You have a 5-liter container with 1.30 x 1024 molecules of ammonia gas (NH3) at STP. How many molecules of argon gas (Ar) are in an identical container at the same temperature and pressure?

1). 2.00 x 1023
molecules
2). 6.02 X 1024
molecules
3). 1.30 X 1024
molecules
4). 6.02 x 1023
molecules

3). 1.30 × 10^(24) molecules

Explanation:

From avogadro's law which state that equal volume of all gases at the same temperature and pressure contain the same number of molecules.

We can relate it to this question as;

V₁/n₁ = V₂/n₂

Where;

V₁ is initial volume

n₁ is initial number of molecules

V₂ is final volume

n₂ is final number of molecules

Thus at STP, we have V₁ = V₂ and as such Plugging in the relevant values gives;

5/(1.30 x 10^(24)) = 5/n₂

n₂ = 1.30 x 10^(24) molecules

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