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) =

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: