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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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1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

Best regards!

8 0
3 years ago
The reaction of Pb+4 and O-2 produces the compound of
vovikov84 [41]

Answer:

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Explanation:

The given species are:

        Pb⁴⁺               O²⁻  

Now, to solve this problem, we use the combining powers which corresponds to the number of electrons usually lost or gained or shared by atoms during the course of a chemical combination.

                                          Pb⁴⁺               O²⁻  

Combining power               4                    2

Exchange of valencies        2                    4

Now the molecular formula is  Pb₂O₄

                               

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