Answer:
The atomic mass of second isotope is 7.016
Explanation:
Given data:
Average Atomic mass of lithium = 6.941 amu
Atomic mass of first isotope = 6.015 amu
Relative abundance of first isotope = 7.49%
Abundance of second isotope = ?
Atomic mass of other isotope = ?
Solution:
Total abundance = 100%
100 - 7.49 = 92.51%
percentage abundance of second isotope = 92.51%
Now we will calculate the mass if second isotope.
Average atomic mass of lithium = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
6.941 = (6.015×7.49)+(x×92.51) /100
6.941 = 45.05235 + (x92.51) / 100
6.941×100 = 45.05235 + (x92.51)
694.1 - 45.05235 = (x92.51)
649.04765 = x
92.51
x = 485.583 /92.51
x = 7.016
The atomic mass of second isotope is 7.016
Answer:
The bond dissociation energy to break 4 bonds in 1 mol of CH is 1644 kJ
Explanation:
Since there are 4 C-H bonds in CH₄, the bond dissociation energy of 1 mol of CH₄ is 4 × bond dissociation energy of one C-H bond.
From the table one mole is C-H bond requires 411 kJ, that is 411 kJ/mol. Therefore, 4 C-H bonds would require 4 × 411 kJ = 1644 kJ
So, the bond dissociation energy to break 4 bonds in 1 mol of CH₄ is 1644 kJ
Answer : The exit temperature of the product is, 
Explanation :
Total heat = Heat lost by liquid + Latent heat of fusion + Heat lost by frozen
where,
Q = Total heat = 6000 kJ
m = mass of product = 15 kg
= specific heat of liquid = 
= latent heat of fusion = 
= specific heat of frozen = 
= initial temperature of liquid = 
= final temperature of liquid = 
= initial temperature of frozen = ?
= final temperature of frozen =
Now put all the given value in the above expression, we get:
![6000kJ=[15kg\times 4kJ/kg^oC\times (10-2)^oC]+[15kg\times 275kJ/kg]+[15kg\times 2.5kJ/kg^oC\times (2-T_3)^oC]](https://tex.z-dn.net/?f=6000kJ%3D%5B15kg%5Ctimes%204kJ%2Fkg%5EoC%5Ctimes%20%2810-2%29%5EoC%5D%2B%5B15kg%5Ctimes%20275kJ%2Fkg%5D%2B%5B15kg%5Ctimes%202.5kJ%2Fkg%5EoC%5Ctimes%20%282-T_3%29%5EoC%5D)
Thus, the exit temperature of the product is,
<span>1 mole of benzene (78g) requires 30.8 kJ/ of heat, so 11.5g will need ..... (it's a proportion calculation.) Temperature does not change at BPt and is not relevant if the temp of the liquid is already at the BPt
ne definition of entropy is qrev/T, where qrev is the heat added in reversible operation (for complicated reasons pertaining to heat as a path function) and T is the temperature at which this is done.
Phase changes are particularly good examples for calculations of changes in entropy, since temperature will not change will the bonds of a state are being broken.
The calculations required boils down to:
1) finding the moles of benzene given from molar mass.
2) multiplying that moles by the heat of vaporization.
3)diving the heat energy required by the temperature of boiling point.</span>
<span>Answer</span>=.000978802802moles H2SO4
How I Got My Answer
<span>Molar mass
</span>H2SO4= 98.079g/mol
What I have
.0960g H2SO4
Equation
.0960g*1mol/98.079g= .000978802802mol H2SO4
