Answer:
112.2L
Explanation:
Volume (V) = 300g
Temperature (T) = 822K
Pressure (P) = 0.9atm
using the ideal gas equation;
Molar gas constant (R) = 
Mole (n) = 
Molar mass of Mercury = 200.59g/mol
= 1.496mol
Now, the volume can be calculated;
V = 
∴Volume of mercury = 
Answer:
m= 4,599.145 g
Explanation:
Let m = mass, d = density and V = volume of the osmium block.
m = d x V
m = 22.610 g/cm3 x (6.70 x 9.20 x 3.3) cm3
m = 4,599.145 g
Answer:
6.05g
Explanation:
The reaction is given as;
Ethane + oxygen --> Carbon dioxide + water
2C2H6 + 7O2 --> 4CO2 + 6H2O
From the reaction above;
2 mol of ethane reacts with 7 mol of oxygen.
To proceed, we have to obtain the limiting reagent,
2,71g of ethane;
Number of moles = Mass / molar mass = 2.71 / 30 = 0.0903 mol
3.8g of oxygen;
Number of moles = Mass / molar mass = 3.8 / 16 = 0.2375 mol
If 0.0903 moles of ethane was used, it would require;
2 = 7
0.0903 = x
x = 0.31605 mol of oxygen needed
This means that oxygen is our limiting reagent.
From the reaction,
7 mol of oxygen yields 4 mol of carbon dioxide
0.2375 yields x?
7 = 4
0.2375 = x
x = 0.1357
Mass = Number of moles * Molar mass = 0.1357 * 44 = 6.05g
1)a. formation of gas
2)b.energy was transferred
3)c.substances that are used up in a reaction
Moles of gas = 0.369
<h3>Further explanation</h3>
Given
P = 2 atm
V = 5.3 L
T = 350 L
Required
moles of gas
Solution
Ideal gas Law
Avogadro's law : at the same temperature and pressure, the ratio of gas volume will be equal to the ratio of gas moles
moles of O₂ = 45% x 0.369 = 0.166
moles of Ar = 12% x 0.369 = 0.044
moles of N = 43% x 0.369 = 0.159
