What mass of ammonia will exert the same pressure as 12.8 mg of hydrogen sulfide (h2s) in the same container under the same conditions?

There will 6.38 mg of NH3 extert the same pressure as 12.8 mg H2S

Explanation:

Step 1: Data given

Mass of hydrogen sulfide (H2S)= 12.8 mg = 0.0128 grams

Molar mass of H2S = 34.1 g/mol

Step 2: Calculate moles of H2S

Moles of H2S = Mass H2S / molar mass H2S

Moles H2S = 0.0128 grams / 34.1 g/mol

Moles H2S = 3.75 * 10^-4

Gases here, exert exacty the same amount of pressure per mole as any other gas.  

The ammonia will exert the same pressure as the hydrogen sulfide when the moles  of ammonia equal the moles of hydrogen sulfide

Step 3: Calculate moles of NH3

Moles of H2S = moles of NH3

moles of NH3 = 3.75 * 10^-4

Step 4: Calculate mass of NH3

Mass of NH3 = moles NH3 * molar mass NH3

Mass of NH3 = 3.75 *10^-4 moles * 17.03 g/mol

Mass of NH3 = 0.00639 grams = 6.38 mg

There will 6.38 mg of NH3 extert the same pressure as 12.8 mg H2S

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b