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andrew11 [14]
3 years ago
11

If a sample of C-14 initially contains 1.5 mmol of C-14, how many millimoles will be left after 2255 years?

Chemistry
1 answer:
CaHeK987 [17]3 years ago
8 0

Answer: 3.2\times 10^{-4} millimoles will be left after 2255 years.

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  

t = age of sample

a = let initial amount of the reactant  

a - x = amount left after decay process  

a)  t_{\frac{1}{2}} of carbon = 5730 years

k}=\frac{0.693}{5730}=1.21\times 10^{-4}years^{-1}

b) millimoles left after 2255 years

2255=\frac{2.303}{1.21\times 10^{-4}}\log\frac{1.5}{a-x}

\log\frac{1.5}{(a-x)}=8.44

\frac{1.5}{(a-x)}=4628.5

(a-x)=3.2\times 10^{-4}millimoles

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A flask contains 6g hydrogen gas and 64 g oxygen at rtp the partial pressure of hydrogen gas in the flask of the total pressure
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Answer:

B.3/5p

Explanation:

For this question, we have to remember <u>"Dalton's Law of Partial Pressures"</u>. This law says that the pressure of the mixture would be equal to the sum of the partial pressure of each gas.

Additionally, we have a <em>proportional relationship between moles and pressure</em>. In other words, more moles indicate more pressure and vice-versa.

P_i=P_t_o_t_a_l*X_i

Where:

P_i=Partial pressure

P_t_o_t_a_l=Total pressure

X_i=mole fraction

With this in mind, we can work with the moles of each compound if we want to analyze the pressure. With the molar mass of each compound we can calculate the moles:

<u>moles of hydrogen gas</u>

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6g~H_2\frac{1~mol~H_2}{2~g~H_2}=~3~mol~H_2

<u>moles of oxygen gas</u>

The molar mass of oxygen gas (O_2) is 32 g/mol, so:

64g~H_2\frac{1~mol~H_2}{32~g~H_2}=~2~mol~O_2

Now, total moles are:

Total moles = 2 + 3 = 5

With this value, we can write the partial pressure expression for each gas:

P_H_2=\frac{3}{5}*P_t_o_t_a_l

P_O_2=\frac{2}{5}*P_t_o_t_a_l

So, the answer would be <u>3/5P</u>.

I hope it helps!

5 0
3 years ago
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