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3 years ago

As part of an investigation, students combined substances in a beaker to observe

Chemistry

1 answer:

aleksklad [387]3 years ago

3 0

Answer:

please mark brainlest and it's Procedure 1: One of the products was a gas that escaped into the air.

Procedure 2: A gas from the air reacted with one of the other reactants

Explanation:

the gas ca evaporate so it would'nt be a or c and b dosent make sense.

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2) A common "rule of thumb" -- for many reactions around room temperature is that the

babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$ and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}$

$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0

3 years ago

Which of the following isotopes is NOT considered to be a major threat to human health from nuclear fallout? a. strontium-90 b.

Nat2105 [25]

Ohhh it’s A) strontium

4 0

3 years ago

In the ground state, an atom of which element has seven valence

sweet-ann [11.9K]

Answer:

Fluorine

Explanation:

Fluorine has 9 total electrons. The first two are in the 1s level, and the remaining electrons are on the outer level of the atom, with 2 in the s level and 5 in the p level. The electron configuration is 1s2 2s2 2p5.

7 0

3 years ago

The half-reaction occurring at the cathode in the balanced reaction shown below is __________?2Ag(s) + 1/2 O2(g) + 2H+(aq) → 2Ag

Shtirlitz [24]

Answer:

Correct choice are C and D (they are both, the same).

Explanation:

Cathode is the positive(+) electrode where a reduction occurs.

Reduction is the chemical reaction where the oxidation state is reduced.

2Ag(s) + 1/2 O2(g) + 2H+(aq) → 2Ag+(aq) + H2O (l)

A. 2H2O (l) → O2 (g) + 4H+ (aq) + 4e-

B. 2Ag (s) → 2Ag+ (aq) + 2e-

C. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)

D. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)

C or D, are ok. They are the same equation.

Oxygen from ground state reduce the oxidation state from 0 to -2

3 0

3 years ago

श्रीरयाफल्सन यसका लागि तियरल प्रतिस्थापन स्तर को के हो

jeka94

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3 0

3 years ago