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3 years ago

would precipitation occur when 500 ml of a 0.02 m solution of agno3 is mixed with 500ml of a 0.001M solution of nacl? explain

1 answer:

3 0

First, write the chemical equation that represents the chemical reation to state the products:

AgNO3 (aq)+ NaCl (aq) → AgCl (s)+ NaNO3 (aq)

The terms (aq) are used to indicate tha the compund is in aqueous solution, whle (s) is used to indicate a solid compound.

You can use the solubility rules or a table of solubilities to check solubilities of Ag Cl and NaNO3 in water. I used a table of solubilities. If you do it you will find that NaNO3 is soluble in water and AgCl is insoluble. Then AgCl will form a precipitate.

Then, the answer is yes, precipitation of AgCl would occur.

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jamie is not sure a new medication will work because it has not had a large test group. Is jamie being creative?

Georgia [21]

Answer:

yes because I wouldn't do it

7 0

3 years ago

Ca(OH)2 (s) precipitates when a 1.0 g sample of CaC2(s) is added to 1.0 L of distilled water at room temperature. If a 0.064 g s

Nina [5.8K]

Answer:

D) Ca(OH)₂ will not precipitate because Q < Ksp

Explanation:

Here we have first a chemical reaction in which Ca(OH)₂ is produced:

CaC₂(s) + H₂O ⇒ Ca(OH)₂ + C₂H₂

Ca(OH)₂ is slightly soluble, and depending on its concentration it may precipitate out of solution.

The solubility product constant for Ca(OH)₂ is:

Ca(OH)₂(s) ⇆ Ca²⁺(aq) + 2OH⁻(aq)

Ksp = [Ca²⁺][OH⁻]²

and the reaction quotient Q:

Q = [Ca²⁺][OH⁻]²

So by comparing Q with Ksp we will be able to determine if a precipitate will form.

From the stoichiometry of the reaction we know the number of moles of hydroxide produced, and since the volume is 1 L the molarity will also be known.

mol Ca(OH)₂ = mol CaC₂( reacted = 0.064 g / 64 g/mol = 0.001 mol Ca(OH)₂

the concentration of ions will be:

[Ca²⁺ ] = 0.001 mol / L 0.001 M

[OH⁻] = 2 x 0.001 M = 0.002 M ( From the coefficient 2 in the equilibrium)

Now we can calculate the reaction quotient.

Q= [Ca²⁺][OH⁻]² = 0.001 x (0.002)² = 4.0 x 10⁻⁹

Q < Ksp since 4.0 x 10⁻⁹ < 8.0 x 10⁻⁸

Therefore no precipitate will form.

The answer that matches is option D

8 0

4 years ago

Paul determines that the hydrogen ion concentration of his unknown solution is 3.60×10^-5 M. what is the pH of this solution?

vivado [14]

Answer:

Explanation:

The pH of a substance can be found by using the formula

$p H = - log[ H^{ + } ]$

where [ H+ ] is the hydrogen ion concentration of the solution

From the question

[ H + ] = 3.60 × 10^-5 M

So the pH is

$pH = - log(3.60 \times {10}^{ - 5} ) \\ =4.44369749923$

We have the final answer as

Hope this helps you

4 0

3 years ago

What is the value of quantum numbers (n, l, ml, ms) of 2p^3?

eduard

Answer:

Explanation:

First digit of the 2p^3 gives you value of n, in this case its = 2, So, n= 2

Second alphabet gives you the value of l,

l=0 =s

l=1 =p

l=3=d

l=4=f

since "p" is the alphabet in 2p^3, so in your case lt shoudlbe = 1 right?

ml= -l to +l , that is -1, 0, +1

Ms= +1/2 or -1/2 alaways remains same foe evrything.

6 0

3 years ago

The following reaction shows the products when sulfuric acid and aluminum hydroxide react.

scoray [572]

Leftover: approximately 11.73 g of sulfuric acid.

<h3>Explanation</h3>

Which reactant is <em>in excess</em>?

The theoretical yield of water from Al(OH)₃ is lower than that from H₂SO₄. As a result,

Al(OH)₃ is the limiting reactant.
H₂SO₄ is in excess.

How many <em>moles</em> of H₂SO₄ is consumed?

Balanced equation:

2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O

Each mole of Al(OH)₃ corresponds to 3/2 moles of H₂SO4. The formula mass of Al(OH)₃ is 78.003 g/mol. There are 15 / 78.003 = 0.19230 moles of Al(OH)₃ in the five grams of Al(OH)₃ available. Al(OH)₃ is in excess, meaning that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.

How many <em>grams</em> of H₂SO₄ is consumed?

The molar mass of H₂SO₄ is 98.076 g.mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.076 = 28.289 g.

How many <em>grams</em> of H₂SO₄ is in excess?

40 grams of sulfuric acid H₂SO₄ is available. 28.289 grams is consumed. The remaining 40 - 28.289 = 11.711 g is in excess. That's closest to the first option: 11.73 g of sulfuric acid.

6 0

3 years ago