1) ideal gas law: p·V = n·R·T.
p - pressure of gas.
V -volume of gas.
n - amount of substance.
R - universal gas constant.
T - temperature of gas.
n₁ = 0,04 mol, V₁ = 0,06 l.
n₂ = 0,07 mol, V₂ = 0,06 · 0,07 ÷ 0,04 = 0,105 l.
2) V₁ = 0,06 l, T₁ = 240,00 K.
T₂ = 340,00 K, V₂ = 340 · 0,06 ÷ 240 = 0,05 l.
<h3><u>Answer;</u></h3>
A dense substance that is hard and incompressible
<h3><u>Explanation</u>;</h3>
- <em><u>Crystals are solids that are made up of particles such as molecules, atoms or ions, </u></em>which are arrange in a highly ordered microscopic structure, forming a crystal lattice which extends in all directions.
- Examples of crystals include<em><u> quartz,</u></em> that is made up of silicon (iV) oxide molecules that are arranged in a lattice,<u><em>table salt</em></u> is another example that is made up of sodium and chloride ions that are arranged in a lattice
- Crystals are very ordered, they may be made up of atoms of the same elements or atoms of different elements such as in the case of table salt.
214.0560
+ 9.3456
=223.4016
The question is incomplete, here is the complete question:
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?
<u>Answer:</u> The rate constant at 324°C is 
<u>Explanation:</u>
To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:
![\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7BK_%7B244%5EoC%7D%7D%29%3D%5Cfrac%7BE_a%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 244°C = 
= equilibrium constant at 324°C = ?
= Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![244^oC=[273+244]K=517K](https://tex.z-dn.net/?f=244%5EoC%3D%5B273%2B244%5DK%3D517K)
= final temperature = ![324^oC=[273+324]K=597K](https://tex.z-dn.net/?f=324%5EoC%3D%5B273%2B324%5DK%3D597K)
Putting values in above equation, we get:
![\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7B6.7%7D%29%3D%5Cfrac%7B71000J%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B517%7D-%5Cfrac%7B1%7D%7B597%7D%5D%5C%5C%5C%5CK_%7B324%5EoC%7D%3D61.29M%5E%7B-1%7Ds%5E%7B-1%7D)
Hence, the rate constant at 324°C is
Answer: The steepness of a ramp affects it by making it easier or harder.
Explanation: It's a bit situational. If you were going up a steep ramp with a heavy load, it will increase the work necessary, whereas if you were going down a ramp, it would decrease the work necessary. If you need this simply put, think about biking up and down a hill. It would be easier going down than up.
