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elena-14-01-66 [18.8K]
3 years ago
11

The addition of which substance will decrease the solubility of AgNO3 in water?

Chemistry
2 answers:
eduard3 years ago
8 0

<u>Answer:</u> The correct answer is Option 3.

<u>Explanation:</u>

Common ion effect is defined as the effect which occurs on equilibrium when a common ion (an ion which is already present in the solution) is added to a solution. This effect decreases the solubility of a solute.

In silver nitrate, the ions present in the solution are:  Ag^+\text{ and }NO_3^-

For the given options:

<u>Option 1:</u> 0.1 M potassium chloride

The ions formed by the ionization of this solution are K^+\text{ and }Cl^-. As, no common ions are present. So, addition of this substance will not decrease the solubility of silver nitrate solution.

<u>Option 2:</u> 0.1 M barium chloride

The ions formed by the ionization of this solution are Ba^{2+}\text{ and }Cl^-. As, no common ions are present. So, addition of this substance will not decrease the solubility of silver nitrate solution.

<u>Option 3:</u> 0.1 M sodium nitrate

The ions formed by the ionization of this solution are Na^+\text{ and }NO_3^-. As, nitrate ions are present which is a common ion. So, addition of this substance will decrease the solubility of silver nitrate solution.

<u>Option 4:</u> 0.1 M sodium carbonate

The ions formed by the ionization of this solution are Na^+\text{ and }CO_3^{2-}. As, no common ions are present. So, addition of this substance will not decrease the solubility of silver nitrate solution.

Hence, the correct answer is Option 3.

sertanlavr [38]3 years ago
5 0
3 is correct because the .1 moles of the NO3 will be better option to decrease the solubility of the Silver Nitrate.
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Identify the Lewis acid and Lewis base from among the reactants in each of the following equations.
Alex

Answer:

Part A

Ag+ is the Lewis acid and NH3 is the Lewis base.

Part B

AlBr3 is the Lewis acid and NH3 is the Lewis base.

Part C

AlCl3 is the Lewis acid and Cl− is the Lewis base.

Explanation:

A Lewis acid is any specie that accepts a lone pair of electrons. Ag^+, AlBr3 and AlCl3 all accepted lone pairs of electrons according to the three chemical reaction equations shown. Hence, they are Lewis acids.

A Lewis base donates a lone pair of electrons. They include neutral molecules having lone pair of electrons such as NH3 or negative ions such as Cl- .

3 0
3 years ago
Pls help!!<br>plz answer correctly!<br>will give the brainliest!<br>Urgent!!​
blagie [28]

a)

A: Copper

B: CuO

C: \mathrm{CuSO_4}

D: $\mathrm{CuCO_3}$

E: $\mathrm{CO_2}$

F: $\mathrm{Cu(NO_3)_2}$

b)

$\mathrm{CuO+ H_2SO_4}\rightarrow \mathrm{CuSO_4 + H_2O}$

c)

$\mathrm{CuCO_3+ 2HNO_3}\rightarrow \mathrm{Cu(NO_3)_2+ CO_2+ H_2O}$

7 0
3 years ago
Read 2 more answers
1. How is melting point of a solid determined?
Oxana [17]

Answer:

Melting points are often used to characterize organic and inorganic crystalline compounds and to ascertain their purity.

Explanation:

4 0
3 years ago
Determine the empirical formula of a compound containing 1.71 g of silicon and 8.63 g of chlorine.
Basile [38]

Answer:

The answer to your question is: SiCl₄

Explanation:

Data

amount of Si      1.71 g

amount of Cl     8.63 g

MW Si = 28 g

MW Cl = 35.5

Process (rule of three)

For Si                                                        For Cl

        28 g of Si ------------------ 1 mol                      35.5 g of Cl --------------- 1 mol

          1.71g of Si  ---------------   x                              8.63 g of Cl --------------  x

         x = 1.71 x 1 / 28 = 0.06 mol                          x = 8.63 x 1 / 35.5 = 0.24 mol

Now, divide both results by the lowest of them.

Si = 0.06 mol / 0.06 = 1 molecule of Si     Cl = 0.24 / 0.06 = 4 molecules of Cl

Finally

                     Si₁ Cl₄ or SiCl₄

8 0
3 years ago
A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre
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Answer:

\mathbf{0.02 M}

Explanation:

\text{So, from the given question:}

\text{EDTA will make complex with} V^{+3} \text{and the remaining EDTA will react with }Ga^{+3}

\text{Hence, the total concentration of} V^{+3} & Ga^{+3} \text{will be equivalent to EDTA concentration.}

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V_{V^{+3}} = 59.0 \ mL

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M_{EDTA} = 0.0680 \ M

M_{V^{+3}} = ???(unknown)

M_{Ga^{+3}} = 0.0400 \ M

V^{+3} + EDTA \to V[EDTA] + EDTA(Excess)  \to^{CoA} \ Ga[EDTA] _{complex}

M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})

0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18

x = \dfrac{1.18}{59}

\mathbf{x =0.02 \ M }

5 0
3 years ago
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