If a gas in a closed container is pressurized from 14.9 atm to 16.5 atm and its original temperature was 33.1 °C, what would the final temperature of the gas be?

339K

Explanation:

Initial pressure P1 = 14.9 atm

Final pressure P2 = 16.5 atm

Original temperature T1 = 33.1 °C

Convert Celsius to Kelvin

(33.1°C + 273 = 306.1K)

Final temperature T2 = ?

Apply the formula for Pressure's law

P1/T1.= P2/T2

14.9atm/306.1K = 16.5atm/ T2

To get T2, cross multiply

14.9 x T2 = 16.5 x 306.1

14.9T2 = 5050.65

T2 = (5050.65/14.9)

T2 = 339K

Thus, the final temperature of the gas would be 339K

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.