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4 years ago

15

If I make a solution by adding water to 7mL of ethanol until the total volume of the solution is 375mL, what’s the percent by vo

lume of ethanol in the solution?

1 answer:

mihalych1998 [28]4 years ago

3 0

(7/375)x100 = 1.866666666 %
= 1.87 (rounded to two decimal places)

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If the equilibrium constant for the reaction A 2B C 5/2 D has a value of 4.0, what is the value of the equilibrium constant for

nirvana33 [79]

Answer:

$K'=\frac{1}{16}$

Explanation:

Hello!

In this case, since the given reaction is:

$A+ 2B \rightleftharpoons C+ \frac{5}{2} D$

Whereas the equilibrium constant is:

$K=\frac{[C][D]^{5/2}}{[A][B]^2} =4.0$

However, the new target reaction reverses and doubles the initial reaction to obtain:

$2C+5D \rightleftharpoons 2A+4B$

Whereas the equilibrium constant is:

$K'=\frac{[A]^2[B]^4}{[C]^2[D]^5}$

Which suggest the following relationship between the equilibrium constants:

$K'=\frac{1}{K^2}$

So we plug in to obtain:

$K'=\frac{1}{4.0^2}\\\\K'=\frac{1}{16}$

Best regards!

8 0

3 years ago

THIS IS EASY BUT FOR SOME REASON MY BRAIN IS DED PLS HELP

Mama L [17]

Answer:

Solid

Explanation:

The particles in the solid matter are held close together and in fixed particles.

6 0

3 years ago

Ammonia react with phosphorus acid to form a compound that contains 28.2% of nitrogen 20.8%, phosphorus 8.1% of hydrogen 42.9%ox

____ [38]

Answer:

Empirical formula will be (NH₄)₃PO₄, which matches the molecular formula

Explanation:

This is the reaction:

NH₃ + H₃PO₄ → 28.2% N, 20.8% P, 8.1% H, 42.9% O

In 100 g of compound we have:

28.2 g N

20.8 g of P

8.1 g of H

42.9 g of O

Now we divide each between the molar mass:

28.2 g / 14 g/mol = 2.01 mol

20.8 g / 30.97 g/mol = 0.671 mol

8.1 g / 1 g/mol = 8.1 mol

42.9 g / 16 g/mol = 2.68 mol

And we divide again between the lowest value of moles

2.01 mol / 0.671 mol → 3

0.671 mol / 0.671 mol → 1

8.1 mol / 0.671 mol → 12

2.68 mol / 0.671 mol → 4

Molecular formula will be: N₃PH₁₂O₄ → (NH₄)₃PO₄

Empirical formula will be (NH₄)₃PO₄, which matches the molecular formula

3 0

4 years ago

Determine the molecular mass ratio of two gases whose rates of effusion have a ratio of 16 : 1.

Usimov [2.4K]

Answer is: <span>the molecular mass ratio of two gases is 1 : 256.
</span>rate of effusion of gas1 : rate of effusion of gas = 16 : 1.<span>
rate of effusion of gas1 = 1/√M(gas1).
rate of effusion of gas2 = 1/√M(gas2).
rate of effusion of gas1 = rate of effusion of gas2 </span>· 16<span>.
</span>1/√M(gas1) = 1/√M(gas2) · 16 /².
<span>1/M(gas1) = 1/M(gas2) </span>· 256.
<span>M(gas1) </span>· 256 = M(gas2).<span>

</span>

8 0

3 years ago

Ammonia reacts with sulfuric acid to produce the important fertilizer, ammonium hydrogen sulfate.

Liula [17]

Answer:

404.8g of (NH4)HSO4 is produced.

Explanation:

Step 1:

Data obtained from the question. This include the following:

Temperature (T) = 10°C = 10°C + 273 = 283K

Pressure (P) = 110KPa = 110/101.325 = 1.09atm

Volume (V) = 75L

Step 2:

Determination of the number of mole of ammonia, NH3.

The number of mole (n) of ammonia, NH3 can be obtained by using the ideal gas equation. This is illustrated below:

Note:

Gas constant (R) = 0.0821atm.L/Kmol

Number of mole (n) =?

PV = nRT

1.09 x 75 = n x 0.0821 x 283

Divide both side by 0.0821 x 283

n = (1.09 x 75) /(0.0821 x 283)

n = 3.52 moles

Step 3:

Determination of the number of mole ammonium hydrogen sulfate produced from the reaction.

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

NH3 + H2SO4 —> (NH4)HSO4

From the balanced equation above,

1 mole of NH3 produced 1 mole of (NH4)HSO4

Therefore, 3.52 moles of NH3 will also produce 3.52 moles of (NH4)HSO4.

Therefore, 3.52 moles of ammonium hydrogen sulfate, (NH4)HSO4 is produced.

Step 4:

Conversion of 3.52 moles of ammonium hydrogen sulfate, (NH4)HSO4 to grams. This is illustrated below:

Molar Mass of (NH4)HSO4 = 14 + (4x1) + 1 + 32 + (16x4) = 115g/mol

Number of mole of (NH4)HSO4 = 3.52 moles

Mass of (NH4)HSO4 =..?

Mass = mole x molar Mass

Mass of (NH4)HSO4 = 3.52 x 115 = 404.8g

Therefore, 404.8g of (NH4)HSO4 is produced.

5 0

3 years ago