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bonufazy [111]
3 years ago
7

A compound is 54.53% c, 9.15% h, and 36.32% o by mass. what is its empirical formula? the molecular mass of the compound is 132

amu. what is the molecular formula?
Chemistry
1 answer:
Yakvenalex [24]3 years ago
6 0
 <span>First - you need the empirical formula. 

So, assume you have 100 g of the compound. 

If so, you'll have 54.53 gram of C, 9.15 g of H and 36.32 g of O. Find the number of moles of each. 

54.53 g C (1 mole C / 12.01 g C) = 4.540 

9.15 g H (1 mole H / 1.008 g H) = 9.077 

36.32 g O (1 mole O / 15.9994 g O) = 2.270 

Take the smallest number found and divide the others by it to get the empirical formula. 

4.540/2.270 = 2. 
9.077/2.270 = 4. 
2.270/2.270 =1. 

So, that gives you the empirical formula of C2H4O. 

Find the weight of this compound. C = 12, H = 1, O = 16. So, C2H4O is 44 amu. 

132/44 = 3. 

So, 3 (C2 H4 O) = C6H12O3 = molecular formula.</span>
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3.2 × 10⁻⁸

Explanation:

Let's consider the solution of magnesium carbonate.

MgCO₃ ⇄ Mg²⁺(aq) + CO₃²⁻(aq)

We can relate the molar solubility (S) with the solubility product (Ksp) using an ICE chart.

         MgCO₃ ⇄ Mg²⁺(aq) + CO₃²⁻(aq)

I                             0                0

C                          +S              +S

E                            S                S

The Ksp is:

Ksp = [Mg²⁺] × [CO₃²⁻] = S × S = S² = (1.8 × 10⁻⁴)² = 3.2 × 10⁻⁸

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Write one common thing between condensation and hydrolysis ?​
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The common thing is the compound water

Explanation:

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Which explanations did you include in your response? Check all that apply. The oxidation number of carbon changes from -1 to +4.
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When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

7 0
2 years ago
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