Answer:
The body with 2.7 
has a larger volume
Explanation:
There is a missespelling in the question. The units of density cannot be g, they usually are
In order to answer this question, we need to use the formula of the density of a body:
For body A:
Mass (M) = 5.0 g
Density (D) = 2.7
2.7 = 5g * 1/V
V1 = 5g / (2.7 )
V1 = 1.85 
For Body B:
Mass (M) = 5.0 g
Density (D) = 8.4
8.4 = 5g * 1/V
V2 = 5g / (8.4 )
V2 = 0.59
So V1 is bigger than V2
Answer:
The energy released in the decay process = 18.63 keV
Explanation:
To solve this question, we have to calculate the binding energy of each isotope and then take the difference.
The mass of Tritium = 3.016049 amu.
So,the binding energy of Tritium = 3.016049 *931.494 MeV
= 2809.43155 MeV.
The mass of Helium 3 = 3.016029 amu.
So, the binding energy of Helium 3 = 3.016029 * 931.494 MeV
= 2809.41292 MeV.
The difference between the binding energy of Tritium and the binding energy of Helium is: 32809.43155 - 2809.412 = 0.01863 MeV
1 MeV = 1000keV.
Thus, 0.01863 MeV = 0.01863*1000keV = 18.63 keV.
So, the energy released in the decay process = 18.63 keV.
Answer: The change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizing compound is dissolved in it is 
Explanation:
Elevation in boiling point:
where,
= boiling point of solution = ?
= boiling point of pure carbon disulfide= 
= boiling point constant =
m = molality
i = Van't Hoff factor = 1 (for non-electrolyte)
= mass of solute = 35.0 g
= mass of solvent (carbon disulphide) = 397.7 g
= molar mass of solute = 70.0 g/mol
Now put all the given values in the above formula, we get:
Therefore, the change in boiling point is 
Answer:
There are twice the molar quantity of nitrogen atoms in nitrous oxide, i.e.
1×10−2⋅mol
Moles of nitrous oxide
=
0.217
⋅
g
44.013
⋅
g
⋅
m
o
l
−
1
=
5
×
10
−
3
⋅
m
o
l
.
Given the composition of
N
2
O
, there are this
2
×
5
×
10
−
3
⋅
m
o
l
of nitrogen atoms.
