Answer:
17.86mL of the HCl solution
Explanation:
The reaction of CaCO₃ with HCl is:
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
The concentration of HCl with a pH of 1.52 is:
pH = 1.52 = -log [H⁺]
[H⁺] = 0.0302M = [HCl]
27.0mg = 0.0270g of CaCO₃ (Molar mass: 100.09g/mol) are:
0.0270g of CaCO₃ ₓ (1mol / 100.09g) = 2.70x10⁻⁴ moles of CaCO₃
Moles of HCl to react completely with these moles of CaCO₃ are:
2.70x10⁻⁴ moles of CaCO₃ ₓ (2 mol HCl / 1 mol CaCO₃) =
5.40x10⁻⁴ moles of HCl
As the concentration of HCl is 0.0302M, volume in 5.40x10⁻⁴ moles is:
5.40x10⁻⁴ moles of HCl * (1L / 0.0302mol) = 0.01786L =
17.86mL of the HCl solutionThe volume in milliliters (mL) of an HCl solution with a pH of 1.52 that can be neutralized by the given CaCO₃ is 17.87 mL
From the question,
We are to determine the volume of HCl that could be neutralized by the given CaCO₃
First, we will write the balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
2HCl + CaCO₃ → CaCl₂ + CO₂ + H₂O
This means
2 moles of HCl is required to neutralize 1 mole of CaCO₃
Now, we will determine the number of moles of CaCO₃ present
Mass of CaCO₃ = 27.0 mg = 0.027 g
Using the formula
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]
Molar mass of CaCO₃ = 100.0869 g/mol
∴ Number of moles of CaCO₃ present = [tex]\frac{0.027}{100.0869}[/tex]
Number of moles of CaCO₃ present = 0.00026977 mole
Since
2 moles of HCl is required to neutralize 1 mole of CaCO₃
Then,
0.00053954 mole of HCl will be required to neutralize the 0.00026977 mole of CaCO₃
∴ 0.00053954 mole of HCl is required to neutralize the CaCO₃
Now, for the volume of HCl solution with a pH of 1.52 required
First,
We will determine the concentration of the HCl
From the given information
pH of the HCl = 1.52
Using the formula
pH = -log[H⁺]
Then,
1.52 = -log[H⁺]
∴ [H⁺] = 10^(-1.52)
[H⁺] = 0.0302 M
∴ The concentration of the HCl is 0.0302 M
Now, for the volume
Using the formula,
[tex]Volume = \frac{Number\ of\ moles}{Concentration}[/tex]
∴ Volume of HCl required = [tex]\frac{0.00053954}{0.0302}[/tex]
Volume of HCl required = 0.01787 L
Volume of HCl required = 17.87 mL
Hence, the volume in milliliters (mL) of an HCl solution with a pH of 1.52 that can be neutralized by the given CaCO₃ is 17.87 mL
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What is T2, if T1= 500 k, v1=10L, V2=8L,P1=600 torr,P2=200 torr?
Answer:
T2 = 133.333°K
Explanation:
Using Combined Gas Laws:
(600 torr)(10L)/500°K = (200 torr)(8L)/x°K
[tex]\frac{600 torr(10L)}{500K} =\frac{200 torr(8L)}{xK}[/tex]
Cross multiply:
x°K (600 torr)(10L) = 500°K(200 torr)(8L)
Divide:
x°K = (500°K(200 torr)(8L))/(600 torr)(10L)
[tex]xK = \frac{500K(200 torr)(8L)}{600 torr(10L)}[/tex]
x = 400/3°K or 133.333°K
When a 21.5-g sample of LiCl was added to 195 g of water at a temperature of 20.00°C in a calorimeter, the temperature increased to 39.26°C. How much heat is involved in the dissolution of the LiCl?
Answer:
[tex]63.09KJ/mol[/tex]
Explanation:
In this case, to calculate the heat of solution (KJ/mol) we have to take into account the mass of water, the specific heat of the water and the temperature change, so:
[tex]m=195~g[/tex]
[tex]c=4.18~J/g^{\circ}C[/tex]
Δ[tex]T=39.26^{\circ}C[/tex]
With this in mind, we can use the equation:
[tex]Q=m*c*T[/tex]
If we plug the values into the equation we will have:
[tex]Q=195~g*4.18~J/g^{\circ}C*39.26^{\circ}C[/tex]
[tex]Q=32000.83~J[/tex]
Now, with the mass value (21.5 g) and the molar mass of LiCl (42.39g/mol) we can calculate the moles of LiCl:
[tex]21.5~g~LiCl\frac{1~mol~LiCl}{42.39~g~LiCl}=0.507~mol~LiCl[/tex]
Now, in the heat of solution, we have KJ/mol units. Therefore, we have to convert from J to KJ:
[tex]32000.83~J\frac{1~KJ}{1000~J}=32~KJ[/tex]
Finally, we can divide by the moles of LiCl:
[tex]\frac{32~KJ}{0.507~mol}=63.09KJ/mol[/tex]
So, for each mole of LiCl, we have 63.09 KJ involved in the dissolution process.
I hope it helps!
The Sun appears to move in the sky because
A) Earth rotates on its axis
B)The sun resolves around Earth
C)the moon revolves around Earth
Answer:
a because the earth rotates on its axis
oxidation number of Se in Se8
Answer:
0.
Explanation:
All elements existing on their own in their free states has an oxidation number of 0.
The oxidation number of Se in Se8 is 0. The element Se in its elemental state has an oxidation number of 0.
Chemistry uses the idea of oxidation number, also known as oxidation state, to describe the charge that an atom seems to have when it forms a compound or ion. It is a notion that aids in our comprehension of how electrons are distributed within a molecule or ion. Oxidation numbers can be used to balance chemical equations and identify the many kinds of chemical reactions that may take place. For a better understanding of the reactivity and bonding of various elements in compounds, they offer insights into the transfer or sharing of electrons between atoms. Se in Se8 has an oxidation number of 0. Se has an oxidation number of 0 when it is in its purest form.
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Need help asap!!! Can someone please help me?
Answer:
The answer is option C.
voltage (V) = Current( I ) × Resistance (R)
V = IR
I = V/ R
V = 10V R = 20Ω
I = 10/20 = 1/2
I = 0.50A
Hope this helps
MnBr2 molecular or ionic:
name:
molecular or ionic
Answer: Manganese(II)bromide
Explanation:
Because it is a chemical compound composed of manganese and bromine
Determine if each statement is True or False. [ Select ] Central atoms with four electron groups will be sp3 hybridized. [ Select ] Hybrid orbitals are delocalized over the entire molecule. [ Select ] The number of hybrid orbitals is equal to the number of atomic orbitals that are blended together. [ Select ] Atoms with a single pi bond and an octet are sp2 hybridized. [ Select ] Sometimes oxygen atoms will be sp3d hybridized in organic molecules. [ Select ] All resonance structures must be considered when assigning hybridization
Answer:
Central atoms with four electron groups will be sp3 hybridized. True
Hybrid orbitals are delocalized over the entire molecule. False
The number of hybrid orbitals is equal to the number of atomic orbitals that are blended together. True
Atoms with a single pi bond and an octet are sp2 hybridized. True
Sometimes oxygen atoms will be sp3d hybridized in organic molecules. False
All resonance structures must be considered when assigning hybridization. False
Explanation:
When a central atom has four electron groups attached to it, then it must be sp3 hybridized. This is the case in ammonia, water, hydrogen sulphide, methane etc.
Hybridization is a valence bond concept while delocalization is a molecular orbital theory concept. Hybridized orbitals are localized on central atoms in a molecule while delocalized orbitals spread across the entire molecule.
According to valence bond theory, the number of atomic orbitals that combined to give hybrid orbitals must be equal to the number of hybrid orbitals formed.
When an atom has a single pi bond and on octet of electrons, then it must be sp2 hybridized. Remember that in ethene for instance, carbon has one pi bond and an octet of electrons.
Oxygen has an empty n=3 level hence it can not have d-orbitals involved in hybridization.
The electron domain geometry of the molecule is considered when assigning hybridization and not the resonance structures. All the resonance structures must have the central atom in the same hybridization state.
The atoms have been resulted in the formation of the covalent bond, by the sharing of the electrons. The sharing of electrons results in the hybrid atomic levels.
Hybridization and Valence bond theoryThe central atom with 4 electrons results in the distribution of electrons in 1 s and 3 p orbitals. Thus, the hybridization of the atom has been [tex]\rm sp^3[/tex].Thus, the given statement is True.
The hybrid orbitals are localized over the bond, while delocalization has been related to MOT. It has been based on the delocalization over the entire molecule.Thus, the given statement is False.
In the valence bond theory, the atomic orbitals combine to form the hybrid orbits for bond sharing.Thus, the given statement is True.
The atom with single pi electrons has fulfilled s orbital and 2 electron sin the p orbital by sharing of bond, Thus, the hybridization has been [tex]\rm sp^2[/tex].
Thus, the given statement is True.
The hybrid orbitals are formed by the combination of the atomic orbital. Oxygen has absence of d orbitals. Thus, it will never form [tex]\rm sp^3d [/tex] hybridization.Thus, the given statement is False.
The hybridization has been assigned based on the central atom and not resonance.Thus, the given statement is False.
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What is science?
O A. Attempting to solve ethical problems through research
O B. Learning about the physical world through observation
O C. Determining a body of knowledge that never changes
O D. Using rules and patterns to predict what humans will do
SUBMIT
Answer:
The answer is B
Explanation:
Answer:
B. Learning about the physical world through observation
Explanation:
A p e x 2021 :) Trust me!
You are provided with a compound fertilizer, 40-15-10.Calculate the quantity of fertilizer to add to a one hectare field supply (a)Nitrogen at 120kg/ha (b)Nitrogen at 90kg/ha (c)Phosphorus at a rate of 60kg/ha (d)Potassium at a rate of 60kg/ha
Answer:
a. 300 Kg of fertilizer
b. 225 Kg of fertilizer
c. 400 Kg of fertilizer
d. 600 Kg of fertilizer
Explanation:
The percentage composition ratio of Nitrogen, Phosphorus and Potassium in a 1 Kg bag of the given fertilizer is 40:15:10.
Therefore a 1 Kg bag contains;
40/100 * 1 Kg = 0.4 Kg of Nitrogen;
15/100 * 1 Kg = 0.15 Kg of phosphorus;
10/100 * 1 Kg = 0.1 Kg of potassium
Quantity of fertilizer required to add to a hectare to supply;
a. Nitrogen at 120 kg/ha = 120/0.4 = 300 Kg of fertilizer
b.. Nitrogen at 90 Kg/ha = 90/0.4 = 225 Kg of fertilizer
c. Phosphorus at a rate of 60 kg/ha = 60/0.15 = 400 Kg of fertilizer
d. Potassium at a rate of 60 kg/ha = 60/0.1 = 600 Kg of fertilizer
Aqueous calcium chloride reacts with aqueous potassium carbonate in a double-displacement reaction. Write a balanced equation to describe this reaction. Include states of matter in your answer. Click in the answer box to open the symbol palett
Answer: [tex]CaCl_2(aq)+K_2CO_3(aq)\rightarrow 2KCl(aq)+CaCO_3(s)[/tex]
Explanation:
A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.
The balanced reaction between aqueous calcium chloride reacts with aqueous potassium carbonate is shown as:
[tex]CaCl_2(aq)+K_2CO_3(aq)\rightarrow 2KCl(aq)+CaCO_3(s)[/tex]
A 1.0 L buffer solution is 0.300 M HC2H3O2 and 0.045 M LiC2H3O2. Which of the following actions will destroy the buffer?
a) Adding 0.050 moles of HC2H3O2
b) Adding 0.075 moles of HCl
c) Adding 0.0500 moles of LiC2H3O2
d) Adding 0.050 moles of NaOH
e) None of the above will destroy the buffer.
Answer:
b) Adding 0.075 moles of HCl
Explanation:
A buffer is defined as the aqueous mixture of a weak acid and its conjugate base or vice versa (Weak base with its conjugate acid).
The buffer of the problem is the acetic acid / lithium acetate.
The addition of any moles of the acid and the conjugate base will not destroy the buffer, just would change the pH of the buffer. Thus, a and c will not destroy the buffer.
The addition of an acid (HCl) or a base (NaOH), produce the following reactions:
HCl + LiC₂H₃O₂ → HC₂H₃O₂ + LiCl
The acid reacts with the conjugate base to produce the weak acid.
And:
NaOH + HC₂H₃O₂ →NaC₂H₃O₂ + H₂O
The base reacts with the weak acid to produce conjugate base.
As the buffer is 1.0L, the moles of the species of the buffer are:
HC₂H₃O₂ = 0.300 moles
LiC₂H₃O₂ = 0.045 moles
The reaction of HCl with LiC₂H₃O₂ consume all LiC₂H₃O₂ -because there are an excess of moles of HCl that react with all LiC₂H₃O₂-
As you will have just HC₂H₃O₂ after the reaction, the addition of b destroy the buffer.
In the other way, 0.0500 moles of NaOH react with the HC₂H₃O₂ but not consuming all HC₂H₃O₂, thus d doesn't destroy the buffer.
what is the modification of the Dalton's atomic theory
Answer:
Dalton's theory had to be modified after mass spectrometry experiments demonstrated that atoms of the same element can have different masses because the number of neutrons can vary for different isotopes of the same element
Consider the three statements below. Which numbered response contains all the statements that are true and no false statements?
I. Hydration is a special case of solvation in which the solvent is water.
II. The oxygen end of water molecules is attracted toward Ca2+ ions.
III. The hydrogen end of water molecules is attracted toward Cl- ions.
a) I, II, and III
b) I and II
c) III
d) I
e) II
Answer:
a) I, II, and III
Explanation:
For the first statement;
Solvation, is the process of attraction and association of molecules of a solvent with molecules or ions of a solute. if the solvent is water, we call this process hydration.
This means the statement is TRUE.
For the second statement;
The negatively-charged side of the water molecules are attracted to positively-charged ions. In the case of water, the oxygen end is the negatively charged side of water. This means the statement is TRUE.
For the third statement;
The positively-charged side of the water molecules are attracted to the negatively-charged chloride ions. In the case of water, the hydrogen end is the positively charged side of water. This means the statement is TRUE.
Going through the options, we can tell that the correct option is option A.
Identify the type of solid for AgCl. Identify the type of solid for AgCl. metallic atomic solid nonbonding atomic solid molecular solid ionic solid networking atomic solid
Silver chloride (AgCl) is a crystalline solid substance that is composed of silver and chloride ions. AgCl is an ionic solid. Thus, option D is correct.
What are ionic solids?Ionic solids are substances that show the properties of solid matter and have ionic, positive, and negative charges in them. They are linked together by the attraction of the opposite charges.
The silver metal in the molecule has a positive charge and the chloride ions are negative in charge making them establish an ionic bond.
The solid molecule is held together by ionic bonds and not the covalent or other metallic bonds. The cations and anions of AgCl are linked together by the electrostatic forces that make their structure appear strong and brittle.
Therefore, AgCl has been known as an ionic solid.
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Una cierta cantidad de gas se encuentra a la presión de 790 mm Hg cuando la temperatura es de 25ºC. Calcula la presión que alcanzará si la temperatura sube hasta los 200ºC.
Respuesta:
1.25 × 10³ mmHg
Explicación:
Presión inicial (P₁): 790 mmHgTemperatura inicial (T₁): 25 °CPresión final (P₂): ?Temperatura final (T₂): 200 °CPaso 1: Convertir las temperaturas a Kelvin
Cuando trabajamos con gases ideales debemos convertir las temperaturas a Kelvin. Usaremos la siguiente ecuación:
K = °C + 273.15
K = 25°C + 273.15 = 298 K
K = 200°C + 273.15 = 473 K
Paso 2: Calcular la presión final
Usaremos la ley de Gay-Lussac.
[tex]\frac{P_1}{T_1} = \frac{P_2}{T_2}\\P_2 = \frac{P_1 \times T_2}{T_1} = \frac{790mmHg \times 473K}{298K} = 1.25 \times 10^{3} mmHg[/tex]
How many ng(nanogram) are in 3.3 mg(milligram)?
Answer:
There are [tex]3.3(10^6)[/tex] nanograms in 3.3 milligrams
Explanation:
The conversion is 1 milligram is equal to [tex]1(10^6)[/tex] nanograms. Use the base 10 decimal system to help you.
For the reaction system, H2(g) + X2(g) <--> 2HX(g), Kc = 24.4 at 300 K. A system made up from these components which is at equilibrium contains 0.150 moles of H2 and 0.600 moles of HX in a 3.00 liter container. Calculate the number of moles of X2(g) present at equilibrium.
Answer:
The number of moles of X2(g) present at equilibrium is 0.0981.
Explanation:
Being:
aA + bB ⇔ cC + dD
the reaction constant Kc is defined as:
[tex]Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }[/tex]
That is, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.
Being the balanced reaction:
H₂(g) + X₂(g) ⇔ 2 HX (g)
the constant Kc is:
[tex]Kc=\frac{[HX]^{2} }{[H_{2} ]*[X_{2} ]}[/tex]
In this case, Kc = 24.4 and being [tex]Concentration=\frac{number of moles of solute}{volume}[/tex]:
[tex][H_{2} ]=\frac{0.150 moles}{3 liters}=0.05 \frac{moles}{liter}[/tex][tex][HX]=\frac{0.600 moles}{3 liters}=0.2 \frac{moles}{liter}[/tex]Replacing in the definition of the constant of Kc:
[tex]24.4=\frac{0.2^{2} }{0.05*[X_{2} ]}[/tex]
Solving:
[tex][X_{2} ]=\frac{0.2^{2} }{0.05*24.4}[/tex]
[X₂]= 0.0327
Applying the definition of concentration [tex][X_{2} ]=\frac{number of moles of X_{2} }{volume}[/tex], and the volume being 3 liters:
[tex]0.0327=\frac{number of moles of X_{2} }{3 liters}[/tex]
Solving:
number of moles of X₂= 0.0327* 3 liters
number of moles of X₂= 0.0981
The number of moles of X2(g) present at equilibrium is 0.0981.
If 25 mL of a HCl solution of unknown concentration was neutralized with 10 mL of a 0.30 M NaOH solution, what was the original concentration of the HCl solution
Answer:
0.12 M
Explanation:
Step 1: Write the balanced equation
NaOH + HCl ⇒ NaCl + H₂O
Step 2: Calculate the reacting moles of NaOH
10 mL of a 0.30 M NaOH solution react.
[tex]0.010L \times \frac{0.30mol}{L} = 3.0 \times 10^{-3} mol[/tex]
Step 3: Calculate the reacting moles of HCl
The molar ratio of NaOH to HCl is 1:1. The reacting moles of HCl are 1/1 × 3.0 × 10⁻³ mol = 3.0 × 10⁻³ mol.
Step 4: Calculate the concentration of HCl
3.0 × 10⁻³ mol of HCl are in 25 mL of solution.
[tex]M = \frac{3.0 \times 10^{-3} mol}{0.025L} = 0.12 M[/tex]
A wave has a frequency of 23.0 Hz and a wavelength of 14.9 m. What is the velocity of the wave?
1.54 m/s
8.1 m/s
37.9 m/s
343 m/s
Answer:
343 m/s
Explanation:
Velocity Formula (Wave): v = fλ
v - velocity
f - frequency
λ - wavelength
We are given f = 23.0 and λ = 14.9. Simply plug it into the formula:
v = 23.0(14.9)
v = 342.7
v ≈ 343
Answer:
343 m/s
Explanation:
There is a formula to calculate the velocity of a wave given the frequency and wavelength.
Velocity = Frequency × Wavelength
v = f λ
v = 23 × 14.9
v = 342.7 ≈ 343
A student was tasked with creating 450.0 mL of a 1.75 M HCl solution from a 4.50 M HCl stock solution.
How much of the stock solution should be transferred in order to conduct the dilution?
A. 1,160 mL
B. 1.16 L
C. 175L
D. 0.175L
Answer:
Option D is correct.
The volume of the 4.50 M stock solution required to be diluted to the specifications given is 0.175 L.
Explanation:
The law of dilution gives that the number of moles in both solutions stay the same. Mathematically, the law is given as
C₁V₁ = C₂V₂
The student wants to make 450 mL of 1.75 M of HCl.
We can find the number of moles of HCl in this required solution.
Number of moles = (Concentration in mol/L) × (Volume in L)
Concentration in mol/L = 1.75 M
Volume in L = (450/1000) = 0.45 L
Number of moles = 1.75 × 0.45 = 0.7875 mole
The stock solution has a concentration of 4.50 M
Volume of Stock solution needed
= (Number of moles)/(Concentration of stock solution in mol/L)
= (0.7875/4.5) = 0.175 L
Hence, the volume of the 4.50 M stock solution required to be diluted to the specifications given is 0.175 L.
Hope this Helps!!!
What is the molarity of a saline solution that contains 0.900g NaCl (58.44 g/mol) dissolved in 100.0 mL of solution?
Hey there!:
Molar mass NaCl = 58.44 g/mol
Number of moles = mass of solute / molar mas
Number of moles = 0.900 / 58.44
Number of moles = 0.0154 moles of NaCl
Volume in liters of solution :
100.0 mL / 1000 => 0.1 L
Therefore:
Molarity = number of moles / volume in liters
Molarity = 0.0154 / 0.1
Molarity = 0.154 M
Hope this helps!
NO(g)+O3(g)⇌NO2(g)+O2(g) The reaction is first order in O3 and second order overall. What is the rate law? View Available Hint(s)
when 1.00g of magnesium is reacted with excess hydrochloric acid, what volume of hydrogen gas will be produced at standard temperature and pressure? statndard temperature and pressure is defined as 0 C and 1 atm. At STP, 1 mole of gas occupies 22.4L.
Answer:
0.93L
Explanation:
Hello,
To solve this question, we need to first of all write down the equation of reaction.
Equation of reaction,
2Mg + 4HCl → 2MgCl₂ + 2H₂
From the equation of reaction, it shows that 2 moles of Mg produces 2 moles of H₂.
Number of moles = mass / molar mass
Mass = number of moles × molar mass
Molar mass of Mg = 24g/mol
Mass = 2 × 24 = 48g
Therefore, 48g of Mg produces (2 × 22.4)L of H₂
48g of Mg = 44.8L of H₂
1g of Mg = xL of H₂
x = (1 × 44.8) / 48
x = 0.93L
1g of Mg will produce 0.93L of H₂
There are five constitutional isomers with the molecular formula C6H14. When treated with chlorine at 300°C, isomer A gives a mixture of two monochlorination products. Under the same conditions, isomer B gives a mixture of five monochlorination products, isomer C gives four monochlorination products, and isomer D gives a mixture of three monochlorination products. From this information, draw the structural formula of isomer D.
Answer:
Pentane or 2,2-dimethylbutane
Explanation:
I've numbered the isomeric hexanes from 1 to 5 and labelled the sets of equivalent hydrogens.
The results are
Isomer 1— three sets of equivalent hydrogens
Isomer 2— five sets of equivalent hydrogens
Isomer 3— four sets of equivalent hydrogens
Isomer 4— two sets of equivalent hydrogens
Isomer 5— three sets of equivalent hydrogens
Each set will give one monochloro substitution product.
4 = A. Two monochloro isomers.
2 = B. Five monochloro isomers.
3 = C. Four monochloro isomers.
Isomers 1 and 5 each give three monochloro isomers.
Thus, we cannot assign Structure D definitively.
D is either pentane or 2,2-dimethylbutane.
A student is conducting their science experiment on the effect of caffeine on
dogs. He has 3 groups of test subjects. The 1st group of dogs receives plain
water. The 2nd group of dogs receives 10 mg of caffeine each, and the 3rd group
receives 50 mg of caffeine each. He will measure their activity levels by
recording how long each dog runs without stopping, after giving them the pills.
What is the dependent variable in this experiment?
A. The amount of dogs tested.
B. The dogs that receive water.
C. The amount of caffeine given.
D. The amount of activity.
Answer:
The amount of activity
Explanation:
For a reaction that follows the general rate law, Rate = k[A][B]2, what will happen to the rate of reaction if the concentration of A is increased by a factor of 5.00?4
Answer:
The rate will increase by a factor of 5.00 too.
Explanation:
Hello,
In this case, given the rate law:
[tex]r=k[A][B]^2[/tex]
Thus, we can notice it is first-order respect to A, for that reason, increasing its concentration by a factor of 5.00, increase the rate by a factor of 5.00 as well.
Moreover, if the concentration of B is increased by the same factor, the rate will increase by a factor of 25.00, since the rate is second-order respect to B (it is squared).
Best regards.
Which state of matter has a definite volume and takes the shape of its container?
liquid
gas
solid
Answer:
solid
Explanation:
liquid can't take the shape of the container because it flows freely and it doesn't have a definite shape and it has space between it e.g water
gas also doesn't shape and e. g air
but solid takes a definite shape because it doesn't have space in between of
Answer:
sold
Explanation:
You start with 6 moles of an acid, 2 moles dissociate into the conjugate base, what is the percent ionization
Answer:
33 %
Explanation:
Step 1: Given data
Initial moles of the acid (nHA(0)): 6 mol
Moles of the conjugate base at equilibrium (nA⁻(eq))
Step 2: Write the balanced generic acid dissociation reaction
HA(aq) ⇄ A⁻(aq) + H⁺(aq)
Step 3: Calculate the percent ionization
We will use the following expression.
[tex]\alpha = \frac{nA^{-}(eq)}{nHA(0)} \times 100 \% = \frac{2mol}{6mol} \times 100 \% = 33 \%[/tex]
Calculate the theoretical percentage of water for the following hydrates.
(a) manganese(II) monohydrate, MnSO4 H2O
(b) manganese(II) tetrahydrate, MnSO4 4H2O
Answer:
(a) [tex]\% H_2O=10.65\%[/tex]
(b) [tex]\% H_2O=32.2[/tex]
Explanation:
Hello.
For this questions we must consider the ratio of the molar mass of water to hydrated compound molar mass as shown below:
(a) In this case, we can consider that inside the manganese (II) sulfate monohydrate, whose molar mass is 169.02 g/mol, there is one water molecule that has a molar mass of 18 g/mol, for which the theoretical percentage of water is:
[tex]\% H_2O=\frac{18g/mol}{169.0g/mol} *100\%\\\\\% H_2O=10.65\%[/tex]
(b) In this case, we can consider that inside the manganese (II) sulfate tetrahydrate, whose molar mass is 223.1 g/mol, there are four water molecules that have a molar mass of 4*18 g/mol, for which the theoretical percentage of water is:
[tex]\% H_2O=\frac{4*18g/mol}{223.1g/mol} *100\%\\\\\% H_2O=32.27\%[/tex]
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When 3.05 moles of CH4 are mixed with 5.03 moles of O2 what is the limiting reactant? How much CO2 will be formed? CH4 + 2O2 → CO2 + 2H2O
Answer:
3.05 moles of Co2
Explanation:
The ratio of Co2 to methane is equal therefore the amount formed is equal