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elena-14-01-66 [18.8K]
3 years ago
8

a factor of 8, it means that A is now only 1/8 of its original concentration. A first order reaction, where A → products, has a

rate constant of 1.56 × 107 s −1 . At some time, a concentration of 1.06 × 10−6 M of species A is introduced into the reactor. How long does it take for the concentration of A to fall by a factor of 8?
Chemistry
1 answer:
Trava [24]3 years ago
4 0

Answer:

It takes 1.32x10⁻⁷s for the concentration of A to fall by a factor of 8

Explanation:

The equation that represents a first-order kinetics is:

Ln ([A] / [A]₀] = -kt

<em>Where [A] is actual concentration, [A]₀ is initial concentration, K is rate constant (For the given problem, 1.57x10⁷s⁻¹ and t is time.</em>

<em />

As you want the time when you have [A] in a factor of 8 = [A] / [A]₀ = 1/8

Replacing:

Ln ([A] / [A]₀] = -kt

Ln (1/8) = -1.57x10⁷s⁻¹*t

t = 1.32x10⁻⁷s

<h3>It takes 1.32x10⁻⁷s for the concentration of A to fall by a factor of 8</h3>

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Which of the following is the correct set up AND answer to convert 6.25 x
BlackZzzverrR [31]

Answer:

Explanation:

How many mols do you have?

1 mol = 6.02 * 10^23 atoms

x mol = 6.25 * 10 ^32 atoms

1/x = 6.02*10^23 / 6.25 * 10^32        Cross multiply

6.02 * 10^23 * x = 1 * 6.25 * 10^32    Divide by 6.02 * 10^23

x = 6.25 * 10*32/ 6.02 ^10^23

x = 1.038 * 10^9 mols which is quite large.

Find the number of grams. (Use the value for copper on your periodic table. I will just use an approximate number.)\

1 mol of copper = 63 grams.

1.038 * 10^9 mols of copper = x

1/1.038 * 10^9 = 63/x         Cross multiply

x = 1.038 * 10^9 * 63

x = 6.54 * 10^10 grams of copper.

7 0
3 years ago
Position always is relative to another object or location <br> options: True False
-Dominant- [34]

Answer:

True

Explanation:because of the motion

4 0
2 years ago
What quantity of copper is deposited by the same quantity of electricity that deposited 9g of aluminum
klio [65]

Answer:

Mass of copper deposited = 31.75 g

Explanation:

According to Faraday's second law of electrolysis, when the same quantity of electricity is passed through different electrolytes, the relative number of moles of the elements deposited are inversely proportional to the charges on the ions of the elements.

From this law, it can be seen that the higher the charge, the lower the number of moles of a given element deposited.

Number of moles of aluminium in 9 g of aluminium = mass / molar mass

Molar mass of aluminium = 27 g

Number of moles of aluminium = 9/27 = 1/3 moles

Charge on aluminium ion = +3

3 moles of electrons will discharge 1 mole of aluminium,

1 mole of electrons will discharge 1/3 moles of aluminium

Number of moles of electrons involved = 1 mole of electrons

Charge on copper ion = +2

1 mole of electrons will discharge 1/2 moles of copper.

Mass of 1/2 moles of copper = number of moles × molar mass of copper

Molar mass of copper = 63.5 g

Mass of copper deposited = 1/2 × 63.5 = 31.75 g

3 0
3 years ago
Hydrochloric acid is a strong acid. Acetic acid is a weak acid. Which statement about hydrochloric acid and acetic acid is corre
vaieri [72.5K]

Answer:

The dissociation constant for hydrochloric acid is greater than the dissociation constant for acetic acid.

Explanation:

yes.

4 0
2 years ago
What are the respective concentrations (M) of Cu2+ and Cl- afforded by dissolving 2.0 g of CuCl2 in water and diluting to 500 mL
Rom4ik [11]

Answer:

M_{Cu^{2+}}=0.030M \\\\M_{Cl^-}=0.060M

Explanation:

Hello.

In this case, the first step is to compute the moles of copper (II) chloride (molar mass: 134.45 g/mol) in 2.0 g as follows:

n_{ClCl_2}=2.0CuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} =0.015molCuCl_2

Thus, since one mole of copper (II) chloride contains 1 mole of copper (its subscript) and 2 moles of chloride (its subscript), those moles are respectively:

n_{Cu^{2+}}=0.015molCuCl_2*\frac{1molCuCl_2}{1molCuCl_2} =0.015molCu^{2+}\\\\n_{Cl^-}=0.015molCuCl_2*\frac{2molCl^-}{1molCuCl_2} =0.030molCl^-

Therefore, the concentrations (in molar units) considering the volume in liters of the solution (0.500 L for 500 mL) are:

M_{Cu^{2+}}=\frac{0.015molCu^{2+}}{0.500L}=0.030M \\\\M_{Cl^-}=\frac{0.030molCu^{2+}}{0.500L}=0.060M

Best regards.

5 0
3 years ago
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