Answer:
29.41% of Calcium and 47.04% of Oxygen
Explanation:
The percent composition of an atom in a molecule is defined as 100 times the ratio between the mass of the atom and the mass of the molecule.
The mass of the molecule of the problem (Ore) is 46.28g. That means the percent composition of Calcium is:
13.61g / 46.28g * 100 = 29.41% of Calcium
And percent composition of Oxygen is:
21.77g / 46.28g * 100 = 47.04% of Oxygen
Answer:
What are the statements please
Explanation:
Answer:
Mass = 141.6 g
Explanation:
Given data:
Mass of Kr in gram = ?
Volume in L = 9.59 L
Temperature = 46.0°C
Pressure = 4.62 atm
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will convert the temperature.
46.0+273 = 319 K
4.62 atm × 9.59 L = n× 0.0821 atm.L/ mol.K ×319 K
44.3 atm.L = n×26.19 atm.L/ mol
n = 44.3 atm.L / 26.19 atm.L/ mol
n = 1.69 mol
Mass in gram:
Mass = number of moles × molar mass
Mass = 1.69 mol × 83.79 g/mol
Mass = 141.6 g
Answer:
0.184 atm
Explanation:
The ideal gas equation is:
PV = nRT
Where<em> P</em> is the pressure, <em>V</em> is the volume, <em>n</em> is the number of moles, <em>R</em> the constant of the gases, and <em>T</em> the temperature.
So, the sample of N₂O₃ will only have its temperature doubled, with the same volume and the same number of moles. Temperature and pressure are directly related, so if one increases the other also increases, then the pressure must double to 0.092 atm.
The decomposition occurs:
N₂O₃(g) ⇄ NO₂(g) + NO(g)
So, 1 mol of N₂O₃ will produce 2 moles of the products (1 of each), the <em>n </em>will double. The volume and the temperature are now constants, and the pressure is directly proportional to the number of moles, so the pressure will double to 0.184 atm.
