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3 years ago

Which variables does equilibrium depend on?

Chemistry

1 answer:

Eddi Din [679]3 years ago

7 0

Answer:

I believe tempurature, though i'm not sure. Give it a shot and tell me what you got!

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Ca(s)+2hno3(aq)→ca(no3)2(aq)+h2(g) identify the oxidizing agent.

aleksandr82 [10.1K]

To find for the oxidizing agent, first let us write the half reactions of this complete chemical reaction:

Ca = Ca2+ + 2e- 
2 H+ + 2e- = H2

The oxidizing agent would be the substance of the element that is reduced. We know that an element is reduced when an electron is added to it. In this case, the element being reduced is H. Therefore the oxidizing agent is HNO3.

Answer:

HNO3

8 0

3 years ago

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Which of the following is an organic moleclue ?

Doss [256]

Answer:

water

Explanation:

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3 years ago

A speed time graph shows that a car moves at 20 m/s for 15s. The car's speed steadily decreases until it comes to a stop at 40s.

ladessa [460]

0s to 15s: constant speed/zero acceleration
15s to 40s: constant gradient, therefore constant deceleration

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2 years ago

Consider the statement: “The temperature of the land is an important factor for the ripening of oranges, because it affects the

Alexxx [7]

Physical Properties: Physical properties can be observed or measured without changing the composition of matter. Physical properties are used to observe and describe matter. so physical changes are the change in temperature of the land and the evaporation of water and change humidity of the air. chemical change is the ripening of the orange

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2 years ago

The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3

9966 [12]

Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol

$k_2=2.3\times 10^8$

$T_2=280\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (280 + 273.15) K = 553.15 K

$T_2=553.15\ K$

$k_1=4.8\times 10^8$

$T_1=376\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (376 + 273.15) K = 649.15 K

$T_1=649.15\ K$

So,

$\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)$

$E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=22.87\ kJ/mol$

The activation energy for this reaction = 23 kJ/mol.

6 0

3 years ago