<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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The answer should be:
KOH (aq) + HCl (aq) --> KCl (aq) + H20 (l).
KOH is a base, because OH can accept a H+.
HCl is an acid because it can donate a H+.
In general, bases are : OH-, and acids are : H+.
![pH=-\log_{10} [H_3O^+] \Rightarrow [H_3O^+]=10^{-pH} \\ \\ pH=2.8 \\ \ [H_3O^+]=10^{-2.8} \approx 1.58 \times 10^{-3}](https://tex.z-dn.net/?f=pH%3D-%5Clog_%7B10%7D%20%5BH_3O%5E%2B%5D%20%5CRightarrow%20%5BH_3O%5E%2B%5D%3D10%5E%7B-pH%7D%20%5C%5C%20%5C%5C%0ApH%3D2.8%20%5C%5C%0A%5C%20%5BH_3O%5E%2B%5D%3D10%5E%7B-2.8%7D%20%5Capprox%201.58%20%5Ctimes%2010%5E%7B-3%7D)
The [H₃O⁺] of the solution is approximately 1.58 × 10⁻³ M.
Answer:
The answer to your question is below:
Explanation:
Having exactly the same data as the previous experiment I think that having the same data as the previous experiment is extremely important but not the most important, for me is the second most important.
Using the same procedure and variables as the previous experiment For me, this is the most importan thing when a scientist is designing an experiment, because if he or she follow exactly the same procedure and variables, then the results will be very close.
Conducting an experiment similar to the previous experiment This characteristic is important but not the most important.
Using the same laboratory that was used in the previous experiment It is not important the laboratory, if the procedure and variables are the same, your experiment must give the same results in whatever laboratory.
Answer:
3) O(g) is an intermediate; 2O3(g)→3O2(g)
Explanation:
The decomposition of ozone to yield oxygen occurs in a sequence of steps. The various non-elementary reactions involved constitute the reaction mechanism. In the sequence of reaction steps O(g) serves as an intermediate.
The overall reaction involves the conversion of two moles of ozone to three moles of oxygen as shown in the answer. Thus the O(g) is merely a reaction intermediate.
