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3 years ago

If aqueous solutions of sodium chloride and mercury(I) nitrate are mixed, which insoluble precipitate is formed?

Chemistry

1 answer:

NeX [460]3 years ago

7 0

Answer:

Hg2^2+(aq) + 2Cl^-(aq) —> Hg2Cl2(s)

Explanation:

The balanced equation for the reaction is given below:

2NaCl(aq) + Hg2(NO3)2(aq) —> 2NaNO3(aq) + Hg2Cl2(s)

Considering the states of each compound in the reaction, we can see that Hg2Cl2 is in solid form meaning it will precipitate out of the solution

In solution the following occurs:

NaCl —> Na+(aq) + Cl-(aq)

Hg2(NO3)2 —> Hg2^2+(aq) + 2NO3^-(aq)

Combining the two equation together, a balanced double displacement reaction occurs as shown below:

2Na+(aq) + 2Cl-(aq) + Hg2^2+(aq) + 2NO3^-(aq) —> 2Na+2NO3^-(aq) + Hg2^2+2Cl-(s)

From the above we can thus right the insoluble precipitate as

Hg2^2+(aq) + 2Cl^-(aq) —> Hg2Cl2(s)

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V1 = l; P1=40Pa; P2= 100 kpa; V2= 1.0 L please help!

blsea [12.9K]

Answer:

V1 = 2500 L

Explanation:

V1P1 = V2P2

x * 40 = 100.000 * 1.0

==> x =2500 L

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3 years ago

How do you distinguish a chemical change from a physical change

kirill [66]

Answer:

One distinguishing feature between physical and chemical changes is that a physical change can be reverse or turn back to its original form, but chemical change cannot be return to its form because the material change.

Explanation:

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3 years ago

What is the mass of oxygen that can be produced from 2.79 moles of lead(ll) nitrate

denis23 [38]

1.38 moles of oxygen

Explanation:

Thermal decomposition of Lead (II) nitrate is shown by the balanced equation below;

2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂

The mole ration of Lead (II) nitrate to oxygen is 2: 1

Therefore 2.76 moles of Lead (II) nitrate will lead to production of? moles of oxygen;

2: 1

2.76: x

Cross-multiply;

2x = 2.76 * 1

x = 2.76 / 2

x = 1.38

8 0

3 years ago

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

Vilka [71]

Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

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

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

$K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}$

$K_c=1.066\aprrox 1.1$

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0

3 years ago

What is the IUPAC name of KNO3<br>

Tasya [4]

Answer: potassium nitrate

hope this helps

4 0

3 years ago

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