The answer to your question is a
Answer:
you need to send us the figure
Explanation:
Answer:
The mass of water = 219.1 grams
Explanation:
Step 1: Data given
Mass of aluminium = 32.5 grams
specific heat capacity aluminium = 0.921 J/g°C
Temperature = 82.4 °C
Temperature of water = 22.3 °C
The final temperature = 24.2 °C
Step 2: Calculate the mass of water
Heat lost = heat gained
Qlost = -Qgained
Qaluminium = -Qwater
Q = m*c*ΔT
m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)
⇒with m(aluminium) = the mass of aluminium = 32.5 grams
⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C
⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C = -58.2 °C
⇒with m(water) = the mass of water = TO BE DETERMINED
⇒with c(water) = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C
32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9
-1742.1 = -7.95m
m = 219.1 grams
The mass of water = 219.1 grams
1. Make a Prediction
2. Fill both beakers with water
3. Dissolve salt in one of the beakers
4. Place both in the freezer and observe
5. Write a report
(Always make the prediction first! That's a hypothesis!)
Answer:
-154KJ/mol
Explanation:
mole of 100ml sample of 0.2M aqueous HCl = Molarity × volume in Liter
= 0.2 × 100 / 1000 ( 1L = 1000 ml) = 0.02 mol and 0.02 mole of HCl solution require 0.02 mole of ammonia according to the mole ratio in the balanced equation.
Heat loss by the reaction = heat gain by calorimeter = mcΔT + 480 J/K
where m is the mass of water = 100g + 100g = 200g since mass of 100ml of water = 100g and it is in both of them and specific heat capacity of water 4.184 J/gK
heat gain by calorimeter = (4.184 × 200 + 480) × 2.34 = 3081.3 J
ΔH per mole = heat loss / number of mole = 3081.3 / 0.02 = 154065.6 = -154KJ/mol
