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3 years ago

Cabr2 will most likely dissolve in which solvent? 1. bi3 2. h2o 3. br2 4. ccl4 5. c8h18

1 answer:

5 0

General principle of solubility is 'like dissolves like'

$CaBr_{2}$ is an ionic compound, wherein the constituent ions ( $Ca^{2+}$ and $Br^{-}$ ) are held by electrostatic forces of interaction.

Such ionic compounds are soluble in polar solvents.

Among the solvent mentioned in question, water ( $H_{2}O$ ) has maximum polarity. Hence, $CaBr_{2}$ is most likely to dissolve in $H_{2}O$

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Questions 18-20 refer to the reaction 2NH3 (g) ↔ N2 (g) + 3H2 (g), and ΔH = 92.2 kJ.

Fittoniya [83]

18. Answer is A

Since the enthalpy of reaction is positive, the forward reaction is an endothermic reaction which means the energy is gained from the surrounding to happen the reaction. If the temperature decreases, according to the Le Chatelier's principle, the system tries to become equilibrium by increasing temperature. Since forward reaction is endothermic (because of the bond breaking), the backward reaction is exothermic (because of the bond making) which releases the energy to the surroundings. This makes the increase of temperature. So if the backward reaction is promoted because of the decrease of temperature, then the concentration of H₂ will decrease.

19. Answer is A.

The reactant side has 2 moles/molecules of reactants and the product side has 4 moles/molecules of products which come from 1 N₂(g) and 3 H₂(g). If the pressure is reduced in the system, according to the Le Chatelier's principle, the system tries to increase the pressure. Hence, forward reaction is promoted because of the higher number of molecules in product side. If the forward reaction is promoted, the concentration of NH₃(g) will decreased.

20. Answer is C.

If the concentration of reactant is increased in the system, according to the Le Chatelier's principle, the system tries to reduce the concentration of that reactant. So if NH₃(g) concentration is increased, then to be equilibrium, the forward reaction will be promoted. Then the concentration of N₂(g) will increase.

4 0

3 years ago

A saturated solution of Pb(IO3)2 in pure water has a lead ion concentration of 5.0 x 10-5 Molar. What is the Ksp value of Pb(IO3

Orlov [11]

Answer:

Option (E) is correct

Explanation:

Solubility equilibrium of $Pb(IO_{3})_{2}$ is given as follows-

$Pb(IO_{3})_{2}\rightleftharpoons Pb^{2+}+2IO_{3}^{-}$

Hence, if solubility of $Pb(IO_{3})_{2}$ is S (M) then-

$[Pb^{2+}]=S(M)$ and $[IO_{3}^{-}]=2S(M)$

Where species under third bracket represent equilibrium concentrations

So, solubility product of $Pb(IO_{3})_{2}$ , $K_{sp}=[Pb^{2+}][IO_{3}^{-}]^{2}$

Here, $[Pb^{2+}]=S(M)=5.0\times 10^{-5}M$

So, $[IO_{3}^{-}]=2S(M)=(2\times 5.0\times 10^{-5})M=1.0\times 10^{-4}M$

So, $K_{sp}=(5.0\times 10^{-5})\times (1.0\times 10^{-4})^{2}=5.0\times 10^{-13}$

Hence option (E) is correct

7 0

3 years ago

Please help in Chemistry!!!! 15 points! I added a picture below...

krek1111 [17]

The answer would be choice B because the energy decreased by 20 J

5 0

3 years ago

Read 2 more answers

A(n) wave carries energy through matter.

denis23 [38]

Answer:

A wave is a disturbance that carries energy from one place to another through matter and space. When we through a stone or a pebble in calm water, then the particles of water moves up and down and this process continues for some time. This implies that there is a disturbance produced in water.

Explanation:

4 0

3 years ago

A sample of 53.0 of carbon dioxide was obtained by heating 1.31 g of calcium carbont. What is the percent yield for this reactio

MaRussiya [10]

Answer:

92.04%

Explanation:

Given:

Mass of CO₂ obtained = 53.0 grams

Mass of calcium carbonate heated = 1.31 grams

Now,

the molar mass of the calcium carbonate = 100.08 grams

The number of moles heated in the problem = Mass / Molar mass

= (1.31 grams) / (100.08 grams/moles)

= 0.013088 moles

now,

1 mol of calcium carbonate yields 1 mol of CO₂

thus,

0.013088 moles of calcium carbonate will yield = 0.013088 mol of CO₂

now,

Theoretical mass of 0.013088 moles of CO₂ will be

= Number of moles × Molar mass of CO₂

= 0.013088 × 44 = 0.5758 grams

Thus, the percent yield for this reaction = $\frac{\textup{Actual yield}}{\textup{Theoretical yield}}\times100$

the percent yield for this reaction = $\frac{0.53}{0.5758}\times100$

the percent yield for this reaction = 92.04%

6 0

3 years ago