Answer:
The volume of water to be added is 0.175 liters of water
Explanation:
The given concentration of the nitric acid = 55% (M/M)
The mass of the nitric acid solution = 100 gm
The concentration solution is to diluted to = 20% (M/M)
The 100 g 55%(M/M) nitric acid solution gives 55g nitric acid in 100 g of solution
Therefore, to have 20% (M/M) nitric acid solution with the 55 g nitric acid, we get
Let "x" represent the volume of the resulting solution, we have;
20% of x = 55 g of nitric acid
∴ 20/100 × x = 55 g
x = 55 g × 100/20 = 275 g
The mass of extra water to be added = The mass of the 20%(M/M) solution solution of nitric acid - The current mass of the 55%(M/M) solution of nitric acid
The mass of extra water to be added = 275 g - 100 g = 175 g
Volume = Mass/Density
The density of water ≈ 1 g/ml
∴ The volume of water to be added that gives 175 g of water = 175 g/(1 g/ml) = 175 ml. = 0.175 l
The volume of water to be added = 0.175 liters of water.
You HAVE TO know that molarity (M) tells you the number of moles of a solute per Liters of solution.
M=mol/L
0.3M = 2.7 mol/ Volume
Volume = 2.7 mol/3.0 L
Volume = 0.9 L
The pressure is 19.3 N/cm².
<em>p</em> = (28.0 lb/1 in²) × (4.45 N/1 lb) × (1 in²/6.45 cm²) = 19.3 N/cm²
Answer:
i think it is a Cobrotoxin
Answer:
1. 31.25 mL
2. 1.98 g/L
3. 0.45 g/mL
Explanation:
For each of the problems, you need to perform unit conversions. You need to use the information given to you to convert to a specific unit.
1. You need volume (mL). You have density (g/mL) and mass (g). Divide mass by density. You will cancel out mL and be left with g.
(50.0 g)/(1.60 g/mL) = 31.25 mL
2. You are given grams and liters. You need to find density with units g/L. This means that you have to divide grams by liters.
(0.891 g)/(0.450 L) = 1.98 g/L
3. You have to find density again but this time with units g/mL. Divide the given mass by the volume.
(10.0 g)/(22.0 mL) = 0.45 g/mL