Answer:
Percent yield of SiC is 77.0%.
Explanation:
Balanced reaction: 
Molar mass of SiC = 40.11 g/mol
Molar mass of 
= 60.08 g/mol
So, 100.0 kg of = 
moles of = 1664 moles of
According to balanced equation, 1 mol of produces 1 mol of SiC
Therefore, 1664 moles of produce 1664 moles of SiC
Mass of 1664 moles of SiC = 
= 66743g = 66.74 kg (4 sig. fig.)
Percent yield of SiC = [(actual yield of SiC)/(theoretical yield of SiC)]
%
= 
%
= 77.0%
The solution needed is prepared as below
by use of the M1V1 =M2 V2 formula where
M1 = 2.25 L
v2 = 1.0M
M2 = 9.0 M
V2 =? l
make V2 the subject of the formula V2 =M1V1/M2
= 2.25 L x 1.0M/9.0 M = 0. 25 L
therefore the solution need 0.25 L of 9.0M H3PO4 and dilute it a final volume of 2.25 l
Explanation:
The end goal of this question is to find density. Density can be found using the following formula:
D=m/v
D=Density
m=mass
v=volume
Therefore we need mass and volume to find density. Mass is already given in the question as 142g. All we have left is volume. When a solid is placed in water in a graduated cylinder the volume goes up. This means if you subtract the volume after the metal was placed from before it was placed, you can find the volume of the metal. Therefore the volume of the metal is 40mL-20mL=20mL. However typically, volume is found in liters so the volume should be .04L-.02L=0.02L. Now that we have volume and mass we can find density using the formula d=m/v. So 142/.02=7,100 grams per liget or 7.1 kilograms per liter. This answer seems unrealistic so I would double check with your teacher about the question.
Hope I helped, 2Trash4U
Answer:
4.96 mol/dm³
Explanation:
From the question,
Mass of NaCl that dissolved in 0.5L of water = 500-346.8 = 153.2 g.
Therefore, 145.2(1/0.5)g of NaCl will dissolve in 1 L of water
mass of NaCl that will dissolve in 1 L of water = 290.4 g/dm³
Molar mass of NaCl = 58.5 g/mol.
Solubility is the amount of substance in mol that will dissolve in 1 L or 1 dm³ Solution.
solubility in (mol/dm³) = solubility in (g/dm³)/molar mass.
solubility in (mol/dm³) = 290.4/58.5
solubility in (mol/dm³) = 4.96 mol/dm³
