the number of students in each group
Answer:
0.97 atm.
Explanation:
From the question given above, the following data were obtained:
Final pressure (P2) = 6.8 atm
Initial temperature (T1) = 40 K
Final temperature (T2) = 280 K
Initial pressure (P1) =?
Thus, we can obtain the initial (original) pressure of the gas as follow:
P1/T1 = P2/T2
P1 /40 = 6.8/280
Cross multiply
P1 × 280 = 40 × 6.8
P1 × 280 = 272
Divide both side by 280
P1 = 272/280
P1 = 0.97 atm
Therefore, the original pressure of the gas is 0.97 atm.
<span>The pointer will be above the zero mark</span>
1 mole of NH42SO4 contains 42 moles of H. We take the 42 moles of H and multiply that times Avogadro's number (6.02x10^23) and get 2.5284x10^25 atoms of hydrogen
22.6 x 10^5 J
‘Heat of vaporization’ is the amount of ‘heat’ required to convert '1g' of mass of a liquid to vapor. Normal “boiling point of water” is 100 C. The heat of vaporization at this temperature is 2260 J/g. It means 2260 J/g of heat is required to convert ‘1g of water’ to 1g of vapor at 100 C. So to convert 1 kg of mass of water or 1000g of water we require 22.6 x 10^5 J of heat.