Which chemicals previously used in products such as refrigerants, aerosol sprays, and cleaning products contributed to the depletion of the ozone layer

Answers

Answer 1

Answer:

Answer:chlorofluorocarbons (CFCs) could deplete Earth's atmospheric ozone layer, which blocks the sun's damaging ultraviolet rays. When the scientists reported their findings in 1974, CFCs were widely used as refrigerant gases and as propellants in aerosol sprays

Explanation:

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When iron reacts with oxygen to form rust, each iron atoma. loses three ions.
c. gains three ions.
b. loses three electrons.
d. gains three electrons.

Answers

Iron rust is the oxidation process that affects iron. Iron reacting with oxygen loses three electrons to form rust. Thus, option b is accurate.

What are oxidation and reduction?

Oxidation is the redox process that raises the oxidationnumber by losing the electron of the species and the reduction will decrease the oxidation state by accepting the electrons.

The iron loses its three electrons when reacts with the atmospheric oxygen in rusting and gets oxidized. The metal oxide is produced as a result of the rust.

Therefore, option b. losing threeelectrons to produce rust is correct.

Learn more about iron rust here:

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When iron reacts with oxygen to form rust, each iron atom loses three electrons. Hope this helps :D

a sample of copper wire with a mass of 1.659 g is heated with excess powdered sulfur until the copper has combined with the sulfur, and all the extra sulfur has been burned off. The final product (a sulfide of copper ) has a mass of 2.080 g. Calculate the empirical formula of the product, and name it.

Answers

Copper mass = 1.659
Copper moles = 1.659 / 64 = 0.0259

Sulfur mass = 2.08 - 1.659 = 0.421
Sulfur moles = 0.421 / 32 = 0.0131

Ratio Copper : Sulfur = 2 : 1

Empirical formula: Cu ₂ S
Name: Copper (I) sulfide

Use electron transfer or electron shift to identify what is oxidized and what is reduced in each reaction :a) 2Na(s) + Br2(l) ----> 2NaBr(s)
b) H2(g) + Cl2(g) ----> 2HCl(g)
c) 2Li(s) + F2(g) ----> 2LiF(s)
d) S(s) + Cl2(g) ----> SCl2(g)
e)N2(g) + 2O2(g) ----> 2NO2(g)
f) Mg(s) +Cu(NO3)2(aq) = Mg(NO3)2(aq) + Cu(s)

For each reaction above, identify the reducing agent and the oxidizing agent

Answers

Answer :

Oxidation-reduction reaction : It is a type of reaction in which oxidation and reduction reaction occur simultaneously.

Oxidation reaction : It is the reaction in which a substance looses its electrons. In the oxidation reaction, the oxidation state of an element increases.

Reduction reaction : It is the reaction in which a substance gains electrons. In the reduction reaction, the oxidation state of an element decreases.

(a) The balanced chemical reactions is,

$2Na(s)+Br_2(l)\rightarrow 2NaBr(s)$

Half reactions of oxidation and reduction are :

Oxidation : $Na\rightarrow Na^(1+)+1e^-$

Reduction : $Br_2+2e^-\rightarrow 2Br^(1-)$

From this we conclude that, 'Na' is oxidized and $'Br_2'$ is reduced in this reaction. The reducing agent is, 'Na' and oxidizing agent is, $'Br_2'$ .

(b) The balanced chemical reactions is,

$H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$

Half reactions of oxidation and reduction are :

Oxidation : $H_2\rightarrow H^(1+)+1e^-$

Reduction : $Cl_2+2e^-\rightarrow 2Cl^(1-)$

From this we conclude that, $'H_2'$ is oxidized and $'Cl_2'$ is reduced in this reaction. The reducing agent is, $'H_2'$ and oxidizing agent is, $'Cl_2'$ .

(c) The balanced chemical reactions is,

$2Li(s)+F_2(g)\rightarrow 2LiF(s)$

Half reactions of oxidation and reduction are :

Oxidation : $Li\rightarrow Li^(1+)+1e^-$

Reduction : $F_2+2e^-\rightarrow 2F^(1-)$

From this we conclude that, 'Li' is oxidized and $'F_2'$ is reduced in this reaction. The reducing agent is, 'Li' and oxidizing agent is, $'F_2'$ .

(d) The balanced chemical reactions is,

$S(s)+Cl_2(g)\rightarrow SCl_2(g)$

Half reactions of oxidation and reduction are :

Oxidation : $S\rightarrow S^(2+)+2e^-$

Reduction : $Cl_2+2e^-\rightarrow 2Cl^(1-)$

From this we conclude that, 'S' is oxidized and $'Cl_2'$ is reduced in this reaction. The reducing agent is, 'S' and oxidizing agent is, $'Cl_2'$ .

(e) The balanced chemical reactions is,

$N_2(g)+2O_2(g)\rightarrow 2NO_2(g)$

Half reactions of oxidation and reduction are :

Oxidation : $N_2\rightarrow N^(4+)+4e^-$

Reduction : $O_2+4e^-\rightarrow 2O^(2-)$

From this we conclude that, $'N_2'$ is oxidized and $'O_2'$ is reduced in this reaction. The reducing agent is, $'N_2'$ and oxidizing agent is, $'O_2'$ .

(f) The balanced chemical reactions is,

$Mg(s)+Cu(NO_3)_2(aq)\rightarrow Mg(NO_3)_2(aq)+Cu(s)$

Half reactions of oxidation and reduction are :

Oxidation : $Mg\rightarrow Mg^(2+)+2e^-$

Reduction : $Cu^(2+)+2e^-\rightarrow Cu$

From this we conclude that, $'Mg'$ is oxidized and $'Cu'$ is reduced in this reaction. The reducing agent is, 'Mg' and oxidizing agent is, 'Cu'.

a) Na is oxidised Br is reduced
b) H is oxidised Cl is reduced
c) Li is oxidised F is reduced
d)S is oxidised Cl is reduced
e) N is oxidised O is reduced
f) Mg is oxidised and N is reduced

Remember: Oxidation= loss and Reduction= gains

For cells in which oxygen gas (O2) crosses the membrane by simple diffusion, what determines the rate at which oxygen enters the cell?

Answers

Explanation:

The diffusion phenomenon is generated by the difference of concentration of a gas between two reference points.

In this case, the two points are: the inside of the cell and the surroundings of it.

The diffusion of oxygen will depend then of the concentration of it in the cell and the concentration in the outside:

$\Delta O_2= Oxygen_(outside) - Oxygen_(cell)$

The bigger this difference, the higher the diffusion rate of oxygen entering the cell.

You fill a rigid steel cylinder that has a volume of 20.0 L with nitrogen gas to a final pressure of 2.00 *10^4 Kpa at 28c How many kilograms of N2 does the cylinder contain