Answer:
D.
Explanation:
I took this test as a freshman and this was the only question of 4 that I can remember
<em>K</em> = 2.4 × 10^(-72)
<em>Step 1</em>. Determine the <em>value of n
</em>
Zn^(2+) + 2e^(-) → Zn
2Cl^(-) → Cl_2 + 2e^(-)
Zn^(2+) + 2Cl^(-) → Zn + Cl_2
∴ <em>n</em> = 2
<em>Step 2</em>. Calculate <em>K</em>
log<em>K</em> = <em>nE</em>°/0.0592 V = [2 × (-2.12 V)]/0.0592 V = -71.62
<em>K</em> = 10^(-71.62) = 2.4 × 10^(-72)
Protons have a mass of 1
Neutrons have a mass of 1
So 13*1 + 14*1 = Mass number 27
Answer: To show the number of atoms present.
Explanation: As in CO² (Carbon dioxide), there is a small 2 next to the symbol "O" (oxygen) to explain that there are two oxygen atoms.
Answer:
mole fraction of N_2 O = 0.330
mole of fraction SF_4 = 0.669
PRESSURE OF N_2 O = 39127.053 Pa
pressure of SF_4 = 792126.36
Total pressure = 118253.413 Pa
Explanation:
Given data:
volume of tank 8 L
Weight of dinitrogen difluoride gas 5.53 g
weight of sulphur hexafluoride gas 17.3 g
Amount of 
amount of 
mole fraction of 
mole of fraction
PV = nRT
P of N_2 O 
mole of SF_4
Total pressure = 39127.053 + 79126.36 = 118253.413 Pa