<u>Answer:</u> The moles of oxygen and carbon dioxide in air is
and
respectively
<u>Explanation:</u>
To calculate the number of moles, we use the equation:

Given mass of atmosphere = 
Average molar mass of atmosphere = 28.96 g/mol
Putting values in above equation, we get:

We know that:
Percent of oxygen in air = 21 %
Percent of carbon dioxide in air = 0.0415 %
Moles of oxygen in air = 
Moles of carbon dioxide in air = 
Hence, the moles of oxygen and carbon dioxide in air is
and
respectively
Answer:
a, and f.
Explanation:
To be deprotonated, the conjugate acid of the base must be weaker than the acid that will react, because the reactions favor the formation of the weakest acid. The pKa value measures the strength of the acid. As higher is the pKa value, as weak is the acid. So, let's identify the conjugate acid and their pKas:
a. NaNH2 will dissociate, and NH2 will gain the proton and forms NH3 as conjugate acid. pKa = 38.0, so it happens.
b. NaOH will dissociate, and OH will gain the proton and forms H2O as conjugate acid. pKa = 14.0, so it doesn't happen.
c. NaC≡N will dissociate, and CN will gain a proton and forms HCN as conjugate acid. pKa = 9.40, so it doesn't happen.
d. NaCH2(CO)N(CH3)2 will dissociate and forms CH3(CO)N(CH3)2 as conjugate acid. pKa = -0.19, so it doesn't happen.
e. H2O must gain one proton and forms H3O+. pKa = -1.7, so it doesn't happen.
f. CH3CH2Li will dissociate, and the acid will be CH3CH3. pKa = 50, so it happens.
Boric acid, H3BO3, in aqueous solution would only give out one H+ ion. As it is also produce OH ion and by hydrolysis it produces one proton. <span>All the boron compounds (BX3) are having only 6 valence electrons in it and should follow the octet rule by taking another electron.</span>
B(OH)3 + 2 H2O → B(OH)4− + H3O
Answer:
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