Part 1. a chemist reacted 15.0 liters of f2 gas with nacl in the laboratory to form cl2 and naf. use the ideal gas law equation to determine the mass of nacl that reacted with f2 at 280. k and 1.50 atm.

f2 + 2nacl → cl2 + 2naf

part 2. explain how you would determine the mass of sodium chloride that can react with the same volume of fluorine gas at stp.

1- 114.54 g.

2- 78.32 g.

Explanation:

Q1:

F₂ + 2NaCl → Cl₂ + 2NaF.

PV = nRT,

where, P is the pressure of the gas (P = 1.0 atm),

V is the volume of the gas (V = 15.0 L),

n is the no. of moles of the gas (n = ??? mole),

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas (T = 280.0 K).

∴ n = PV/RT = (1.5 atm)(15.0 L) / (0.082 L/atm/mol.K)(280.0 K) = 0.98 mol.

Using cross multiplication:

1.0 mole of F₂ reacts with → 2.0 moles of NaCl, from the stichiometry.

0.98 mole of F₂ reacts with → ??? moles of NaCl.

∴ The no, of moles of reacted NaCl = (2.0 mole)(0.98 mole)/(1.0 mole) = 1.96 mol.

∴ The mass of NaCl = n x molar mass of NaCl = (1.96 mol)(58.44 g/mol) = 114.54 g.

Q2:

∴ n = PV/RT = (1.0 atm)(15.0 L) / (0.082 L/atm/mol.K)(273.0 K) = 0.67 mol.

Using cross multiplication:

1.0 mole of F₂ reacts with → 2.0 moles of NaCl, from the stichiometry.

0.67 mole of F₂ reacts with → ??? moles of NaCl.

∴ The no, of moles of reacted NaCl = (2.0 mole)(0.67 mole)/(1.0 mole) = 1.34 mol.

∴ The mass of NaCl = n x molar mass of NaCl = (1.34 mol)(58.44 g/mol) = 78.32 g.

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)