What mass of barium nitrate is dissolved in 21.29 mL of a 3.38 M solution?

18.81 g

Explanation:

We are given;

We are required to determine the mass of Barium nitrate

Step 1: Determine the number of moles of Barium nitrate

Number of moles = Molarity × Volume

                             = 3.38 M × 0.02129 L

                             = 0.07196 moles

Step 2: Determine the mass of Barium nitrate

We know that;

Mass = Number of moles × Molar mass

Molar mass of Barium Nitrate is 261.337 g/mol

Therefore;

Mass of barium nitrate = 0.07196 moles × 261.337 g/mol

                                     = 18.8058 g

                                     = 18.81 g

Thus, the mass of barium nitrate dissolved is 18.81 g

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

b. examine and describe cells