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3 years ago

When a reversible reaction is at equilibrium, the concentration of products and the concentration of reactants must be

Chemistry

1 answer:

ivolga24 [154]3 years ago

7 0

Answer:

Constant

Explanation:

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How many iron atoms are in 7.890 g of iron?

zhenek [66]

Answer:

see explanation

Explanation:

#Atoms = (mass/atomic wt) x 6.02 x 10²³

mass is grams
atomic weight is grams/mole
6.02 x 10²³ is atoms/mole

8 0

2 years ago

Calculate the concentration (M) of sodium ions in a solution made by diluting 25.0 mL of a 0.765 M solution of sodium sulfide to

Yuri [45]

Given:

Concentration of Na₂S = 0.765 M

Volume of Na₂S solution = 25.0 ml = 0.025 L

Final total volume = 225 ml = 0.225 L

To determine:

The concentration of Na+ ions

Explanation:

# moles of Na₂S = 0.765 * 0.025 = 0.0191 moles

Based on the atomic stoichiometry of Na₂S-

# moles of Na⁺ ions = 2*0.0191 = 0.0382 moles

conc of Na⁺ ions = 0.0382 moles/ 0.225 L = 0.169 M

Ans: [Na⁺] = 0.169 M

5 0

3 years ago

Read 2 more answers

For the reaction ag2s(s) ⇌ 2 ag+ (aq) + s2- (aq), keq = 2.4 × 10-4 and the equilibrium concentration of silver ion is [ag+] = 2.

Juliette [100K]

Answer : The concentration of silver ion is, $3.8\times 10^{-3}M$

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq)+S^{2-}(aq)$

The expression of $K$ will be,

$K_{eq}=[Ag^+]^2[S^{2-}]$

$2.4\times 10^{-4}=(2.5\times 110^{-1})^2[S^{2-}]$

$[S^{2-}]=3.8\times 10^{-3}M$

Therefore, the concentration of silver ion is, $3.8\times 10^{-3}M$

4 0

3 years ago

Read 2 more answers

How you get a peer to like you?

netineya [11]

Answer:

talk to them.

Explanation:

some people you can never get to like you but by talking to all types of people you find the people you can relate to and when you do it should feel natural for you.

3 0

3 years ago

The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL water. Because these com

Elanso [62]

Answer:

(a) $Ksp=4.50x10^{-7}$

(b) $Ksp=1.55x10^{-6}$

(d) $Ksp=1.05x10^{-22}$

Explanation:

Hello,

In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:

(a) $BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)$

$Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}$

In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:

$Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L} )^2\\\\Ksp=4.50x10^{-7}$

(B) $Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)$

$Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}$

In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:

$Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}$

$Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}$

In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:

$Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}$

(D) $La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)$

$Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}$

In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:

$Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}$

Best regards.

7 0

4 years ago