Answer:
454.3 g.
Explanation:
1.0 mol of CaO liberates → – 64.8 kJ.
??? mol of CaO liberates → - 525 kJ.
∴ The no. of moles needed = (1.0 mol)(- 525 kJ)/(- 64.8 kJ) = 8.1 mol.
<em>∴ The no. of grams of CaO needed = no. of moles x molar mass</em> = (8.1 mol)(56.077 g/mol) = <em>454.3 g.</em>
<u>Answer:</u>
211.9 J
<u>Explanation:</u>
The molecules of water release heat during the transition of water vapor to liquid water, but the temperature of the water does not change with it.
The amount of heat released can be represented by the formula:
where 
= heat energy, 
= mass of water and 
= latent heat of evaporation.
The latent heat of evaporation for water is 
and the mass of the water is 
.
The amount of heat released in this process is:
211.9 J
10 x 70 = 100 x Part, or
700 = 100 x Part
Now, divide by 100 and get the answer:
Part = 700 / 100 = <span>7</span>
The final temperature of the water will be 31.2 °C... i don’t know the second one sorry :(
Answer:
2.57 g of H₂
Solution:
The Balance Chemical Equation is as follow,
N₂ + 3 H₂ → 2 NH₃
According to Balance equation,
34.06 g (2 moles) NH₃ is produced by = 6.04 g (3 moles) of H₂
So,
14.51 g of NH₃ will be produced by = X g of H₂
Solving for X,
X = (14.51 g × 6.04 g) ÷ 34.06 g
X = 2.57 g of H₂
