I think a type B fire extinguisher should be on board a vessel with a permanently installed fuel tank.
All vessels are required to have a type B fire extinguisher on board if one or more of the following conditions exists. That is; Inboard engine, vessel length of 26 feet or longer, enclosed living spaces, closed storage compartments in which flammable or combustible materials may be stored, permanently installed fuel tanks, and also closed compartments where portable fuel tanks may be stored.
Answer:

Explanation:
The molar mass is the mass of a substance in grams per mole.
To find it, add the mass of each element in the compound. These masses can be found on the Periodic Table.
The compound given is:

The compound has 1 Ca (calcium) and 2 Cl (chlorine).
Mass of Calcium
- The molar mass of calcium is 40.08 g/mol
- There is only one atom of Calcium in CaCl₂, so the number above is what we will use.
Mass of Chlorine
- The molar mass of chlorine is 35.45 g/mol
- There are two atoms of chlorine in CaCl₂, therefore we need to multiply the molar mass by 2.
- 35.45 * 2= 70.9 g/mol
Molar Mass of CaCl₂
- Now, to find the molar mass, add the molar mass of 1 calcium and 2 chlorine.
- 40.08 g/mol + 70.9 g/mol =110.98 g/mol
The molar mass of CaCl₂ is <u>110.98 grams per mole. </u>
Density= mass/volume
= 100/25
density = 4g/ml
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.