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3 years ago

Help plzzzzzzzzzz ASAP!!!

Chemistry

2 answers:

Reil [10]3 years ago

8 0

1st one light bouncesfrom objects saneangleand intensity that the object received it

lord [1]3 years ago

8 0

First one is the right one

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The number of nitrogen atoms in one mole of nitrogen gas are...

9966 [12]

Explanation:

The number of nitrogen atoms in one mole of nitrogen gas are 6.02214179×1023 nitrogen atoms. 

Hope this helps...

3 0

3 years ago

In regard to protective actions for explosive devices, the area where the blast originates is referred to as ___________ perimet

lianna [129]

This is to fill in the answer to the question.

The area where the blast originates is referred to as Scene Perimeter/Isolation Zone. This whole area is dangerous for people since it can contain harmful gasses or falling debris depending on the environment of the blast.

7 0

3 years ago

Compare the density of 25g of copper to 80g of copper?

tiny-mole [99]

Answer:

Explanation:

25g is less dense than 80g therefore will mostly float. if you subtract 80-25 that will leave you will 55 as the difference

7 0

3 years ago

What are the functions of ammonia in the solvay process

enot [183]

The ammonia gas is absorbed in the concentrated brine to produce aqueous sodium chloride and aqueous ammonia. This ammoniation process is exothermic, so energy is released as heat. The ammonia tower eventually needs to be cooled.

7 0

3 years ago

If a temperature increase from 10.0 ∘C to 22.0 ∘C doubles the rate constant for a reaction, what is the value of the activation

maria [59]

The activation energy barrier is 40.1 kJ·mol⁻¹

Use the Arrhenius equation

$\ln( \frac{k_2 }{k_1 }) = (\frac{E_{a} }{R })(\frac{ 1}{T_1} - \frac{1 }{T_2 })\\$

$\ln( \frac{2k }{k}) = (\frac{E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1} })(\frac{ 1}{\text{283.15 K}} - \frac{1 }{\text{295.15 K }})\\$

$\ln2 = (\frac{ E_{a} }{\text{8.314 J} \cdot \text{K}^{-1} \text{mol}^{-1}}) \times 1.436 \times10^{-4}\\$

$\ln2 = E_{a} \times 1.727 \times 10^{-5} \text{ mol} \cdot \text{J}^{-1}$

$E_{a} = \frac{\ln2 }{ 1.727 \times10^{-5}\text{ mol} \cdot \text{J}^{-1}}\\$

$E_{a} = \text{40 100 J}\cdot\text{mol}^{-1} = \textbf{40.1 kJ}\cdot \textbf{mol}^{-1}$

3 0

3 years ago