Liquids and solids are left out of the equilibrium constant expression because their concentrations remain constant during reactions. what is the molarity concentration of liquid water at 25.0 latex: ^\circ ∘c given that its density is 0.997 g/ml at that temperature?
a.23.5 m
b.0.997 m
c.0.180 m
d.0.156 m
e.55.3 m

55.3 M

Explanation:

Given temperature = 25°C

Density = 0.997 g/mL

Based on density the mass of water per L = 997 grams

The molar mass of water = 18

The moles of water =

The molarity is defined as the moles of water per litre.

The moles = 55.3

Volume = 1 L

So molarity = 55.3 /1 = 55.3 M

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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