Suppose 25 L to be 25 kg water (almost, -0.2% at room temperature)
then divide 25,000 g by the moalr mass of H2O being 2+16=18 g/mole
you get = 1385 moles (because 1 mole = 18 grams)
By filling in this formula with SI units of the symbols R, T, p, V you'll find the number of moles for ALL possible (ideal) gases according to Ideal Gas Law (Boyle's):
n = p*V / R*T or p*V = n*R*T n= number of moles of any gas => 25 mol (your 'known')p= pressure of (any) gas (Pa) => 1.01325*105 Pa (=1 atm) (Standard)
V= Volume of that gas (m3) => ? (to be calculated)
R= universal Gas constant (J.mol-1.K-1 )=> 8.3145 (universal constant)
T= absolute temperature (K) => 273.15 K (Standard)
With the questioned figures:
V = nRT/p = [25*8.3145*273.15] / [1.01325*105] = 0.56035 (m3) = 560 litres for 25 moles of Ar (or any) gas at (Standard)TP
1 mole of gas at STP occupies 22.4Lt. of space. So 25 mole will occupie 22.4*25=560lt.
1 mole = 22.414 liters
So, 5 moles = 112.07 liters
1,67.1024 argon atoms is equal to 2,773 moles.
Molarity = moles of solute/Liters of solution 3.42 M NaOH = 1.3 moles NaOH/Liters NaOH Liters NaOH = 1.3 moles NaOH/3.42 M NaOH = 0.38 Liters
Molarity = moles of solute/Liters of solution ( 50.0 ml = 0.05 Liters ) 0.552 M KCl = moles/.0.05 liters = 0.0276 moles of KCl
152
8,4 liters of nitrous oxide at STP contain 2,65 moles.
0.25 moles, approx.
moles = mass in grams / atomic weight So moles in 37.9 g or argon = 37.9 / 39.948 = 0.948 moles
Approx 0.223 moles.
1,67.1024 argon atoms is equal to 2,773 moles.
Moles = weight in g / atomic weight. So moles in 24.7 g of Ar = 24.7 / 39.948 = 0.62 moles
Have: 607gAr Need: Moles of Argon From the periodic table we know that there are 39.948gAr per every 1 mole of Argon. 607g/39.948 = your answer.
1.7x10^-18
55.64
1 mole occupies 22.414 liters So, 1.84 moles will occupy 41.242 liters
Molarity = moles of solute/Liters of solution 3.42 M NaOH = 1.3 moles NaOH/Liters NaOH Liters NaOH = 1.3 moles NaOH/3.42 M NaOH = 0.38 Liters
Molarity = moles of solute/Liters of solution (40 ml = 0.04 Liters) algebraically manipulated, Moles of solute = Liters of solution * Molarity Moles HCl = (0.04 Liters)(0.035 M) = 0.0014 moles HCl ==============
Molarity = moles of solute/liters of solution ( 15ml = 0.015 liters ) 2.9 M NaOH = moles NaOH/0.015 liters = 0.0435 moles of NaOH