Explanation:
<u>Moles is denoted by given mass divided by the molecular mass , </u>
Hence ,
n = w / m
n = moles ,
w = given mass ,
m = molecular mass .
For example ,
For a compound X ,
The given mass i.e. w = 20 g
and the molecular mass ,i.e. , m = 10 g / mol
Then the moles can easily be calculated by using the above formula ,
n = w / m
n = 20 g / 10 g/mol = 2 mol
Hence , answer = 2 mol.
The number of energy levels to which an electron can jump depends on the amount of energy the electron possesses. Each energy level has a specific amount of energy an electron needs to have before it can be in there. So, if an electron doesn't have enough energy to be in that energy level then it won't jump to that higher level.
Volume of the tank is 5.5 litres.
Explanation:
mass of the CO2 is given 8.6 grams
Pressure of the gas is 89 Kilopascal which is 0.8762 atm
Temperature of the gas is 29 degrees ( 0 degrees +273.5= K) so (29+273)
R = gas constant 0.0821 liter atmosphere per kelvin)
FROM THE IDEAL GAS LAW
PV=nRT ( P Pressure, V Volume, n is number of moles of gas, R gas constant, Temperature in Kelvin)
no of moles = mass/atomic mass
= 8.6/44
= 0.195 moles
now putting the values in equation
V=nRT/P
= 0.195*0.0821*302/ 0.8762
= 5.5 litres.
As the carbon dioxide gas occupies the volume os the tank hence volume of tank is 5.5 litres.
Answer: Theoretical Yield = 0.2952 g
Percentage Yield = 75.3%
Explanation:
Calculation of limiting reactant:
n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol
pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol
- n-trans-cinnamic acid is the limiting reactant
The molar ratio according to the equation mentioned is equals to 1:1
The brominated product moles is also = 9.584*10⁻⁴ mol
Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)
= (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g
Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952
= 75.3%
Answer:

Explanation:
Data:
50/50 ethylene glycol (EG):water
V = 4.70 gal
ρ(EG) = 1.11 g/mL
ρ(water) = 0.988 g/mL
Calculations:
The formula for the boiling point elevation ΔTb is

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.
1. Moles of EG

2. Kilograms of water

3. Molal concentration of EG

4. Increase in boiling point

5. Boiling point