Answer:
The correct answer is "1.0100".
Explanation:
Let the volume of mixture be 100 ml.
then,
The volume of DMSO will be 10 mL as well as that of water will be 90 mL.
DMSO will be:
= 
= 
The total mass of mixture will be:
= 
= 
Density of mixture will be:
= 
= 
= 
hence,
Specific gravity of mixture will be:
= 
= 
= 
We are given with a compound, Methane (CH4), with a molar
mass of 0.893 mol sample. We are tasked to solve for it's corresponding mass in
g. We need to solve first the molecular weight of Methane, that is
C=12 g/mol
H=1g/mol
CH4= 12 g/mol +1(4) g/mol = 16 g/mol
With 0.893 mol sample, its corresponding mass is
g CH4= 0.893 mol x 16g/mol =14.288 g
Therefore, the mass of methane is 14.288 g
Answer : Option B) The period 2 element would be more reactive because the attractive force of protons is stronger when electrons are attracted to a closer electron shell.
Explanation : The reactivity of the Periods decreases as we go from left to right across a period. The farther to the left and down the periodic chart we go, the easier it is for electrons to be donated or taken away, resulting in higher reactivities of the elements. The attractive force of the protons is found to be stronger when electrons are found to be attracted to a closer electron shell.
Number of moles of CO2 =
Mass /Ar
= 50.2 / (12 + 32)
1.14 mols
For every 1 mol of gas, there will be
24000 cm^3 of gas
Vol. = 1.14 x 24 dm^3
= 27.36 dm^3
The answer is 200 g.
If the molar mass of CaCl2 is 110.98 g/mol, this means there are 110.98 g in 1 L of 1 M solution.
Let's find how many g of CaCl2 are present in 0.720 M.
110.98 g : 1 M = x : 0.720 M
x = 110.98 g * 0.720 M : 1 M
x = 79.90 g
So there are 79.90 g in 0.720 M. In other words, in 1 L of 0.720 M solution there will be 79.90 g.
Now, we need to prepare ten beakers with 250 mL of solutions:
10 * 250 mL = 2500 mL = 2.5 L
79.90 g : 1 L = x : 2.5 L
x = 79.90 g * 2.5 L : 1 L
x = 199.75 g ≈ 200 g
