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11Alexandr11 [23.1K]

3 years ago

8

How can a chemical reaction be sped up? Select from the drop-down menu to correctly complete the statement. Expose more of the r

eactant by increasing the

1 answer:

Alexus [3.1K]3 years ago

5 0

Answer:

dont know the options I wish I could help sorry

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Q1. Calculate the amount of copper produced in 1.0 hour when aqueous CuBr2 solution was electrolyzed by using a current of 4.50

OLga [1]

Answer:

$\boxed{\text{Q1. 3.6 g; Q2. 0.2 A}}$

Explanation:

Q1. Mass of Cu

(a) Write the equation for the half-reaction.

Cu²⁺ + 2e⁻ ⟶ Cu

The number of electrons transferred (z) is 2 mol per mole of Cu.

(b) Calculate the number of coulombs

q = It

$\text{t} = \text{1.0 h} \times \dfrac{\text{3600 s}}{\text{1 h}} = \text{3600 s}\\\\q = \text{3 C/s} \times \text{ 3600 s} = \textbf{10 800 C}$

(c) Mass of Cu

We can summarize Faraday's laws of electrolysis as

$\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\& = &\dfrac{10 800 \times 63.55}{2 \times 96 485}\\\\& = & \textbf{3.6 g}\\\end{array}\\\text{The mass of Cu produced is $\boxed{\textbf{3.6 g}}$}$

Note: The answer can have only two significant figures because that is all you gave for the time.

Q2. Current used

(a) Write the equation for the half-reaction.

Ag⁺ + e⁻ ⟶ Ag

The number of electrons transferred (z) is 1 mol per mole of Ag.

(a) Calculate q

$\begin{array}{rcl}m &=& \dfrac{qM}{zF}\\\\2.00& = &\dfrac{q \times 107.87}{1 \times 96 485}\\\\q &=& \dfrac{2.00 \times 96485}{107.87}\\\\& = & \textbf{1789 C}\\\end{array}$

(b) Calculate the current

t = 3 h = 3 × 3600 s = 10 800 s

$\begin{array}{rcl}q&=& It\\1789 & = & I \times 10800\\I & = & \dfrac{1789}{10800}\\\\& = & \textbf{0.2 A}\\\end{array}\\\text{The current used was $\large \boxed{\textbf{0.2 A}}$}$

Note: The answer can have only one significant figure because that is all you gave for the time.

4 0

3 years ago

A surfactant used for cleaning is called a(n):_________.

Hoochie [10]

I’m not really sure but I wanna think I te catiónic

5 0

3 years ago

The rate constant for the oxidation of nitric oxide by ozone is 2 x 10^14 molecule cm s, whereas that for the competing reaction

andreev551 [17]

Answer:

The NO + O3 is the dominant reaction.

Explanation:

First of all, let's convert to molecules/cm³;

For O3;

O3 at 40 ppb in atm= 4 x 10^(-8) atm and from ideal gas law PV = nRT or simplify n/V = P/RT

Thus, plugging in the relevant values to get;

n/V = [4 x 10^(-8)]/(0.0821 x 298) = 1.636 x 10^(-9)

So, n/V = 1.636 x 10^(-9) = (1.635 x 10-9 mol L-1)(6.02 x10^(23) molec/mol)(L/1000 cm3) =

9.84 x 10^(11) molecules/cm³

But from the question, NO has 2 moles, and thus concentration is;

2 x 9.84 x 10^(11) = 1.968 x 10^(12) molec/cm³

For O2;

Following the same pattern for O3, we obtain;

(0.21 atm)/[(0.0821 L atm mol-1 K-1)(298K)] = 5.167 x 1018 molecules/cm³

Now, for NO and O3 reaction the rate is; k[NO] [O3]

Thus rate;

= (2 x 10^(-14)cm³/molec.s)( 9.84 x 10^(11)molec/cm³)(1.968 x 10^(12) molec/cm³) = 3.9 molec/cm³.s

For 2NO + O2 → 2NO2 reaction, rate = k[NO]2 [O2]

Thus, rate;

= (2 x 10^(-38) cm^(6)/molec².s )( 1.968 x 10^(12) molec/cm³) ²

(5.167 x 1018 molec/cm³)

= 40,000 molec/cm³.s

Observing the two rates, it's clear that the NO + O3 is the dominant reaction.

6 0

3 years ago

BRAINLIESTTT ASAP!! PLEASE HELP ME :)

fredd [130]

Answer:

Is B, it is B because you don't need arrows on it.

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I am lonely I need anime weebs

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Answer:

IM HEREEEEEEEEEEEEEEEEE

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