Rolf prepares four solutions using different solutes as shown in the table below. Solution
Preparation
Solubility of the solutes
A
Rolf added 20 g of solute to 100 g The solute in solution A has a
of water and mixed the solution solubility of 37 g/100 g H20 at 20°C!
B
Rolf added 10 g of solute to 100 g The solute in solution B has a
of water and mixed the solution. solubility of 10 g/100 g H20 at 20°C.
С
Rolf added 65 g of solute to 100g The solute in solution C has a
of water, mixed the solution, and
then heated the solution,
solubility of 32 g/100 g H20 at 20°C
D
Rolf added 7 g of solute to 100 g
of water, mixed the solution, and
then heated the solution
The solute in solution D has a
solubility of 4 g/100 g H20 at 20°C.
Which solution is unsaturated?

by cartels in other countries, shipped by air, mail, or land. i cant really see your answer choices : /

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

answer; natural resources;

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