Answer:
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Explanation:
<u>Step 1: </u>Data given
mass of water = 300 grams
initial temperature = 10°C
final temperature = 50°C
Temperature rise = 50 °C - 10 °C = 40 °C
Specific heat capacity of water = 4.184 J/g °C
<u>Step 2:</u> Calculate the heat
Q = m*c*ΔT
Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)
Q = 50208 Joule = 50.2 kJ
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
The answer is d I'm pretty sure because the second one and the first one don't make sense but I'm not 100 percent positive about it not being d ..... sorry if it's wrong.
I am unable to see the attachment pm me and i can help you
Answer:
The answer to your question is <u>111 g of CaCl₂</u>
Explanation:
Reaction
2HCl + CaCO₃ ⇒ CaCl₂ + CO₂ + H₂O
Process
1.- Calculate the molecular mass of Calcium carbonate and calcium chloride
CaCO₃ = (1 x 40) + (1 x 12) + ((16 x 3) = 100 g
CaCl₂ = (1 x 40) + (35.5 x 2) = 111 g
2.- Calculate the amount of calcium chloride produced using proportions.
The proportion CaCO₃ to CaCl₂ is 1 : 1.
100 g of CaCO₃ ------------- 111 g of CaCl₂
Then 111g of CaCl₂ will be produced.
