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Aneli [31]
3 years ago
9

How many molecules are in 3.5 moles of H2O A. 2.107x10^24 B. 3.87x10^24 C. 21.07 B. 1.72x10^23

Chemistry
1 answer:
Pani-rosa [81]3 years ago
8 0

Answer:

A. 2.107x10^24

Explanation:

1 mole is defined as the particles containing in 12g of 12-C. That is 6.022x10²³ particles.

When you say 1 mole of atoms means 6.022x10²³ atoms, 1 mole of molecules means 6.022x10²³ molecules, etc.

Now, 3.5 moles of H₂O are:

3.5mol * (6.022x10²³ molecules / 1mol) = 2.107x10²⁴ molecules of H₂O.

Thus, right answer is:

<h3>A. 2.107x10^24</h3>
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2) A common "rule of thumb" -- for many reactions around room temperature is that the
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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.)

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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.

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