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emmainna [20.7K]
3 years ago
9

Ammonium sulfate is used as a fertilizer to supply nitrogen to crops. Write the equation for the disassociation of this compound

when it dissolves in water and include states in your answer.
Chemistry
1 answer:
aksik [14]3 years ago
5 0

Answer:

(NH₄)₂(SO₄)+H₂O⇒(NH₄)₂O+H₂(SO₄)

Explanation:

(NH₄)⁺(SO₄)²⁻+H₂O⇒(NH₄)⁺O²⁻+H₂(SO₄)²⁻

(NH₄)₂(SO₄)+H₂O⇒(NH₄)₂O+H₂(SO₄)

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Select the most likely product for this reaction:
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The answer is B- LiNO3(aq)+H2O(I)

Explanation:

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Similarities between pluton and pegmatite
snow_lady [41]
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7 0
3 years ago
Balance the following half reaction in basic conditions. Then, indicate the coefficients for H2O and OH– for the balanced half r
Ugo [173]

Answer:

The ballance half reactions are:

Mg²⁺  + 2e⁻ → Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

Coefficients for H2O and OH– are 3 for H₂O (in products side) and 6 for OH⁻ (in reactants side)

Explanation:

Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)

Let's see the oxidations number.

As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.

Mg in reactants, acts with +2, so the oxidation number has decreased.

This is the reduction, so it has gained electrons.

Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.

Let's take a look to half reactions:

Mg²⁺  + 2e⁻ → Mg

Si  → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺  + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺  + 4e⁻ → 2Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

2Mg²⁺  + 4e⁻  + 6OH⁻ + Si  → 2Mg  +  SiO₃²⁻ + 4e⁻ + 3 H₂O

7 0
3 years ago
Use the following balanced reaction to solve:
Naily [24]

Answer:  60.7 g of PH_3 will be formed.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}    

\text{Moles of} H_2=\frac{60L}{22.4L}=2.68moles

The balanced chemical reaction is

P_4(s)+6H_2(g)\rightarrow 4PH_3(g)

H_2 is the limiting reagent as it limits the formation of product and P_4 is the excess reagent.

According to stoichiometry :

6 moles of H_2 produce = 4 moles of PH_3

Thus 2.68 moles of H_2 will produce=\frac{4}{6}\times 2.68=1.79moles  of PH_3

Mass of PH_3=moles\times {\text {Molar mass}}=1.79moles\times 33.9g/mol=60.7g

Thus 60.7 g of PH_3 will be formed by reactiong 60 L of hydrogen gas with an excess of P_4

3 0
3 years ago
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