Answer:
Weather is describes as the term of the temperature, humidity, wind velocity, precipitation, and barometric pressure. It usually happens in the troposphere, or lower atmosphere. (Hope this helped!)
Explanation:
The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m
Answer:
A) Carbon dioxide and Oxygen
Explanation:
Percentage of gases in the Planet's atmosphere:
- Carbon dioxide =<u> </u><u>4</u><u> </u><u>%</u><u> </u>
- Nitrogen =<u> </u><u>72</u><u> </u><u>%</u><u> </u>
- Oxygen =<u> </u><u>24</u><u> </u><u>%</u><u> </u>
Percentage of gases in the Earth's atmosphere:
- Carbon dioxide =<u> </u><u>0.036</u><u> </u><u>%</u><u> </u><em>(</em><em>traces</em><em>)</em>
- Nitrogen =<u> </u><u>78</u><u> </u><u>%</u><u> </u>
- Oxygen =<u> </u><u>21</u><u> </u><u>%</u><u> </u>
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On observing the percentage composition of the atmospheres of the two Planets, we get:
- The newly discovered planet has more percentage of <u>Oxygen</u> than The Earth.
- Percentage of <u>Carbon dioxide</u> is more in the planet than in the Earth
- <em>Nitrogen</em> is <u>less</u> in the planet than in the Earth.
___________________
Answer:
What's asked is the name of the gases that are in higher amounts in the atmosphere of the newly discover planet
Therefore,
<h3>A) Carbon dioxide and Oxygen</h3>
Answer:
94.4g/mol is molar mass of the unknown
Explanation:
Based on the freezing point depression equation:
ΔT = Kf*m*i
<em>Where ΔT is the depression in freezing point (1.87°C)</em>
<em>Kf is freezing point depression constant of water (1.86°Ckg/mol)</em>
<em>And i is Van't Hoff factor (1 for nonelectrolyte solutes)</em>
<em />
Replacing:
1.87°C = 1.86°CKg/mol*m*i
1.005mol/kg solvent = m
Using the mass of the solvent we can find the oles of the nonelectrolyte:
1.005mol/kg solvent * 0.4764kg = 0.479moles
Molar mass is defined as the ratio between mass of a substance in grams and moles, that is:
45.2g / 0.479mol =
<h3>94.4g/mol is molar mass of the unknown</h3>