Because there are a lot of transition metals.
X=240 g NaCl salt dissolves in solution.
Example: If we add 68 g sugar and 272 g water to 160 g solution having concentration 20 %, find final concentration of this solution.
Solution:
Mass of solution is 160 g before addition sugar and water.
100 g solution includes 20 g sugar
160 g solution includes X g sugar
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X=32 g sugar
Mass of solute after addition=32 + 68=100 g sugar
Mass of solution after addition=272 +68 + 160=500 g
500 g solution includes 100 g sugar
100 g solution includes X g sugar
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X= 20 % is concentration of final solution.
The reported number of moles will be too low since the additional amount of oxygen is not included. It may be 6 drops but the molarity of the solution is 6M. this meas that for every ml of the JNO3 solution there is 6 moles of JNO3. In every mole of <span>JNO3 there are 3 moles of oxygen. So in every drop of oxygen there are 18 moles of oxygen. </span>
84.24 g of water (H₂O)
Explanation:
We have the following chemical reaction:
2 H₂O → 2 H₂ + O₂
Now we calculate the number of moles of products.
number of moles = mass / molar weight
number of moles of H₂ = 50 / 2 = 25 moles
number of moles of O₂ = 75 / 32 = 2.34 moles
We see from the chemical reaction that for every 2 moles of H₂ produced there are 1 mole of O₂ produces for every 25 moles of H₂ produced there are 12.5 moles of O₂ but we only have 2.35 moles of O₂ available. The O₂ will be the limiting quantity from which we devise the following reasoning:
if 2 moles of H₂O produces 1 mole of O₂
then X moles of H₂O produces 2.34 mole of O₂
X = (2 × 2.34) / 1 = 4.68 moles of H₂O
mass = number of moles × molar weight
mass of H₂O = 4.68 × 18 = 84.24 g
Learn more about:
limiting reactant
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