Dinitrogen monoxide gas is collected at-3.0 °c in an evacuated flask with a measured volume of 5.0 l. when all the gas has been collected, the press the flask is measured to be 0.100 atm . calculate the mass and number of moles of dinitrogen monoxide gas that were collected. round your answer to 2 significant digits. mass: mole: nol explanation check 2019 mcgraw-hill education all rights reserved

0.971 grams

Explanation:

Given:

Temperature = 3.0° C = 3 + 273 = 276 K

Volume, V = 5.0 L

Pressure, P = 0.100 atm

Now, from the relation

PV = nRT

where,

n is the number of moles,

R is the ideal gas constant  = 0.082057 L atm/mol.K

thus,

0.1 × 5 = n ×  0.082057 × 276

or

n = 0.022 moles

Also,

Molar mass of the Dinitrogen monoxide gas (N₂O)

= 2 × Molar mass of nitrogen + 1 × Molar mass of oxygen

= 2 × 14 + 16 = 44 grams/mol

Therefore, Mass of 0.022 moles of N₂O = 0.022 × 44 = 0.971 grams

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

it takes on an electronical energy