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3 years ago

Calculate the pH of a solution with [OH−]=1.3×10−2M. (Hint: Begin by using Kw to find [H3O+].)

Chemistry

1 answer:

Nookie1986 [14]3 years ago

3 0

Answer:

12.11.

Explanation:

∵ pOH = - log[OH⁻]

∴ pOH = - log(1.3 x 10⁻²) = 1.886.

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 1.886 = 12.11.

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Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.45 M of reagent

sladkih [1.3K]

Answer:

$\large\boxed{\large\boxed{0.0014M/s}}$

Explanation:

From the table, first find the order of reaction, then find the rate constant, write the rate equation, and, finally, subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

1. Table

Trial [A] (M) [B] (M) [C] (M) Initial rate (M/s)

1 0.20 0.20 0.20 6.0×10⁻⁵

2 0.20 0.20 0.60 1.8×10⁻⁴

3 0.40 0.20 0.20 2.4×10⁻⁴

4 0.40 0.40 0.20 2.4×10⁻⁴

2. Orders

a) From trials 3 and 4 you learn that the initial concentration of B does not change the change teh rate of the reaction. Hence the order with respect to [B] is 0.

b) From trials 1 and 2 you learn that when the concentration of C is tripled the rate of reaction is also tripled:

0.60 / 0.2 = 3, and

1.8×10⁻⁴ / 6.0×10⁻⁵ = 3

Hence, the order with respect to C is 1.

c) From trials 1 and 3 you get:

0.40/0.2 = 2

2.4×10⁻⁴ / 6.0×10⁻⁵ = 4

Which means that when the concentration of A is doubled, the rate of the reaction is quadruplicated. Hence, the order of the reaction with respect to A is 2.

3. Rate equation

Ther orders are:

$a=2\\\\b=0\\\\c=1$

Hence the rate is:

$rate=k[A]^a{B}^b[C]^c\\ \\ rate=k[A]^2[B]^0[C]^1=k[A]^2C$

4. Rate constant, k

You can use any trial to find the value of the constant, k

Using trial 1:

$6.0\times 10^{-5}M/s=k(0.20M)^2(0.20M)\\ \\ k=\frac{6.0\times 10^{-5}M/s}{(0.20M)^2(0.20M)}=0.0075M^{-2}s^{-1}$

5. Rate law:

$rate=k[A]^2C=0.0075[A]^2[C]$

6. Substitute

Subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

$rate=0.0075M^{-2}s^{-1}[A]^2[C]=0.0075M^{-2}s^{-1}[0.45M]^2[0.9M]$

$r=0.00136688M/s\approx 0.0014M/s$

7 0

3 years ago

4. Froath floatation process is used for*

lbvjy [14]

Answer:

Explanation:

Froath flotaion process is based on thie wttebality properteis like hydrophillicity and hydrophobicity.

Incase of sulphide ore processing,

The metallic sulphide particles of ore are preferentially wetted by oil and the gangue particles by water. By doing thise you are sperating required sulphide ore particels from unwanted gaunge , by doing this you are incresing the concentation of Sulfide in ore by removing the not required gaunge particles.

Hope this helps you to decide your annswer

5 0

3 years ago

It get hotter towards the center

STatiana [176]

Answer:

earth

Explanation:

8 0

3 years ago

Determine % yield if a student obtains 45 g of product in an experiment and the theoretical amount is determined to be 50 g

Katen [24]

Percentage yield=(actual yield/theoretical yield) x 100%
= (45/50) x 100%
= 90%

8 0

4 years ago

Calculate the number of moles in 3.01 x 10²² atoms of calcium.

Lera25 [3.4K]

Answer:

Using dimensional analysis:

3.01x1022 molecules CO2 x 1 mol CO2/6.02x1023 molecules x 44. g CO2/mole = 2.20 g CO2

Explanation:

8 0

3 years ago