Answer:
The answer is definitely D
Explanation:
2 elections will fill the first energy level
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
Answer:
Lithium hydroxide is a base.
Carbon dioxide is the anhydride of the carbonic acid, H₂CO₃.
Therefore, the reaction awaited is a typical neutralization reaction with the formation of a salt and water.
2LiOH + CO₂ → Li₂CO₃ + H₂O
So, 2*20 = 40 moles of LiOH react with 20 moles of CO₂.
Molar Mass of LiOH = 23.95 g/mol
So, 40 * 23.95 = 958 g
