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S_A_V [24]
3 years ago
5

Two liquids are analyzed and found to both be 85.7% carbon and 14.3% hydrogen. At 750 mmHg and 150 C, both are gases. At these c

onditions, an 800.0 mL bottle will hold 1.60 g of Compound A, while the same bottle will hold 2.22 g of Compound B. Find the molecular formulas of both compounds A and B. (Hint: use percent composition to find the empirical formula, and then use the Ideal gas law to find moles of each.) Enter your answer for both molecular formulas like this: XY, X2Y
Chemistry
1 answer:
vodomira [7]3 years ago
8 0

Answer:

Molecular formula A: C₅H₁₀

Molecular formula B: C₇H₁₄

Explanation:

It is possible to obtain empirical formula of compounds using percent composition, thus:

C: 85.7% × (1mol / 12.01g) = 7.136 moles C

H: 14.3% × (1mol / 1.01g) = 14.158 moles H

Mole ratio of H:C is:

14.158mol / 7.136mol = 2

That means in compounds A and B you have 2 hydrogens per atom of carbon and empirical formila is:

CH₂

Using PV = nRT, moles of A and B are:

<em>Where P is pressure (750mmHg / 760 = 0.987atm), V is volume (0.8000L), R is gas constant (0.082atmL/molK), and T is temperature (150°C +273.15 = 423.15K)</em>

Moles A and B: n = PV / RT

n = 0.987atm×0.8000L / 0.082atmL/molK×423.15K

n = 0.0228 moles of A and B.

Using the mass of A and B it is possible to find molar mass of each compound:

A = 1.60g / 0.0228mol = 70.31g/mol

B = 2.22g / 0.0228mol = 97.56g/mol

As empirical formula of both compounds is CH₂, (molar mass = 14.03g/mol). Molecular formula of compounds is:

A = 70.31g/mol / 14.03g/mol = 5 → Molecular formula: 5×CH₂ = <em>C₅H₁₀</em>

B = 97.56g/mol / 14.03g/mol = 7 → Molecular formula: 7×CH₂ = <em>C₇H₁₄</em>

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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
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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\%

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A community consists of all the organisms in an ecosystem that belong to the same species.
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Plesse help asap!!!​
Naily [24]

Answer:

4.4×10² cm³

Explanation:

From the question given above, the following data were obtained:

Diameter (d) = 68.3 mm

Height (h) = 0.120 m

Volume (V) =?

Next, we shall convert the diameter (i.e 68.3 mm) to cm.

This can be obtained as follow:

10 mm = 1 cm

Therefore

68.3 mm = 68.3 mm / 10 mm × 1 cm

68.3 mm = 6.83 cm

Therefore, the diameter 68.3 mm is equivalent 6.83 cm.

Next, we shall convert the height (i.e 0.120 m) to cm. This can be obtained as follow:

1 m = 100 cm

Therefore,

0.120 m = 0.120 m/ 1 m × 100 cm

0.120 m = 12 cm

Therefore, the height 0.120 m is equivalent 12 cm.

Next, we shall determine the radius of the cylinder. This can be obtained as follow:

Radius (r) is simply half of a diameter i.e

Radius (r) = Diameter (d) /2

r = d/2

Diameter (d) = 6.83 cm

Radius (r) =?

r = d/2

r = 6.83/2

r = 3.415 cm

Finally, we shall determine the volume of the cylinder as follow:

Radius (r) = 3.415 cm

Height (h) = 12 cm

Volume (V) =?

Pi (π) = 3.14

V = πr²h

V = 3.14 × (3.415) ² × 12

V = 440 cm³

V = 4.4×10² cm³

Therefore, the volume of the cylinder is 4.4×10² cm³

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