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3 years ago

Calculate the molar solubility of CaSO4. Ksp = 7.10 x 10-5

Chemistry

2 answers:

zubka84 [21]3 years ago

4 0

Multiple or divide them

Aliun [14]3 years ago

4 0

Answer: The solubility of calcium sulfate is $8.43\times 10^{-3}M$

Explanation:

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The chemical equation for the ionization of calcium sulfate follows:

$CaSO_4(aq.)\rightleftharpoons Ca^{2+}(aq.)+SO_4^{2-}(aq.)$

s s

The expression of $K_{sp}$ for above equation follows:

$K_{sp}=s\times s$

We are given:

$K_{sp}=7.10\times 10^{-5}$

Putting values in above expression, we get:

$7.10\times 10^{-5}=s\times s\\\\s=\sqrt{7.10\times 10^{-5}}=8.43\times 10^{-3}M$

Hence, the solubility of calcium sulfate is $8.43\times 10^{-3}M$

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Answer:

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Codeine C18H21NO3 is a weak organic base. A 5.0 x 10^-3 M solution of codeine has a pH of 9.95. Calculate the value of Kb for th

arsen [322]

Answer:

Kb = $4.45\times10^{-7}\ mol/L$

$p^{Kb}=6.35$

Explanation:

For a weak organic base, the formula to find $p^{OH}$ is given by:

$p^{OH}=p^{K_b}+\log c$

where c is the concentration of base.

Here c= $5\times10^{-3}\ M$

$p^{H}=9.95\\p^{OH}=14-p^{H}=14-9.95=4.05$

Substituting the above values in the formula,we get:

$p^{k_b}=p^{OH}-\log c\\p^{k_b}=4.05-\log (5\times10^{-3})\\p^{K_b}=6.35\\K_b=$antilog 6.35=4.45\times10^{-7}\ mol/L$

Hence:

Kb = $4.45\times10^{-7}\ mol/L$

$p^{Kb}=6.35$

7 0

2 years ago

How many grams of silver oxide are needed to react with 7.9 g of hydrochloric acid produce silver chloide and water?

Anastaziya [24]

Answer:

25.11 g.

Explanation:

It is clear from the balanced equation:

Ag₂O + 2HCl → 2AgCl + H₂O.

that 1.0 mole of Ag₂O reacts with 2.0 moles of HCl to produce 2.0 mole of AgCl and 1.0 moles of H₂O.

7.8 g of HCl reacts with excess Ag₂O. To calculate the no. of grams of Ag₂O that reacted, we should calculate the no. of moles of HCl:

no. of moles of HCl = mass/atomic mass = (7.9 g)/(36.46 g/mol) = 0.2167 mol.

From the balanced equation; every 1.0 mol of Ag₂O reacts with 2 moles of HCl.

∴ 0.2167 mol of HCl will react with (0.2617 mol / 2 = 0.1083 mol) of Ag₂O.

∴ The mass of reacted Ag₂O = no. of moles x molar mass = (0.1083 mol)(231.735 g/mol) = 25.11 g.

7 0

3 years ago

A. 14.8

Scilla [17]

Answer:

Explanation:

Given data:

Initial volume = 2.5 L

Initial temperature = 300 k

Final temperature = 80 k

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 2.5 L × 80 K / 300 k

V₂ = 200 L.K / 300 K

V₂ = 0.67 L

Given data:

Initial volume = 752 mL

Initial temperature = 25.0°C (25+273 = 298 K)

Final temperature = 50.0°C(50+273 = 323 K)

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 752 mL × 323 K / 298 k

V₂ = 242896 mL.K / 298 K

V₂ = 815.1 mL

Given data:

Initial volume = 2.75 L

Initial temperature = 20°C (20+273 = 293 K)

Final temperature = ?

Final volume = 2.46 L

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁/V₁

T₂ = 2.46 L × 293 K / 2.75 L

T₂ = 720.78 L.K / 2.75 L

T₂ = 262.1 K

Given data:

Initial volume = 1500 L

Initial temperature = 5°C (5+273 = 278 K)

Final temperature = 30 °C(30+273 = 303 K)

Final volume of heated air = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 1500 L × 303 K / 278 k

V₂ = 454500 L.K / 278 K

V₂ = 1634.89 L

Given data:

Initial volume = 15.5 L

Initial temperature = 20°C (20+273 = 293 K)

Final temperature = 7.0 °C(7.0+273 = 280 K)

Final volume of balloon = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 15.5 L × 280 K / 293 k

V₂ = 4340 L.K / 293 K

V₂ = 14.8 L

Given data:

Initial volume = 150 mL

Initial temperature = 23.5°C (23.5+273 = 296.5 K)

Final temperature = 72.5 °C(72.5+273 = 345.5 K)

Final volume of balloon = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁

V₂ = 150 mL × 345.5 K / 296.5 k

V₂ = 51825 mL.K / 296.5 K

V₂ = 174.79 mL

6 0

3 years ago