The Atomic Mass of Copper is 63.5 grams One mole of any element has a mass equal to the atomic mass.
0.75 grams of Cu = x moles of Cu
63.5 grams of Cu = 1 mole of Cu
Set up a proportion and solve for x
Divide
0.75 / 63.5 = x /1
0.75 ÷ 63.5 = x
Number of moles = (# of grams) ÷ (molar mass)
15.45 g Cu / 63.546 g = 0.24313 mol
Read more: How_do_you_convert_from_grams_to_moles_and_also_from_moles_to_grams
For this you need the atomic mass of Cu. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel.
180 grams Cu / (63.5 grams) = 2.83 moles Cu
3 moles Cu (63.55 grams Cu/1mole Cu)
= 190.65 grams
If you work out the relative formula mass, which is one mole, then you find 0.75 of that answer.
150 grams copper (1 mole Cu/63.55 grams)(6.022 X 10^23/1 mole Cu)(1 mole Cu/6.022 X 10^23)
= 2.36 moles of copper atoms
1 mole Cu = 63.546g
75.92g Cu x 1mol Cu/63.546g Cu = 1.195 moles Cu
Considering the molar mass of water is 18 g, there are 100 moles in 1800 grams of water.
0.125 Molar solution! Molarity = moles of solute/Liters of solution Algebraically manipulated, Moles of copper sulfate = 2.50 Liters * 0.125 M = 0.313 moles copper sulfate needed ===========================
How many MOLES of sodium nitrate are present in 2.85 grams of this compound ?
5.00 moles H x 1 mole C2H4O2/4 moles H = 1.25 moles of C2H4O2 present.
There are 10 moles present in 585 g of sodium chloride.
The formula for converting is: atoms ÷ Avogadro's constant = # moles (1.8 × 1023) ÷ (6.02 × 1023)= 0.299 moles Cu
Only one mole of copper.
2.83 moles
1020g
Approx. 10e-17 moles.
0.01362 moles.
5 moles
0.125 Molar solution! Molarity = moles of solute/Liters of solution Algebraically manipulated, Moles of copper sulfate = 2.50 Liters * 0.125 M = 0.313 moles copper sulfate needed ===========================
The answer is 0,615 moles.
The answer is 8,33 moles.
How many moles are present in 92 grams of ethyl alcohol?
How many MOLES of sodium nitrate are present in 2.85 grams of this compound ?
The answer is 14,93 moles.