Answer:
13.687 grams of salt should be added
The total grams of 8.6% of brine solution produced is 852.687g
Explanation:
Solution mass= 962g
Salt= 6.2%
Water = 93.8%
962 gram of water is made up of:
902.356g ( due to vaporization which reduces mass)
= 902.356 - 123
= 779. 356g of water
59.644g of salt.
If we add x gram of salt for making brine solution up to 8.6%
=(59. 644g + x)g.of salt
% salt = Mass of Salt / Total mass of solution
= 0.086= 59.644 + x / 779.356 + 59.644 + x
= 59.644 + x / 839 + x
x= 13.687 g of salt
Grams of 8.6% brine solution will be:
Gram of water + total gram of salt added to form 8.6% brine solution.
= 779.356g +59.644g + 13.687g
= 852.687g
The total grams of 8.6% of brine solution produced is 852.687g
Dilution<span> is when you decrease the concentration of a </span>solution<span> by adding a solvent. As a result, if you want to </span>dilute<span> salt water, just add water. ... Add more solute until it quits dissolving. That point at which a solute quits dissolving is the point at which it's </span>saturated<span>.</span>
The correct reaction equation is:
Answer:
b) 1 mole of water is produced for every mole of carbon dioxide produced.
Explanation: <u>CONVERT EVERYTHING TO MOLES OR VOLUME, THEN COMPARE IT WITH THE COMPOUND'S STOICHIOMETRY IN CHEMICAL EQUATION.</u>
a) <u>22.4 L of 
gas</u> is produced only when <u>
L of 
</u> is reacted with 22.4 L of 
. So it is wrong.
b) Since in the chemical equation the stoichiometric coefficient of 
and 
are same so the number of moles or volume of each of them will be same whatever the amount of reactants taken. <u>Therefore it is correct option.</u>
c) 
molecules is equal 1 mole of 
if produced then 3 moles of is required, which is not given in the option. So it is wrong.
d) 54 g of water or 3 moles of (<em>Molecular Weight of water is 18 g</em>) is produced when 3 moles of is used but in this option only one mole of is given. So it is wrong.
The answers would be organisms in water, water, and air.
Answer:
180.56 Kilo joules of energy is removed in the form of heat when 1.00 kg of freon-11 is evaporated.
Explanation:
Molar mass of freon-11 = 137.35 g/mol
Enthalpy of vaporization of freon-11= 
Mass of freon-11 evaporated = 1.00 kg = 1000 g
Moles of freon-11 evaporated = 
Energy in the form of heat removed when 1.00 kg of freon-11 gets evaporated:
