<em>m Na₂CO₃: 23g×2 + 12g + 16g×3 = 106 g/mol</em>
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1 mol ------- 106g
X ------------ 10,6g
X = 10,6/106
<u>X = 0,1 mol Na₂CO₃</u>
Answer:
The molarity of urea in this solution is 6.39 M.
Explanation:
Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is
To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.
Our first step is to calculate the moles of urea in 100 grams of the solution,
using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is
60.06 g/mol ÷ 37.2 g = 0.619 mol
Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.
1.032 g/mL ÷ 100 g = 96.9 mL
This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.
0.619 mol/96.9 mL × 1000 mL= 6.39 M
Therefore, the molarity of the solution is 6.39 M.
Answer:
60 moles of NaF
Explanation:
The balanced equation for the reaction is given below:
Al(NO3)3 + 3NaF —> 3NaNO3 + AlF3
From the balanced equation above,
3 moles of NaF reacted to produce 1 mole of AlF3.
Therefore, Xmol of NaF will react to produce 20 moles of AlF3 i.e
Xmol of NaF = 3 x 20
Xmol of NaF = 60 moles
Therefore, 60 moles of NaF are required to produce 20 moles of AlF3.
The anwser is atoms are destroyed
<span>The correct answer is that an ionic bond forms between charged particles. To form this bond, the particles transfer valence electrons (those in the outermost orbit). Specifically, in ionic bonding, the metal atom loses its electrons (thus becoming positive) and the nonmetal atom gains electrons (thus becoming negative).</span>