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meriva
3 years ago
9

Convert 1.25 atm to bar.

Chemistry
1 answer:
Tasya [4]3 years ago
8 0

Answer: 1.27 bar

Explanation:

1 atm = 1.01325 bar

1.25 atm = Z (let Z be the unknown value)

To get the value of Z, cross multiply

Z x 1 atm = 1.25 atm x 1.01325 bar

1 atm•Z = 1.2665625 atm•bar

To get the value of Z, divide both sides by 1 atm

1 atm•Z/1 atm = 1.2665625 atm•bar/1atm

Z = 1.2665625 bar

(Round up Z to the nearest hundredth as 1.27 bar)

Thus, 1.25 atm when coverted gives 1.27 bar

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3 years ago
The rate constant, k, for a reaction is 0.0354 sec1 at 40°C. Calculate the rate constant for the
deff fn [24]

Answer:

The rate constant of the reaction at 125˚ is 0.3115 \ \text{sec}^{-1}.

Explanation:

The Arrhenius equation is a simple equation that describes the dependent relationship between temperature and the rate constant of a chemical reaction. The Arrhenius equation is written mathematically as

                                                  k \ = \ Ae^{\displaystyle\frac{-E_{a}}{RT}}

                                               \ln k \ = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT}

where k is the rate constant, E_{a} represents the activation energy of the chemical reaction, R is the gas constant, T is the temperature, and A is the frequency factor.

The frequency factor, A, is a constant that is derived experimentally and numerically that describes the frequency of molecular collisions and their orientation which varies slightly with temperature but this can be assumed to be constant across a small range of temperatures.

Consider that the rate constant be k_{1} at an initial temperature T_{1} and the rate constant k_{2} at a final temperature T_{2}, thus

                         \ln k_{2} \ - \ \ln k_{1} = \ \ln A \ - \ \displaystyle\frac{E_{a}}{RT_{2}} \ - \ \left(\ln A \ - \ \displaystyle\frac{E_{a}}{RT_{1}}\right) \\ \\ \\ \rule{0.62cm}{0cm} \ln \left(\displaystyle\frac{k_{2}}{k_{1}}\right) \ = \ \displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)

                                         \rule{1.62cm}{0cm} \displaystyle\frac{k_{2}}{k_{1}} \ = \ e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)} \\ \\ \\ \rule{1.62cm}{0cm} k_{2} \ = \ k_{1}e^{\displaystyle\frac{E_{a}}{R}\left(\displaystyle\frac{1}{T_{1}} \ - \ \displaystyle\frac{1}{T_{2}} \right)}

Given that E_{a} \ = \ 26.5 \ \ \text{kJ/mol}, R \ = \ 8.3145 \ \ \text{J mol}^{-1} \ \text{K}^{-1}, T_{1} \ = \ \left(40 \ + \ 273\right) \ K, T_{2} \ = \ \left(125 \ + \ 273\right) \ K, and k_{1} \ = \ 0.0354 \ \ \text{sec}^{-1}, therefore,

           k_{2} \ = \ \left(0.0354 \ \ \text{sec}^{-1}\right)e^{\displaystyle\frac{26500 \ \text{J mol}^{-1}}{8.3145 \ \text{J mol}^{-1} \ \text{K}^{-1}}\left(\displaystyle\frac{1}{313 \ \text{K}} \ - \ \displaystyle\frac{1}{398 \ \text{K}} \right)} \\ \\ \\ k_{2} \ = \ 0.3115 \ \ \text{sec}^{-1}                      

8 0
2 years ago
2. A chemical reaction occurs when substances interact and their atoms
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Pyruvate is converted to acetyl-CoA in the reaction given below:

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1 molecule of carbon dioxide is eliminated from 1 molecule of pyruvate.

Also,  

2 molecules of carbon dioxide is eliminated from 2 molecules of pyruvate (as glucose on glycolysis yields 2 molecules of pyruvate).

Also, acetyl-CoA further goes into the citric acid cycle and produces 2 molecules of carbon dioxide.

Thus pyruvate produces total 3 molecules of CO₂  and hence glucose produces 6 molecules of CO₂ (as glucose on glycolysis yields 2 molecules of pyruvate)

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Temperature comes under interval scale , because interval scale has no zero point.

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Therefore , Temperature is an example of a quantitative variable.

Hence, the correct answer is "quantitative variable"

5 0
3 years ago
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