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:
0.97 kg, 1.0kg
Explanation:
As you have mentioned in the question, to convert kilo-gram to grams, this conversion is used, 1 kilogram = 1,000 grams
Initially, when one has to convert from one unit to another, there is a lot of confusion. To avoid any such, multiply and divide the number with units to which it has to be converted.
Here,
970 g = 
This makes no difference to the number.
970 g = 
As, 1 kg= 1,000 g
970 g = 
970 g = 0.97 kg
Rounding this off to the nearest tenth of a kilogram.
The tenths place is immediately to the right of the decimal point.
Here, 9 is at the tenth's place. Round the tenth's place up if the digit at hundredth's place is greater than or equal to 5. Since, 7 > 5, up the number at tenth's place.
So, answer is 1.0 kg
<span>it is limited to situations that involve aqueous solutions of specific compounds
I hope this helps</span>
Enzymes affect the rate of the reaction in both the forward and reverse directions; the reaction proceeds faster because less energy is required for molecules to react when they collide. Thus, the rate constant (k) increases. Figure 3: Lowering the Activation Energy of a Reaction by a Catalyst.
<em>Best of luck,</em>
<em>-Squeak</em>
