Name two compounds in unpolluted air?

An atom has 13 protons and 14 neutrons. What is its mass number?

Hitman42 [59]

Protons have a mass of 1
Neutrons have a mass of 1

So 13*1 + 14*1 = Mass number 27

3 0

3 years ago

Calculate the number of grams in 4.56 x 1026 atoms of sodium phosphate. Be sure to balance the charges of sodium phosphate. Help

zalisa [80]

Answer:

124225.91 g of Na₃PO₄

Explanation:

From the question given above, the following data were obtained:

Number of atoms of Na₃PO₄ = 4.56×10²⁶ atoms

Mass of Na₃PO₄ =?

From Avogadro's hypothesis,

6.02×10²³ atoms = 1 mole of Na₃PO₄

Next, we shall determine the mass of 1 mole of Na₃PO₄. This can be obtained as follow:

1 mole of Na₃PO₄ = (23×3) + 31 + (16×4)

= 69 + 31 + 64

= 164 g

Thus,

6.02×10²³ atoms = 164 g of Na₃PO₄

Finally, we shall determine the mass of Na₃PO₄ that contains 4.56×10²⁶ atoms. This can be obtained as follow:

6.02×10²³ atoms = 164 g of Na₃PO₄

Therefore,

4.56×10²⁶ atoms = (4.56×10²⁶ × 164)/6.02×10²³

4.56×10²⁶ atoms = 124225.91 g of Na₃PO₄

Therefore, 124225.91 g of Na₃PO₄ contains 4.56×10²⁶ atoms

4 0

2 years ago

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers

When a -COOH group replaces a hydrogen in a hydrocarbon the result is a/an

soldier1979 [14.2K]

Answer:

drugs

Explanation:

4 0

3 years ago

when the orbitals or energy levels of an atom are full, it is considered to be stable. which atom in the ground state has a stab

Sonbull [250]

Answer:

Noble gases, atoms which have complete filled outer orbits.

Explanation:

SInce, you have given no options elements like Neon, Helium,etc

7 0

2 years ago