Answer:
It's well Explained below.
Explanation:
First of Excess product of CaCO_3 would be produced due to the fact that there would not be enough CaCl_2 to react with Na_2•CO_3. The main purpose of having stoichiometric quantities is for us to know the correct amount or near the correct amount of each reactant in order to create a product that will be close to the theoretical amount and thus have a higher percent yield.
Draw structural formulas for the major organic product(s) of the reaction shown below.
• You do not have to consider stereochemistry.
If no reaction occurs, draw the organic starting material.
Remember to include all of the formal charges on the atoms of any nitro groups.
Answer:
3-bromobenzoic acid
Explanation:
In this case, we have to remember that the [tex]Br_2/FeBr_3[/tex] is a reaction in which we add Br into the molecule an electrophilic aromatic substitution. Additionally, we have a carboxylic acid in the benzene. This carboxylic acid is an ortho director because is a deactivating group (it removes electrons from the benzene ring). With this in mind, a "Br" atom would be added in an ortho position respect to the COOH group and we will obtain 3-bromobenzoic acid.
See figure 1.
I hope it helps!
To create 3-bromobenzoic acid, a "Br" atom would be placed at an orthogonal position to the COOH group according to electrophilic aromatic substitution.
Electrophilic aromatic substitution is a type of organic reaction in which an atom or group in an aromatic ring is substituted with an electrophile. It is a fundamental reaction in aromatic chemistry that happens due to the aromatic system's high electron density.
It is an electrophilic aromatic substitution process in which Br is incorporated into the molecule. In addition, the benzene contains a carboxylic acid. Because it removes electrons from the benzene ring, this carboxylic acid functions as an ortho director. To create 3-bromobenzoic acid, a "Br" atom would be placed at an orthogonal position to the COOH group. The product is seen in the photographs below.
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What are the correct formulas and coefficients for the products of the following double-replacement reaction? RbOH + H3PO4→
Answer:
3 RbOH + H₃PO₄ → Rb₃PO₄ + 3 H₂O
Explanation:
Let's consider the double-replacement reaction between rubidium hydroxide and phosphoric acid to form rubidium phosphate and water. The cation rubidium replaces the cation hydrogen and the anion hydroxyl replaces the anion phosphate. The balanced chemical reaction is:
3 RbOH + H₃PO₄ → Rb₃PO₄ + 3 H₂O
a soluation of acetone in water has a molarity of 2.422M and a density of 0.970 g/mL. Calculate the mole fraction
Answer:
[tex]x_{acetone}=7.970x10^{-3}[/tex]
Explanation:
Hello,
In this case, for the given molarity, we can assume a volume of 1 L of solution, to obtain the following moles of acetone:
[tex]n=0.422mol/L*1L=0.422mol[/tex]
Then, with the density of solution, we can compute the mass of the solution for the selected 1-L volume basis:
[tex]m_{solution}=1L*\frac{1000mL}{1L}*\frac{0.970g}{1mL}=970g[/tex]
After that, we compute the mass of water in the solution, considering the mass of acetone (molar mass = 58.08 g/mol):
[tex]m_{H_2O}=970g-0.422molAcetone*\frac{58.08g\ Acetone}{1mol\ Acetone} =945.49gH_2O[/tex]
Next, the moles of water:
[tex]n_{H_2O}=945.49g*\frac{1molH_2O}{18gH_2O} =52.53molH_2O[/tex]
Finally, the mole fraction:
[tex]x_{acetone}=\frac{n_{acetone}}{n_{acetone}+n_{H_2O}}=\frac{0.422mol}{0.422mol+52.53mol}\\ \\x_{acetone}=7.970x10^{-3}[/tex]
Regards.
The mathematics of combining quantum theory with wave motion of atomic particles is known as _____.
Combining quantum theory with wave motion of atomic particles is: Wave Mechanics
If 50 ml of 1.00 M of H2SO4 and 50 ml of 2.0 M KOH are mixed what is the concentration of the resulting solutes?
Answer: [H2SO4] = 0.5M;
[KOH] = 1M
Explanation: Molarity is the solution concentration defined by:
molarity = [tex]\frac{mol}{L}[/tex] or M
To determine the concentration of the mixture, find how many mols of each compound there are in the mixture:
50 mL = 0.05L
H2SO4
1 mol/L * 0.05L = 0.05mol
KOH
2mol/L * 0.05L = 0.1 mol
The mixture has a total volume of:
V = 50 + 50 = 100 mL = 0.1 L
The concentration of the resullting solutes:
[H2SO4] = [tex]\frac{0.05}{0.1}[/tex] = 0.5 M
[KOH] = [tex]\frac{0.1}{0.1}[/tex] = 1 M
Concentration of H2SO4 is 0.5M while for KOH is 1M.
Cual es la diferencia entre agua pesada y agua ligera a) el agua pesada contiene mas minerales que el agua ligera b) el agua ligera es liquida mientras el agua pesada es solida c) el agua ligera es agua purificada y el agua pesada es agua contaminada d) el agua pesada contiene mas elementos estearato de sodio
Answer:
d) El agua pesada contiene mas elementos
Explanation:
La diferencia fundamental entre el agua pesada y el agua ligera es que la primera tiene una proporción mayor de deuterio que la segunda. El deuterio es un ión del hidrógeno que tiene un peso atómico mayor que el hidrógeno común y corriente. Por ende, la opción D ofrece la mejor aproximación.
Answer:
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Explanation:ki
What is the molarity of a solution that is 7.00% by mass magnesium sulfate and has a density of 1.071 g/mL?
Answer:
0.623 M
Explanation:
Step 1: Given data
Percent by mass (%m/m): 7.00 %Density of the solution (ρ): 1.071 g/mLMolar mass of magnesium sulfate: 120.37 g/molStep 2: Calculate the percent by volume (%m/v)
We will use the following expression.
[tex]\%m/v = \%m/m \times \rho = 7.00\% \times 1.071g/mL = 7.50g\%mL[/tex]
Step 3: Calculate the molarity
7.50 g of magnesium sulfate are dissolved in 100 mL of the solution. The molarity is:
[tex]M = \frac{7.50g}{120.37g/mol \times 0.100L } = 0.623 M[/tex]
Explain with examples following characteristics of chemical reactions: a. Change of colour b. Evolution of gas c. Change of smell d. Change of state
Answer:
Explanation:
a. change of colour:
A chemical reaction rearranges the constituent atoms of the reactants to create different substances as products. The products have different molecular structures than the reactants. Different atoms and molecules radiate different colours of light. Hence, there usually is a change in colour during a chemical reaction.
Eg: copper reactions with the elements
b. Evolution of gas:
A gas evolution reaction is a chemical reaction in which one of the end products is a gas such as oxygen or carbon dioxide.
Eg: ammonium hydroxide breaks down to water and ammonia gas.
c. Change of smell :
Production of an Odor Some chemical changes produce new smells. ... The formation of gas bubbles is another indicator that a chemical change may have occured.
Eg: The chemical change that occurs when an egg is rotting produces the smell of sulfur.
d. Change of state:
A chemical reaction is a process in which one or more substances, also called reactants, are converted to one or more different substances, known as products.
Eg: candle wax (solid) melts initially to produce molten wax (liquid)
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what is the value of the equilibrium constant at 500k for a chemical equilivrium that has a delta h value of 250kj mol and s value of 48 j mol k
Kinetic energy and gravitational potential energy are both forms of which type
of energy?
A. Internal energy
B. Mechanical energy
C. Potential energy
D. Thermal energy
Answer:
C. Potential energy
Explanation:
Kinetic energy and gravitational potential energy are both forms of potential energy. Potential energy is stored energy, when an object is not in motion it has stored energy. When an object is an motion it has kinetic energy. An object posses gravitational potential energy when it is above or below the zero height.
Identify the particle that must receive 2 electrons to acquire a charge of +1. a) K b) Fe2+ c) O2- d) Nee) Al3+ (URGENT) Needs to be done in 30 mins
Answer:
E) Al³⁺
Explanation:
A reaction involving a gain of electrons is known as a reduction reaction because the oxidation number of the species gaining the electron is reduced.
In the given question, the oxidation number (charge) of particle accepting two electrons will decrease by 2. From the given options;
A. K is a neutral atom with oxidation number of 0. If is accepts two electrons, its oxidation number becomes -2.
K + 2e⁻ ----> K⁻²
B) Fe²⁺ has a charge of +2. If it accepts two electrons, its charge comes 0.
Fe⁺ + 2e⁻ ----> Fe
C) O²⁻ has a charge of -2. if it accepts two electrons, it will have a charge of -4.
O²⁻ + 2e⁻ ----> O⁴⁻
D) Ne has a charge of zero. If it accepts two electrons, its charge becomes -2.
Ne + 2e⁻ ----> Ne²⁻
E) Al³⁺ has a charge of +3. If it gains two electrons, its charge becomes +1.
Al³⁺ + 2e⁻ ----> Al⁺
Which of the following would be useful for converting g/mol to g/L?
A. Mass percent
B. Avogadro's number
C. Molarity
D. Molar mass
Answer:
Molarity
Explanation:
The conversion of g/mol to g/L molarity can be used. Thus, option C is correct.
The g/mol has been the amount of solute present in a mole. The g/mol has been the molecular weight of the compound.
The g/L has been the mass of solute present in a L of solution. The g/L has the unit for density.
Molarity has been the moles of solute present in the liter of solution. It has been given as mol/L.
The product of g/mol and g/L gives the value of mol/L. Thus, to convert g/mol to g/L molarity can be used. Thus, option C is correct.
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Identify a reaction of sugars, starch, or cellulose to produce ethanol and carbon dioxide halogenation polymerization
Answer:
Fermentation
Explanation:
Fermentation is the general term used to describe the process by which sugars such as glucose, starch or cellulose are converted to ethanol and carbon (iv) oxide. It is anaerobic process meaning that it occurs in the absence of air or in very low oxygen concentrations.
Yeast and other microorganisms ferment glucose into ethanol and carbon (iv) oxide with the help of the enzyme zymase. Polysaccharides such as starch and cellulose are first broken down into glucose by enzymes such as diastases, maltase and cellulase, before it is then converted into ethanol and carbon (iv) oxide.
The equation for the conversion of glucose to ethanol and carbon (iv) oxide is as follows:
C₆H₁₂O₆(aq) -----> 2C₂H₅OH(aq) + 2CO₂(g)
Considering the steps involved in dissolution, which of the following do you expect to speed up a dissolution process?
A. Sweeping all the solute particles into a pile within the solvent.
B. Stirring the solution vigorously.
C. Grinding the solute down into tiny particles.
D. Gently heating the solution.
Answer:
C. Grinding the solute down into tiny particles.
Explanation:
The dissolution of a solute has something to do with particle size. The size of solute particles usually determines how quickly a solute dissolves in a solvent. When large solute particles are introduced into the solvent, the large solute particles do not easily interact with solvent particles hence preventing easy dissolution in the solvent.
However, when the solute is ground into tiny particles, smaller solute particles interact more effectively with solvent particles hence dissolution is faster.
Therefore, tiny solute particles will dissolve faster in a solvent than a lump of solute. Summarily, small particle size enhances dissolution of a solute in the appropriate solvent.
Answer: stirring the solution vigorously
Grinding the solute down into tiny particles
gently heating the solution
Explanation:
A dissolution will proceed more readily when heated . Breaking up the solute as much as possible will aid in overcoming the solute-solute interaction, as will stirring the solution
Write the empirical formula
Answer:
[tex]Pb(CO_{3})_{2} \\Pb(NO_{3})_{4} \\FeCO_{3}\\Fe(NO_{3})_{2}[/tex]
Explanation:
[tex]Pb^{4+}(CO_{3}^{2-})_{2} --->Pb(CO_{3})_{2} \\Pb^{4+} (NO_{3}^{-})_{4} --->Pb(NO_{3})_{4} \\Fe^{2+} CO_{3}^{2-} --->FeCO_{3}\\Fe^{2+} (NO_{3}^{-})_{2}--->Fe(NO_{3})_{2}[/tex]
What is the [OH-] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution
Answer: The [tex][OH^-][/tex] of a solution is [tex]10^{-12}[/tex] M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n = moles of solute
[tex]V_s[/tex] = volume of solution in ml
moles of [tex]HCl[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{0.0912g}{36.5g/mol}=0.0025mol[/tex]
Now put all the given values in the formula of molality, we get
[tex]Molarity=\frac{0.0025\times 1000}{250}=0.01[/tex]
pH or pOH is the measure of acidity or alkalinity of a solution.
[tex]HCl\rightarrow H^++Cl^{-}[/tex]
According to stoichiometry,
1 mole of [tex]HCl[/tex] gives 1 mole of [tex]H^+[/tex]
Thus [tex]0.01[/tex] moles of [tex]HCl[/tex] gives =[tex]\frac{1}{1}\times 0.01=0.01[/tex] moles of [tex]H^+[/tex]
Putting in the values:
[tex][H^+][OH^-]=10^{-14}[/tex]
[tex][0.01][OH^-]=10^{-14}[/tex]
[tex][OH^-]=10^{-12}[/tex]
Thus the [tex][OH^-][/tex] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution is [tex]10^{-12}[/tex] M
The [OH-] of a solution is [tex]10^{12}[/tex] M.
What is Molarity?Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
M = n/ V..................(1)
where,
n = moles of solute
V = volume of solution in ml
Calculation for number of moles:
Moles of HCl = 0.0912 g/ 36.5 g/mol = 0.0025 mol
On substituting the values in equation 1:
M = n/ V
M= 0.0025*1000 / 250
M=0.01 M
pH or pOH is the measure of acidity or alkalinity of a solution.
[tex]HCl---- > H^++Cl^-[/tex]
According to stoichiometry,
1 mole of HCl gives 1 mole of [tex]H^+[/tex]
Thus, 0.01 moles of HCl gives = 1 / 1 *0.01 = 0.01 mole of [tex]H^+[/tex]
On adding the values:
[tex][H^+][OH^-]=10^{14}\\\\(0.01)[OH^-]=10^{-14}\\\\OH^-=10^{-12}[/tex]
Thus, the [OH-] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution is [tex]10^{-12}[/tex] M.
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The force that opposes drag and is powered by combustion reactions in the
engine is
Answer:
Thrust.
Explanation:
hope this helps you :)
Answer:
thrust
Explanation:
A silver cube with an edge length of 2.42 cm and a gold cube with an edge length of 2.75 cm are both heated to 85.4 ∘C and placed in 112.0 mL of water at 20.5 ∘C . What is the final temperature of the water when thermal equilibrium is reached?
Answer:
Explanation:
Volume of silver cube = 2.42³ = 14.17 cm³
mass of silver cube = volume x density
= 14.17 x 10.49 = 148.64 gm
Volume of gold cube = 2.75³ = 20.8 cm³
mass of gold cube = 20.8 x 19.3 = 401.44 gm
specific heat of silver and gold are .24 and .129 J /g°C
mass of 112 mL water = 112 g
Heat absorbed = heat lost = mass x specific heat x temperature fall or rise
Heat lost by metals
= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )
= (35.67 + 51.78 ) x ( 85.4 - T )
87.45 x ( 85.4 - T )
= 7468.23 - 87.45 T
Heat gained by water
= 112 x 1 x ( T - 20.5 )
= 112 T - 2296
Heat lost = heat gained
7468.23 - 87.45 T = 112 T - 2296
199.45 T = 9764.23
T = 48.95° C
Use the reaction I2(s) I2(g), H = 62.4 kJ/mol, S = 0.145 kJ/(molK)
At what temperature is the reaction at equilibrium?
A.157K
B.430K
C.0.002K
D.62K
Answer: B. 430 K
Explanation:
According to Gibb's equation:
[tex]\Delta G=\Delta H-T\Delta S[/tex]
[tex]\Delta G[/tex] = Gibbs free energy
[tex]\Delta H[/tex] = enthalpy change = +62.4 kJ/mol
[tex]\Delta S[/tex] = entropy change = +0.145 kJ/molK
T = temperature in Kelvin
[tex]\Delta G[/tex] = +ve, reaction is non spontaneous
[tex]\Delta G[/tex] = -ve, reaction is spontaneous
[tex]\Delta G[/tex] = 0, reaction is in equilibrium
[tex]\Delta H-T\Delta S=0[/tex] for reaction to be spontaneous
[tex]T=\frac{\Delta H}{\Delta S}[/tex]
[tex]T=\frac{62.4kJ/mol}{0.145kJ/molK}=430K[/tex]
Thus the Reaction is spontaneous when temperature is 430 K.
Answer:
430 K
Explanation:
i just took the test on a pex :)
need this asap , help please
Answer:
Path A-B-D involves a catalyst and is slower than A-C-D
Explanation:
The diagram above illustrates both the catalyzed path and the uncatalyzed path of a chemical reaction.
The catalysed path is the path expressed with broken lines and the uncatalyzed path is the path expressed with thick small line as shown in the diagram above.
The catalyzed path has a higher activation energy than the uncatalyzed path.
Therefore, the catalyzed path will be slower that the uncatalyzed path because, the catalyzed path will require a higher energy to overcome the activation energy in order for the reaction to proceed to product.
On the other hand, the uncatalyzed path has a lower activation energy and a lesser amount of energy is needed to overcome it in order for the reaction to proceed to product.
what bonding is similar to ionic bonding, except there are no high-electronegativity atosms present to accept any electrons that the present atoms are willing to donate.
Answer:
Metallic bonding
Explanation:
Ionic bonding involves the transfer of electrons from a highly electropositive metal to a highly electronegative nonmetal.
The metallic bond is somewhat similar to the ionic bond since there are also charged positive metal ions. The only difference is that there isn't any electronegative element that accepts the electrons.
In a metallic bond, the positively charged metal ions are bound together by a sea of mobile electrons. The attractive force between the metal ions and the mobile electrons hold the metallic crystal lattice together.
A 3.00-g sample of an alloy (containing only Pb and Sn) was dissolved in nitric acid (HNO3). Sulfuric acid was added to this solution, which precipitated 2.93 g of PbSO4. Assuming that all of the lead was precipitated, what is the percentage of Sn in the sample? (molar mass of PbSO4 = 303.3 g/mol)
Answer:
33.3% of Sn in the sample
Explanation:
The addition of SO₄⁻ ions produce the selective precipitation of Pb²⁺ to produce PbSO₄.
Moles of PbSO₄ (molar mass 303.26g/mol) in 2.93g are:
2.93g ₓ (1mol / 303.26) = 9.66x10⁻³ moles PbSO₄ = Moles Pb²⁺.
As molar mass of Pb is 207.2g/mol, mass in 9.66x10⁻³ moles of Pb²⁺ is:
9.66x10⁻³ moles of Pb²⁺ ₓ (207.2g / mol) = 2.00g of Pb²⁺
As mass of the sample is 3.00g, mass of Sn²⁺ is 3.00g - 2.00g = 1.00g
And the percentage of Sn in the sample is:
1.00g / 3.00g ₓ 100 =
33.3% of Sn in the sampleThe lock and key model and the induced fit model are two models of enzyme action explaining both the specificity and the catalytic activity of enzymes. Indicate whether each statement is part of the lock and key model, the induced fit model, or is common to both models.
a. Enzyme conformation changes when it binds the substrate so the active site fits the substrate
b. Substrate binds to the enzyme at the active site, forming an enzyme-substrate complex
c. Enzyme active site has a rigid structure complementary
d. Substrate binds to the enzyme through noncovalent interactions
Answer:
"The active site of the enzyme has a complementary rigid structure" belongs to the key and lock system
"The conformation of the enzyme changes when it binds to the substrate so that the active site conforms to the substrate." belongs to the induced fit system.
"The substrate binds to the enzyme at the active site, forming an enzyme-substrate complex" belongs to both, that is, the key and lock system and the induced fit system.
"The substrate binds to the enzyme through non-covalent interactions" can belong to both enzyme systems.
Explanation:
Enzymatic key and lock systems bear this name because the enzyme at its site of union with the substrate has an ideal shape so that its fit is perfect, similar to a headbreaker, so once they are joined they are not It can bind another substrate to the enzyme, since they are generally associated with strong chemical bonds.
The shape of the enzyme's active site is a negative of what the shape of the substrate would be.
On the other hand, in the mechanism or enzyme system of induced adjustment, the enzyme has an active site that is where it binds with the substrate and another site where another chemical component binds, which when this chemical component binds this enzyme changes its morphology and becomes "active" to bond with your substrate.
This happens a lot in the inactive enzymes that are usually activated in digestive processes since the fact that these enzymes are constantly active would be dangerous, therefore the body takes the induced enzyme system as a control mechanism, where a molecule or chemical compound induces change morphological of an enzyme by means of the allosteric union so that it joins its substrate and catalyzes or analyzes it, depending on the enzymatic character of the enzyme.
A compound consisting of atoms of small atomic mass is more likely to require what
Answer:
a lower temperature to liquefy
Explanation:
g Suppose you are titrating an acid of unknown concentration with a standardized base. At the beginning of the titration, you read the base titrant volume as 1.94 mL. After running the titration and reaching the endpoint, you read the base titrant volume as 23.82 mL. What volume of base was required for the titration
Answer:
21.88mL is the volume of base required for the titration.
Explanation:
For an acid-base titration trying to find the concentration of an acid, you must add a known quantity of the acid and titrate it with an standarized base.
If you know the moles of base you add to the acid solution, these moles are equal to moles of acid.
In the buret of the titration, initial volume is 1.94mL and final volume is 23.82mL. The volume you are adding is the difference between initial and final volume, that is:
23.82mL - 1.94mL
21.88mL is the volume of base required for the titration.
A chemist observed an unknown Balmer Series decay through an emission of 410 nm. Using the experimental wavelength, determine the energy levels transition involved in the
emitted wavelength.
Answer:
Option D is correct.
n = 6 to n = 2
Explanation:
Like all waves emitted from the movement of electrons from one energy level to another, the wavelength (λ) is given by the equation involving Rydberg's constant
(1/λ) = Rₕ [(1/n₂²) - (1/n₁²)]
where Rₕ = 10973731.57 m⁻¹ = (1.0974 × 10⁷) m⁻¹
n₂ = principal quantum number corresponding to the final energy level of the electron = 2 (For Balmer Series)
n₁ = principal quantum number corresponding to the final energy level of the electron = ?
λ = 410 nm = (410 × 10⁻⁹) m
(1/λ) = (2.439 × 10⁶) m⁻¹
2.439 × 10⁶ = (1.0974 × 10⁷) [(1/2²) - (1/n₁²)]
0.25 - (1/n₁²) = (2.439 × 10⁶) ÷ (1.0974 × 10⁷) = 0.2222602562
(1/n₁²) = 0.25 - 0.2222602562 = 0.0277397438
n₁² = (1/0.0277397438) = 36.05
n₁ = 6
Hope this Helps!!!
A balloon filled with helium has a volume of 4.5 × 103 L at 25°C. What volume will the balloon occupy at 50°C if the pressure surrounding the balloon remains constant?
Answer:
[tex]V_2 = 4.87 * 10^3[/tex]
Explanation:
This question is an illustration of ideal Gas Law;
The given parameters are as follows;
Initial Temperature = 25C
Initial Volume = 4.5 * 10³L
Required
Calculate the volume when temperature is 50C
NB: Pressure remains constant;
Ideal Gas Law states that;
[tex]PV = nRT[/tex]
The question states that the pressure is constant; this implies that the constant in the above formula are P, R and n
Divide both sides by PT
[tex]\frac{PV}{PT} = \frac{nRT}{PT}[/tex]
[tex]\frac{V}{T} = \frac{nR}{P}[/tex]
Represent [tex]\frac{nR}{P}[/tex] with k
[tex]\frac{V}{T} = k[/tex]
[tex]k = \frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
At this point, we can solve for the required parameter using the following;
[tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
Where V1 and V2 represent the initial & final volume and T1 and T2 represent the initial and final temperature;
From the given parameters;
V1 = 4.5 * 10³L
T1 = 25C
T2 = 50C
Convert temperatures to degree kelvin
V1 = 4.5 * 10³L
T1 = 25 +273 = 298K
T2 = 50 + 273 = 323K
Substitute values for V1, T1 and T2 in [tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
[tex]\frac{4.5 * 10^3}{298} = \frac{V_2}{323}[/tex]
Multiply both sides by 323
[tex]323 * \frac{4.5 * 10^3}{298} = \frac{V_2}{323} * 323[/tex]
[tex]323 * \frac{4.5 * 10^3}{298} = V_2[/tex]
[tex]V_2 = 323 * \frac{4.5 * 10^3}{298}[/tex]
[tex]V_2 = \frac{323 * 4.5 * 10^3}{298}[/tex]
[tex]V_2 = \frac{1453.5 * 10^3}{298}[/tex]
[tex]V_2 = 4.87 * 10^3[/tex]
Hence, the final volume at 50C is [tex]V_2 = 4.87 * 10^3[/tex]
A solid white substance A is heated strongly in the absence of air. It decomposes to form a new white substance B and a gas C. The gas has exactly the same properties as the product obtained when carbon is burned in an excess of oxygen. Based on these observations, can we determine whether solids A and B and the gas C are elements or compounds?
Answer:
A, B and C are compounds
Explanation:
First of all, I need to establish that when carbon is burnt in excess oxygen, carbon dioxide is obtained as shown by this equation; C(s) + O2(g) ----> CO2(g).
Looking at the presentation in the question, A was said to be heated strongly and it decomposed to B and C. Only a compound can decompose when heated. Elements can not decompose on heating. Secondly, compounds usually decompose to give the same compounds that combined to form them. Compounds hardly decompose into their constituent elements.
Again from the information provided, the compound A is a white solid. This is likely to be CaCO3. It decomposes to give another white solid. This may be CaO and the gas was identified as CO2.
Hence;
CaCO3(s)--------> CaO(s) + CO2(g)
Given that the molar mass of NAOH is 40.00G/MOL, what mass of NAOH is needed to make 2.500 L of a 2.000 M NaOH Solution?
Answer:
200g
Explanation:
n = CV
n = mass/molar mass
mass/molar mass = CV
mass/40 = 2 x 2.5
mass/40 = 5
mass = 5x 40
mass = 200g
A sample of oxygen is collected over water at a total pressure of 692.2 mmHg at 17°C. The vapor pressure of water at 17°C is 14.5 mmHg. The partial pressure of the O2
Answer:
677.7 mmHg
Explanation:
The first empirical study on the behaviour of a mixture of gases was carried out by John Dalton. He established the effects of mixing gases at different pressures in the same vessel.
Dalton's law states that,the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the individual gases present in the mixture of gases. When a gas is collected over water, the gas also contains some water vapour. The partial pressure of the gas will now be given as; total pressure of gas mixture - saturated vapour pressure of water (SVP) at that temperature.
Given that;
Total pressure of gas mixture = 692.2 mmHg
SVP of water at 17°C = 14.5 mmHg
Therefore, partial pressure of oxygen = 692.2-14.5
Partial pressure of oxygen = 677.7 mmHg