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

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.