Answer: 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether
Explanation:
First we have to calculate the moles of diethyl ether
As, 1 mole of diethyl ether require heat = 26.5 kJ
So, 1.01 moles of diethyl ether require heat = 
Thus 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether
Nitrogen — 78 percent.
Oxygen — 21 percent.
Argon — 0.93 percent.
Carbon dioxide — 0.04 percent.
Trace amounts of neon, helium, methane, krypton and hydrogen, as well as water vapor.
I’m pretty sure it’s D because energy measures the kinetic, thermal, electrical, chemical, nuclear and other forms.
Answer:
48.75 g of AgCl
11.60 g of H₂S
Solution:
The Balance Chemical Equation is as follow,
Ag₂S + HCl → AgCl + H₂S
<u>Calculate amount of AgCl produced</u><u>,</u>
According to equation,
247.8 g (1 mol) of Ag₂S produces = 143.32 g (1 mol) of AgCl
So,
84.3 g of Ag₂S will produce = X g of AgCl
Solving for X,
X = (84.3 g × 143.32 g) ÷ 247.8 g
X = 48.75 g of AgCl
<u>Calculate amount of H</u><u>₂</u><u>S produced</u><u>,</u>
According to equation,
247.8 g (1 mol) of Ag₂S produces = 34.1 g (1 mol) of H₂S
So,
84.3 g of Ag₂S will produce = X g of H₂S
Solving for X,
X = (84.3 g × 34.1 g) ÷ 247.8 g
X = 11.60 g of H₂S
Rock formed by heat and pressure
