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tekilochka [14]
3 years ago
13

What is log of 52????

Chemistry
2 answers:
irinina [24]3 years ago
8 0

1.71600334       Hope Im right                

erik [133]3 years ago
3 0
1.716003344 <--- this the log of 52!
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At standard temperature and pressure (0 ∘C and 1.00 atm ), 1.00 mol of an ideal gas occupies a volume of 22.4 L . What volume wo
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Taking into account the Charles's law, the same amount of gas at the same pressure and 65 ∘C would occupy a volume of 27.73 L.

<h3>Charles's Law</h3>

Charles's Law consists of the relationship that exists between the volume and the temperature of a certain quantity of ideal gas, at a constant pressure.

Volume is directly proportional to the temperature of the gas: if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.

Mathematically, Charles's law is a law that says that the quotient that exists between the volume and the temperature will always have the same value:

V÷ T= k

Considering an initial state 1 and a final state 2, it is satisfied:

V1÷ T1= V2÷ T2

<h3>Volume at 65°C</h3>

In this case, you know:

  • V1= 22.4 L
  • T1= 0 C= 273 K
  • V2= ?
  • T2= 65 C= 338 K

Replacing in Charles's law:

22.4 L÷ 273 K= V2÷ 338 K

Solving:

(22.4 L÷ 273 K) ×338 K= V2

<u><em>V2= 27.73 L</em></u>

Finally, the same amount of gas at the same pressure and 65 ∘C would occupy a volume of 27.73 L.

Learn more about Charles's law:

brainly.com/question/4147359

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Explanation:

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Explain in terms of charge distribution why a molecule of butanamide is polar
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The butanamide molecule has an asymmetrical distribution of charge.

The molecule has uneven charge distribution<span />

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The decomposition of hydrogen peroxide follows first order kinetics and has a rate constant of 2.54 x 10-4 s-1 at a certain temp
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Decomposition of hydrogen peroxide is following first order kinetics.

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

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k = rate constant = 2.54\times 10^{-4}s^{-1}

t = time taken for decay process = 855 s

[A_o] = initial amount of the reactant = ?

[A] = amount left after decay process =  0.321 M

Putting values in above equation, we get:

2.54\times 10^{-4}s^{-1}=\frac{2.303}{855s}\log \frac{[A_o]}{0.321}

[A_o]=0.399M

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What happens to the length of a year for each planet as you get farther away from the sun
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