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mr Goodwill [35]
3 years ago
11

How many total atoms are in 0.560 g of P2O5

Chemistry
1 answer:
ioda3 years ago
7 0

Answer:

16.499 × 10∧ 21 atoms

Explanation:

Given data:

mass of P2O5= 0.560 g

number of atoms= ?

first of all we will calculate the molar mass of P2O5:

P = 2×31 g/mol = 62 g/mol

O = 5× 16 = 80 g/mol

molar mass of P2O5 = 142 g/mol

Noe we will find the moles of 0.560 g P2O5:

moles = mass / molar mass

moles = 0.560 g/ 142 g/mol

moles = 0.0039 mol

now we will find the atoms present in 0.0039 moles:

0.0039 × 6.02 × 10∧ 23 molecules

2.357 × 10∧ 21 molecules

P2O5 consist of 7 atoms:

2.357 × 10∧ 21  × 7 = 16.499 × 10∧ 21 atoms

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Answer:

The volume of first and second compound are 9.15 ml and 10.85 ml.

Explanation:

Given that,

Density of graphite = 2.25 g/cm³

Volume of mixture = 20.0 mL

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Density of second compound = 2.890 g/ml

Let the volume of first mixture = x

The volume of second mixture = (20-x)....(I)

We need to calculate the volume of first compound

Using formula of  density of mixture

\rho=\dfrac{V_{1}\rho_{1}+V_{2}\rho_{2}}{V_{1}+V_{2}}

Where, V_{1} = volume of first compound

V_{2} = volume of second compound

\rho_{1} =density of first compound

\rho_{1} = density of first compound

Put the volume into the formula

2.25=\dfrac{x\times1.492+(20-x)\times2.890}{x+20-x}

45=1.492x+57.8-2.890x

45-57.8=1.492x-2.890x

12.8=1.398x

x=\dfrac{12.8}{1.398}

x=9.15\ ml

We need to calculate the volume of second compound

Using equation (I)

V_{2}=20-x

Put the value of x

V_{2}=20-9.15

V_{2}=10.85\ ml

Hence, The volume of first and second compound are 9.15 ml and 10.85 ml.

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What is the structure of a fluorine atom?
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How many grams of nickel (ni, 58.69 g/mol) are in 1.57 moles of nickel?
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3 0
4 years ago
Read 2 more answers
Find percent yield:
saveliy_v [14]

<u>Answer:</u> The percent yield of the reaction is 91.8 %

<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 B_5H_9 :</u>

Given mass of B_5H_9 = 4.0 g

Molar mass of B_5H_9 = 63.12 g/mol

Putting values in equation 1, we get:

\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol

The chemical equation for the reaction of B_5H_9 and oxygen gas follows:

2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of B_2H_5

So, 0.3125 moles of oxygen gas will react with = \frac{2}{12}\times 0.3125=0.052mol of B_2H_5

As, given amount of B_2H_5 is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of B_2O_3

So, 0.3125 moles of oxygen gas will produce = \frac{5}{12}\times 0.3125=0.130moles of water

Now, calculating the mass of B_2O_3 from equation 1, we get:

Molar mass of B_2O_3 = 69.93 g/mol

Moles of B_2O_3 = 0.130 moles

Putting values in equation 1, we get:

0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g

To calculate the percentage yield of B_2O_3, we use the equation:

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

Experimental yield of B_2O_3 = 8.32 g

Theoretical yield of B_2O_3 = 9.052 g

Putting values in above equation, we get:

\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%

Hence, the percent yield of the reaction is 91.8 %

6 0
4 years ago
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