This depends on:
- if it is an anhydrous or hydrated salt
- if it is a salt of Cu(I) or Cu(II)
For CuSO4(anh.) the answer is 0,00364 moles.
400 grams of nickel sulphate (anhydrous) is equivalent to 2,58 moles.
The molar mass of Potassium Sulphate (K2SO4) is 174.26 g/mol. Therefore, the mass of 5 moles of Potassium Sulphate would be 5 moles x 174.26 g/mol = 871.3 grams.
To determine the number of electrons required to deposit 6.35 grams of copper, you need to first calculate the moles of copper deposited using the molar mass of copper. Then, use Faraday's constant (1 mol of electrons = 1 Faraday) to convert moles of copper to the number of electrons. Finally, multiply the number of moles of electrons by Avogadro's number to get the total number of electrons required.
To completely replace silver in the solution with copper, you would need an equal number of moles of copper to the moles of silver present. Calculate the moles of silver in the solution using the concentration and volume given. Then use the mole ratio between copper and silver to determine the moles of copper needed, and convert this to grams.
To convert grams into atoms, you have to convert them into moles first. Get the molar mass and multiply it by the number of moles to get the atoms.
2.83 moles
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.2068 grams Cu / (63.5 grams) = 32.6 moles Cu
Using the given masses, calculate the moles of each reactant. The limiting reactant will be the one that produces the least amount of product, which is copper. Convert the moles of copper to moles of copper (II) oxide using the balanced chemical equation. Then, convert the moles of copper (II) oxide to grams to find the mass produced.
The answer is 1,54 moles.
The Atomic weight of Copper is 63.55, which is the number of grams in one mole of Copper [Cu].So, the weight in g of 0.252 mol of copper is calculated as 0.252 x 63.55 = 16.0 g Cu
Multiply the number of moles by the molecular weight.
6.066 into 10(up 23 )