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IRISSAK [1]
4 years ago
5

Use electron transfer or electron shift to identify what is oxidized and what is reduced in each reaction :

Chemistry
2 answers:
olga2289 [7]4 years ago
5 0
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
ss7ja [257]4 years ago
4 0

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'.

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35.2 J of heat is
ruslelena [56]

The specific heat : c = 0.306 J/g K

<h3>Further explanation</h3>

Given

Heat = 35.2 J

Mass = 16 g

Temperature difference : 7.2 K =

Required

The specific heat

Solution

Heat can be calculated using the formula:  

Q = mc∆T  

Q = heat, J  

m = mass, g  

c = specific heat, joules / g ° C  

∆T = temperature difference, ° C / K  

Input the value :

c = Q / m.∆T  

c = 35.2 / 16 x 7.2

c = 0.306 J/g K

7 0
3 years ago
If an ideal gas at a constant temperature is initially at a pressure of 3.8 atm and is then allowed to expand to a volume of 5.6
BabaBlast [244]

Answer:

The most common example is the molar volume of a gas at STP (Standard Temperature and Pressure), which is equal to 22.4 L for 1 mole of any ideal gas at a temperature equal to 273.15 K and a pressure equal to 1.00 atm.If an ideal gas at a constant temperature is initially at a pressure of 3.8 atm and is then allowed to expand to a volume of 5.6 L and a pressure of 2.1 - 18914… ... of 5.6 L and a pressure of 2.1 atm, what is the initial volume of the gas? ... An ideal gas is at a pressure of 1.4 atm and has a volume of 3 L.

Explanation:

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7 0
3 years ago
What is an example of a neutral mutation? a frog born with white skin a person's inability to break down sugar a red flower with
Naya [18.7K]

Answer:

I think a red flower with one part colored yellow

Explanation:

8 0
3 years ago
When a 0.4500 g sample of impure potassium chloride was dissolved in water and treated with an excess ofsilver nitrate, 0.8402 g
Svetllana [295]

Answer:

97.78% KCl in the original sample

Explanation:

5 0
3 years ago
Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e
igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em>  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

Explanation:

1) Moles of NaCl ,n_1=1 mol

Mass of water = m= 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute(NaCl)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p=17.19 Torr

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose ,n_1=1mol

Mass of water = m  = 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p'

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute ( glucose)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p'=17.19 Torr

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

3 0
3 years ago
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