Answer:
Round to the number of significant figures in the original question. However, if you're going to proceed with further calculations using this mass, it's best not to round, as rounding will cause your answer to be less precise.
Explanation:
1245 liters because you move the decimal place over to the right six times
Answer:
(a) the mass of the water is 3704 g
(b) the mass of the water is 199, 285.7 g
Explanation:
Given;
Quantity of heat, H= 8.37 x 10⁶ J
Part (a) mass of water (as sweat) need to evaporate to cool that person off
Latent heat of vaporization of water, Lvap. = 2.26 x 10⁶ J/kg
H = m x Lvap.

mass in gram ⇒ 3.704 kg x 1000g = 3704 g
Part (b) quantity of water raised from 25.0 °C to 35.0 °C by 8.37 x 10⁶ J
specific heat capacity of water, C, 4200 J/kg.°C
H = mcΔθ
where;
Δθ is the change in temperature = 35 - 25 = 10°C

mass in gram ⇒ 199.2857 kg x 1000 g = 199285.7 g
Out of the five major mass extinction, one is Cretaceous-Tertiary extinction (K-T extinction). It happened 66 million years ago marked by the end of the Cretinous period.
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.