100. cal of heat are added to 18.0 g of ethanol (0.581 cal/g °C) originally at 23 °C. The final temperature is .

Final temperature is 32.56 °C

Explanation:

The specific heat of a substance is the amount of heat required to raise the temperature of 1g of the substance by 1°C.

The following equation/formula is used;

Q = m × c × ΔT

Where; Q= amount of heat supplied

(cal)

M= mass of ethanol (g)

C= specific heat of ethanol

(cal/g °C)

ΔT= change in temperature (°C)

i.e. (final temperature - initial

temperature)

According to the question, Q= 100 calories (cal), M= 18g, C= 0.581 cal/g °C, initial temperature = 23°C, final temperature = ?

Hence, we insert our values into the equation;

Q = m × c × ΔT

ΔT = Q/mc

(Final T - Initial T) = Q/mc

(Final T - 23) = 100/ 18 × 0.581

(Final T - 23) = 100/10.458

Final T - 23 = 9.562

Final T = 23 + 9.562

Final T = 32.562

Hence, the final temperature of ethanol is 32.56°C

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"