All categories

2 years ago

How many ammonium ions and how many sulfate ions are present in a 0.310 mol sample of nh42so4?

1 answer:

3 0

$1.87* 10^{23}$ sulphate ions and $1.87* 10^{24}$ ammonium ions are present in a 0.310 mole sample of (NH₄)₂SO₄ .

The International System of Units uses the mole (symbol: mol) as the unit of material quantity (SI). How many elementary entities of a particular substance are present in an object or sample is determined by the quantity of that material.

There are precisely 6.022140761023 elementary entities in a mole. An elementary entity can be an atom, a molecule, an ion, an ion pair, or a subatomic particle like an electron depending on what the substance is.

How this is calculated?

The subscripts next to each atom allow you to find the number of moles of each element in the compound.
Then, in every mole of (NH₄)₂SO₄ there are:

2 moles of ammonium (NH₄) ions and 1 mole of sulfate (SO₄) ions.

Number of sulphate ions:
$0.310 moles of( NH_{4})_{2} SO_{4} *\frac{ 1 sulphte mole}{1mole of( NH_{4})_{2}SO_{4} }*\frac{6.023*10^{23}}{ 1 sulphte mole} =1.87*10^{23}$

Number of ammonium ions:

$0.310 moles of( NH_{4})_{2} SO_{4} *\frac{ 10 ammonium mole}{1mole of( NH_{4})_{2}SO_{4} }*\frac{6.023*10^{23}}{ 1 ammonium mole} =1.87*10^{24}$

To know more about moles, refer:

brainly.com/question/19964502

#SPJ4

You might be interested in

Write a balanced chemical equation, including states of matter, for the combustion of gaseous benzene, c6h6.

snow_tiger [21]

Almost all hydrocarbon 'burn' reactions involve oxygen; it's by far the most reactive substance in air.

Hydrocarbon combustions always involve 
[some hydrocarbon] + oxygen --> carbon dioxide + steam. 

C6H6(l) + O2 (g)--> CO2 (g)+ H2O (g)

Balance carbon, six on each side: 
C6H6(l) + O2 (g)--> 6CO2 (g)+ H2O (g)

Balance hydrogen, six on each side: 
C6H6(l) + O2 (g)--> 6CO2(g) + 3H2O (g)

Now, we have fifteen oxygens on the right and O2 on the left. 
Two ways to deal with that. We can use a fraction: 
C6H6 (l)+ (15/2)O2 (g)--> 6CO2 (g)+ 3H2O (g)

Or, if you prefer to have whole number coefficients, double everything 
to get rid of the fraction: 
2C6H6 (l)+ 15O2 (g)--> 12CO2 (g)+ 6H2O (g)

With the SATP states thrown in... 
C6H6(l) + (15/2)O2(g) --> 6CO2(g) + 3H2O(g)

4 0

3 years ago

Who earned the nobel prize twice for their work with radioactivity

lord [1]

Answer:

Marie Curie, née Maria Salomea Skłodowska, (born November 7, 1867, Warsaw, Congress Kingdom of Poland, Russian Empire—died July 4, 1934, near Sallanches, France), Polish-born French physicist, famous for her work on radioactivity and twice a winner of the Nobel Prize.

Explanation:

pls give brainlest almost lvled up.

7 0

3 years ago

Read 2 more answers

Write a caption for a photo of a polluted city.

arlik [135]

Answer:

Explanation:

this is a picture of a city that we have made worse everyday

5 0

3 years ago

Now you will solve the same problem as above, but using the quadratic formula instead of iterations, to show that the same value

Dafna1 [17]

Answer:

a: 1

b: 4.5x10⁻⁴

c: 1.125x10⁻⁶

[H₃O⁺] = 0.000859M

Explanation:

As HNO₂ is a weak acid, its equilibrium in water is:

HNO₂(aq) + H₂O(l) ⇄ H₃O⁺(aq) + NO₂⁻(aq)

Equilibrium constant, ka, is defined as:

ka = 4.5x10⁻⁴ = [H₃O⁺] [NO₂⁻] / [HNO₂] (1)

Equilibrium concentration of each specie are:

[HNO₂] = 0.00250M - x

[H₃O⁺] = x

[NO₂⁻] = x

Replacing in (1):

4.5x10⁻⁴ = x × x / 0.00250M - x

1.125x10⁻⁶ - 4.5x10⁻⁴x = x²

0 = x² + 4.5x10⁻⁴x - 1.125x10⁻⁶

As the quadratic equation is ax² + bx + c = 0

Coefficients are:

a: 1

b: 4.5x10⁻⁴

c: 1.125x10⁻⁶

Now, solving quadratic equation:

x = -0.0013 → False answer, there is no negative concentrations.

x = 0.000859

As [H₃O⁺] = x; [H₃O⁺] = 0.000859M

I hope it helps!

6 0

3 years ago

What do you notice about the distance between the terrestrial planets compared with the spacing between the gas giants

LekaFEV [45]

Answer:

The spacing between the terrestrial planets is less than that between the gas planets.

Explanation:

Since the terrestrial planets which are Mercury, Venus, Earth and Mars have a distance of 0.4 AU, 0.7 A.U, 1.0 AU and 1.5 A.U respectively from while the gas planets which are Jupiter, Saturn, Uranus and Neptune have a distance of 5.2 AU, 9.6 AU, 19.6 AU and 30 AU respectively from the sun, we can see that the spacing between the terrestrial planets is much smaller than the spacing between the gas planets.

For example, the spacing between Mercury and Venus is 0.7 AU - 0.4 AU = 0.3 AU while that between Earth and Venus is 1.0 AU - 0.7 AU = 0.3 AU.

Also, the spacing between Earth and Mars is 1.5 AU - 1.0 AU = 0.5 AU.

Also too, the spacing between the gas planets is shown below the spacing between Jupiter and Saturn is 9.6 AU - 5.2 AU = 4.4 AU while that between Uranus and Saturn is 19.6 AU - 9.6 AU = 10.0 AU.

Also, the spacing between Neptune and Uranus is 30 AU - 19.6 AU = 10.4 AU.

Thus, we can see that the spacing between the terrestrial planets is less than that between the gas planets.

Note that AU means astronomical unit and 1 AU is the distance between the Earth and the Sun. 1 AU = 1.50 × 10⁸ km

8 0

3 years ago