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3 years ago

Which situation is appropriate for using rate laws?

Chemistry

2 answers:

alukav5142 [94]3 years ago

8 0

The answer is: To see how fast hydrogen peroxide decomposes into water and oxygen.
Hope I helped god bless U ;)

Nastasia [14]3 years ago

3 0

Answer:

to see how fast hydrogen peroxide decomposes into water and oxygen

Explanation:

Let’s consider a chemical reaction,

A -> B

Rate law is given by the equation:

r = k [A]

where r is the rate of a reaction, k is the rate constant and [A] is the molar concentration of the reactant A. The rate of the reaction, r measures how fast the reactant gets converted to products.

Therefore, the rate of the reaction will indicate how fast hydrogen peroxide decomposes into water and oxygen.

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KBr(aq) + NaCl(aq) =

torisob [31]

It’s ionic
KCl(at)
Potassium chloride

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3 years ago

Which of the following describes a characteristic of water that helps it in dissolving a wide range of chemicals and biomolecule

Papessa [141]

<h2>Hey there!</h2>

<h3>The correct option is (A) It has a partial negative charge on oxygen and a partial positive charge on hydrogen.</h3>

<h3>☆ Explanation:</h3>

¤ As water has the ability to form hydrogen bonds which makes it an excellent solvent.

¤ For this ability of water it can dissolve many different kinds of molecules.

<h2>Hope it helps </h2>

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1 year ago

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

Vilka [71]

Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

$K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}$

$K_c=1.066\aprrox 1.1$

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0

2 years ago

What occurs in electroplating?

mixas84 [53]

I think it’s the first one. Electroplating means you put on a layer of metal on something.

5 0

2 years ago

Determine the volume of atmospheric air (at 14 lb/in.^2) needed to fill your bike tires (assuming it holds 500 mL of air) to the

Fiesta28 [93]

Boyle Law says “the pressure of fixed amount of ideal gas which is at constant temperature is inversely proportional to its volume".<span>

P = 1/V

<span>Where, P is pressure of the ideal gas and V is volume of the ideal gas.</span>

<span>For two situations, this law can be added as;
P</span>₁V₁ = P₂V₂<span>

</span><span>14 lb/in² x V₁ = 70 lb/in² x 500 mL</span><span>
</span><span>V₁ = 2500 mL</span><span>

Hence, the needed volume of atmospheric air = 2500 mL

<span>Here, we made two </span>assumptions. They are,
1. The atmospheric air acts as ideal gas.
2. Temperature is a constant.

<span>We didn't convert the units to SI units since converting volume and pressure are products of two numbers, they will cut off. </span></span></span>

3 0

3 years ago