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FrozenT [24]
2 years ago
13

Combustion analysis of an unknown compound containing only carbon and hydrogen produced 0.2845 g of co2 and 0.1451 g of h2o. wha

t is the empirical formula of the compound?
Chemistry
1 answer:
Serjik [45]2 years ago
7 0
CxHy + (x+0.25)O₂ → xCO₂ + 0.5yH₂O

m(CO₂)/{xM(CO₂)}=m(H₂O)/{0.5yM(H₂O)}

0.2845/{44.01x}=0.1451/{9.01y}

x/y=0.4=2:5

C₂H₅ is the empirical formula 
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andre [41]

1.always listen to teacher

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Compare Solution X to Solution A, Solution B, and Solution C. Beaker with a zoom view showing eight shperes, four groups of part
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Answer:

SOLUTION A

Explanation:

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A sample of an unknown compound with a mass of 0.847 g has the following composition: 50.51 % fluorine and 49.49 % iron. When th
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Answer: 0,4278g of F and 0,4191g of Fe

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For Fluorine (F)

0,847g * 0,5051 = 0,4278g of F

For iron (Fe)

0,847 * 0,4949 = 0,4191g of Fe

This is determined because even when the compound is decomposed, due to conservative law of mass, the decomposition process do not affect the amount of matter, so the mass of the elements remain even if they are separated from the original molecule.

At the end, the sum of the elements masses should be the total mass of the compound.

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Hydrochloric acid (75.0 mL of 0.250 M) is added to 225.0 mL of 0.0550 M Ba(OH)2 solution. What is the concentration of the exces
Shalnov [3]

Answer:  The concentration of excess [OH^-] in solution is 0.017 M.

Explanation:

1. Molarity=\frac{moles}{\text {Volume in L}}

moles of HCl=Molarity\times {\text {Volume in L}}=0.250\times 0.075=0.019moles

1 mole of HCl give = 1 mole of H^+

Thus 0.019 moles of HCl give = 0.019 mole of H^+

2. moles of Ba(OH)_2=Molarity\times {\text {Volume in L}}=0.0550\times 0.225=0.012moles

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1 mole of Ba(OH)_2 gives = 2 moles of OH^-

Thus 0.012 moles of Ba(OH)_2 give = 2 \times 0.012=0.024 moles of OH^-

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As 1 mole of H^+ neutralize 1 mole of OH^-

0.019 mole of H^+ will neutralize 0.019 mole of OH^-

Thus (0.024-0.019)= 0.005 moles of OH^- will be left.

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AnnZ [28]

Answer : The correct options are,

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