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AfilCa [17]
3 years ago
10

A compound is made of two or more different atoms joined together. which image would best fit

Chemistry
1 answer:
emmasim [6.3K]3 years ago
8 0

Answer:

Where is your images?

Explanation:

cant solve the prblm without ur images

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What is the concentration of each ion in a solution that is prepared by dissolving 5.00 g of ammonium chloride in enough water t
sukhopar [10]

The concentration of each ion in the solution of ammonium chloride is:

  • NH₄⁺ = 0.1886 M
  • Cl⁻ = 0.1886 M

To solve this problem, the formulas and the procedures that we have to use are:

  • M = n(solute)/v(solution) L
  • n = m / MW
  • MW= ∑ AWT

Where:

  • M= molarity
  • n = moles
  • m = mass
  • v = volume
  • MW = molecular weight
  • AWT = atomic weight

Information about the problem:

  • m(NH₄Cl) = 5 g
  • v(solution) = 500 ml
  • AWT (N)= 14 g/mol
  • AWT (H)= 1 g/mol
  • AWT (Cl) = 35 g/mol

Converting the volume units from (ml) to (L) we have:

v(solution) = 500 ml * (1 L/1000 ml)

v(solution) = 0,50 L

We calculate the moles of the NH₄Cl from the MW:

MW = ∑ AWT

MW (NH₄Cl)= AWT (N) + AWT (H)*4 + AWT (Cl)

MW (NH₄Cl)= 14 g/mol +( 1 g/mol * 4) + 35 g/mol

MW (NH₄Cl)= 14 g/mol + 4 g/mol + 35 g/mol

MW (NH₄Cl)= 53 g/mol

Having the MW we calculate the moles of NH₄Cl:

n(NH₄Cl) = m(NH₄Cl) / MW(NH₄Cl)

n(NH₄Cl) = m(H2SO4) / MW (H2SO4)

n(NH₄Cl) =  5 g / 53 g/mol

n(NH₄Cl) = 0.0943 mol

Applying the molarity formula, we get:

M(NH₄Cl) = n(NH₄Cl)/v(solution) L

M(NH₄Cl) = 0.0943 mol / 0,50 L

M(NH₄Cl) = 0.1886 M

There are 0.1886 moles of NH₄Cl per liter of solution.

Let's recognize that 1 mol NH₄Cl contains:

  • 1 mol NH₄⁺
  • 1 mol Cl⁻

The concentration of each ion is thus:

(1)*(0.1886 M) = 0.1886 M NH₄⁺

(1)*(0.1886 M) = 0.1886 M Cl⁻

<h3>What is a solution?</h3>

In chemistry a solution is known as a homogeneous mixture of two or more components called:

  • Solvent
  • Solute

Learn more about chemical solution at: brainly.com/question/13182946 and brainly.com/question/25326161

#SPJ4

4 0
2 years ago
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