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 : The excess reactant in the combustion of methane in opem atmosphere is
molecule.
Solution : Given,
Mass of methane = 23 g
Molar mass of methane = 16.04 g/mole
The Net balanced chemical reaction for combustion of methane is,

First we have to calculate the moles of methane.
=
= 1.434 moles
From the above chemical reaction, we conclude that
1 mole of methane react with the 2 moles of oxygen
and 1.434 moles of methane react to give
moles of oxygen
The Moles of oxygen = 2.868 moles
Now we conclude that the moles of oxygen are more than the moles of methane.
Therefore, the excess reactant in the combustion of methane in open atmosphere is
molecule.
Yes everything looks good to me
Answer:
Explanation:
D. When the truck driver slammed on the brakes, all the boxes in the back
of the truck slid forward.