Write the net ionic equation for the reaction that takes places between aqueous copper and nitrate.

Determine the mass of carbon iv oxide ,produced on burning 104g of ethyne

mars1129 [50]

163 grams (3 sig. fig.).

<h3>Explanation</h3>

Formula of <em>carbon(IV) oxide</em> (a.k.a. carbon dioxide): $\text{CO}_2$ .
Molar mass of $\text{CO}_2$ : $\underbrace{12.01}_{\text{C}} + 2\times\underbrace{16.00}_{\text{O}}=44.01\;\text{g}\cdot\text{mol}^{-1}$ .

Formula of ethyne: structural $\text{H}-\text{C}\equiv\text{C}-\text{H}$ or molecular $\text{C}_2\text{H}_2$ .
Molar mass of $\text{C}_2\text{H}_2$ : $2\times\underbrace{12.01}_{\text{C}}+2 \times\underbrace{16.00}_{\text{O}} = 56.02\;\text{g}\cdot\text{mol}^{-1}$ .

All carbon atoms in that 104 grams of ethyne will end up in $\text{CO}_2$ . Number of moles of molecules in 104 grams of ethyne:

$n = \dfrac{m}{M} = \dfrac{104}{56.02} = 1.85648\;\text{mol}$ .

There are two carbon atoms in each ethyne molecule. Number of carbon atoms in that many ethyne molecules:

$n(\text{C}) = 2\;n(\text{C}_2\text{H}_2) = 3.71296\;\text{mol}$ .

There are one carbon atom in each $\text{CO}_2$ molecule. In case oxygen is in excess, all those carbon atoms from that 104 grams of ethyne will make $n(\text{CO}_2) = n(\text{C}) =3.71296\;\text{mol}$ of $\text{CO}_2$ .

Mass of all those $\text{CO}_2$ molecules:

$m = n\cdot M = 163\;\text{g}$ . (3 sig. fig. as in the mass of ethyne.)

7 0

3 years ago

How much heat is required to vaporize 43.9 g of acetone at its boiling point?

bogdanovich [222]

The heat required to vaporize 43.9 g of acetone at its boiling point is calculated as below

the heat of vaporization of acetone at its boiling point is 29.1 kj/mole

find the moles of acetone = mass/molar mass
= 43.9g /58 g/mol =0.757 moles

heat (Q) = moles x heat of vaporization

= 29.1 kj/mole x 0.757 moles = 22.03 kj

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3 years ago

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Which condition increases the number of collisions between reactant molecules in a given volume?

Nezavi [6.7K]

Answer:

Increasing Surface Area

Explanation:

A greater surface area (meaning more, smaller particles) allows for more opportunity for particles to collide. On the other hand, decreasing temperature and removing a catalyst would only decrease the number of collisions, and the clumping option doesn't make much sense. Hope this helps!

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3 years ago

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What is the molarity of a solution prepared by dissolving 10.7 g NaI in 0.250 L?0.0714

hjlf

Answer:0.286M

Explanation:

10.7g / 149.89g/mol×0.25L

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3 years ago

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For each of the following reactions, identify the missing reactant(s) or products(s) and then balance the resulting equation. No

Pachacha [2.7K]

Answer:

A. The reactants are= Li and O2

The balanced equation is:

4Li + O2 —> 2Li2O

B. The products are: MgCl2 and O2

The balanced equation is:

Mg(ClO3)2 —> MgCl2 + 3O2

C. The products are: Ca(NO3)2 and H2O

The balanced equation is:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 +

2H2O

D. The products are: CO2 and H2O

The balanced equation is:

C5H12 + 8O2 —> 5CO2 + 6H2O

Explanation:

A. ____ —> Li2O

The reactants are Li and O2. Thus the equation is given below:

Li + O2 —> Li2O

Thus the equation is balanced as follow:

There are 2 atoms of O on the left side and 1 atom on the right side. It can be balance by putting 2 in front of Li2O as shown below:

Li + O2 —> 2Li2O

Now, we have 4 atoms of Li on the right side and 1 atom on the left. It can be balance by putting 4 in front of Li as shown below:

4Li + O2 —> 2Li2O

Now the equation is balanced

B. Mg(ClO3)2 —> __

The products are MgCl2 and O2

The equation is given below:

Mg(ClO3)2 —> MgCl2 + O2

The equation can be balance as follow:

There are 6 atoms of O on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of O2 as shown below:

Mg(ClO3)2 —> MgCl2 + 3O2

Now the equation is balanced

C. HNO3 + Ca(OH)2 —>___

The products are: Ca(NO3)2 and H2O

The equation is given below:

HNO3 + Ca(OH)2 —> Ca(NO3)2 + H2O

The equation is balanced as follow:

There are 2 atoms of NO3 on the right side and 1 atom on the left. It can be balance by putting 2 in front of HNO3 as shown below:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 + H2O

There are a total of 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of H2O as shown below:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 + 2H2O

Now the equation is balanced

D. C5H12 + O2 —>__

The products are: CO2 and H2O

The equation is given below:

C5H12 + O2 —> CO2 + H2O

The equation can be balance as follow:

There are 5 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 5 in front of CO2 as shown below:

C5H12 + O2 —> 5CO2 + H2O

There are 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H2O as shown below:

C5H12 + O2 —> 5CO2 + 6H2O

Now, there are a total of 16 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 8 in front of O2 as shown below:

C5H12 + 8O2 —> 5CO2 + 6H2O

Now we can see that the equation is balanced.

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3 years ago

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