Answer:
1.332 g.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- At the same T and P and constant V (1.0 L), different gases have the same no. of moles (n):
<em>∴ (n) of CO₂ = (n) of C₂H₆</em>
<em></em>
∵ n = mass/molar mass
<em>∴ (mass/molar mass) of CO₂ = (mass/molar mass) of C₂H₆</em>
mass of CO₂ = 1.95 g, molar mass of CO₂ = 44.01 g/mol.
mass of C₂H₆ = ??? g, molar mass of C₂H₆ = 30.07 g/mol.
<em>∴ mass of C₂H₆ = [(mass/molar mass) of CO₂]*(molar mass) of C₂H₆</em> = [(1.95 g / 44.01 g/mol)] * (30.07 g/mol) =<em> 1.332 g.</em>
<em></em>
Answer:
The molarity of urea in this solution is 6.39 M.
Explanation:
Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is
To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.
Our first step is to calculate the moles of urea in 100 grams of the solution,
using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is
60.06 g/mol ÷ 37.2 g = 0.619 mol
Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.
1.032 g/mL ÷ 100 g = 96.9 mL
This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.
0.619 mol/96.9 mL × 1000 mL= 6.39 M
Therefore, the molarity of the solution is 6.39 M.
Answer:
true im pretttyyy sure
Explanation:
bcuz the stronger the intermolecular forces the higher the boiling point :3
So this can explain the density of a person very clearly.
Answer:
covalent bond
Explanation:
a covalent bond forms when electrons are shared between two nonmetals
plz mark as the brainliest