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3 years ago

Question 3. A batch chemical reactor achieves a reduction in

Chemistry

1 answer:

kotykmax [81]3 years ago

4 0

Answer:

Rate constant for zero-order kinetics: 1, 58 [mg/L.s]

Rate constant for first-order kinetics: 0,05 [1/s]

Explanation:

The reaction order is the relationship between the concentration of species and the rate of the reaction. The rate law is as follows:

$r = k [A]^{x} [B]^{y}$

where:

[A] is the concentration of species A,
x is the order with respect to species A.
[B] is the concentration of species B,
y is the order with respect to species B

k is the rate constant

The concentration time equation gives the concentration of reactants and products as a function of time. To obtain this equation we have to integrate de velocity law:

$v(t) = -\frac{d[A]}{dt} = k [A]^{n}$

For the kinetics of zero-order, the rate is apparently independent of the reactant concentration.

Rate Law: rate = k

Concentration-time Equation: [A]=[A]o - kt

where

k: rate constant [M/s]

[A]: concentration in the time t [M]
[A]o: initial concentration [M]
t: elapsed reaction time [s]

For first-order kinetics, we have:

Rate Law: rate= k[A]

Concentration -Time Equation: ln[A]=ln[A]o - kt

where:

K: rate constant [1/s]
ln[A]: natural logarithm of the concentration in the time t [M]
ln[A]o: natural logarithm of the initial concentration [M]
t: elapsed reaction time [s]

To solve the problem, wee have the following data:

[A]o = 100 mg/L

[A] = 5 mg/L

t = 1 hour = 60 s

As we don't know the molar mass of the compound A, we can't convert the used concentration unit (mg/L) to molar concentration (M). So we'll solve the problem using mg/L as the concentration unit.

Zero-order kinetics

we use: [A]=[A]o - Kt

we replace the data: 5 = 100 - K (60)

we clear K: K = [100 - 5 ] (mg/L) /60 (s) = 1, 583 [mg/L.s]

First-order kinetics

we use: ln[A]=ln[A]o - Kt

we replace the data: ln(5) = ln(100) - K (60)

we clear K: K = [ln(100) - ln(5)] /60 (s) = 0,05 [1/s]

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A pharmaceutical company is making a large volume of nitrous oxide (NO). They predict they will be able to make a maximum amount

iragen [17]

Answer:

The answer is "Option B"

Explanation:

From the query, the following knowledge is derived:

Yield in percentage = 47%

Performance of theory = 4860 g

Actual yield Rate =?

The percentage return is defined simply by the ratio between both the real return as well as the conceptual return multiplied by the 100. It's also represented as numerically:

$Rate = \frac{Existing \ Rate} {Theoretical \ Rate} \times 100$

Now We can obtain the percent yield as followed using the above formula:

$\text{Yield in percentage}= \frac{Actual \ yield \ Rate} {Theorical \ Rate} \times 100$

$47\% = \frac{Actual \ yield \ Rate}{4860}$

The value of the Actual yield Rate = $47\% \times 4860$

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3 years ago

What is Alkaline and Acid in full definition

Harrizon [31]

Answer:

The solution is neither acidic or basic. An acid is a substance that donates hydrogen ions. ... Because the base "soaks up" hydrogen ions, the result is a solution with more hydroxide ions than hydrogen ions. This kind of solution is alkaline. Acidity and alkalinity are measured with a logarithmic scale called pH.

Explanation:

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3 years ago

A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

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aleksley [76]

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3 years ago

What is the simplest method of determining the composition of a compound?

kaheart [24]

Answer:

Chemical formulas tell you how many atoms of each element are in a compound, and empirical formulas tell you the simplest or most reduced ratio of elements in a compound. If a compound's chemical formula cannot be reduced anymore, then the empirical formula is the same as the chemical formula.

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2 years ago