Potassium 23.5g/39.0983g/mol = 0.601mol
The Ratio of reactants is 2 to 1 so (0.601mol)/2 = 0.3005mol
Therefore 0.3005mol of F2 is needed to find liters use
formula V = nRT/P (V)Volume = 22.41L
(T)Temperature = 273K or 0.0 Celsius
(P)Pressure = 1.0atm
<span>(R)value is always .08206 with atm n = 0.3005moles
(273)(.08206)(0.3005)/1 = V V = 6.7319 Liters</span>
Answer:
Mass = 0.37 g
Explanation:
Given data:
Number of moles of sulfur = 11.9 mol
Mass of sulfur in 11.9 mol = ?
Molar mass of sulfur = 32.06 g
Solution:
Number of moles = mass/molar mass
by putting values,
11.9 mol = mass/ 32.06 g/mol
Mass = 11.9 mol × 32.06 g/mol
Mass = 0.37 g
The answer would be D. because when an acid and base mix the start to cancel each other out causing it to neutralize
Answer:
19 g
Explanation:
Data Given:
Sodium Chloride (table salt) = 50 g
Amount of sodium (Na) = ?
Solution:
Molecular weight calculation:
NaCl = 23 + 35.5
NaCl = 58.5 g/mol
Mass contributed by Sodium = 23 g
calculate the mole percent composition of sodium (Na) in sodium Chloride.
Since the percentage of compound is 100
So,
Percent of sodium (Na) = 23 / 58.5 x 100
Percent of sodium (Na) = 39.3 %
It means that for ever gram of sodium chloride there is 0.393 g of Na is present.
So,
for the 50 grams of table salt (NaCl) the mass of Na will be
mass of sodium (Na) = 0.393 x 50 g
mass of sodium (Na) = 19 g
