The answer for the following problem is mentioned below.
- <u><em>Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule </em></u>
Explanation:
Given:
no of moles of the oxygen gas = 0.692
Also given:
2 HgO → 2 Hg + 
where,
HgO represents mercuric oxide
Hg represents mercury
represents oxygen
To calculate:
Molar mass of HgO:
Molar mass of HgO = 216 grams
molar mass of mercury (Hg) = 200 grams
molar mass of oxygen (O) =16 grams
HgO = 200 +16 = 216 grams
We know;
2×216 grams of HgO → 1 mole of oxygen molecule
? → 0.692 moles of oxygen molecule
= 
= 298.944 grams of HgO
<u><em>Therefore 298.44 grams of mercuric oxide is needed to produce 0.692 moles of oxygen molecule </em></u>
<u />
It is avogrado number. One molecue of magnesium has 6.023 x 10^23 atoms
Explanation:
Initial Pressure = 24 lb in-2
Initial Temperature = –5 o C = 268 K (Converting to kelvin temperature)
Final Pressure = ?
Final Temperature = 35 o C = 308 K (Converting to kelvin temperature)
No Change in Volume.
From Gay Lusaac's law; pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.
P1T1 = P2T2
P2 = P1T1 / T2
P2 = 24 * 268 / 308 = 20.88 lb in-2
There would be a drop in pressure as the temperature increases. Appropriate measures should b taken by regularly gauging the pressure of the tire.
Answer:
397 L
Explanation:
Recall the ideal gas law:
If temperature and pressure stays constant, we can rearrange all constant variables onto one side of the equation:
The left-hand side is simply some constant. Hence, we can write that:
Substitute in known values:
Solving for <em>V</em>₂ yields:
In conclusion, 13.15 moles of argon will occupy 397* L under the same temperature and pressure.
(Assuming 100 L has three significant figures.)
