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Ira Lisetskai [31]
3 years ago
7

Which type of chemical reaction does this word equation exemplify: iron metal plus copper (II) sulfate solution yields iron (II)

sulfate solution and copper metal?
Chemistry
2 answers:
olchik [2.2K]3 years ago
4 0

Answer: Single replacement

Fe+CuSO_4\rightarrow Cu+FeSO_4

Explanation:

Single replacement reaction is a chemical reaction in which more reactive element displaces the less reactive element from its salt solution.

Iron being more reactive than copper would displace copper from its salt solution and thus will generate copper and iron sulphate.

The chemical equation representing iron metal plus copper (II) sulfate solution yields iron (II) sulfate solution and copper metal is:

Fe+CuSO_4\rightarrow Cu+FeSO_4

yan [13]3 years ago
3 0

Answer:

Synthesis

Explanation:

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What mass of hydrogen sulfide, H2S, will completely react with 2.00 moles of silver nitrate, AgNO3?
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Answer:

34g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

H2S + 2AgNO3 —> 2HNO3 + Ag2S

Next, we shall determine the number of mole of H2S required to react with 2 moles of AgNO3.

This is illustrated below:

From the balanced equation above,

We can see that 1 mole of H2S is required to react completely with 2 moles of AgNO3.

Finally, we shall convert 1 mole of H2S to grams. This is shown below:

Number of mole H2S = 1 mole

Molar mass of H2S = (2x1) + 32 = 34g/mol

Mass = number of mole x molar Mass

Mass of H2S = 1 x 34

Mass of H2S = 34g

Therefore, 34g of H2S is needed to react with 2 moles of AgNO3.

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What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )
yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution :  Given,

Concentration (c) = 0.150 M

Acid dissociation constant = k_a=4.5\times 10^{-4}

The equilibrium reaction for dissociation of HNO_2 (weak acid) is,

                           HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+

initially conc.         c                       0         0

At eqm.              c(1-\alpha)                c\alpha        c\alpha

First we have to calculate the concentration of value of dissociation constant (\alpha ).

Formula used :

k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}

Now put all the given values in this formula ,we get the value of dissociation constant (\alpha ).

4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}

4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2

0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0

By solving the terms, we get

\alpha=0.0533

No we have to calculate the concentration of hydronium ion or hydrogen ion.

[H^+]=c\alpha=0.150\times 0.0533=0.007995 M

Now we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.007995 M)

pH=2.097\approx 2.1

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0
3 years ago
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