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3 years ago

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Given the following equation formats, which form matches a single replacement

1 answer:

Ber [7]3 years ago

7 0

Answer:

The answer is c. A + BC → AC + B

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If 11.0 g of ccl3f is enclosed in a 1.1 −l container, will any liquid be present?if so, what mass of liquid?

Anastasy [175]

Considering that CCL3F gas behave like an ideal gas then we can use the Ideal Gas Law
<span>PV = nRT, however is an approximation and not the only way to resolve this problem with the given data..So,at the end of the solution I am posting some sources for further understanding and a expanded point of view. </span>

<span>Data: P= 856torr, T = 300K, V= 1.1L, R = 62.36 L Torr / KMol </span>

<span>Solving and substituting in the Gas equation for n = PV / RT = (856)(1.1L) /( 62.36)(300) = 0.05 Mol. This RESULT is of any gas. To tie it up to our gas we need to look for its molecular weight:MW of CCL3F = 137.7 gm/mol. </span>

<span>Then : 0.05x 137.5 = 6.88gm of vapor </span>

<span>If we sustract the vapor weight from the TOTAL weight of liquid we have: 11.5gm - 6.88gm = 4.62 gm of liquid.d</span>

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3 years ago

What have chemists done to help pople conserve energy

Ainat [17]

Chemists have developed insulation.

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3 years ago

How much time would it take for 5.2 x 10^5 atoms of fermium-253 (half-life 3 days) to decay to 6.5 x 104 atoms?

Jlenok [28]

Answer:

9 days, or 3 half-lives

Explanation:

5.2x10^5=520000

6.5x10^4=65000

65000/520000=1/8, or 3 half-lives

3x3=9 days

I'm not the greatest at Chem but this seems more like math than Chem :)

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3 years ago

A gas is located inside a closed container. if the gas is allowed to occupy more volume inside the container, what will happen t

nataly862011 [7]

Pressure in the container will decrease.

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3 years ago

A reactant decomposes with a half-life of 29.5 s when its initial concentration is 0.229 M. When the initial concentration is 0.

Dmitrij [34]

Answer :

The order of reaction is, 0 (zero order reaction).

The value of rate constant is, $0.00388Ms^{-1}$

Explanation :

Half life : It is defined as the time in which the concentration of a reactant is reduced to half of its original value.

The general expression of half-life for nth order is:

$t_{1/2}\propto \frac{1}{[A_o]^{n-1}}$

or,

$\frac{t_{1/2}_1}{t_{1/2}_2}=\frac{[A_2]^{n-1}}{[A_1]^{n-1}}$

or,

$n=\left(\frac{\log\frac{(t_{1/2})_1}{(t_{1/2})_2}}{\log\frac{(A)_2}{(A)_1}}\right )+1$ .............(1)

where,

$t_{1/2}$ = half-life of the reaction

n = order of reaction

[A] = concentration

As we are given:

Initial concentration of A = 0.229 M

Final concentration of A = 0.639 M

Initial half-life of the reaction = 29.5 s

Final half-life of the reaction = 82.3 s

Now put all the given values in the above formula 1, we get:

$n=\left (\frac{\log \frac{29.5}{82.3}}{\log\frac{0.639}{0.229}}\right )+1$

$n=0.000196\approx 0$

Thus, the order of reaction is, 0 (zero order reaction).

Now we have to determine the rate constant.

To calculate the rate constant for zero order the expression will be:

$t_{1/2}=\frac{[A_o]}{2k}$

When,

$t_{1/2}$ = 29.5 s

$[A_o]$ = 0.229 M

$29.5s=\frac{0.229M}{2k}$

$k=0.00388Ms^{-1}$

Thus, the value of rate constant is, $0.00388Ms^{-1}$

4 0

3 years ago