Which equation is set up correctly to determine the volume of a 3.2 mole sample of oxygen gas at 50°c and 101.325 kpa? a. `"v" = (("3.2 mol")("8.314 l" * "kpa/k" * "mol")("323 k"))/("101.325 kpa")` b. `"v" = (("3.2 mol")("8.314 l" * "kpa/k" * "mol")("50"^@"c"))/("1 atm")` c. `"v" = (("8.314 l" * "kpa/k" * "mol")("50"^@"c"))/(("3.2 mol")("101.325 kpa"))` d. `"v" = (("3.2 mol")("8.314 l" * "kpa/k" * "mol"))/(("1 atm")("323 k"))`

Given that the molar mass of nano3  is 85.00  g/mol, what mass of nano3  is needed to make 4.50  l of a 1.50  m nano3solution? use  .6.75 g18.9 g255 g574 g

City a and city b had two different temperatures on a particular day. on that day, four times the temperature of city a was 8â° c more than 3 times the temperature of city b. the temperature of city a minus twice the temperature of city b was â’3â° c. what was the temperature of city a and city b on that day? city a was 5â° c, and city b was 4â° c. city a was 3â° c, and city b was â’1â° c. city a was 8â° c, and city b was â’3â° c. city a was 5â° c, and city b was â’5â° c.

Explanation for basic character?