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3 years ago

Chemistry!! Please help asap!! Will mark brainiliest!!

Chemistry

1 answer:

Whitepunk [10]3 years ago

5 0

<h2>Answer:</h2><h3>Part 1. </h3>

Option B is correct option.

The half-reaction 2MnO2 + H2O + 2e- Mn2O3 is missing OH- ions.

Explanation:

Full equation:

2MnO2 + H2O + 2e- → Mn2O3 + 2OH-

The option B which is Mg is stronger reducing agent than Ag is correct option.

Explanation:

Equation:

Mg(s) + 2Ag+ (aq) → Mg2+ (aq) + 2Ag(s)

According to equation Mg converts to Mg+2 which means it gives to electron to reduce Ag. So it act as an reducing agent.

The correct option is B. Which is 5, 1, 8, 5, 1, 4.

Explanation:

Full equation :

5 Fe²⁺ (aq) + MnO₄⁻ (aq) + 8 H⁺ (aq) --> 5 Fe³⁺ (aq) + Mn²⁺ (aq) + 4 H₂O (l)

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Calculate the mass of magnesium carbonate ( MgCO3), in grams, required to produce 110.0 g of carbon dioxide using the following

bearhunter [10]

Answer:

$210.7~g~MgCO_3$

Explanation:

We have to start with the <u>reaction</u>:

$MgCO_3~->~MgO~+~CO_2$

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the <u>number of moles</u> that we have in the 110.0 g of carbon dioxide, to this, we have to know the <u>atomic mass of each atom</u>:

C: 12 g/mol

O: 16 g/mol

Mg: 23.3 g/mol

If we take into account the number of atoms in the formula, we can calculate the <u>molar mass</u> of carbon dioxide:

$(12*1)+(16*2)=44~g/mol$

In other words: $1~mol~CO_2=~44~g~CO_2$ . With this in mind, we can calculate the moles:

$110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2$

Now, the <u>molar ratio</u> between carbon dioxide and magnesium carbonate is 1:1, so:

$2.5~mol~CO_2=2.5~mol~MgCO_3$

With the molar mass of $MgCO_3$ ( $(23.3*1)+(12*1)+(16*3)=84.3~g/mol$ . With this in mind, we can calculate the <u>grams of magnesium carbonate</u>:

$2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3$

I hope it helps!

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Answer:

I think it's A

Explanation:

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