A 5.20 mol sample of solid A was placed in a sealed 1.00 L container and allowed to decompose into gaseous B and C. The concentration of B steadily increased until it reached 1.30 M, where it remained constant. A(s)↽−−⇀B(g)+C(g) A ( s ) ↽ − − ⇀ B ( g ) + C ( g ) Then, the container volume was doubled and equilibrium was re‑established. How many moles of A A remain?

2.60 moles of A remaining.

Explanation:

According to  Le Chatelier's principle, the equilibrium would shift if the volume, concentration, pressure, or temperature changes.

In this question, we were told that  the volume doubles, that implies that we would have to double the molarity of B/ C (since B=C.)

However, it is obvious and clear from the given equation of the reaction that  A is  solid in it's activity = 1. Hence, it is then ignored.

So doubling B would be 1.30 M × 2 = 2.60 M

i.e 2.60 M moles of A was consumed.

Now;  the number of moles of A remaining  is 5.20 - 2.60 = 2.60 moles of A remaining.

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

a polar bond is formed when electrons are shared unevenly between atoms