Divide that my the molar mass which is 23 so 1.4087 g
Answer:
147.2g
Explanation:
The full solution can be found in the image attached. Graham's law was applied to the problem. The rate of diffusion of a gas is inversely proportional to its molar mass or vapour density. Molar mass= 2vapour density
Answer: See below
Explanation:
1. a) 0.15 moles calcium carbonate (15g/100g/mole)
b) 0.15 moles CaO (molar ratio of CaO to CaCO3 is 1:1)
c) 8.4 grams CaO (0.15 moles)*(56 grams/mole)
2. a) 0.274 moles Na2O (17g/62 grams/mole)
b) 46.6 grams NaNO3 (2 moles NaNO3/1 mole Na2O)*(0.274 moles Na2O)*(85 g/mole NaNO3)
Explanation:
The pH of a weak base falls somewhere between 7 and 10.
382.85 Celsius is the temperature does 0.750 moles of an ideal gas occupy a volume of 35.9 L at 114 kPa.
Explanation:
Given data:
number of moles of the gas = 0.75 moles
volume of the gas = 35.9 liters
pressure of the gas = 114 KPa or 1.125 atm
R = 0.0821 latm/moleK
temperature of the gas T = ?
The equation used to calculate temperature from above data is ideal gas law equation.
the equation is :
PV = nRT
T = 
Putting the values in the above rewritten equation:
T = 
T = 655.9 K
To convert kelvin into celsius, formula used is
K = 273.15+ C
putting the values in the equation
C = 656 - 273.15
= 382.85 Celsius