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3 years ago

Which of the following results in a chemical change in matter?

Chemistry

1 answer:

kodGreya [7K]3 years ago

7 0

You have a raw egg you put it in a hot pan boom cooked egg most chemical changes are nonreversibal

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Just as the depletion of stratospheric ozone today threatens life on Earth today, its accumulation was one of the crucial proces

inessss [21]

Answer:

(a) rate = -(1/3) Δ[O₂]/Δt = +(1/2) Δ[O₃]/Δt

(b) Δ[O₃]/Δt = 1.07x10⁻⁵ mol/Ls

Explanation:

By definition, the reaction rate for a chemical reaction can be expressed by the decrease in the concentration of reactants or the increase in the concentration of products:

aX → bY (1)

$rate= -\frac{1}{a} \frac{\Delta[X]}{ \Delta t} = +\frac{1}{b} \frac{\Delta[Y]}{ \Delta t}$

where, a and b are the coefficients of de reactant X and product Y, respectively.

(a) Based on the definition above, we can express the rate of reaction (2) as follows:

3O₂(g) → 2O₃(g) (2)

$rate = -\frac{1}{3} \frac{\Delta[O_{2}]}{\Delta t} = +\frac{1}{2} \frac{ \Delta[O_{3}]}{ \Delta t}$ (3)

(b) From the rate of disappearance of O₂ in equation (3), we can find the rate of appearance of O₃:

$rate = +\frac{1}{2} \frac{\Delta[O_{3}]}{ \Delta t} = -\frac{1}{3} \frac{\Delta[O_{2}]}{ \Delta t}$

$+\frac{\Delta[O_{3}]}{ \Delta t} = -\frac{2}{3} \frac{\Delta[-1.61 \cdot 10^{-5}]}{ \Delta t}$

$\frac{\Delta[O_{3}]}{ \Delta t} = 1.07 \cdot 10^{-5} \frac{mol}{Ls}$

So the rate of appearance of O₃ is 1.07x10⁻⁵ mol/Ls.

Have a nice day!

6 0

4 years ago

If a neutral atom of beryllium has 4 protons and 5 neutrons, what is the mass number of the atom?

Fynjy0 [20]

Answer:

Explanation:

mass number =protons+neutrons

4 0

4 years ago

Write .. hot copper metal reacts with chlorine gas to form solid green copper chloride

Flauer [41]

Answer:

Cu + Cl2 → CuCl

To happen the reaction, copper metal is heated.

5 0

3 years ago

A pressure cooker contains 5.68 liters of air at a temperature of 394 K. If the absolute pressure of the air in the pressure coo

Arturiano [62]

Answer:

Explanation:

We shall find out volume of air at NTP or at 273 K and 10⁵ Pa ( 1 atm )

Let it be V₂

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

$\frac{2\times 10^5\times 5.68}{394} =\frac{10^5\times V_2}{273}$

V₂ = 7.87 litres

22.4 litres of any gas is equivalent to 1 mole

7.87 litres of air will be equivalent to

7.87 / 22.4 moles

= .35 moles .

7 0

3 years ago

How many moles of NaCl are equivalent to 15.6g NaCl

aev [14]

Hey there!:

Molar mass NaCl = 58.44 g/mol

Number of moles:

n = mass of solute / molar mass

n = 15.6 / 58.44

n = 0.2669 moles of NaCl

hope this helps!

7 0

3 years ago

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