Answer:
The freezing point will be 
Explanation:
The depression in freezing point is a colligative property.
It is related to molality as:
Where
Kf= 
the molality is calculated as:
Depression in freezing point = 
The new freezing point = 
<span>Answer:
.01 moles of D to .005 moles of L ~ so, .01+.005 = .015 total; using this total value, divide the portions of D and L.
so .01/.015 to .005/.015 ~ 67% D to 33% L.
And thus, the enantiomer excess will be 34%.</span>
Momentum = Mass x Velocity
80kg Runner: 80 x 2.45 = 196 Kg m/s
65kg Runner: 65 x 3= 195 Kg m/s
The 80kg runner has a greater momentum
Answer:
7.46 g
Explanation:
From the balanced equation, 2 moles of Mg is required for 2 moles of MgO.
The mole ratio is 1:1
mole = mass/molar mass
mole of 4.50 g Mg = 4.50/24.3 = 0.185 mole
0.185 mole Mg will tiled 0.185 MgO
Hence, theoretical yield of MgO in g
mass = mole x molar mass
0.185 x 40.3 = 7.46 g
<span>BaCl2+Na2SO4---->BaSO4+2NaCl
There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent?
"First convert grams into moles"
1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2
1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4
(7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2
"From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl"
The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.</span>
