H = 1 amu
P = 31 amu
O = 16 amu
Therefore:
H3PO4 = 1 x 3 + 31 + 16 x 4 => 98 u
hope this helps!
2.a)R b)R c)L d)L e)R f)L g)L
Answer:
Mass = 1.33 g
Explanation:
Given data:
Mass of argon required = ?
Volume of bulb = 0.745 L
Temperature and pressure = standard
Solution:
We will calculate the number of moles of argon first.
Formula:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
By putting values,
1 atm ×0.745 L = n × 0.0821 atm.L/mol.K× 273.15 K
0.745 atm. L = n × 22.43 atm.L/mol
n = 0.745 atm. L / 22.43 atm.L/mol
n = 0.0332 mol
Mass of argon:
Mass = number of moles × molar mass
Mass = 0.0332 mol × 39.95 g/mol
Mass = 1.33 g
Is bubble chamber one of your choices? Bubble chamber sounds like a good fit for the question.
