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NikAS [45]
3 years ago
10

The solubility of Cr(NO3)3⋅9H2O in water is 208 g per 100 g of water at 15 ∘C. A solution of Cr(NO3)3⋅9H2O in water at 35 ∘C is

formed by dissolving 310 g in 100 g water. When this solution is slowly cooled to 15 ∘C, no precipitate forms. At equilibrium, what mass of crystals do you expect to form?
Chemistry
1 answer:
Zinaida [17]3 years ago
5 0

Answer:

102g of crystals

Explanation:

When the Cr(NO₃)₃⋅9H₂O is dissolved in water at 15°C, the maximum mass that water will dissolve in the equilibrium is 208 g per 100g of water. When you heat the water, this mass will increases.

In this problem, at 35°C the water dissolves 310g in 100g of water, as in the equilibrium at 15°C the maximum mass is 208g, the mass of crystals that will form is:

310g - 208g = <em>102g of crystals</em>

<em>-Crystals are the Cr(NO₃)₃⋅9H₂O that is not dissolved-.</em>

I hope it helps!

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Answer:

Kp = 0.049

Explanation:

The equilibrium in question is;

2 SO₂ (g)  +  O₂ (g)   ⇄ 2 SO₃ (g)  

Kp = p SO₃² / ( p SO₂² x p O₂ )

The initial pressures are given, so lets set up the ICE table for the equilibrium:

atm        SO₂         O₂          SO₃

I              3.3        0.79           0

C              -2x           -x          2x

E             3.3 - 2x    0.79 - x    2x

We are told 2x = partial pressure of SO₃ is 0.47 atm at equilibrium, so we can determine the partial pressures of  SO₂ and O₂ as follows:

p SO₂  = 3.3 -0.47 atm = 2.83 atm

p O₂ = 0.79 - (0.47/2) atm = .56 atm

Now we can calculate Kp:

Kp = 0.47² /[ ( 2.83 )² x 0.56 ] = 0.049 ( rounded to 2 significant figures )

Note that we have extra data in this problem we did not need since once we setup the ICE table for the equilibrium we realize we have all the information needed to solve the question.

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What is the wavelength, in nm, of the light photon emitted by a hydrogen atom when an electron goes from n
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Explanation:

8 0
3 years ago
A chemical reaction can theoretically produce 137.5 grams of product, but in actuality 112.9 grams are
Butoxors [25]
<h3>Answer:</h3>

82.11%

<h3>Explanation:</h3>

We are given;

  • Theoretical mass of the product is 137.5 g
  • Actual mass of the product is 112.9 g

We are supposed to calculate the percentage yield

  • We need to know how percentage yield is calculated;
  • To calculate the percentage yield we get the ratio of the actual mass to theoretical mass and express it as a percentage.

Thus;

% yield = (Actual mass ÷ Experimental mass) × 100%

            = (112.9 g ÷ 137.5 g) × 100%

            = 82.11%

Therefore, the percentage yield of the product is 82.11 %

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3 years ago
Assuming the same temperature and pressure for each gas, how many milliliters of carbon dioxide are produced from 16.0 mL of CO?
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Answer:

V_{CO_2}=16.0mL

Explanation:

Hello,

In this case, given that the same temperature and pressure is given for all the gases, we can notice that 16.0 mL are related with two moles of carbon monoxide by means of the Avogadro's law which allows us to understand the volume-moles relationship as a directly proportional relationship. In such a way, since in the chemical reaction:

2CO(g)+O_2(g)\rightarrow 2CO_2(g)

We notice two moles of carbon monoxide yield two moles of carbon dioxide, therefore we have the relationship:

n_{CO}V_{CO}=n_{CO_2}V_{CO_2}

Thus, solving for the yielded volume of carbon dioxide we obtain:

V_{CO_2}=\frac{n_{CO}V_{CO}}{n_{CO_2}} =\frac{2mol*16.0mL}{2mol}\\ \\V_{CO_2}=16.0mL

Best regards.

4 0
3 years ago
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