Answer:
B
Explanation:
Shows actual sacrifice in order to train and be the best.
<u>Given:</u>
Mass of solvent water = 4.50 kg
Freezing point of the solution = -11 C
Freezing point depression constant = 1.86 C/m
<u>To determine:</u>
Moles of methanol to be added
<u>Explanation:</u>
The freezing point depression ΔTf is related to the molality m through the constant kf, as follows:
ΔTf = kf*m
where ΔTf = Freezing point of pure solvent (water) - Freezing pt of solution
ΔTf = 0 C - (-11.0 C) = 11.0 C
m = molality = moles of methanol/kg of water = moles of methanol/4.50 kg
11.0 = 1.86 * moles of methanol/4.50
moles of methanol = 26.613 moles
Ans: Thus around 26.6 moles of methanol should be added to 4.50 kg of water.
Answer:
Explanation:
Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts with oxygen. What is the limiting reactant in this experiment?
Mg + O2 → MgO (unbalanced)
first, balance the equation
2Mg +O2-------> 2MgO
two magnesium atoms react with one diatomic oxygen molecule
there is a 1:1 ratio of magnesium to oxygen atoms
but we have 2 moles of magnesium atoms and 2X5 = 10 moles of oxygen atoms
the lesser magnesium LIMITS the amount of product we can make, so it is the LIMITING REAGENT.
Answer:
The first one is 425g of copper
Explanation:
