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aleksley [76]
3 years ago
12

The overall change in enthalpy of a reaction depends only on the reactants and the products. the path that is taken from the rea

ctants to the products. the reactants and the catalysts only. the reactants, the products, and the catalysts.
Chemistry
2 answers:
vfiekz [6]3 years ago
8 0

Answer:

Here is the complete statement: The overall change in enthalpy of a reaction depend ONLY ON REACTANTS AND THE PRODUCTS.

Explanation:

Change in enthalpy of a chemical reaction refers to the heat change that is associated with a chemical change when the pressure is held constant. Generally, chemical reaction involves the breaking of chemical bonds in the reactants and the formation of new chemical bonds in the products. The breaking and forming of bonds involve heat change; the amount of heat that is given off and absorb in each chemical reaction depend on the type of reactants and products that are involved in the chemical reaction.

galina1969 [7]3 years ago
3 0

Answer: A. the reactants and the products

Explanation:

i just took the test edg2020

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What is the standard state of oxygen?
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Consider the reaction of gaseous hydrogen with gaseous oxygen to produce gaseous water. Given that the first picture represents
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The question is incomplete. There's missing the image, which is shown below.

Answer:

Volume of O₂ = 6 L, volume of mixture: 18 L, volume of H₂O = 12 L, molecule volume of H₂O = 0.667 molecule/L

Explanation:

The reaction between hydrogen gas and oxygen gas to form water is:

2H₂(g) + O₂(g) → 2H₂O(g)

So, for 1 mol of O₂ is necessary 2 moles of H₂ form 2 moles of H₂O. As the images below there's 8 molecules of H₂, 4 molecules of O₂, 12 molecules in the mixture, and 8 molecules of H₂O. Thus, there are stoichiometric values.

All the images are at the same temperature and pressure, so, by the ideal gas law:

PV= nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

The number of moles and molecules are related, so let's substitute it in the equation. For the H₂:

P*12 = 8*RT

RT/P = 12/8 = 1.5

Thus, for O₂:

PV= nRT

V = n*(RT/P)

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6 0
2 years ago
Consider the second-order reaction:
kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

Explanation:

2HI(g)\rightarrow H_2(g)+I_2(g)


Rate Law: k[HI]^2


Rate constant of the reaction = k = 6.4\times 10^{-9} L/mol s

Order of the reaction = 2

Initial rate of reaction = R=1.6\times 10^{-7} Mol/L s

Initial concentration of HI =[A_o]

1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2

[A_o]=5 mol/L

Final concentration of HI after t = [A]

t = 4.53\times 10^{10} s

Integrated rate law for second order kinetics is given by:

\frac{1}{[A]}=kt+\frac{1}{[A_o]}

\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}

[A]=0.00345 mol/L

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

5 0
3 years ago
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