NaOH+HCl-> NaCl+H2O
1 mole of NaOH
1 mole of HCl.
To calculate volume of NaOH
CaVa/CbVb= Na/Nb
Where Ca=2M
Cb=1M
Va=200cm³
Vb=xcm³
Substitute into the equation.
2×200/1×Vb=1/1
400/Vb=1/1
Cross multiply
Vb×1=400×1
Vb=400cm³
To calculate the mass of sodium chloride, NaCl from the neutralization rxn.
Mole of NaCl=1
Molar mass of NaCl= 23+35.5=58.5
Mass=xgrammes.
Mass of NaCl=Number of moles × Molar mass.
Substitute
Mass of NaCl= 1×58.5
=58.5g
This is what I could come up with.
Answer:
16.6 g of Al are produced in the reaction of 82.4 g of AlCl₃
Explanation:
Let's see the decomposition reaction:
2AlCl₃ → 2Al + 3Cl₂
2 moles of aluminum chloride decompose to 2 moles of solid Al and 3 moles of chlorine gas.
We determine the moles of salt:
82.4 g . 1mol/ 133.34g = 0.618 moles
Ratio is 2:2. 2 moles of salt, can produce 2 moles of Al
Then, 0.618 moles of salt must produce 0.618 moles of Al.
Let's convert the moles to mass → 0.618 mol . 26.98g /mol = 16.6 g
Answer:
B. two rounds of cell division
