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CaHeK987 [17]
3 years ago
13

Calculate the mass of nitrogen dissolved at room temperature in an 95.0 LL home aquarium. Assume a total pressure of 1.0 atmatm

and a mole fraction for nitrogen of 0.78.
Chemistry
1 answer:
harina [27]3 years ago
8 0

Answer:

86.3 g  of N₂ are in the room

Explanation:

First of all we need the pressure from the N₂ in order to apply the Ideal Gases Law and determine, the moles of gas that are contained in the room.

We apply the mole fraction:

Mole fraction N₂ = N₂ pressure / Total pressure

0.78 . 1 atm = 0.78 atm → N₂ pressure

Room temperature → 20°C → 20°C + 273 = 293K

Let's replace data: 0.78 atm . 95L = n . 0.082 . 293K

(0.78 atm . 95L) /0.082 . 293K = n

3.08 moles = n

Let's convert the moles to mass → 3.08 mol . 28g /1mol = 86.3 g  

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Calculate the mole fraction of each component in a solution with 6.87 g of sodium chloride (NaCl) dissolved in 65.2 g of water.
zheka24 [161]

Answer:

mole fraction of NaCl = 0.03145.

mole fraction of water = 0.9686.

Explanation:

  • Mole fraction is an expression of the concentration of a solution or mixture.
  • It is equal to the moles of one component divided by the total moles in the solution or mixture.
  • The summation of mole fraction of all mixture components = 1.

mole fraction of NaCl = (no. of moles of NaCl) / (total no. of moles).

<em>no. of moles of NaCl = mass/molar mass </em>= (6.87 g)/(58.44 g/mol) = 0.1176 mol.

<em>no. of moles of water = mass/molar mass</em> = (65.2 g)/(18.0 g/mol) = <em>3.622 mol.</em>

<em></em>

∴ mole fraction of NaCl = (no. of moles of NaCl) / (total no. of moles) = (0.1176 mol)/(0.1176 mol + 3.622 mol) = 0.03145.

<em>∵ mole fraction of NaCl + mole fraction of water = 1.0.</em>

∴ mole fraction of water = 1.0 - mole fraction of NaCl = 1.0 - 0.03145 = 0.9686.

7 0
3 years ago
Differentiate between a precipitate and an aqueous solution
antiseptic1488 [7]

Answer:

A precipate is a solid while an aqueous solution is liquid.

Explanation:

A precipitate is a solid which separates after a chemical reaction occurs. It is the solid product of the reaction.

An aqueous solution is formed when a substance is dissolved in water.

6 0
3 years ago
The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×
Anuta_ua [19.1K]

Answer:

Kc=~1.49x10^3^4}

Explanation:

We have the reactions:

A: N_2_(_g_) + O_2_(_g_)  2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9

Our <u>target reaction</u> is:

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)

We have NO_(_g_) as a reactive in the target reaction and  NO_(_g_) is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

Then if we add reactions A and B we can obtain the target reaction, so:

A: 2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}

B: 2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9

For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.

4NO_(_g_)  N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}

Kc=~1.49x10^+^3^4}

3 0
3 years ago
What is the percent yield of NaCl if 31g of CuCl2 reacts with excess NaNo3 to produce 21.2g of NaCl
Valentin [98]
The answer is 78.7% yield.
3 0
4 years ago
Under what conditions does the molar volume of a gas decrease
nydimaria [60]

Answer:

Gases are compressible, meaning that when put under high pressure, the particles are forced closer to one another. This decreases the amount of empty space and reduces the volume of the gas. Gas volume is also affected by temperature. When a gas is heated, its molecules move faster and the gas expands.

4 0
3 years ago
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