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Aleonysh [2.5K]
3 years ago
9

How many moles are in 1.5 x 10^20 molecules of NH3?

Chemistry
1 answer:
Keith_Richards [23]3 years ago
5 0
<span>Because the question is asking moles of NH3, the compound, any subscripts are irrelevant. It only wants to know how many moles of NH3 molecules, not individual atoms.

Therefore, we can simply convert to moles. 1.5x10^23/6.022x10^23 = .249 moles of NH3.

(If it were to ask moles of Hydrogen, for example, you would multiply the answer by 3, because there are 3 atoms of Hydrogen per one molecule of NH3. But this only asks for moles of the entire compound).  

hope you have a great day! :)
</span>
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When the concentrations of CO2 and H2CO3 are both horizontal lines then the rate of the forward reaction is the same as the rate of the reverse reaction.

<h3>What is rate of reaction?</h3>

The term rate of reaction refers to how fast or slow a reaction proceeds. Recall that the rate  of reaction is measured from the rate of disappearance of reactants or the rate of appearance of products.

When the [CO2] and [H2CO3 ] are both horizontal lines, the rate of the forward reaction is the same as the rate of the reverse reaction.

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Answer:

1-1) NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

1-2) 0.5 mole of CO2

2-1) 2C4H10 + 13O2 --> 8CO2 + 10H2O

2-2) 4 mol CO2

Explanation:

<u>Question 1</u>

NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

<em>To balance the equation, count the number of atoms on both sides of the equation</em>

(1 Na, 1+3+1H, 1+1+1C, 3+2Oxygen) --> (1 Na, 1+1+1C, 3+2H, 2+1+2Oxygen)

<em>Combining the pluses will give you the following</em>

(1 Na, 5H, 3C, 5Oxygen) --> (1 Na, 3C, 5H, 5Oxygen)

<em>Both sides are the same, therefore the chemical equation is balanced (originally). </em>

From the equation, we can see that <u>1 mole of NaHCO3</u> produces <u>1 mole of CO2</u>.

So that means <u>0.5 mole of NaHCO3</u> would produce <u>0.5 mole of CO2</u>.

<u>Question 2</u>

C4H10 + O2 --> CO2 + H2O

<em>Again, count the number of atoms on both sides of the equation</em>

(4C, 10H, 2O) --> (1C, 2H, 3O)     <em>This time left does not equal right side</em>

<em>You now need to find </em><u><em>factors </em></u><em>that can make both sides equal. </em>

C4H10 + O2 --> <u>4</u>CO2 + H2O    <em>Now the C is balanced, let's recount </em>

<em>(4C, 10H, 2Oxygen) --> (4C, 8+1Oxygen, 2H)      H&O is still not balanced</em>

C4H10 + O2 --> 4CO2 + <u>5</u>H2O    <em>Now the H is balanced, let's recount</em>

<em>(4C, 10H, 2Oxygen) --> (4C, 8+5Oxygen, 10H)      O is still not balanced</em>

C4H10 + (<u>13/2</u>)O2 --> 4CO2 + 5H2O    <em>Now the O is balanced</em>

<em>(4C, 10H, 13Oxygen) --> (4C, 13Oxygen, 10H)</em>

<em>But because 13/2 is a fraction, we want to eliminate that by multiplying every reactant and product by 2 (the denominator).</em>

<u>2</u>C4H10 + <u>13</u>O2 --> <u>8</u>CO2 + <u>10</u>H2O    Now it's completely balanced!

<em>(8C, 20H, 28Oxygen) --> (8C, 28Oxygen, 20H)     Yayy! It's balanced.</em>

Now, 2 mol C4H10 produces 8 mol CO2.

So 1 mol C4H10 produces 4 mol CO2.

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