Answer:
4804.5 g of SO₂ are needed to the reaction
Explanation:
The reaction to produce sulfuric acid is:
2SO₂ + O₂ + 2H₂O → 2H₂SO₄
Ratio is 1:2. 1 mol of oxygen needs 2 moles of sulfur dioxide in order to react. We can propose this rule of three.
If 1 mol of O₂ react to 2 moles of SO₂
Then, 37.50 moles of O₂ will react with (37.5 . 2) /1 = 75 moles of SO₂
We convert the moles to mass, to know the answer:
75 mol . 64.06 g / 1 mol = 4804.5 g of SO₂
Answer:
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Answer:
This question is incomplete, here's the complete question:
<em><u>"Suppose 0.0842g of potassium sulfate is dissolved in 50.mL of a 52.0mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Round your answer to 2 significant digits."</u></em>
Explanation:
Reaction :-
K2SO4 + Na2CrO4 ------> K2CrO4 + Na2SO4
Mass of K2SO4 = 0.0842 g, Molar mass of K2SO4 = 174.26 g/mol
Number of moles of K2SO4 = 0.0842 g / 174.26 g/mol = 0.000483 mol
Concentration of Na2CrO4 = 52.0 mM = 52.0 * 10^-3 M = 0.052 mol/L
Volume of Na2CrO4 solution = 50.0 ml = 50 L / 1000 = 0.05 L
Number of moles of Na2CrO4 = 0.05 L * 0.052 mol/L = 0.0026 mol
Since number of moles of K2SO4 is smaller than number of moles Na2CrO4, so 0.000483 mol of K2SO4 will react with 0.000483 mol of Na2CrO4 will produce 0.000483 mol of K2CrO4.
0.000483 mol of K2CrO4 will dissociate into 2* 0.000483 mol of K^+
Final concentration of potassium cation
= (2*0.000483 mol) / 0.05 L = 0.02 mol/L = 0.02 M
Protons and neutrons are in the center of the atom, making up the nucleus. The charge on the proton and electron are exactly the same size but opposite. Neutrons have no charge.
Answer:
Explanation:
Pro – Low carbon. Unlike traditional fossil fuels like coal, nuclear power does not produce greenhouse gas emissions like methane and CO2. ...
Con – If it goes wrong… ...
Pro – Not intermittent. ...
Con – Nuclear waste. ...
Pro – Cheap to run. ...
Con – Expensive to build
