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zheka24 [161]
3 years ago
6

If the volume occupied by 0.500 mol of nitrogen gas at 0°C is 11.2 L, then the volume occupied by 2.00 mol of nitrogen gas at th

e same temperature and pressure will be:
A. the data given is not sufficient to determine the volume
B. 22.4 L
C. 44.8 L
D. the same as that occupied by 0.500 mol of nitrogen gas
Chemistry
1 answer:
Paraphin [41]3 years ago
5 0
Avagadros law states that volume of gas is directly proportional to number of moles of gas at constant pressure and temperature.
\frac{V1}{n1} =  \frac{V2}{n2}
where V -volume , n - number of moles 
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation 
substituting these values in the equation 
\frac{11.2 L}{0.500 mol} =  \frac{V}{2.00 mol}
V = 44.8 L 
answer is C. 44.8 L 
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Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp
gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})

Therefore

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T1= 21+273.15

T1= 294.15K

T2= 200  

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in  j/mol R Ea is

Ea=35.5*1000 j/mol R

ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

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In conclusion, rate constant

K2=2.67sec^{-1}

Read more about rate constant

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3 years ago
50.0 ml of 0.010m naoh was titrated with 0.50m hcl using a dropper pipet. if the average drop from the pipet has a volume of 0.0
creativ13 [48]

25 drops of acid is required to neutralize the 50.0 ml of 0.010m of NaOH in the experiment.

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We can use the titration formula;

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VA = volume of acid

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VB = 50.0 ml

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VA = ?

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Substituting values;

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