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Vera_Pavlovna [14]
3 years ago
13

What concentration of so32– is in equilibrium with ag2so3(s) and 9.50 × 10-3 m ag ? the ksp of ag2so3 can be found here?

Chemistry
1 answer:
satela [25.4K]3 years ago
8 0
Answer is: concentration of SO₃² is 1.66·10⁻¹⁰ M.
Reaction of dissociation: Ag₂SO₃(s) → 2Ag⁺(aq) + SO₃²⁻(aq)<span>.
Ksp for silver sulfite is 1.5</span>·10⁻¹⁴.
c(Ag⁺) = 9.50·10⁻³ M; equilibrium concentration of silver cations.
Ksp = c(Ag⁺)² · c(SO₃²⁻).
c(SO₃²⁻) = Ksp ÷ c(Ag⁺)².
c(SO₃²⁻) = 1.5·10⁻¹⁴ M³ ÷ (9.50·10⁻³ M)².
c(SO₃²⁻) = 1.5·10⁻¹⁴ M³ ÷ 9.025·10⁻⁵ M².
c(SO₃²⁻) = 1.66·10⁻¹⁰ M; equilibrium concentration of sulfite anions.
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2 years ago
An experiment reveals that 125.0 grams of an unknown metal increases in temperature from 22.0 oC to 43.6 oC upon absorbing 640 j
nydimaria [60]

Answer:

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Explanation:

                      Amount of energy required by known amount of a substance to raise its temperature by one degree is called specific heat capacity.

The equation used for this problem is as follow,

                                                 Q  =  m Cp ΔT   ----- (1)

Where;

           Q  =  Heat  =  640 J

           m  =  mass  =  125 g

           Cp  =  Specific Heat Capacity  =  <u>??</u>

           ΔT  =  Change in Temperature  =  43.6 °C  -  22 °C  =  21.6 °C

Solving eq. 1 for Cp,

                                Cp  =  Q / m ΔT

Putting values,

                                Cp  =  640 J / (125 g × 21.6 °C)

                                Cp  =  0.237 J.g⁻¹.°C⁻¹

3 0
2 years ago
Need help setting the problem up
Sveta_85 [38]

Answer:

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Explanation:

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Volume (V) = 12L

Pressure = 5.6 atm

Temperature (T) = 205K

Gas constant (R) = 0.08206 atm.L/Kmol

Number of mole (n) =?

Using the ideal gas equation: PV = nRT, the number of mole of the gas can be obtained as follow

PV = nRT

5.6 x 12 = n x 0.08206 x 205

Divide both side by 0.08206 x 205

n = (5.6 x 12)/(0.08206 x 205)

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5 0
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sertanlavr [38]

Answer:

Reaction A and B are unfavorable.

Explanation:

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ΔG°=ΔH°-TΔS°

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ΔH° =  enthalpy of reaction

T = temperature of eh reaction

ΔS° = Entropy change

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According to given information in the question:

Reaction A and B are non spontaneous as their Gibbs free energy value is positive.hence both are unfavorable.

3 0
3 years ago
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