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rodikova [14]
3 years ago
8

Which of the following equations is balanced for charge and mass?

Chemistry
2 answers:
Nataliya [291]3 years ago
8 0
<span>B. S⁰(s) + 2H⁺ + 2e⁻ –--> H2S⁰(g)

by mass: 1 S  and 2 H ----> 2 H and 1S  True.
by charge : 0 +(2*(+1)) + 2*(-1) = 0, 0+2-2=0, 0 = 0  True.</span>
mamaluj [8]3 years ago
4 0

 

The following equation(option B) is balanced for charge and mass

A. S(s) + 2H+ → H2S(g) + 2e–


B. S(s) + 2H+  + 2e–  →  H2S(g)


C. S(s) + 2e- →H2S(g)


D. S(s) + 2H+→ H2S(g)


S(s) + 2H+  + 2e–  →  H2S(g)



here S is one on either sides

H is two on either sides, hence mass is balanced.


And charge is also balanced,


as  2H+ has +2 charge

and 2e- has -2 charge,thus

total charge on left side is +2-2 = 0

and charge on right side is also zero here or no charge on either sides.

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Magnesium carbonate decomposes on heating to form magnesium oxide and carbon dioxide as shown.
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In the reaction C7H16+_______O2→ 8H2O + 7CO2, what coefficient should be placed in front og O2 to balence the reaction
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Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates. The phosphate ion forms in
avanturin [10]
<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
</span>
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I think it would be solubility but I’m not sure

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