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3 years ago

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The solubility of silver sulfate (Ag2SO4), in moles per liter, can be expressed in terms of the resulting ion concentrations. Wh

ich relationship is correct?

1 answer:

lisabon 2012 [21]3 years ago

8 0

Answer:

Solubility = [SO₄²⁻]

Explanation:

Solubility of Ag₂SO₄ can be determined by its ksp equilibrium:

Ag₂SO₄(s) ⇄ SO₄²⁻ + 2Ag⁺

<em>Where 1 mole of silver sulfate dissolves producing 1 mole of sulfate ion and 2 moles of silver ion.</em>

Solubility is defined as the amount of solid that can be dissolved per liter of solution.

The moles of Ag₂SO₄ dissolved are equal to moles of SO₄²⁻.

That means:

<h3>Solubility = [SO₄²⁻]</h3>

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calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w

kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is $-22Jmole^-^1k^-^1$

(b) The normal boiling point of water (J·K−1·mol−1) is $-109Jmole^-^1K^-^1$

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole

Explanation:

Lets calculate

(a) - General equation -

$(\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p$ = $-5_m(\beta )+5_m(\alpha )$ = $-\frac{\Delta H}{T}$

$\alpha ,\beta$ → phases

ΔH → enthalpy of transition

T → temperature transition

$(\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p$ = $= -\frac{\Delta_fH}{T_f}$

= $\frac{-6.008kJ/mole}{273.15K}$ ( $\Delta_fH$ is the enthalpy of fusion of water)

= $-22Jmole^-^1k^-^1$

(b) $(\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}$

= $\frac{40.656kJ/mole}{373.15K}$ ( $\Delta_v_a_p_o_u_rH$ is the enthalpy of vaporization)

= $-109Jmole^-^1K^-^1$

(c) $\Delta\mu =\Delta\mu(l)-\Delta\mu(s)$ = $-S_m\DeltaT$

$[\mu(l-5$ ° $C)-\mu(l,0$ ° $C)]$ = $[\mu(s-5$ ° $C)-\mu(s,0$ ° $C)]$ $=-S_m$ ΔT

$\mu(l,-5$ ° $C)-\mu(s,-5$ ° $C)=-Sm\DeltaT [\mu(l,0$

$\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)$

= 109J/mole

6 0

2 years ago

Consider the equation: 2NO(0) - N.04(). Using ONLY the information given by the equation which of the following

ratelena [41]

Answer:

By increasing the pressure, the molar concentration of N2O4 will increase

Explanation:

We have the equation 2NO2 ⇔ N2O4

This equation is reversible and exotherm. By <u>decreasing the temperature</u>, the reaction will produce more energy, so the reaction will move to the right. But a lower temperature also lowers the rate of the process, so, the temperature is set at a compromise value that allows N2O4 to be made at a reasonable rate with an equilibrium concentration that is not too unfavorable

So <u>increasing the temperature</u> will shift the equilibrium to the left. The equilibrium shifts in the direction that consumes energy.

If we d<u>ecrease the concentration of NO2</u>, the equilibrium will shift to the left, resulting in forming more reactants.

To increase the molar concentration of the product N2O4, we have to <u>increase the pressure</u> of the system.

NO2 takes up more space than N2O4, so increasing the pressure would allow the reactant to collide more form more product.

By increasing the pressure, the molar concentration of N2O4 will increase

7 0

3 years ago

Read 2 more answers

Explain why the Amazon River does not create deltas?

jeka57 [31]

Answer:

The ocean currents are too strong by the Amazon River to form deltas.

Explanation:

The Atlantic has sufficient wave and tidal energy to carry most of the Amazon's sediments out to sea, thus the Amazon does not form a true delta. The great deltas of the world are all in relatively protected bodies of water, while the Amazon empties directly into the turbulent Atlantic.

4 0

2 years ago

Find the total surface area of the rectangular prism in the figure. A) 174 cm^ 2 B) 522c * m ^ 2; 348c * m ^ 2 D) 432c * m ^ 2

WARRIOR [948]

Surface area: 348 m^2

8 0

2 years ago

Read 2 more answers

a solution with a transmittance of 0.44 is analyzed in a spectrophotometer with 6% stray light. calculate the absorbance reporte

IgorLugansk [536]

The absorbance reported by the defective instrument was 0.3933.

Absorbance A = - log₁₀ T

Tm = transmittance measured by spectrophotometer

Tm = 0.44

Absorbance reported in this equipment = -log₁₀ (0.44) = 0.35654

True absorbance can be calculated by true transmittance, Tm = T+S(α-T)

S = fraction of stray light = 6%= 6/100 = 0.06

α= 1, ideal case

T = true transmittance of the sample

Tm = T+S(α-T)

now, T= Tm-S/ 1-S = 0.44-0.06/ 1-0.06 = 0.404233

therefore, actual reading measured is A = -log₁₀ T = -log₁₀ (0.404233)

i.e; 0.3933

To know more about transmittance click here:

brainly.com/question/17088180

#SPJ4

3 0

1 year ago