What is the percent of nitrogen in dinitrogen pentoxide? A 74.1

B 62.8

C 44.4

D 25.9

D.) 25.9%

Explanation:

Dinitrogen pentoxide chemical formular : N2O5

Calculating the molar mass of N2O5

Atomic mass of nitrogen(N) = 14

Atomic mass of oxygen(O) = 16

Therefore molar mass :

N2O5 = (2 × 14) + (5 × 16) = 28 + 80 = 108g/mol

Percentage amount of elements in N205:

NITROGEN (N) :

(Mass of nitrogen / molar mass of N2O5) × 100%

MASS OF NITROGEN = (N2) = 2 × 14 = 28

PERCENT OF NITROGEN : (28/108) × 100%

0.259259 × 100%

= 25.925%

= 25.9%

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