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polet [3.4K]
3 years ago
8

write a net ionic equation for the reaction that occurs when aqueous solutions of perchloric acid and potassium hydroxide are co

mbined.
Chemistry
1 answer:
BaLLatris [955]3 years ago
3 0

Answer:

H⁺(aq) + OH⁻(aq) ⇒ H₂O(l)

Explanation:

Let's consider the molecular equation that occurs when aqueous solutions of perchloric acid and potassium hydroxide are combined. This is a neutralization reaction.

HClO₄(aq) + KOH(aq) ⇒ KClO₄(aq) + H₂O(l)

The complete ionic equation includes all the ions and molecular species.

H⁺(aq) + ClO₄⁻(aq) + K⁺(aq) + OH⁻(aq) ⇒ K⁺(aq) + ClO₄⁻(aq) + H₂O(l)

The net ionic equation includes only the ions that participate in the reaction and the molecular species.

H⁺(aq) + OH⁻(aq) ⇒ H₂O(l)

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A number of valence electrons can be determined by looking at the... *
inessss [21]

Answer:

The main group number for an element can be found from its column on the periodic table.

Explanation:

For example, carbon is in group 4 and has 4 valence electrons. Oxygen is in group 6 and has 6 valence electrons.

8 0
3 years ago
The solid fuel in the booster stage of the space shuttle is a mix- ture of ammonium perchlorate and aluminum powder. Upon igni-
nataly862011 [7]

<u>Answer:</u>

<u>For A:</u> The mass of aluminium required will be 183 g

<u>For B:</u> The mass of alumina produced will be 6.63\times 10^6g

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For a:</u>

Given mass of NH_4ClO_4 = 1.325 kg = 1325 g   (Conversion factor:  1 kg = 1000 g)

Molar mass of NH_4ClO_4 = 117.50 g/mol

Putting values in equation 1, we get:

\text{Moles of }NH_4ClO_4=\frac{1325g}{117.50g/mol}=11.28mol

For the given chemical reaction:

6NH_4ClO_4(s)+10Al(s)\rightarrow 5Al_2O_3(s)+3N_2(g)+6HCl(g)+9H_2O(g)

By stoichiometry of the reaction:

6 moles of NH_4ClO_4 reacts with 10 moles of aluminium

So, 11.28 moles of NH_4ClO_4 will react with = \frac{10}{6}\times 11.28=6.77mol of aluminium

Now, calculating the mass of aluminium by using equation 1, we get:

Molar mass of aluminium = 27.00 g/mol

Moles of aluminium = 6.77 moles

Putting values in equation 1, we get:

6.77mol=\frac{\text{Mass of aluminium}}{27.00g/mol}\\\\\text{Mass of aluminium}=(6.77mol\times 27.00g/mol)=183g

Hence, the mass of aluminium required will be 183 g

  • <u>For b:</u>

Given mass of aluminium = 3.500\times 10^3=3.5\times 10^6g   (Conversion factor:  1 kg = 1000 g)

Molar mass of aluminium = 27.00 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium}=\frac{3.5\times 10^6g}{27.00g/mol}=1.3\times 10^5mol

For the given chemical reaction:

6NH_4ClO_4(s)+10Al(s)\rightarrow 5Al_2O_3(s)+3N_2(g)+6HCl(g)+9H_2O(g)

By stoichiometry of the reaction:

10 moles of aluminium produces 5 moles of alumina

So, 1.3\times 10^5mol of aluminium will react with = \frac{5}{10}\times 1.3\times 10^5=6.5\times 10^4mol of alumina

Now, calculating the mass of alumina by using equation 1, we get:

Molar mass of alumina = 101.96 g/mol

Moles of alumina = 6.5\times 10^4mol

Putting values in equation 1, we get:

6.5\times 10^4mol=\frac{\text{Mass of alumina}}{101.96g/mol}\\\\\text{Mass of alumina}=(6.5\times 10^4mol\times 101.96g/mol)=6.63\times 10^6g

Hence, the mass of alumina produced will be 6.63\times 10^6g

6 0
3 years ago
If 10 grams of aluminum chloride are reacted how many atoms of aluminum will be produced​
LekaFEV [45]

Answer:

26.3 Yield

Explanation:

2AICI-3+ 2AI + 3CI-2

2.10g

           =  26.3 Yield

7.98g

6 0
3 years ago
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Composition, and a c<span>hemical reaction.</span>
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B. Number of protons
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