Part a what is the concentration the k+ in 0.15 m the k2s?

express her answer come one decimal place and include the proper units. 0.3 m submithintsmy answersgive upreview component correct part b if cacl2 is dissolved in water, what have the right to be said about the concentration the the ca2+ ion? if is liquified in water, what can be said around the concentration of the ion? it has actually the same concentration together the cl− ion. The concentration is half that that the cl− ion. That is concentration is twice that that the cl− ion. Its concentration is one-third the of the cl− ion. Submithintsmy answersgive upreview component correct component c a scientist desires to do a equipment of tribasic sodium phosphate, na3po4, for a laboratory experiment. How plenty of grams of na3po4 will certainly be necessary to create 650. Ml that a solution that has actually a concentration of na+ ion of 1.30 m ?


Stokiometry in Chemistry studies around chemical reactions greatly emphasizing quantitative, such together the calculate of volume, mass, amount, i beg your pardon is regarded the number of ions, molecules, elements, etc.

You are watching: What is the concentration of k+ in 0.15 m of k2s?

Reaction equations space chemical formulas for reagents and also product substances

Reaction coefficients room numbers in the chemistry formula of substances associated in the reaction equation. Reaction coefficients are beneficial for balancing reagents and also products.

The reaction coefficient shows the ratio of the number of moles or molecule of the reacting substance

The ionization reaction is the reaction of the decomposition the a substance into its ions when the problem is liquified in water.

Molarity (M)

Molarity shows the number of moles of solute in every 1 liter the solution.

M = n/V


M = Molarity

n = variety of moles of solute

V = volume of the solution


1) price is: the concentration that potassium cations (K⁺) is 0.3 M.

Balanced chemical reaction (dissociation) the potassium sulfide in the water:

K₂S(aq) → 2K⁺(aq) + S²⁻(aq).

c(K₂S) = 0.15 M.

From well balanced reaction: n(K₂S) : n(K⁺) = 1 : 2, because volume that the systems is constant: c(K⁺) = 2 · c(K₂S).

c(K⁺) = 0.3 M; concentration the potassium cations.

2) prize is: that concentration is fifty percent that of the Cl⁻ ion.

Balanced chemical reaction (dissociation) the calcium chloride (CaCl₂) in the water:

CaCl₂(aq) → Ca²⁺(aq) + 2Cl⁻(aq).

From balanced reaction: n(Ca²⁺) : n(Cl⁻) = 1 : 2, because volume of the systems is constant: c(Ca²⁺) = c(Cl⁻) ÷ 2.

n is amount of the substance.

3) price is: 46.176 grams that Na₃PO₄ will certainly be needed.

Balanced chemical reaction (dissociation) of salt phosphate (Na₃PO₄) in the water:

Na₃PO₄(aq) → 3Na⁺(aq) + PO₄³⁻(aq).

V = 650 mL ÷ 1000 mL.

V = 0.650 L; volume of the solution.

c(Na⁺) = 1.3 M; concentratium of salt cations.

From balanced reaction: n(Na₃PO₄) : n(Na⁺) = 1 : 3, since volume the the systems is constant: c(Na₃PO₄) = c(Na⁺) ÷ 3.

c(Na₃PO₄) = 1.3 M ÷ 3.

c(Na₃PO₄) = 0.43 M.

n(Na₃PO₄) = c(Na₃PO₄) · V(solution).

n(Na₃PO₄) = 0.43 M · 0.65 L.

n(Na₃PO₄) = 0.281 mol.

m(Na₃PO₄) = n(Na₃PO₄) · M(Na₃PO₄).

m(Na₃PO₄) = 0.281 mol · 163.94 g/mol.

m(Na₃PO₄) = 46.176 g.

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