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1 year ago

What is the mass of a 32.5-ml sample of ethanol? the density of ethanol is 0.789 g/ml.

Chemistry

1 answer:

Anon25 [30]1 year ago

6 0

Answer:

25.6 grams of ethanol in 32.5 ml of the substance.

Explanation:

The density of ethanol can be used as a conversion factor. We know that:

Ethanol is 0.789 g/ml. That becomes a conversion factor that we may write as (0.789 g/ml) and then convert either grams ethaanol or ml ethanol into the other unit. We are given volume of ethanol: 32.5 ml.

Note that when we multiply (0.789 g/ml) by (32.5 ml) the ml unit cancels, leaving just grams. That's what we want, so:

(0.789 g/ml)(32.5 ml) = 25.6 grams of ethanol in 32.5 ml of the substance.

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Answer:

So the volume will be 2.33 L

Explanation:

The reaction for the combustion is:

2 C₄H₁₀ (g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (l)

mass of butane to moles (mass / molar mass)

1.4 g / 58 g/mol

= 0.024 moles

2 moles of butane can produce 8 moles of carbon dioxide

0.024 moles of butane must produce (0.024 × 8) /2

= 0.096 moles of CO₂

Now we apply the Ideal Gases Law to find out the volume formed.

P . V = n . R . T

p = 1atm

n = 0.096 mol

R = 0.082 L.atm/mol.K

T = 273 + 23 = 296K

V = ?

1atm × V = 0.096 mol × 0.082 L.atm/mol.K × 296K

V = 0.096 mol × 0.082 L.atm/mol.K × 296K / 1atm

= 2.33 L

So the volume will be 2.33 L

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3 years ago

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Which polymer is a biopolymer?

KATRIN_1 [288]

C) poly lactic acidosis

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2 years ago

Hydrochloric acid reacts with sodium hydroxide to form water and sodium chloride. Hydrochloric acid is an extremely acidic, clea

olya-2409 [2.1K]

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I think it is a).

Explanation:

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2 years ago

The formation of SO3 from SO2 and O2 is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for

jeka57 [31]

Answer:

118.22 atm

Explanation:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

KP = 0.13 = $\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}$

Where p(SO₃) is the partial pressure of SO₃, p(SO₂) is the partial pressure of SO₂ and p(O₂) is the partial pressure of O₂.

With 2.00 mol SO₂ and 2.00 mol O₂ if there was a 100% yield of SO₃, then 2 moles of SO₃ would be produced and 1.00 mol of O₂ would remain.
With a 71.0% yield, there are only 2*0.71 = 1.42 mol SO₃, the moles of SO₂ that didn't react would be 2 - 1.42 = 0.58; and the moles of O₂ that didn't react would be 2 - 1.42/2 = 1.29.

The total number of moles is 1.42 + 0.58 + 1.29 = 3.29. With that value we can calculate the molar fraction (X) of each component:

XSO₂ = 0.58/3.29 = 0.176

XO₂ = 1.29/3.29 = 0.392

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The partial pressure of each gas is equal to the total pressure (PT) multiplied by the molar fraction of each component.

p(SO₂) = 0.176 * PT

p(O₂) = 0.392 * PT

p(SO₃) = 0.432 * PT

Rewriting KP and solving for PT:

$\frac{p(SO_{3})^{2}}{p(SO_{2})^{2}p(O_{2})}=0.13\\\frac{(0.432*P_{T})^{2}}{(0.176*P_{T})^{2}(0.392*P_{T})} =0.13\\\frac{0.1866*P_{T}^{2}}{0.0121*P_{T}^{3}} =0.13\\\frac{15.369}{P_{T}}=0.13\\P_{T}=118.22 atm$

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3 years ago

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shusha [124]

Answer:

Correct option is

5 liters of CH

(g)NO

at STP

No. of molecules=

22.4

mol=

22.4

×N

molecules

A) 5ℊ of H

(g)

No. of moles=

mol=

×N

molecules

B) 5l of CH

(g)

No. of moles of CH

22.4

mol=

22.4

molecules

C) 5 mol of O

=5N

molecules

D) 5×10

molecules of CO

(g)

Molecules of 5l NO

(g) at STP=5l of CH

(g) molecules at STP

Therefore, option B is correct.

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