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3 years ago

Kc is 1.67 x 10^20 at 25 °C for the formation of iron(III) oxalate complex ion:

Chemistry

1 answer:

Liula [17]3 years ago

7 0

Answer:

$n_{C_2O_4^{2-}}=1.59mol C_2O_4^{2-}$

Explanation:

Hello there!

In this case, according to the given information and chemical equation, it turns out possible for us to calculate the moles of C2O4^2- by firstly setting up the equilibrium expression:

$Kc=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[Fe^{3+}][C_2O_4^{2-}]^3}$

However, according to the question, we just need to apply the given 1:3 mole ratio in the chemical reaction, of iron (III) ions to oxalate ions to obtain:

$n_{C_2O_4^{2-}}=0.53molFe^{3+}*\frac{3molC_2O_4^{2-}}{1molFe^{3+}}=1.59mol C_2O_4^{2-}$

Regards!

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The decomposition reaction of N2O5 in carbon tetrachloride is 2N2O5−→−4NO2+O2. The rate law is first order in N2O5. At 64 °C the

Mademuasel [1]

Answer:

(a) rate = 4.82 x 10⁻³s⁻¹ [N2O5]

(b) rate = 1.16 x 10⁻⁴ M/s

(d) rate = 5.80 x 10⁻⁵ M/s

Explanation:

We are told the rate law is first order in N₂O₅, and its rate constant is 4.82 x 10⁻³s⁻¹ . This means the rate is proportional to the molar concentration of N₂O₅, so

(a) rate = k [N2O5] = 4.82 x 10⁻³s⁻¹ x [N2O5]

(b) rate = 4.82×10⁻³s⁻¹ x 0.0240 M = 1.16 x 10⁻⁴ M/s

(c) Since the reaction is first order if the concentration of N₂O₅ is double the rate will double too: 2 x 1.16 x 10⁻⁴ M/s = 2.32 x 10⁻⁴ M/s

(d) Again since the reaction is halved to 0.0120 M, the rate will be halved to

1.16 x 10⁻⁴ M/s / 2 = 5.80 x 10⁻⁵ M/s

3 0

3 years ago

Read 2 more answers

Match the chemical Equation with his correct type

Alex

Answer:

The answer to your question is given below

Explanation:

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From the above equation, we can see that a single compound H2O produces two elements H2 and O2. This is simply called a decomposition reaction.

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From the above equation, we can see that Cl replaces S in ZnS to produce ZnCl2 and S replaces Cl in HCl to produce H2S. This is simply called double displacement reaction.

5. B. H2 + Br2 —> 2HBr => B. Synthesis.

From the above equation, we can see that two element H2 and Br2 combine to produce a single compound HBr. This is simply called a synthesis reaction.

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3 years ago

If a substance changes its shape and size, is it a physical or chemical change? - chemical - physical - both

Alex

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3 years ago

Which of the following is an example of a polymer?

svetlana [45]

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3 years ago

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate

vovangra [49]

Answer:

K2 = 61.2 M^-1.S^-1

Explanation:

We complete the question fully:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0kJ/mol . If the rate constant of this reaction is 6.7M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

Answer is as follows:

The question asks us to calculate the value of the rate constant at a certain temperature, given that it is at a particular value for a particular temperature. We solve the question as follows:

According to Arrhenius equation, the relationship between temperature and activation energy is as follows:

k = Ae^-(Ea/RT)

where, k = rate constant

A = pre-exponential factor

Ea = activation energy

R = gas constant

T = temperature in kelvin

From the equation, the following was derived for a double temperature problem:

ln(k2/k1) = (-Ea/R) * (1/T1 - 1/T2)

We list out the parameters as follows:

T1= (244 + 273.15) K = 517.15 K

T2= (324+ 273.15) K =597.15 K

K1 = 6.7 , K2 = ?

R = 8.314 J/mol K

Ea = 71.0 kJ/mol = 71000 J/mol

Putting the given values into the above formula as follows:

ln(k2/6.7) = (-71000/8.314) * (1/517.15 - 1/597.15)

lnk2 - 1.902 = 8539.8 * 0.000259

lnK2 = 1.902 + 2.21

lnK2 = 4.114

K2 = e^(4.114)

K2 = 61.2

Hence, K2 = 61.2 (M.S)^-1

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4 years ago

Read 2 more answers