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ANEK [815]
3 years ago
12

What is the number of molecules present in 1.12 dm^3 of nitrogen gas at STP​

Chemistry
1 answer:
Anuta_ua [19.1K]3 years ago
4 0

Answer:

\huge\boxed{\sf No.\ of\ molecules = 3 * 10\²\² \ molecules}

Explanation:

<u>Given Data:</u>

Volume = v = 1.12 dm³ = 1.12 L

Density of nitrogen at STP = D = 1.25 g / L

Molar mass = M = 14 * 2 = 28 g / mol

Avogadro's Number = \tt{N_{A}} = 6.023 * 10²³ mol⁻¹

<u>Required:</u>

No. of molecules = ?

<u>Formula:</u>

\tt{No. \ of \ molecules = \frac{Density * Volume}{Molar\ Mass} * N_{A}}

<u>Solution:</u>

No. of molecules = (1.25*1.12) / 28 * (6.023 * 10²³)

No. of molecules = ( 1.4 / 28 ) * 6.023 * 10²³

No. of molecules = 0.05 * 6.023 * 10²³

No. of molecules = 0.3 * 10²³

No. of molecules = 3 * 10²² molecules

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
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7 0
3 years ago
You have 15.42g of C2H6. How many moles of H2O can be made?
Amiraneli [1.4K]

<u>Answer:</u> The moles of water produced are 1.54 moles.

<u>Explanation:</u>

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

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of ethane = 15.42 g

Molar mass of ethane = 30.07 g/mol

Putting values in above equation, we get:

\text{Moles of ethane}=\frac{15.42g}{30.07g/mol}=0.513mol

The chemical equation for the combustion of ethane follows:

2C_2H_6+5O_2\rightarrow 4CO2+6H_2O

By Stoichiometry of the reaction:

2 moles of ethane produces 6 moles of water

So, 0.513 moles of ethane will produce = \frac{6}{2}\times 0.513=1.54mol of water

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4 0
3 years ago
Information-
tekilochka [14]

Answer:

So first thing to do in these types of problems is write out your chemical reaction and balance it:

Mg + O2 --> MgO

Then you need to start thinking about moles of Magnesium for moles of Magnesium Oxide. Based on the above equation 1 mole of Magnesium is needed to make one mole of Magnesium Oxide.

To get moles of magnesium you need to take the grams you started with (.418) and convert to moles by dividing by molecular weight of Mg (24.305), this gives you .0172 moles of Mg.

The theoretical yield would be the assumption that 100% of the magnesium will be converted into Magnesium Oxide, so you would get, based on the first equation, .0172 mol of MgO. Multiplying this by the molecular weight of MgO (24.305+16) gives us .693 g of MgO.

The percent yield is what you actually got in the experiment, and for this you subtract off the total mass from the crucible mass, or 27.374 - 26.687, which gives .66 g of MgO obtained.

Percent yield is acutal/theoretical, .66/.693, or 95.24%.

I'll let you do the same for the second trial, and average percent yield is just an average of the two trials percent yield.

Hope this helps.

6 0
3 years ago
10 In a bond between an atom of carbon and an atom of fluorine, the fluorine atom has a
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