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Alina [70]
3 years ago
9

If 1.546 g of copper was used by a student at the start of the lab, and 0.732 g of copper were obtained at

Chemistry
1 answer:
mash [69]3 years ago
8 0

Answer: Percent recovery is 47.34 %

Explanation:

Percent yield is defined as the ratio of experimental yiled to theoretical yield in terms of percentage.

{\text{ percent yield}}=\frac{\text{amount recovered}}{\text{total amount}}\times 100

Putting in the values we get:

{\text{ percent yield}}=\frac{0.732}{1.546}\times 100=47.34\%

Therefore, the percent recovery is 47.34 %

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What is the mass in grams of 1.8 mol of Magnesium Sulfide?
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101.466000

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Give the complete ionic equation for the reaction (if any) that occurs when aqueous solutions of MgSO3 and HI are mixed. Give th
Sunny_sXe [5.5K]

Answer:

Mg²⁺(aq) + SO₃²⁻(aq) + 2 H⁺(aq) + 2 I⁻(aq) ⇄ Mg²⁺(aq) + 2I⁻(aq) + H₂O(l) + SO₂(g)

Explanation:

<em>Give the complete ionic equation for the reaction (if any) that occurs when aqueous solutions of MgSO₃ and HI are mixed.</em>

When MgSO₃ reacts with HI they experience a double displacement reaction, in which the cations and anions of each compound are exchanged, forming H₂SO₃ and MgI₂. At the same time, H₂SO₃ tends to decompose to H₂O and SO₂. The complete molecular equation is:

MgSO₃(aq) + 2 HI(aq) ⇄ MgI₂(aq) + H₂O(l) + SO₂(g)

In the complete ionic equation, species with ionic bonds dissociate into ions.

Mg²⁺(aq) + SO₃²⁻(aq) + 2 H⁺(aq) + 2 I⁻(aq) ⇄ Mg²⁺(aq) + 2I⁻(aq) + H₂O(l) + SO₂(g)

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3 years ago
Bohr's planetary of the atom stated that
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3 0
3 years ago
When 1.95 g of co(no3)2 is dissolved in 0.350 l of 0.220 m koh, what are [ co2+], [ co(oh)42−], and [ oh−] if kf of co(oh)42− =
Airida [17]
Mass of Co(NO₃)₂ = 1.95 g
V KOH = 0.350 L
[KOH] = 0.220 M
Kf = 5.0 x 10⁹
molar mass of Co(NO₃)₂ = 182.943 g/mol
so [Co(NO₃)₂] = 1.95 / (0.350 * 182.943) = 0.03045 M
[Co²⁺] = 0.03045 M
[OH⁻] = 0.22 M
chemical reaction:
             Co²⁺(aq) + 4 OH⁻    ⇄      Co(OH)₄²⁻
I (M)      0.03045      0.22                   0
C (M)   - 0.03045   - 4 (0.03045)      0.03045
E (M)       - x         0.22 - 4(0.03045)   0.03045
                              = 0.0982
Kf  = [Co(OH)₄²⁻] / [Co⁺²][OH⁻]⁴
5.0 x 10⁹ = (0.03045) / x (0.0982)⁴
x = 6.5489 x 10⁻⁸
at equilibrium:
[Co²⁺] = 6.54 x 10⁻⁸
[OH⁻] = 0.0982 M
[Co(OH)₄²⁻] = 0.03045 M



4 0
3 years ago
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