Let's say that you were given the following problem:
A gas that exerts a pressure of 151 mm Hg in a
container with a volume of L will exert a
pressure of 166 mm Hg when transferred to a
container with a volume of 0.532 L.
Let's begin by assigning our variables. Let's say
that P1 = 151 mm Hg. What is V2?

V2~0.4839M

Explanation:

We're going to use Boyles law to answer the question.

Boyle's law:

P1V1=P2V2

P1=151mmHg

P2=166mmHg

V1=0.532L

V2=?

V2=(P1 x V1)/P2

V2=(151 x 0.532)/166

V2~0.4839M

Hope it helps:)

oh it is true if thats what you are asking

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