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vfiekz [6]
4 years ago
12

The concentration of reactants or products at any time during the reaction can be calculated from the integrated rate law.

Chemistry
1 answer:
pashok25 [27]4 years ago
5 0

Answer:

True

Explanation:

Due to the fact that it is an equation that expresses the concentrations of reactants or products as a function of time, meaning that it describes the change in concentration of reactant(s) per change in time.

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How does a liquid substance change into gas? ​
Afina-wow [57]

Answer: heat

Explanation:

Is that heat boils the liquid and the liquid eventually evaporated into gas

3 0
3 years ago
Which of the following are true statements about equilibrium systems? For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(
Grace [21]

Answer:

The first, third and fourth statements are correct.

Explanation:

1) For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(s) + CO2(g) adding more CaCO3 will shift the equilibrium to the right.

⇒ Le Chatellier says As the CaCO3 concentration is increased, the system will attempt to undo that concentration change by shifting the balance to the right. <u>This statement is true.</u>

<u />

2) For the following reaction at equilibrium: CaCO3(s)⇌ CaO(s) + CO2(g) increasing the total pressure by adding Ar(g) will shift the equilibrium to the right.

⇒ Le chatellier says that if we increase the pressure, the equilibrium will shift to the side with the least number of particles.

Since the molar densities of CaO and CaCO3 are constant, they don't appear in the equilibrium expression. This is why only changes to the pressure (concentration) of CO2 affect the position of the equilibrium.

If the pressure in the container is increased by adding an inert or non-reacting gas, nothing happens to the amounts of CO2, CaO or CaCO3. The added gas won't affect the partial pressure of CO2. <u>This statement is false. </u>

3)For the following reaction at equilibrium: 2 H2(g) + O2(g) ⇌ 2 H2O(g) the equilibrium will shift to the left if the volume is doubled.

⇒ Le Chatellier says if we increase the pressure, the equilibrium will shift to the side with the most particles.

In this case we have 2 moles of H2 and 1 mole of O2 on the left side and 2 mole of H2O on the right side. This means on the left side are more particles. So the equilibrium will shift to the left, so <u>this statement is true.</u>

4) For the following reaction at equilibrium: H2(g) + F2(g) ⇌ 2HF(g) removing H2 will increase the amount of F2 present once equilibrium is reestablished. Increasing the temperature of an endothermic reaction shifts the equilibrium position to the right.

⇒ Le chatellier says if H2 will be removed (this means the left side will get less particles) so the equilibrium will shift to the left, to increase the amount of F2.

⇒Le chatelier says if we increase the temperature of an exotherm reaction , there will be less energy released. The equilibrium will shift to the side of the reactants (the left side).

If we increase the temperature of an endotherm reaction, the equilibrium will shift to the side of the products (the right side). <u>This statement is true.</u>

4 0
4 years ago
Hurry!!
pishuonlain [190]

Answer:

condensation is the answer

6 0
3 years ago
Read 2 more answers
N204(0) + 2NO2(g)
user100 [1]

setup 1 : to the right

setup 2 : equilibrium

setup 3 : to the left

<h3>Further explanation</h3>

The reaction quotient (Q) : determine a reaction has reached equilibrium

For reaction :

aA+bB⇔cC+dD

\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}

Comparing Q with K( the equilibrium constant) :

K is the product of ions in an equilibrium saturated state  

Q is the product of the ion ions from the reacting substance  

Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)

Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium

Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)

Keq = 6.16 x 10⁻³

Q for reaction N₂O₄(0) ⇒ 2NO₂(g)

\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}

Setup 1 :

\tt Q=\dfrac{0.0064^2}{0.098}=0.000418=4.18\times 10^{-4}

Q<K⇒The reaction moved to the right (products)

Setup 2 :

\tt Q=\dfrac{0.0304^2}{0.15}=0.00616=6.16\times 10^{-3}

Q=K⇒the system at equilibrium

Setup 3 :

\tt Q=\dfrac{0.230^2}{0.420}=0.126

Q>K⇒The reaction moved to the left (reactants)

8 0
3 years ago
Read 2 more answers
Chemical equation=CH4 + O2 = CO2 + H2O.
tankabanditka [31]

Answer:

1.The substance(s) to the left of the arrow in a chemical equation are called reactants. A reactant is a substance that is present at the start of a chemical reaction. The substance(s) to the right of the arrow are called products . A product is a substance that is present at the end of a chemical reaction.The combustion of methane or octane is exothermic; it releases energy. CH4 + 2 O2 → CO2 + 2 H2O + energy The energies of the products are lower than the energiies of the reactants.

Explanation:

4 0
3 years ago
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