Adding 1.56 g of K2SO4 to 6.00 mL of water at 16.2ºC causes the temperature of the solution to drop by 7.70ºC. How many grams of NaOH (ΔHsoln = –44.3 kJ/mol) would you need to add to raise the temperature back to 16.2ºC?

You need to add 0.243g of NaOH to raise the temperature back to 16.2°C

Explanation:

Using the equation:

Q = C*m*ΔT

Where Q is heat

C is specific heat

m is mass

and ΔT is change in temperature

We can find the heat required to increase the temperature of the solution back to 16.2°C:

Assuming specific heat of the solution of water + K2SO4 = Specific heat of water:

C = 4.184J/g°C

m = 1.56g + 6.00g = 7.56g

ΔT = 16.2°C - 7.70°C = 8.50°C

Q = 4.184J/g°C * 7.56g * 8.50°C

Q = 268.86J = 0.269kJ of heat are required

As this heat is obtained from the dissolution of NaOH:

0.269kJ * (1mol NaOH / 44.3kJ) = 0.00607 moles of NaOH are required

In grams -Molar mass NaOH: 40g/mol-:

0.00607 moles NaOH * (40g / mol) =

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i