Resistance of a material being scratched in known as: Hardness
Answer:
the final volume of the gas is = 1311.5 mL
Explanation:
Given that:
a sample gas has an initial volume of 61.5 mL
The workdone = 130.1 J
Pressure = 783 torr
The objective is to determine the final volume of the gas.
Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.
Converting the external pressure to atm ; we have
External Pressure :
The workdone W = V
The change in volume ΔV=
ΔV =
ΔV =
ΔV = 1.25 L
ΔV = 1250 mL
Recall that the initial volume = 61.5 mL
The change in volume V is
multiply through by (-), we have:
= 1250 mL + 61.5 mL
= 1311.5 mL
∴ the final volume of the gas is = 1311.5 mL
Atmospheric
pressure<span>, sometimes also called barometric pressure, is the pressure exerted by the weight of air in
the </span>atmosphere of Earth<span> (or that of another planet)</span>
1 atm is equivalent to = 101325
Pa
= 760 mmHg
= 760 torr
= 1.01325 bar
So 1.23 atm is equal to
= 124629.8 Pa
= 934.8 mmHg
= 934.8 torr
<span>= 1.2462 bar</span>
Answer:
The average atomic weight = 121.7598 amu
Explanation:
The average atomic weight of natural occurring antimony can be calculated as follows :
To calculate the average atomic mass the percentage abundance must be converted to decimal.
121 Sb has a percentage abundance of 57.21%, the decimal format will be
57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .
123 Sb has a percentage abundance of 42.79%, the decimal format will be
42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .
Next step is multiplying the fractional abundance to it masses
121 Sb = 0.5721 × 120.904 = 69.169178400
123 Sb = 0.4279 × 122.904 = 52.590621600
The final step is adding the value to get the average atomic weight.
69.169178400 + 52.590621600 = 121.7598 amu