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shepuryov [24]
3 years ago
13

An aspirin tablet contains 325 mg acetylsalicylic acid. How many acetylsalicylic acid molecules does it contain?

Chemistry
1 answer:
lubasha [3.4K]3 years ago
4 0
The formula of acetylsalicylic acid is C9H8O4. And the gram formula mass of the molecule is 180 g/mol. So there is 325/180 = 1.8 mol. There is 6.02*10^23 molecules per mol. So the answer is 1.08 * 10^24 molecules.
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Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine. CCl4(g) + (1/2)O
madam [21]

The question is incomplete, here is the complete question:

Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine.

CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c=4.4\times 10^9 at 1,000 K

Calculate Kc for the reaction 2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)

<u>Answer:</u> The value of K_c' for the final reaction is 1.936\times 10^{19}

<u>Explanation:</u>

The given chemical equations follows:

CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c

We need to calculate the equilibrium constant for the equation, which is:

2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)

As, the final reaction is the twice of the initial equation. So, the equilibrium constant for the final reaction will be the square of the initial equilibrium constant.

The value of equilibrium constant for net reaction is:

K_c'=(K_c)^2

We are given:

K_c=4.4\times 10^9

Putting values in above equation, we get:

K_c'=(4.4\times 10^9)^2=1.936\times 10^{19}

Hence, the value of K_c' for the final reaction is 1.936\times 10^{19}

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3 years ago
Which of the following is an example of point-source pollution? A. Salt from roads B. Water from a sewage treatment plant C. Fer
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3 years ago
An electrochemical cell is composed of pure nickel and pure iron electrodes immersed in solutions of their divalent ions at room
Andrej [43]

Answer:

0.758 V.

Explanation:

Hello!

In this case, case when we include the effect of concentration on an electrochemical cell, we need to consider the Nerst equation at 25 °C:

E=E\°-\frac{0.0591}{n} log(Q)

Whereas n stands for the number of moles of transferred electrons and Q the reaction quotient relating the concentration of the oxidized species over the concentration of the reduced species. In such a way, we can write the undergoing half-reactions in the cell, considering the iron's one is reversed because it has the most positive standard potential so it tends to reduction:

Fe^{2+}+2e^-\rightarrow Fe^0\ \ \ E\°=0.440V\\\\Ni^0\rightarrow Ni^{2+}+2e^-\ \ \ E\°=-0.250V

It means that the concentration of the oxidized species is 0.002 M (that of nickel), that of the reduced species is 0.40 M and there are two moles of transferred electrons; therefore, the generated potential turns out:

E=(0.440V+0.250V)-\frac{0.0591}{2} log(\frac{0.002M}{0.40M} )\\\\E=0.758V

Beat regards!

8 0
3 years ago
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