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Nutka1998 [239]
2 years ago
10

When calcium carbonate reacts with hydrochloric acid, calcium ions are transferred to the solution. Write only one net ionic equ

ation for this reaction:
Chemistry
1 answer:
nika2105 [10]2 years ago
6 0

Answer:

In conclusion, the net ionic equation for the reaction between calcium carbonate and hydrochloric acid is CaCO3 solid plus two H+ aqueous react to form Ca2+ aqueous plus CO2 gas plus H2O liquid.

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3 years ago
Why estuaries are an important habitat for many marine organisms in comparison to the open ocean?
Alexxandr [17]

Answer:

There are more nutrients available in estuaries. There are more nutrients available in estuaries.

Explanation:

5 0
3 years ago
Science question down below
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3 years ago
For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi
natka813 [3]

<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<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 Iron:</u>

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:  

\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol

For the given chemical reaction:

2Fe(s)+O_2(g)\rightarrow 2FeO(s)

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = \frac{2}{2}\times 0.0771=0.0771mol of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:  

0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g

To calculate the percentage yield of iron (ii) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0
3 years ago
When electrolyzing copper (ll) chloride, what reaction takes place at the anode? What reaction takes place at the cathode?
klemol [59]

Answer:

Copper ions are reduced into copper atoms.

Cu²⁺₍aq₎ + 2e⁻ →  Cu₍s₎

Explanation:

During electrolysis,  the positive H⁺ and Cu⁺ ions move to the negative cathode and negative OH⁻ and Cl⁻ ions move to the positive anode.

At cathode, copper ions are preferentially discharged due to the low electromotive force required to discharge them compared to the hydrogen ion. The copper ions gain the two electrons lost by the chloride ions when the are discharged. (2 Cl⁻₍aq₎ → Cl₂₍g₎ + 2e⁻)

Thus the half equation is as follows:

Cu²⁺₍aq₎ + 2e⁻ →  Cu₍s₎

5 0
3 years ago
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