Answer: The volume of 0.640 grams of
gas at Standard Temperature and Pressure (STP) is 0.449 L.
Explanation:
Given: Mass of
gas = 0.640 g
Pressure = 1.0 atm
Temperature = 273 K
As number of moles is the mass of substance divided by its molar mass.
So, moles of
(molar mass = 32.0 g/mol) is as follows.

Now, ideal gas equation is used to calculate the volume as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.

Thus, we can conclude that the volume of 0.640 grams of
gas at Standard Temperature and Pressure (STP) is 0.449 L.
Just multiply 151 x 0.0001
That will equal:
0.00151 km or if you want to round it then it would be 0.02km
Answer:
Q = 90,000 J
Explanation:
Given data:
Mass skillet = 2000 g
Specific heat capacity = 0.450 J/g.°C
Energy required to raise temperature = ?
Initial temperature = 25°C
Final temperature = 125°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 125°C - 25°C
ΔT = 100°C
Q = 2000 g × 0.450 J/g.°C × 100°C
Q = 90,000 J