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4 years ago

What is the empirical formula for the compound P4O6??

Chemistry

1 answer:

777dan777 [17]4 years ago

5 0

The empirical formula is P₂O₃

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An element has an average atomic mass of 1.008 amu. It consists of two isotopes , one having a mass of 1.007 amu, and one having

dimulka [17.4K]

Answer:

The most abundant isotope is 1.007 amu.

Explanation:

Given data:

Average atomic mass = 1.008 amu

Mass of first isotope = 1.007 amu

Mass of 2nd isotope = 2.014 amu

Most abundant isotope = ?

Solution:

First of all we will set the fraction for both isotopes

X for the isotopes having mass 2.014 amu

1-x for isotopes having mass 1.007 amu

The average atomic mass is 1.008 amu

we will use the following equation,

2.014x + 1.007 (1-x) = 1.008

2.014x + 1.007 - 1.007 x = 1.008

2.014x - 1.007x = 1.008 - 1.007

1.007 x = 0.001

x= 0.001/ 1.007

x= 0.0009

0.0009 × 100 = 0.09 %

0.09 % is abundance of isotope having mass 2.014 amu because we solve the fraction x.

now we will calculate the abundance of second isotope.

(1-x)

1-0.0009 = 0.9991

0.9991 × 100= 99.91%

6 0

3 years ago

Which statement most likely describes the relationship between ethylene and polyethylene?

Musya8 [376]

Answer:

<em><u>B. Ethylene can be chemically changed to become polyethylene.</u></em>

Explanation:

You can make polyethylene by purifying a quantity of ethylene.

7 0

3 years ago

Read 2 more answers

What are some differences between a human , a dog a frog and a fish name 3

AleksandrR [38]

Well, human bodies differ from all those animals.

7 0

4 years ago

Read 2 more answers

Write the balanced NET ionic equation for the reaction when aqueous Cs₃PO₄ and aqueous AgNO₃ are mixed in solution to form solid

Maurinko [17]

Answer:

joe

Explanation:

mama

3 0

3 years ago

Calculate the moles of O atoms in 0.658 g of Mg(NO3)2.

lara31 [8.8K]

Answer:

There are 0.0267 moles of oxygen in 0.658 grams of magnesium nitrate.

Explanation:

Mass of magnesium nitrate = m = 0.658 g

Molar mass of magnesium nitrate = M = 148 g/mol

Moles of magnesium nitrate = n

$n=\frac{m}{M}=\frac{0.658 g}{148 g/mol}=0.004446 mol$

1 mole of magnesium nitrate has 6 moles of oxygen atoms. Then 0.004446 moles magnesium nitrate will have :

$6\times 0.004446 mol=0.026676 mol\approx 0.0267$

There are 0.0267 moles of oxygen in 0.658 grams of magnesium nitrate.

8 0

4 years ago