Answer:
56.2÷6.02×10^23
=9.34×10^23
Explanation:
Divide the given mass of the atom by the mass of an Atom (the avogadro's constant) to find the number of atoms in the given mass.
<span>5.5×10−2M in calcium chloride and 8.0×10−2M in magnesium nitrate.
What mass of sodium phosphate must be added to 1.5L of this solution to completely eliminate the hard water ion
1) Content of Ca (2+) ions
Calcium chloride = CaCl2
Ionization equation: CaCl2 ---> Ca (2+) + 2 Cl (-)
=> Molar ratios: 1 mol of CaCl2 : 1 mol Ca(2+) : 2 mol Cl(-)
Calculate the number of moles of CaCl2 in 1.5 liters of 5.5 * 10^-2 M solution
M = n / V => n = M*V = 5.5 * 10^ -2 M * 1.5 l = 0.0825 mol CaCl2
=> 0.0825 mol Ca(2+)
2) Number of phosphate ions needed to react with 0.0825 mol Ca(2+)
formula of phospahte ion: PO4 (3-)
molar ratio: 2PO4(3-) + 3Ca(2+) = Ca3 (PO4)2
Proportion: 2 mol PO4(3-) / 3 mol Ca(2+) = x / 0.0825 mol Ca(2+)
=> x = 0.0825 coml Ca(2+) * 2 mol PO4(3-) / 3 mol Ca(2+) = 0.055 mol PO4(3-)
3) Content of Mg(2+) ions
Ionization equation: Mg (NO3)2 ----> Mg(2+) + 2 NO3 (-)
Molar ratios: 1 mol Mg(NO3)2 : 1 mol Mg(2+) + 2 mol NO3(-)
number of moles of Mg(NO3)2 in 1.5 liter of 8.0 * 10^-2 M solution
n = M * V = 8.0 * 10^ -2 M * 1.5 liter = 0.12 moles Mg(NO3)2
ions of Mg(2+) = 0.12 mol Mg(NO3)2 * 1 mol Mg(2+) / mol Mg(NO3)2 = 0.12 mol Mg(2+)
4) Number of phosphate ions needed to react with 0.12 mol Mg(2+)
2PO4(3-) + 3Mg(2+) = Mg3(PO4)2
=> 2 mol PO4(3-) / 3 mol Mg(2+) = x / 0.12 mol Mg(2+)
=> x = 0.12 * 2/3 mol PO4(3-) = 0.16 mol PO4(3-)
5) Total number of moles of PO4(3-)
0.055 mol + 0.16 mol = 0.215 mol
6) Sodium phosphate
Sodium phosphate = Na3(PO4)
Na3PO4 ---> 3Na(+) + PO4(3-)
=> 1 mol Na3PO4 : 1 mol PO4(3-)
=> 0.215 mol PO4(3-) : 0.215 mol Na3PO4
mass in grams = number of moles * molar mass
molar mass of Na3 PO4 = 3*23 g/mol + 31 g/mol + 4*16 g/mol = 164 g/mol
=> mass in grams = 0.215 mol * 164 g/mol = 35.26 g
Answer: 35.26 g of sodium phosphate
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Answer:
B, H, E, G, C, D, A, F
Explanation:
The ones closest to the bottom are the oldest
