (CH3COO-).
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The ground state of He atom is nondegenerate. The first excited state has degeneracy 3. The energy difference is 19.82eV. Suppose the system obeys semiclassical distribution and T = 1 × 10ªK. What is the ratio between numbers the molecules at the ground state and at the first excited state?
The ratio between the numbers of molecules in the ground state and the first excited state at T = 1 × 10 K, when the system obeys semiclassical distribution is 1:3.
The given information is as follows:
The ground state of He atom is nondegenerateThe first excited state has degeneracy 3. The energy difference is 19.82 eVT = 1 × 10 K .We are required to find the ratio of the number of molecules in the ground state and the first excited state. Let N1 and N2 be the number of molecules in the ground state and first excited state, respectively.For a non-degenerate state, the population density of the state is given by the Boltzmann distribution.
Therefore, the ratio between the population density of the ground state and first excited state is given by:
[tex]$$\frac{N_1}{N_2}=\frac{g_1}{g_2}\exp\left(-\frac{E_2-E_1}{k_BT}\right)$$[/tex]
Where g1, g2 are the degeneracy of states 1 and 2,
E1, E2 are the energy of states 1 and 2,
k is the Boltzmann constant,
T is the temperature.
In this case,
E2 − E1 = 19.82 eV,
g1 = 1, g2 = 3,
k = 8.62 × 10−5 eV/K,
T = 1 × [tex]10 K.$$ \frac{N_1}{N_2}[/tex]
=[tex]\frac{1}{3}\exp\left(-\frac{19.82 eV}{8.62 × 10^{−5} eV/K \cdot 1 × 10 K}\right)[/tex]
=[tex]1:3 $$[/tex]
Therefore, the required ratio between the numbers of molecules in the ground state and the first excited state is 1:3.
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1) Draw the reaction of glucose with the Benedict's reagent. 2) Draw the hydrolysis reaction of sucrose, structures and names.
1) The reaction of glucose with Benedict's reagent involves the reduction of copper ions, resulting in a color change from blue to orange-red.
2) The hydrolysis reaction of sucrose involves the breakdown of the glycosidic bond, forming glucose and fructose.
The Hydrolysis reaction of sucrose:
Sucrose + H2O → Glucose + Fructose
1.The reaction of glucose with Benedict's reagent is a common test for the presence of reducing sugars. Benedict's reagent consists of copper(II) sulfate, sodium carbonate, and sodium citrate. When glucose is added to Benedict's reagent and heated, the copper(II) ions in the reagent are reduced to copper(I) oxide, resulting in a color change. Initially, the solution is blue due to the presence of copper(II) ions. However, when heated in the presence of glucose, the copper(II) ions are reduced to copper(I) ions, which form orange-red precipitates of copper(I) oxide. This color change indicates the presence of glucose or other reducing sugars.
2.Sucrose is a disaccharide composed of glucose and fructose units joined together by a glycosidic bond. The hydrolysis of sucrose occurs when it reacts with water, breaking the glycosidic bond and yielding glucose and fructose as individual monosaccharides. The reaction is catalyzed by an enzyme called sucrase or invertase. The hydrolysis reaction can be represented as follows:
Sucrose + H2O → Glucose + Fructose
In this reaction, a water molecule is consumed, and the glycosidic bond between glucose and fructose is broken. As a result, sucrose is converted into its constituent monosaccharides, glucose, and fructose. This hydrolysis reaction is important for the digestion and metabolism of sucrose in organisms.
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What is the expected boiling point of a solution prepared by
dissolving 7.27 g of sodium bromide (NaBr) in 74.7 g of water (H
2O)? For water, T b = 100.00
oC and K b = 0.512 oC
m –1.
a.
100.66o
The expected boiling point of the solution prepared by dissolving sodium bromide (NaBr) in water can be calculated using the equation: ΔTb = Kbm, 7.27 g of sodium bromide in 74.7 g of water is approximately 100.49 degrees C.
To calculate the molality, we need to determine the moles of solute and the mass of the solvent. The molar mass of NaBr is 102.9 g/mol, so the moles of NaBr can be calculated as 7.27 g / 102.9 g/mol = 0.0707 mol. The molality (m) is defined as moles of solute per kilogram of solvent, so we need to convert the mass of water to kilograms: 74.7 g / 1000 = 0.0747 kg. Therefore, the molality is 0.0707 mol / 0.0747 kg = 0.946 m.
Substituting the values into the boiling point elevation equation, we have ΔTb = (0.512 degrees C/m) * (0.946 m) = 0.485 degrees C.
The boiling point of pure water is 100.00 degrees C. Adding the boiling point elevation to the boiling point of pure water gives us the expected boiling point of the solution: 100.00 degrees C + 0.485 degrees C = 100.485 degrees C.
Therefore, the expected boiling point of the solution prepared by dissolving 7.27 g of sodium bromide in 74.7 g of water is approximately 100.49 degrees C.
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3. A cantilever spring is composed of six graduated leaves, each 51 mm wide and 9.50 mm thick. Estimate the maximum induced stress and the deflection if loaded with 7,117 N at the tip. The length of the spring is 686 mm.
The maximum induced stress is 28.4 y/N mm² and the deflection is 2.5 mm.
Width (W) = 51 mm
Thickness (t) = 9.50 mm
Load (P) = 7,117 N
Length (L) = 686 mm
For the maximum induced stress and the deflection if loaded with 7,117 N at the tip. The formula for the deflection of the cantilever spring is given by: y = (PL³)/(3EI), where
y = deflection,
P = load,
L = length,
E = Young's modulus of elasticity,
I = moment of inertia of cross-section.
The moment of inertia of the rectangular cross-section of the cantilever spring is given by: I = (1/12)wt³
Let's calculate the moment of inertia,I = (1/12)wt³= (1/12)×(51 × 9.50³) mm⁴
= 91.9 × 10⁶ mm⁴
The Young's modulus of elasticity of spring steel is 200 GPa = 200 × 10⁹ N/mm²
Maximum induced stress is given by the relation,σ = Py/IAfter substituting the values,σ = (P×L×y)/(4I)
Maximum induced stress,σ = (P×L×y)/(4I)
= (7,117 × 686 × y)/(4 × 91.9 × 10⁶)= 28.4 y/Nmm² The maximum induced stress is 28.4 y/N mm².
The deflection of the cantilever spring,
y = (PL³)/(3EI)
= (7,117 × 686³)/(3 × 200 × 10⁹ × 91.9 × 10⁶)
= 2.5 mm
The deflection of the cantilever spring is 2.5 mm.
Therefore, the maximum induced stress is 28.4 y/N mm² and the deflection is 2.5 mm.
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Amanda dissolves 0.64 moles of NaCl in 343 ml of solution. What is the concentration of NaCl in the solution using the molarity method? Remember that 1,000 mL = 1 Liter
Amanda dissolves 0.64 moles of NaCl in 343 ml of solution, the concentration of NaCl in the solution using the molarity method is 1.866.
The Molarity of a solution is defined as the number of moles of solute present in 1 litre of a solution.
here, given that 0.64 moles of solute is dissolved in 343 ml of solution
we know that
1000 ml = 1 l solution
1 ml= 0.001 l solution
343 ml = 0.343 l solution
therefore, molarity
= no of moles of solute/ volume of the solution in L
= 0.64 moles/ 0.343 l
= 1.866
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Consider the following system at equilibrium where AH = -108 kJ, and Ke = 77.5, at 6.00 × 10² K: CO(g) + Cl₂(g) COC1₂ (9) If the temperature on the equilibrium system is suddenly increased: The
1) If the temperature is increased, Kc would decrease
The value of Qc would decrease
2) The reaction must run in the reverse direction to regain equilibrium
The concentration of chlorine would increase.
What is the equilibrium position?When the equilibrium constant, Kc, decreases, the reaction quotient, Qc, tends to increase.
The reaction quotient (Qc) is calculated in the same way as the equilibrium constant (Kc), but it is determined using concentrations of reactants and products at any given point in the reaction, not just at equilibrium.
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Carbon monoxide and chlorine gas react to form phosgene: CO(g) + Cl₂ (g) = COC12 (g) Kp: = 3.10 at 700 K Part A If a reaction mixture initially contains 174 torr of CO and 211 torr of C12, what is t
Given: Partial pressure of CO = 174 torr Partial pressure of Cl2 = 211 torr Temperature T = 700 K Phosgene = COCl2Gaseous reactions and their equilibrium constants are related by Kp = (pCOCl2) / (pCO * pCl2)Where pCOCl2, pCO, pCl2 are the partial pressures of COCl2, CO, and Cl2 respectively.
Kp = 3.10 = (pCOCl2) / (174) (211)For the given initial partial pressures of CO and Cl2, the reaction will move in which direction Let's assume x be the partial pressure of COCl2 formed at equilibrium, then the partial pressures of CO and Cl2 would change to (174 - x) and (211 - x) respectively.
As per the reaction, 1 mole of CO reacts with 1 mole of Cl2 to produce 1 mole of COCl2. Therefore, the moles of CO and Cl2 consumed will be equal to x, and the moles of COCl2 formed will be equal to x. The total moles of gases before reaction = (174 + 211) / 760 = 0.404After the reaction, the total moles of gases = (174 - x + 211 - x + x) / 760 = (385 - x) / 760At equilibrium, the value of Kp can be given as:
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v) Consider a combined gas-steam power plant. Water for the steam cycle is heated in a well- insulated heat exchanger by the exhaust gases that enter at 800 K at a rate of 30 kg/s and leave at 400 K.
The problem describes a combined gas-steam power plant where water is heated in a well-insulated heat exchanger using exhaust gases. The exhaust gases enter the heat exchanger at 800 K with a mass flow rate of 30 kg/s and exit at 400 K.
In a combined gas-steam power plant, the exhaust gases from the gas turbine are utilized to generate steam in a heat exchanger. In this case, the heat exchanger is well-insulated, meaning there is no heat transfer to the surroundings.
The heat transfer in the heat exchanger can be analyzed using the energy balance equation:
Q = m * c * (T2 - T1)
Where:
Q is the heat transfer
m is the mass flow rate of the exhaust gases
c is the specific heat capacity of the exhaust gases
T1 is the initial temperature of the exhaust gases
T2 is the final temperature of the exhaust gases
Given the values of the mass flow rate and the initial and final temperatures of the exhaust gases, the heat transfer in the heat exchanger can be calculated.
The purpose of heating the water in the heat exchanger is to generate steam for the steam cycle of the power plant. The hot water is then used to drive a steam turbine, which is connected to a generator to produce electricity.
Overall, the heat exchanger in the combined gas-steam power plant allows for efficient utilization of waste heat from the gas turbine to generate additional power through the steam cycle, enhancing the overall energy efficiency of the power plant.
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A 140.0-mLmL
solution contains 2.40 gg
of sodium benzoate and 2.53 gg
of benzoic acid. Calculate the pHpH
of the solution. For benzoic acid, Ka=6.5×10−5Ka=6.5×10−5.
Express your answer
The pH of the solution can be calculated using the Henderson-Hasselbalch equation and the given information. The pH of the solution is approximately 3.60.
To calculate the pH of the solution, we need to consider the dissociation of benzoic acid (C6H5COOH) in water. Benzoic acid is a weak acid, so it partially dissociates into its conjugate base, benzoate ion (C6H5COO-), and releases a proton (H+).
Given:
Amount of sodium benzoate (C6H5COONa) = 2.40 g
Amount of benzoic acid (C6H5COOH) = 2.53 g
Ka for benzoic acid = 6.5 × 10^(-5)
First, we need to calculate the concentrations of benzoate ion and benzoic acid in the solution. The molar mass of sodium benzoate (C6H5COONa) is 144.11 g/mol, and the molar mass of benzoic acid (C6H5COOH) is 122.12 g/mol.
Concentration of benzoate ion (C6H5COO-) = (2.40 g / 144.11 g/mol) / 0.140 L
Concentration of benzoic acid (C6H5COOH) = (2.53 g / 122.12 g/mol) / 0.140 L
Next, we can calculate the ratio of benzoate ion to benzoic acid (base/acid) using their concentrations. This ratio is essential for the Henderson-Hasselbalch equation.
Ratio = [C6H5COO-] / [C6H5COOH]
Finally, we can use the Henderson-Hasselbalch equation to calculate the pH of the solution:
pH = pKa + log10(Ratio)
pKa is the negative logarithm of the acid dissociation constant (Ka), which is given as 6.5 × 10^(-5).
By substituting the values into the equation, we can determine the pH of the solution, which is approximately 3.60.
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Determine the number of protons, neutrons and
electrons for Al+3:
Group of answer choices
Protons = 13, neutrons = 27, electrons = 10
Protons = 13, neutrons = 14, electrons = 13
Protons = 13, neu
The number of protons, neutrons, and electrons in Al+3 is: Protons = 13, neutrons = 14, electrons = 10.
Aluminum (Al) is a chemical element with atomic number 13. The atomic number represents the number of protons in an atom. Since Al+3 indicates the presence of a positive charge of +3, it means that the atom has lost three electrons.
1. Protons: The atomic number of aluminum is 13, so it has 13 protons.
2. Electrons: Al+3 indicates a positive charge of +3, which means the atom has lost three electrons. Since neutral aluminum has 13 electrons, subtracting three gives us 13 - 3 = 10 electrons in Al+3.
3. Neutrons: To determine the number of neutrons, we subtract the number of protons (13) from the atomic mass of aluminum. The atomic mass of aluminum is approximately 27 (rounded to the nearest whole number). Therefore, the number of neutrons is 27 - 13 = 14.
In summary, Al+3 has 13 protons, 14 neutrons, and 10 electrons.
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Determine the number of phases, components, and the variance
(degrees of freedom) for the following:
6. Determine the number of phases, components, and the variance (degrees of freedom) for the following: (16 pts) (a) A solution made from water, NaCl and methanol. (b) A solid mixture containing powde
To determine the number of phases, components, and variance (degrees of freedom) for the given systems, we need to analyze the number and types of substances present in each system.
(a) A solution made from water, NaCl, and methanol: In this system, we have three substances present - water, NaCl, and methanol. Each substance is a component. The number of phases depends on the conditions of the system.
If the solution is homogeneous and uniform, it will be a single phase. The variance, or degrees of freedom, can be determined using the Gibbs phase rule, which states that variance = number of components - number of phases + 2. In this case, the number of phases and components is 3, so the variance will be 2.
(b) A solid mixture containing powdered substances: In this system, we have a solid mixture composed of different powdered substances. The number of components will depend on the number of distinct substances present in the mixture. Each distinct substance will be considered a component. The number of phases will depend on the physical properties and arrangement of the mixture. If the mixture is homogeneous, it will be a single phase. The variance can be calculated using the Gibbs phase rule as mentioned above.
By analyzing the composition and properties of each system, we can determine the number of phases, components, and variance (degrees of freedom) for the given systems.
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Among the following pairs of substances, the one that does not
serve as a buffer system is:
Select one:
a.CH3NH2, CH3NH3Cl
b.H2CO3, NaHCO3
c.HBr, KBr
d.HOBr, KOBr
e. KH2PO4, K2HPO4
Among the given pairs of substances, the one that does not serve as a buffer system is c. HBr, KBr
A buffer system consists of a weak acid and its conjugate base or a weak base and its conjugate acid. It helps maintain the pH of a solution by resisting changes in pH when small amounts of acid or base are added. In order to serve as a buffer system, the pair of substances should consist of a weak acid and its conjugate base or a weak base and its conjugate acid.
Option c, HBr and KBr, does not represent a buffer system. HBr is a strong acid (hydrobromic acid) and KBr is a salt of a strong acid and a strong base (potassium bromide). Both substances do not possess the properties required for a buffer system, as they do not consist of a weak acid and its conjugate base or a weak base and its conjugate acid.
On the other hand, options a, b, d, and e represent buffer systems. Option a consists of the weak base CH3NH2 and its conjugate acid CH3NH3Cl. Option b consists of the weak acid H2CO3 (carbonic acid) and its conjugate base NaHCO3 (sodium bicarbonate). Option d consists of the weak acid HOBr (hypobromous acid) and its conjugate base KOBr. Option e consists of the weak acid KH2PO4 (monopotassium phosphate) and its conjugate base K2HPO4 (dipotassium phosphate).
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O A patient is seen in the emergency room with the following values pH: 7.32 (normal 7.37-742) PCO,: 35 mmHg (normal: 35-42 mmHg) HCO, 20 mEqL (normal: 22-28 mEq/L) Which of the following acid base conditions is the patient most likely experiencing? A Compensated respiratory alkalosis • Compensated respiratory acidosis Compensated metabolic acidosis Compensated metabolic alkalnsis C
Based on the given values, the patient is most likely experiencing compensated metabolic acidosis.
The pH value of 7.32 indicates acidemia, as it is below the normal range of 7.37-7.42. The P[tex]CO_{2}[/tex] value of 35 mmHg falls within the normal range of 35-42 mmHg, suggesting that the respiratory system is adequately compensating for the acid-base disturbance. However, the [tex]HCO_{3}[/tex]- value of 20 mEq/L is below the normal range of 22-28 mEq/L, indicating a primary decrease in bicarbonate levels.
Compensated metabolic acidosis occurs when the body compensates for a primary decrease in bicarbonate levels by decreasing the partial pressure of carbon dioxide (P[tex]CO_{2}[/tex]) through increased ventilation. This helps to restore the acid-base balance by reducing the concentration of carbonic acid.
In this case, the patient's P[tex]CO_{2}[/tex] value is within the normal range, indicating appropriate compensation by the respiratory system to decrease the P[tex]CO_{2}[/tex] levels. However, the [tex]HCO_{3}[/tex]- value is below the normal range, indicating a primary metabolic acidosis. The compensatory decrease in P[tex]CO_{2}[/tex] indicates that the respiratory system is trying to correct the acid-base disturbance.
Therefore, the patient is most likely experiencing compensated metabolic acidosis.
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How many coulombs of charge are transferred by a current of \( 3.84 \) A running for 204 seconds? Include 3 significant figures in your answer.
The charge transferred by a current of 3.84 A running for 204 seconds is 784.9 Coulombs (C).Note: The answer has 3 significant figures.
The charge that is transferred by a current of 3.84 A running for 204 seconds is 784.9 Coulombs (C).How many coulombs of charge are transferred by a current of 3.84 A running for 204 seconds?The charge that is transferred is given by the formula Q = I × t
Where Q is the charge, I is the current, and t is the time.
The current is 3.84 AThe time for which the current runs is 204 seconds.
Substituting the values, we have;
Q = I × t
= 3.84 A × 204 s
= 783.36 C
≈ 784.9 C
Therefore, the charge transferred by a current of 3.84 A running for 204 seconds is 784.9 Coulombs (C).Note: The answer has 3 significant figures.
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complete the following table
Questions: I. Hydrocarbons 1. Physical state at RT 11. III. IV. V. VI. Appearance Color Odor Solubility in conc. H₂SO4 Ignition Test Baeyer's test (theoretical visible result) Bromine Test (theoreti
The table provided contains various questions related to hydrocarbons, including physical state at room temperature, appearance, color, odor, solubility in concentrated sulfuric acid, ignition test, Baeyer's test, and bromine test.
To complete the table, we need specific information for each hydrocarbon being considered. Each hydrocarbon has its own properties, and the answers to the questions will vary depending on the specific hydrocarbon.
Physical state at RT: This refers to whether the hydrocarbon is a gas, liquid, or solid at room temperature. The answer can vary depending on the number of carbon atoms and the molecular structure of the hydrocarbon.
Appearance: This refers to the visual appearance of the hydrocarbon, such as its physical form (e.g., clear liquid, crystalline solid, etc.).
Color: This refers to the color of the hydrocarbon, if applicable. Some hydrocarbons may be colorless, while others may have a distinct color.
Odor: This refers to the smell or odor associated with the hydrocarbon. Different hydrocarbons can have different odors, ranging from odorless to pungent or characteristic smells.
Solubility in conc. H₂SO₄: This refers to the ability of the hydrocarbon to dissolve in concentrated sulfuric acid. The solubility can vary depending on the nature of the hydrocarbon.
Ignition Test: This test determines whether the hydrocarbon can undergo combustion or burn when exposed to an ignition source, such as a flame.
Baeyer's Test: Baeyer's test is used to detect the presence of unsaturation (double or triple bonds) in a hydrocarbon. The result can be a color change or the formation of a precipitate.
Bromine Test: The bromine test is used to determine the presence of unsaturation in a hydrocarbon. Bromine reacts with unsaturated hydrocarbons, resulting in a color change or the formation of a precipitate.
To complete the table accurately, specific hydrocarbons need to be provided, as the answers will vary depending on the hydrocarbon in question.
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Identify the structure of elements. Consider the following elements in their stable forms: 1. Hydrogen 2. Fluorine 3. Carbon 4. Lithium Then for each of the following questions, enter the number liste
Hydrogen and fluorine exist as diatomic molecules (H₂ and F₂, respectively), while carbon and lithium exist as individual atoms (C and Li, respectively) in their stable forms.
The structures of the given elements in their stable forms are as follows:
1. Hydrogen: H₂ (diatomic molecule)
2. Fluorine: F₂ (diatomic molecule)
3. Carbon: C (individual atom)
4. Lithium: Li (individual atom)
Hydrogen exists as a diatomic molecule, meaning two hydrogen atoms combine to form H₂. Fluorine also exists as a diatomic molecule, where two fluorine atoms combine to form F₂.
Both hydrogen and fluorine readily form stable diatomic molecules due to the sharing of electrons through covalent bonds. Carbon exists as an individual atom and is the basis of organic chemistry.
It can form various compounds due to its ability to form covalent bonds with other elements, including itself. Carbon forms stable bonds with multiple atoms, allowing for the formation of complex organic molecules.
Lithium exists an individual atom, forming a stable monatomic species. It belongs to the alkali metal group and readily loses its outermost electron to form a cation with a +1 charge.
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Describe the process of anodizing and give an example (at least one) of a material that can be effectively anodized.
Anodizing is an electrochemical process that enhances the surface of a metal, typically aluminum, by forming an oxide layer on its surface. The process involves creating a controlled oxidation reaction, which results in the formation of a durable and protective coating.
The anodizing process involves:
Preparation: The metal, usually aluminum, is thoroughly cleaned to remove any dirt, grease, or oxide layers from its surface. This can be achieved through various cleaning methods, including degreasing, etching, or chemical cleaning.
Pre-Treatment: After cleaning, the metal is typically immersed in an alkaline solution to remove any remaining impurities and to promote adhesion of the oxide layer. This step prepares the metal surface for anodizing.
Anodizing: The cleaned metal is immersed in an electrolyte solution, usually containing sulfuric acid.
The metal acts as the anode (positive electrode) in the electrolyte bath, while a cathode (negative electrode) is also present.
When an electric current is passed through the circuit, oxidation occurs at the metal's surface, forming a layer of aluminum oxide.
The anodizing process can be controlled to achieve different thicknesses of the oxide layer, which impacts the final appearance and properties of the anodized metal.
Thicker coatings provide increased protection but may alter the metal's visual appearance.
Coloring (optional): If desired, the anodized surface can be colored using various methods.
This can be achieved by introducing dyes into the electrolyte solution during anodizing or by post-anodizing processes such as dyeing or electrolytic coloring.
The coloration is absorbed into the porous anodic oxide layer, resulting in a range of decorative options.
Sealing: After anodizing, the porous oxide layer is often sealed to enhance its corrosion resistance, improve durability, and prevent color fading.
Sealing is typically accomplished by immersing the anodized metal in a hot water bath or by using other sealing agents.
Example:
Aluminum is widely used in various industries due to its lightweight, corrosion resistance, and versatility.
Anodized aluminum offers additional benefits such as increased durability, improved scratch resistance, and the ability to apply decorative finishes.
It is commonly utilized in architectural applications, aerospace components, consumer goods, electronics, automotive parts, and more.
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Use the References to access important values if needed for this question. What is the binding energy in kJ/mol nucleons for lanthanum-139? kJ/mol nucleons 57 H+ 82 n ¹39 La The required masses (g/mo
The binding energy of lanthanum-139 can be calculated using the mass defect and the Einstein's mass-energy equivalence principle (E = mc²).
To calculate the binding energy per nucleon for Lanthanum-139, we need to use the mass defect and convert it into energy using Einstein's mass-energy equation (E = mc^2).
The binding energy is the energy required to completely separate all the nucleons in the nucleus.
Given:
Mass of proton (H+): 1.007825 u
Mass of neutron (n): 1.008665 u
Mass of Lanthanum-139 (La): 138.906 u
First, we need to calculate the total mass of the nucleons (protons and neutrons) in Lanthanum-139:
Mass of nucleons = (57 * mass of proton) + (82 * mass of neutron)
Mass of nucleons = (57 * 1.007825 u) + (82 * 1.008665 u)
Next, we calculate the mass defect, which is the difference between the actual mass of Lanthanum-139 and the mass of its constituent nucleons:
Mass defect = mass of nucleons - mass of Lanthanum-139
Finally, we can convert the mass defect into energy using Einstein's equation:
Binding energy = Mass defect * c^2
where c is the speed of light (3.00 x 10^8 m/s).
Let's perform the calculations:
Mass of nucleons = (57 * 1.007825 u) + (82 * 1.008665 u) = 141.126955 u
Mass defect = 141.126955 u - 138.906 u = 2.220955 u
Binding energy = (2.220955 u) * (1.66053906660 x 10^-27 kg/u) * (3.00 x 10^8 m/s)^2
Convert the binding energy from Joules to kilojoules and divide by the number of nucleons in Lanthanum-139 (139 nucleons) to get the binding energy per nucleon in kJ/mol nucleons.
Finally, we can calculate the binding energy per nucleon:
Binding energy per nucleon = (Binding energy * 1 kJ / 1000 J) / 139
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If
445 g of N2O and H2O decomposes to N2O and H2O , how many grams of
N2O are formed?
If
445g of NH4NO3 decomposes to N2O and H2O, how many grams of N2O are
formed?
In both cases, the question is asking for the grams of [tex]N_2O[/tex] formed when a certain amount of substance decomposes.
In the first case, when [tex]N_2O[/tex] and H2O decompose to form [tex]N_2O[/tex], we need to determine the molar ratio between [tex]N_2O[/tex] and the decomposing substance. Once we have the ratio, we can calculate the moles of [tex]N_2O[/tex] formed by dividing the given mass of [tex]N_2O[/tex] by its molar mass.
Finally, we convert the moles of [tex]N_2O[/tex] to grams using its molar mass. In the second case, when [tex]NH_4NO_3[/tex] decomposes to form [tex]N_2O[/tex] and H2O, we follow a similar procedure.
We first determine the molar ratio between [tex]NH_4NO_3[/tex] and [tex]N_2O[/tex]. Then, we calculate the moles of [tex]N_2O[/tex] formed by dividing the given mass of [tex]NH_4NO_3[/tex] by its molar mass. Finally, we convert the moles of [tex]N_2O[/tex] to grams using the molar mass of [tex]N_2O[/tex].
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Calculate the change in enthalpy as 1 kg of nitrogen is heated from 1,000°K to 1,500°K, assuming the nitrogen is an ideal gas at a constant pressure. The temperature dependent specific heat of nitrogen is Cp=39.06 512.79T^(−1.5)+1072.7T^(−2)−820.4T^(−3) where Cp is in kJ/kg mol- K, and T is in K. A. 19,524 kJ B. 697.3 kJ C. 953.26 kJ D. 0 kJ
The change in enthalpy as 1 kg of nitrogen is heated from 1,000°K to 1,500°K, assuming the nitrogen is an ideal gas at a constant pressure, is 953.26 kJ (Option C).
To calculate the change in enthalpy, we can use the formula ΔH = ∫Cp dT, where ΔH is the change in enthalpy and Cp is the specific heat capacity at constant pressure. Given that Cp = 39.06 + 512.79T^(-1.5) + 1072.7T^(-2) - 820.4T^(-3), we can substitute this expression into the integral.
Evaluating the integral ∫Cp dT over the temperature range from 1,000°K to 1,500°K, we obtain the change in enthalpy:
ΔH = ∫[39.06 + 512.79T^(-1.5) + 1072.7T^(-2) - 820.4T^(-3)] dT
Evaluating this integral using the limits 1,000°K and 1,500°K, we find that the change in enthalpy is approximately 953.26 kJ. Therefore, the correct answer is option C.
The integral calculation involves solving the antiderivatives of the given equation, which can be complex. The specific form of the equation and the integration limits are essential for obtaining an accurate result.
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how to prepare 2L contain
(40%v/v) of .4 N (Ca(OH)2) and (30%v/v) of .4M (Ca(OH)2 with (30%
distuled water ) hint (Mw=74.1 g/mol) after that calculate final
concentration in final solution 2L
Two liters of 0.4M (Ca(OH)₂) will be required to prepare 2L of 30%v/v of 0.4M ((Ca(OH)₂)) with 30% distilled water and the final concentration of the solution is 0.4M.
To prepare 2L of a solution that is 40%v/v of 0.4 N ((Ca(OH)₂)) and 30%v/v of 0.4M (Ca(OH)₂) with 30% distilled water and calculate the final concentration in the final solution, the following steps should be followed:
1: Calculate the number of moles of Ca(OH)₂ that will be required to prepare 2L of 40%v/v of 0.4 N (Ca(OH)₂)
.Volume of solution = 2L
Percentage volume of Ca(OH)2 = 40%v/v
Let the volume of Ca(OH)2 required = V L
Then:V × 0.4 N = (2 - V) × 0 N → 0.4V = 0 → V = 0L
This shows that 0L of 0.4 N (Ca(OH)₂) will be required to prepare 2L of 40%v/v of 0.4 N (Ca(OH)₂).
2: Calculate the number of moles of Ca(OH)₂ that will be required to prepare 2L of 30%v/v of 0.4M (Ca(OH)₂) with 30% distilled water.
Volume of solution = 2L
Percentage volume of Ca(OH)₂ = 30%v/v
Let the volume of Ca(OH)2 required = V L Then:
V × 0.4M = (2 - V) × 0 N → 0.4V = 0.8 → V = 2L
Therefore, 2L of 0.4M (Ca(OH)₂) will be required to prepare 2L of 30%v/v of 0.4M (Ca(OH)₂) with 30% distilled water.
3: Calculate the volume of distilled water required to make up the 30%v/v of 0.4M (Ca(OH)₂) solution.
Volume of 0.4M (Ca(OH)₂) = 2L
Concentration of 0.4M (Ca(OH)₂) = 0.4M
Therefore, number of moles of 0.4M (Ca(OH)₂) = 0.4 × 2 = 0.8 mol
Then:0.3V = 2 - 0.8 → V = 4L
Therefore, 4L of distilled water will be required to make up the 30%v/v of 0.4M (Ca(OH)₂) solution.
4: Calculate the final concentration of the solution.Final volume of solution = 2L
Total number of moles of Ca(OH)₂ = Number of moles from 0.4M (Ca(OH)₂) + Number of moles from 0.4 N (Ca(OH)₂)
Number of moles from 0.4M (Ca(OH)₂) = 0.4 × 2 = 0.8 mol
Number of moles from 0.4 N (Ca(OH)₂) = 0.4 × 0 × 2 = 0 mol
Therefore, total number of moles of Ca(OH)₂ = 0.8 mol
Volume of solution = 2L
Therefore, final concentration of the solution = (Total number of moles of Ca(OH)₂ / Volume of solution) = 0.8 / 2 = 0.4 M
Thus, the final concentration of the solution is 0.4M.
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please answer all, follow instructions, and give the letter of
the answer only. thank you
PART 2- LONG EXAM 3 41. What is the first step in the general mechanism for nucleophilic acyl substitution? A) Protonation of the carbonyl B) Removal of an a-proton C) Addition of the nucleophile to t
The first step in the general mechanism for nucleophilic acyl substitution is the c) addition of the nucleophile to the carbonyl carbon.
Nucleophilic acyl substitution is a reaction that involves the replacement of a leaving group (often an acyl group) with a nucleophile. The general mechanism for this reaction proceeds through several steps. The first step is the addition of the nucleophile to the carbonyl carbon of the acyl group.
In this step, the lone pair of electrons on the nucleophile attacks the electrophilic carbonyl carbon, resulting in the formation of a tetrahedral intermediate. This addition step is typically facilitated by a Lewis base catalyst or a basic medium.
The nucleophile can be a wide range of species, such as a negatively charged ion, a neutral molecule with a lone pair of electrons, or even a metal complex.
After the addition of the nucleophile, subsequent steps in the mechanism involve proton transfer, rearrangement, and/or elimination of the leaving group to form the final product. However, it is important to note that the first step, the addition of the nucleophile to the carbonyl carbon, sets the stage for the subsequent reactions and determines the stereochemistry and radiochemistry of the product.
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Close Problem Use and interpret standard heats of formation. (a) Write the balanced chemical equation that represents the standard heat of formation of \( \operatorname{CO}(\mathrm{g}) \) at \( 298 \m
The standard heat of formation of CO(g) at 298 K is given by the equation:
ΔHf°(CO(g)) = ΔHf°(CO(g))
The standard heat of formation (ΔHf°) is the enthalpy change that occurs when one mole of a compound is formed from its constituent elements, with all substances in their standard states at a given temperature (usually 298 K) and at a pressure of 1 bar.
To determine the standard heat of formation of CO(g), we need to subtract the sum of the standard heats of formation of the reactants from the sum of the standard heats of formation of the products.
The standard heat of formation of an element in its standard state is defined as zero. Therefore, the standard heat of formation of carbon (C(s)) and oxygen (O2(g)) is zero.
The standard heat of formation of CO(g) can be calculated using the following equation:
ΔHf°(CO(g)) = ΣnΔHf°(products) - ΣmΔHf°(reactants)
Where:
ΔHf°(CO(g)) is the standard heat of formation of CO(g),
n and m are the stoichiometric coefficients of the products and reactants, respectively,
ΔHf°(products) is the sum of the standard heats of formation of the products, and
ΔHf°(reactants) is the sum of the standard heats of formation of the reactants.
For the given equation C(s) + 0.5O2(g) → CO(g), the stoichiometric coefficients are 1 for CO(g), 1 for C(s), and 0.5 for O2(g).
Plugging in the values:
ΔHf°(CO(g)) = [ΔHf°(CO(g))] - [ΔHf°(C(s))] - [ΔHf°(O2(g))]
Since ΔHf°(C(s)) and ΔHf°(O2(g)) are both zero, the equation simplifies to:
ΔHf°(CO(g)) = [ΔHf°(CO(g))] - 0 - 0
Therefore, the standard heat of formation of CO(g) at 298 K is equal to the standard heat of formation of CO(g) itself.
The standard heat of formation of CO(g) at 298 K is given by the equation: ΔHf°(CO(g)) = ΔHf°(CO(g))
This means that the standard heat of formation of CO(g) at 298 K is equal to itself, and it does not require any calculation as it is a defined value.
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Question 19 Which of the following does NOT describe an exothermic chemical reaction? O The temperature of the surroundings increases. O Heat is released by the reaction The potential energy of the re
An exothermic chemical reaction is characterized by the release of heat, but it does not necessarily involve an increase in the temperature of the surroundings. Therefore the correct option is A. The temperature of the surroundings increase.
An exothermic chemical reaction is a reaction that releases heat energy to the surroundings. During an exothermic reaction, the potential energy of the reactants is higher than the potential energy of the products. As a result, the excess energy is released in the form of heat. This release of heat can lead to an increase in the temperature of the surroundings if the heat is not efficiently transferred or dissipated.
However, it is important to note that an increase in the temperature of the surroundings is not a defining characteristic of an exothermic reaction. In some cases, the released heat may be quickly transferred to the surroundings, resulting in a negligible change in temperature. Therefore, an exothermic reaction can occur without a noticeable increase in the temperature of the surroundings.
The key aspect of an exothermic reaction is the release of heat energy, which can be detected through the measurement of temperature changes or by observing other indicators of heat release, such as light emission or an increase in the reaction rate.
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What are the dissociation products when methanoic acid is mixed
with water?
Group of answer choices:
a. Methanoate ion and hydronium (H3O+)
b. Methanoic acid and hydroxide (OH-)
c. Methanoic acid and
The dissociation products when methanoic acid (formic acid) is mixed with water are a. Methanoate ion (HCOO-) and hydronium ion (H3O+).
Methanoic acid, also known as formic acid (HCOOH), is a weak acid. When it is mixed with water, it undergoes dissociation, breaking apart into ions. The dissociation reaction can be represented as follows:
HCOOH + H2O ⇌ HCOO- + H3O+
The products of the dissociation are the methanoate ion (HCOO-) and the hydronium ion (H3O+). Here's an explanation of each dissociation product:
a. Methanoate ion (HCOO-): This is the conjugate base of methanoic acid. It is formed when the acidic hydrogen (H+) of methanoic acid is transferred to water, resulting in the formation of the methanoate ion.
b. Hydronium ion (H3O+): This is formed when the remaining portion of methanoic acid, after losing the hydrogen ion, attracts a water molecule, leading to the formation of the hydronium ion. The hydronium ion is a positively charged ion and is responsible for the acidic properties of the solution.
Therefore, the correct answer is option a. Methanoate ion and hydronium (H3O+), as these are the dissociation products when methanoic acid is mixed with water. The other options, b. Methanoic acid and hydroxide (OH-), c. Methanoic acid and hydronium (H3O+), and d. Methanoate ion and hydroxide (OH-), are not the correct dissociation products for this reaction.
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