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Scilla [17]
4 years ago
11

Suppose you have a solution that might contain or all of the following cations: Cu2+, Ag+, Ba2+, and Mn2+. Addition of HBr cause

s a precipitate to form. After the precipitate is filtered off, H2SO4 is added to the supernate another precipitate forms. This precipitate is filtered off, and a solution of NaOH is added to the supernatant liquid until it is strongly alkaline. No precipitate is formed. Which ions are present in each of the precipitates? Which cations are not present in the original solution?
Chemistry
2 answers:
Aneli [31]4 years ago
8 0
Bat2 is the cations for your question
timofeeve [1]4 years ago
3 0

Answer:

Both Ba2+ and Ag+ are present.

Neither Cu2+ nor Mn^2+ are present.

Explanation:

Hello,

At first, the addition of HBr promotes the precipitation of AgBr as it is largely insoluble in water. Next, the addition of H2SO4 promotes the precipitation of BaSO4 as it is largely insoluble in water as well.

Finally, as no precipitate was formed due to the addition of sodium hydroxide and both Cu(OH)2 and Mn(OH)2, one concludes they aren't present into the solution.

Best regards.

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At a certain temperature, 0.800 mol SO 3 is placed in a 1.50 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) At equil
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Answer : The value of equilibrium constant (K) is, 0.004

Explanation :

First we have to calculate the concentration of SO_3\text{ and }O_2

\text{Concentration of }SO_3=\frac{\text{Moles of }SO_3}{\text{Volume of solution}}=\frac{0.800mol}{1.50L}=1.2M

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Now we have to calculate the value of equilibrium constant (K).

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                       2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)

Initial conc.       1.2             0               0

At eqm.          (1.2-2x)        2x               x

As we are given:

Concentration of O_2 at equilibrium = x = 0.1 M

The expression for equilibrium constant is:

K_c=\frac{[SO_2]^2[O_2]}{[SO_3]^2}

Now put all the given values in this expression, we get:

K_c=\frac{(2x)^2\times (x)}{(1.2-2x)^2}

K_c=\frac{(2\times 0.1)^2\times (0.1)}{(1.2-2\times 0.1)^2}

K_c=0.004

Thus, the value of equilibrium constant (K) is, 0.004

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If 0.896 g of a gas occupies a 250 mL flask at 20°C and 760 mm Hg of pressure, what is the molar mass of the gas?​
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Answer:

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Explanation:

Before you can find the molar mass, you first need to calculate the number of moles of the gas. To find this value, you need to use the Ideal Gas Law:

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In this equation,

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-----> n = moles

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After you convert the volume from mL to L and the temperature from Celsius to Kelvin, you can use the equation to find the moles.

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n = ? moles

PV = nRT

(760 mmHg)(0.250 L) = n(62.36 L*mmHg/mol*K)(293.15 K)

190 = n(18280.834)

0.0104 = n

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?grams = 86.2                                               <----- Divide both sides by 0.0104

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