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ladessa [460]
3 years ago
10

What does lead ii acetate and hydrogen sulfide produce? and what type of reaction is it?

Chemistry
1 answer:
kati45 [8]3 years ago
5 0
When Lead (II) acetate and Hydrogen sulfide react, they form Lead sulfide and Acetic acid. The reaction is a reduction-oxidation (redox) reaction.
The balanced chemical reaction is this:
Pb(C2H3O2)2 + H2S --> PbS + 2C2H4O2
And the net ionic reaction is this:
Pb2+ + S2- --> PbS
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Density is the ratio of mass to volume. Another way to say this is
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Answer:

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Please Help, will give 30 points! The following data was collected when a reaction was performed experimentally in the laborator
Sedaia [141]

Answer:

9 moles of NaNO3.

Explanation:

The balanced equation for the reaction is given below:

Al(NO3)3 + 3NaCl —> 3NaNO3 + AlCl3

Next, we shall determine the number of mole of Al(NO3)3 and NaCl that reacted and the number of mole of NaNO3 produced from the balanced equation.

From the balanced equation above,

1 mole of Al(NO3)3 reacted with 3 moles of NaCl to produce 3 moles of NaNO3.

Next, we shall determine the limiting reactant.

The limiting reactant can be obtained as follow:

From the balanced equation above,

1 mole of Al(NO3)3 reacted with 3 moles of NaCl.

Therefore, 4 moles of Al(NO3)3 will react with = (4 x 3)/1 = 12 moles of NaCl.

From the calculations made above, we can see that it will take a higher amount i.e 12 moles than what was given i.e 9 moles of NaCl to react completely with 4 moles of Al(NO3)3.

Therefore, NaCl is the limiting reactant and Al(NO3)3 is the excess reactant.

Finally, we shall determine the maximum amount of NaNO3 produced from the reaction.

In this case, the limiting reactant will be used as it will produce the maximum amount of NaNO3 since all of it is consumed by the reaction.

The limit reactant is NaCl and the maximum amount of NaNO3 produced can be obtained as follow:

From the balanced equation above,

3 moles of NaCl reacted to produce 3 moles of NaNO3.

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From the calculations made above, the maximum amount of NaNO3 produced is 9 moles

6 0
3 years ago
The equilibrium constant for the reaction
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Answer: The concentrations of Cl_2 at equilibrium is 0.023 M

Explanation:

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Volume of solution = 1 L

Initial concentration of Cl_2 = \frac{0.14mol}{1L}=0.14M

The given balanced equilibrium reaction is,

                            COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)

Initial conc.           0.14 M           0 M       0M    

At eqm. conc.     (0.14-x) M        (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}

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

4.63\times 10^{-3}=\frac{x)^2}{(0.14-x)}

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A molecule of glucose is comprised of 6 atoms of carbon, 12 atoms of hydrogen, and 6 atoms of oxygen.
Tcecarenko [31]
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