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How many atoms of phosphorus are in 4.5 grams of tetraphosphorus decoxide?

Alexxx [7]

Answer:

In 4.5 grams of tetraphosphorus decoxide we have 3.85 * 10^22 phosphorus atoms

Explanation:

Step 1: Data given

tetraphosphorus decoxide = P4O10

Molar mass of P4O10 = 283.89 g/mol

Mass of P4O10 = 4.5 grams

Number of Avogadro = 6.022 * 10^23 / mol

Step 2: Calculate moles of P4O10

Moles P4O10 = mass P4O10 / molar mass P4O10

Moles P4O10 = 4.5 grams / 283.89 g/mol

Moles = 0.016 moles

Step 3: Calculate moles of P

For 1 mol P4O10 we have 4 moles of phosphorus

For 0.016 moles P4O10 we have 4*0.016 = 0.064 moles P

Step 4: Calculate number of P atoms

Number of P atoms = moles P * number of Avogadro

Number of P atoms = 0.064 moles * 6.022*10^23

Number of P atoms = 3.85 * 10^22 atoms

In 4.5 grams of tetraphosphorus decoxide we have 3.85 * 10^22 phosphorus atoms

3 0

3 years ago

Manganese forms several oxides when combined with oxygen. One of the oxides (Oxide 1) contains 63.2% of Mn and another oxide (Ox

Nina [5.8K]

Explanation:

Defining law of definite proportions, it states that when two elements form more than one compound, the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers.

A. One of the oxides (Oxide 1) contains 63.2% of Mn.

Mass of the oxide = 100g

Mass of Mn = 63.2 g

Mass of O = 100 - 63.2

= 36.8 g

Ratio of Mn to O = 63.2/36.8

= 1.72

Another oxide (Oxide 2) contains 77.5% Mn.

Mass of oxide = 100 g

Mass of Mn = 77.5 g

Mass of O = 100 - 77.5

= 22.5 g

Ratio of Mn to O = 77.5/22.5

= 3.44

Therefore, the ratio of the masses of Mn and O in Oxide 1 and Oxide 2 is in the ratio 1.72 : 3.44, which is also 1 : 2. So the law of multiple proportions is obeyed.

B.

Oxide 1

Mass of Mn per 1 g of O = mass of Mn/mass of O

= 77.5/22.5

= 3.44 g/g of Oxygen.

Oxide 2

Mass of Mn per 1 g of O = mass of Mn/mass of O

= 77.5/22.5

= 3.44 g/g of Oxygen.

3 0

3 years ago

n-Butane (C4H10) is burned with stoichiometric amount of oxygen. Determine the mole fraction of carbon dioxide and water in the

Fudgin [204]

Answer:

See details below

Explanation:

The balanced reaction equation is given below:

$2C_{4} H_{10}$ + $13O_{2}$ → $8CO_{2}$ + $10H_{2} O$

Mole fraction of CO2 to H20

= 8/10 = $\frac{4}{5}$

Mole ratio of C4H10 to CO2 is 2:8 = 1:4

1 mole of n-butane - 38.12 g

4 moles - ?

= 152.48g fuel consumed.

8 0

3 years ago

How is the mass of carbon related to the mole?

Anuta_ua [19.1K]

Answer:

Simply put, you can go from moles to grams and vice versa by using the mass of 1 mole of that substance, i.e its molar mass. For example, the molar mass of carbon is 12.011 g/mol. This means that 1 mole of carbon, or 6.022⋅1023 atoms of carbon, weigh 12.011 g.

Explanation:

4 0

3 years ago

If 2.0 g of copper(II) chloride react with excess sodium nitrate, what mass of sodium chloride is formed in this double replacem

cluponka [151]

Taking into account the reaction stoichiometry, 1.729 grams of NaCl is formed.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CuCl₂ + 2 NaNO₃ → Cu(NO₃)₂ + 2 NaCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CuCl₂: 1 mole
NaNO₃: 2 moles
Cu(NO₃)₂ : 1 mole
NaCl: 2 moles

The molar mass of the compounds is:

CuCl₂: 134.44 g/mole
NaNO₃: 85 g/mole
Cu(NO₃)₂ : 187.54 g/mole
NaCl: 58.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

CuCl₂: 1 mole ×134.44 g/mole= 134.44 grams
NaNO₃: 2 moles ×85 g/mole= 170 grams
Cu(NO₃)₂ : 1 mole ×187.54 g/mole= 187.54 grams
NaCl: 2 moles ×58.45 g/mole= 116.9 grams

<h3>Mass of NaCl formed</h3>

The following rule of three can be applied: If by stoichiometry of the reaction 134.44 grams of CuCl₂ form 116.9 grams of NaCl, 2 grams of CuCl₂ form how much mass of NaCl?

$mass of NaCl=\frac{2 grams of CuCl_{2}x116.9 grams of NaCl }{134.44grams of CuCl_{2}}$

<u><em>mass of NaCl= 1.739 grams</em></u>

Finally, 1.729 grams of NaCl is formed.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

6 0

2 years ago