Answer : The mass of 
precipitate produced will be, 9.681 grams.
Explanation : Given,
Molarity of NaI = 0.210 M
Volume of solution = 0.2 L
Molar mass of = 461.01 g/mole
First we have to calculate the moles of 
.
Now we have to calculate the moles of .
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 moles of react to give 1 mole of
So, 0.042 moles of react to give 
moles of
Now we have to calculate the mass of .
Therefore, the mass of precipitate produced will be, 9.681 grams.
Answer:
0.100 M AlCl₃
Explanation:
The variation of boiling point by the addition of a nonvolatile solute is called ebullioscopy, and the temperature variation is calculated by:
ΔT = W.i
Where W = nsolute/msolvent, and i is the Van't Hoff factor. Because all the substances have the same molarity, n is equal for all of them.
i = final particles/initial particles
C₆H₁₂O₆ don't dissociate, so final particles = initial particles => i = 1;
AlCl₃ dissociates at Al⁺³ and 3Cl⁻, so has 4 final particles and 1 initial particle, i = 4/1 = 4;
NaCl dissociates at Na⁺ and Cl⁻ so has 2 final particles and 1 initial particle, i = 2/1 = 2;
MgCl₂ dissociates at Mg⁺² and 2Cl⁻, so has 3 final particles and 1 initial particle, i = 3/1 = 3.
So, the solution with AlCl₃ will have the highest ΔT, and because of that the highest boiling point.
Instead, they are created when magma cools and hardens —> the temperature contributes as the magma would be at extreme heat while it need a cooler temperature to harden and mold
Sodium phosphide is the inorganic compound with the formula Na3P.
Answer:
The answer is: 18 moles and 1341, 72 grams of KCl
Explanation:
The molarity is defined as the moles of solute ( in this case KCl) in 1 liter of solution:
1L solution-----3 moles of KCl
6L solution-----x= (6L solutionx 3 moles of KCl)/1 L solution= <em>18 moles of KCl</em>
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We calculate the weight of 1 mol of KCl:
Weight 1 mol KCl= Weight K + Weight Cl= 39,09 g + 35, 45 g=74, 54 g/mol
1 mol KCl----- 74, 54 g
18 mol KCl----x= (18 mol KCl x 74, 54 g)/1 mol KCl=<em>1341, 72 g</em>
