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3 years ago

What is true when a reaction has reached equilibrium?

Chemistry

2 answers:

goldfiish [28.3K]3 years ago

7 0

Answer:

In a chemical equilibrium, the forward and reverse reactions occur at equal rates, and the concentrations of products and reactants remain constant. A catalyst speeds up the rate of a chemical reaction, but has no effect upon the equilibrium position for that reaction.

Explanation:

SSSSS [86.1K]3 years ago

3 0

Answer:

The Answer is A. The reaction rate is equal in both directions.

Explanation:

My explanation is don't listen to the guy above me. The reaction is constant therefore it goes both ways.

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A 17.9 mL sample of a 0.445 M aqueous hypochlorous acid solution is titrated with a 0.374 M aqueous sodium hydroxide solution. W

In-s [12.5K]

Answer:

the pH at the start of the titration is 3

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3 years ago

The entropy of an exothermic reaction decreases. This reaction will be spontaneous under which of the following temperatures?

kakasveta [241]

Answer:Low temperatures

Explanation:

∆G= ∆H-T∆S

If ∆H is negative (exothermic reaction), then in order to maintain ∆G<0 which is the condition for spontaneity; T must decrease. This is because, decrease in T will keep the difference of ∆H and T∆S at a negative value in order to satisfy the above stated condition for spontaneity.

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3 years ago

Read 2 more answers

What is the percent composition of hydrogen in NH4HCO3?

Readme [11.4K]

To find the Percent Composition of an atom, you use this formula:
Mass of element in the compound you're studying on ( in this case it's 5 since there is 5 Hydrogens) over the mass of the compound (which is here 79), Multiplied by 100 since you want a percent.
So we get:
$\frac{5}{79} * 100$
So you get about:
$0.063 * 100$
$6.3$

So, the percent composition of Hydrogen in NH4HCO3 is 6.3%

Hope this Helps! :D

7 0

3 years ago

Read 2 more answers

Calculate the work done when an ideal gas expands isothermally and reversibly in a piston and cylinder assembly for expansion of

pantera1 [17]

Answer:

$W=5743.1077\ J$

Explanation:

The expression for the work done is:

$W=RT \ln \left( \dfrac{P_1}{P_2} \right)$

Where,

W is the amount of work done by the gas

R is Gas constant having value = 8.314 J / K mol

T is the temperature

P₁ is the initial pressure

P₂ is the final pressure

Given that:

T = 300 K

P₁ = 10 bar

P₂ = 1 bar

Applying in the equation as:

$W=8.314\times 300 \ln \left( \dfrac{10}{1} \right)$

$W=300\times \:8.314\ln \left(10\right)$

$W=2.30258\times \:2494.2$

$W=5743.1077\ J$

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2 years ago

The incredible catalytic power of enzymes can perhaps best be appreciated by imagining how challenging life would be without jus

ioda

Answer:

t = 7.58 * 10¹⁹ seconds

Explanation:

First order rate constant is given as,

k = (2.303 /t) log [A₀] /[Aₙ]

where [A₀] is the initial concentraion of the reactant; [Aₙ] is the concentration of the reactant at time, <em>t</em>

[A₀] = 615 calories;

[Aₙ] = 615 - 480 = 135 calories

k = 2.00 * 10⁻²⁰ sec⁻¹

substituting the values in the equation of the rate constant;

2.00 * 10⁻²⁰ sec⁻¹ = (2.303/t) log (615/135)

(2.00 * 10⁻²⁰ sec⁻¹) / log (615/135) = (2.303/t)

t = 2.303 / 3.037 * 10⁻²⁰

t = 7.58 * 10¹⁹ seconds

8 0

2 years ago