Correct answer: a. releasing CO2 that dissolves and forms acid in the oceans
The fuels used in automobiles release gases like carbon dioxide, carbon monoxide, oxides of nitrogen and sulfur. Carbon dioxide when dissolved in water forms carbonic acid. So, when the usage of cars is high, these emissions of carbon-dioxide into the atmosphere increase and this leads to the lowering of pH of the oceans as the carbon dioxide present in higher amounts in to atmosphere diffuses into the oceanic waters and form carbonic acid which makes the ocean slightly acidic.
<span>100.
ppb of chcl3 in drinking water means 100 g of CHCl3 in 1,000,0000,000 g of water
Molarity, M
M = number of moles of solute / volume of solution in liters
number of moles of solute = mass of CHCl3 / molar mass of CHCl3
molar mass of CHCl3 = 119.37 g/mol
number of moles of solute = 100 g / 119.37 g/mol = 0.838 mol
using density of water = 1 g/ ml => 1,000,000,000 g = 1,000,000 liters
M = 0.838 / 1,000,000 = 8.38 * 10^ - 7 M <----- answer
Molality, m
m = number of moles of solute / kg of solvent
number of moles of solute = 0.838
kg of solvent = kg of water = 1,000,000 kg
m = 0.838 moles / 1,000,000 kg = 8.38 * 10^ - 7 m <----- answer
mole fraction of solute, X solute
X solute = number of moles of solute / number of moles of solution
number of moles of solute = 0.838
number of moles of solution = number of moles of solute + number of moles of solvent
number of moles of solvent = mass of water / molar mass of water = 1,000,000,000 g / 18.01528 g/mol = 55,508,435 moles
number of moles of solution = 0.838 moles + 55,508,435 moles = 55,508,436 moles
X solute = 0.838 / 55,508,435 = 1.51 * 10 ^ - 8 <------ answer
mass percent, %
% = (mass of solute / mass of solution) * 100 = (100g / 1,000,000,100 g) * 100 =
% = 10 ^ - 6 % <------- answer
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The balanced chemical reaction is written as:
4Al + 3O2 = 2Al2O3
To determine the mass of oxygen gas that would react with the given amount of aluminum metal, we use the initial amount and relate this amount to the ratio of the substances from the chemical reaction. We do as follows:
moles Al = 16.4 g ( 1 mol / 26.98 g ) = 0.61 mol Al
moles O2 = 0.61 mol Al ( 3 mol O2 / 4 mol Al ) = 0.46 mol O2
mass O2 = 0.46 mol O2 ( 32.0 g / mol ) = 14.59 g O2
Therefore, to completely react 16.4 grams of aluminum metal we need a minimum of 14.59 grams of oxygen gas.
