19 What is the number of moles of nitrogen gas which react with 18 g of magnesium to form magnesium nitride compound ? [Mg = 24,N=14] a) 0.25 mol 0 b) 0.5 mol C) 1 mol d) 2 mol

0.5 mol of nitrogen gas.

Explanation:

18g of Magnesium in mol is equal to 18g/24 gmol-1 = 0.75 mol and 7g of nitrogen in mol is equal to 7g/14 gmol-1 = 0.5 mol,

Thus, When 18g of magnesium reacts with 7g of nitrogen it implies 0.75 mol of magnesium which reacts with 0.5 mol of nitrogen. :)

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

1.43 x 10^-11 j the guy commented the awnser just making sure u see it .