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4 years ago

If the anode compartment contains [so2−4]= 1.30 m in equilibrium with pbso4(s, what is the ksp of pbso4?

Chemistry

2 answers:

Lena [83]4 years ago

8 0

Answer:

The solubility product lead sulfate at anode is 1.69.

Explanation:

$PbSO_4\rightleftharpoons Pb^{2+}+SO_4^{2-}$

S S

Concentration of sulfate ion at anode:

$[SO_4^{2-}]=S=1.30 M$

1 mole of lead sulfate dissociates into 1 mol of lead ion and 1 mole of sulfate ion.So.

$SO_4^{2-}=[Pb^{2+}]=S$

The solubility product of the lead sulfate is:

$K_{sp}=S\times S=S^2=(1.30 M)^2=1.69$

The solubility product lead sulfate at anode is 1.69.

notsponge [240]4 years ago

7 0

The anode compartment contains [SO2-4] with a concentration of 1.30 m, this is in equilibrium with PbSO4 (s). The dissociation of PbSO4 is:

PbSO4 + H2O ---> Pb 2+ + SO4 2-

The Ksp of PbSO4:

Ksp = [Pb2+] [SO4-2]

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Need help !!!!! ASAP

Ksivusya [100]

<h2>Hello!</h2>

The answer is:

We have that there were produced 0.120 moles of $CO_{2}$

$n=0.120mol$

We are asked to calculate the number of moles of the given gas, also, we are given the volume, the temperature and the pressure of the gas, we can calculate the approximate volume using The Ideal Gas Law.

The Ideal Gas Law is based on Boyle's Law, Gay-Lussac's Law, Charles's Law, and Avogadro's Law, and it's described by the following equation:

$PV=nRT$

Where,

P is the pressure of the gas.

V is the volume of the gas.

n is the number of moles of the gas.

T is the absolute temperature of the gas (Kelvin).

R is the ideal gas constant (to work with pressure in mmHg), which is equal to:

$R=62.363\frac{mmHg.L}{mol.K}$

We must remember that the The Ideal Gas Law equation works with absolute temperatures (K), so, if we are given relative temperatures such as Celsius degrees or Fahrenheit degrees, we need to convert it to Kelvin before we proceed to work with the equation.

We can convert from Celsius degrees to Kelvin using the following formula:

$Temperature(K)=Temperature(C\°) + 273K$

So, we are given the following information:

$Pressure=760mmHg\\Volume=2.965L\\Temperature=25.5C\°=25.5+273K=298.5K$

Now, isolating the number of moles, and substituting the given information, we have:

$PV=nRT$

$n=\frac{PV}{RT}$

$n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}$

$n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}\\\\n=\frac{2242mmHg.L}{18615.355\frac{mmHg.L}{mol.}}\\\\n=0.120mole$

Hence, we have that there were produced 0.120 moles of $CO_{2}$

$n=0.120mol$

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3 years ago

The second-order rate constants for the reaction of oxygen atoms witharomatic hydrocarbons have been measured (R. Atkinson and J

Pepsi [2]

Answer:

A = 1,13x10¹⁰

Ea = 16,7 kJ/mol

Explanation:

Using Arrhenius law:

ln k = -Ea/R × 1/T + ln(A)

You can graph ln rate constant in x vs 1/T in y to obtain slope: -Ea/R and intercept is ln(A).

Using the values you will obtain:

y = -2006,9 x +23,147

As R = 8,314472x10⁻³ kJ/molK:

-Ea/8,314472x10⁻³ kJ/molK = -2006,9 K⁻¹

<em>Ea = 16,7 kJ/mol</em>

Pre-exponential factor is:

ln A = 23,147

A = e^23,147

I hope it helps!

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