All categories

3 years ago

Calculate the molarity of 0.850 mol of Na2S in 1.70 L of solution.

1 answer:

5 0

Molarity is simply the ratio of the number of moles of a substance over the total volume of the solution. Assuming that the addition of 0.850 moles does not change the overall volume of 1.70 L, therefore molarity is simply:

Molarity = 0.850 moles / 1.70 L = 0.5 moles / L = 0.5 M

You might be interested in

Give the spectator ions for the reaction that occurs when aqueous solutions of H 2SO 4 and KOH are mixed.

Sever21 [200]

Answer:

The spectator ions is: $K^+$ and $SO^{2-}_4$

Explanation:

The equation of reaction between H₂ SO₄ and KOH is:

$H_2SO_{4(aq)} + 2KOH _{aq} \to K_2SO_{4(aq)} +2H_2O _{(l)}$

Rewriting this equation as ionic;

$[2H^{+} + SO^{2-}_4 + 2K^+ +2OH^- \to 2K^+ SO_4^{2-} + 2H_2O ]$

Spectators ions are ions present on both sides of the ionic equation by the same quantity but do not take part in the net reaction.

5 0

3 years ago

If all the deer disappeared from this community, which change would be most

butalik [34]

Answer:c

Explanation:

4 0

3 years ago

Read 2 more answers

What is the molecular shape and bond angle at the indicated carbon?

aliina [53]

Answer:

bond angles of 120 degrees.

7 0

3 years ago

An amino acid is usually more soluble in aqueous solvent at pH extremes than it is at a pH near the isolelectric point of the am

gogolik [260]

Answer:

Option B. At pH extremes, the amino acid molecules mostly carry a net charge, thus increasing their solubility in polar solvent.

C. At very low or very high pH, the amino acid molecules have increased charge, thus form more salt bonds with water solvent molecules.

Explanation:

4 0

3 years ago

"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"

Andru [333]

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

pKa NH₃/NH₄⁺

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = 4.74

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

Moles NH₃

2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃

H-H equation:

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 = log [0.400 moles] / [NH₄Cl]

0.0087 = [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

<h3>245.66g of NH₄Cl is the mass we need to add to obtain the desire pH</h3>

3 0

4 years ago