However when the solute dissolves in the water creating the solution, the volume increases. So 3 mol of solute in 2 L of water creates more than 2 L of solution.

The correct method for making a 3 mol/L solution would be to place some water into a two liter volume container. Dissolve all 3 mol of the solute into the water. Then add water to the 2 L mark. Now there is 3 mol of solute and 2 L of solution.

Explanation: I hope this helps XDDDD

8 0

3 years ago

When a pure substance was heated a chemical reaction occurred, and a student obtained another solid and a gas, each of which was

bija089 [108]

The original solid is not an element

8 0

4 years ago

180 N 3) 40 kg I 140 N Net Force

IgorC [24]

Yoo your andwrr would be 2945

5 0

3 years ago

9. 6.59 g of ammonia (NH3, 17.031 g/mol) react completely in the presence of excess carbon dioxide (CO2, 44.01 g/mol) according

Katarina [22]

Answer:

The percent yield for the reaction is 63.7%

Explanation:

This problem can be solved, from different rules of three.

The reaction is:

2NH₃(g) + CO₂(g) → CH₄N₂O(s) + H₂O(l)

2 moles of ammonia, produce 1 mol of urea.

Mass / molar mass = moles

6.59g / 17.031 g/m = 0.387 moles

As ratio is 2:1, I will produce half a mole of urea.

(0.387 m . 1m)/ 2m = 0.193 m

Let's find out the mass.

Mass of urea = Moles of urea . molar mass of urea.

Mass = 0.193m . 60.056 g/m = 11.6 g

If the percent yield for the reaction was 100 %, we make 11.6 g, but I only obtained 7.40 g so let's calculate the new percent yield.

11.6 g ____ 100 %

7.40 g ___ (7.40g / 11.6g) .100 = 63.7 %

6 0

4 years ago