In ionic bonding, atoms SHARE ELECTRONS
1 g/L ------- 0.001 g/mL
150 g/L ----- ?
150 x 0.001 / 1
= 0.15 g/mL
Answer C
Answer:
The heat capacity of the calorimeter is 5.11 J/g°C
Explanation:
Step 1: Given data
50.0 mL of water with temperature of 80.0 °C
Specific heat capacity of water = 4.184 J/g°C
Consider the density of water = 1g/mL
50.0 mL of water in a calorimeter at 20.0 °C
Final temperature = 47.0 °C
Step 2: Calculate specific heat capacity of the water in calorimeter
Q = Q(cal) + Q(water)
Q(cal) = mass * C(cal) * ΔT
Qwater = mass * Cwater * ΔT
Qcal = -Qwater
mass(cal) * C(cal) * ΔT(cal) = mass(water) * C(water) * ΔT(water)
50 grams * C(cal) * (47.0 - 20.0) =- 50grams * 4.184 J/g°C * (47-80)
1350 * C(cal) = 6903.6
C(cal) = 5.11 J/g°C
The heat capacity of the calorimeter is 5.11 J/g°C
Answer : A
Explanation:
Answer A is correct I just did the assignment.
