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 1

Answer:

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'.

Answer 2

Answer: 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

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Determine the number of neutrons in an atom of Si-29

Answers

15 as 29 (atomic mass-number of neutrons and protons) - 14 (atomic number-number of protons)

The symbol for hydronium ion concentration is

Answers

The symbol for hydronium ion concentration is H+. There are quite a few relationships between [H+] and [OH−] ions. And because there is a large range of number between 10 to 10-15 M, the pH is used. pH = -log[H+] and pOH = -log[OH−]. In aqueous solutions, [H+ ][OH- ] = 10-14. From here we can derive the values of each concentration.

Answer:

H+

Explanation:

An bhfuil an freagra.

Carbon-14 has a half-life of 5,730 y. How much of a 144g sample of carbon-14 will remain after 1.719x10 ^4 y.

Answers

To determine how much of a 144g sample of carbon-14 will remain after 1.719 x 10^4 years, you can use the formula for exponential decay:

\[N(t) = N_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{T}}\]

Where:

- $N(t)$ is the remaining amount after time $t$.

- $N_0$ is the initial amount.

- $t$ is the time that has passed.

- $T$ is the half-life.

In this case, $N_0$ is 144g, $t$ is 1.719 x 10^4 years, and $T$ is the half-life of carbon-14, which is 5,730 years.

Plug these values into the formula:

\[N(t) = 144g \cdot \left(\frac{1}{2}\right)^{\frac{1.719 \times 10^4\text{ years}}{5,730\text{ years}}}\]

Now, calculate:

\[N(t) = 144g \cdot \left(\frac{1}{2}\right)^{\frac{3}{2}}\]

\[N(t) = 144g \cdot \left(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\right)\]

\[N(t) = 144g \cdot \frac{1}{8}\]

Now, multiply 144g by 1/8 to find the remaining amount:

\[N(t) = \frac{144g}{8} = 18g\]

So, after 1.719 x 10^4 years, only 18g of the 144g sample of carbon-14 will remain.

What is the pressure of 64 grams of O2 in a 1L container at a temperature of 10°C?

Answers

So PV=nRT
P=?

We know that:
-V=1L
-R=8.314kPa•L/mol•K
-T=10°C (we need to converse °C in K so 10+273) =283K
For n, we know that we have 64g of O2 and were looking for the quantity of mole in these 64g. 1 mole of O2 is 15.999g so 64g would be 4.0mol ((64g • 1mol)/15.999g) so n=4.0

P•1L= 4.0mol • 8.314kPa•L/mol•K • 283K
P • 1L = 9411
P = 9411/1L
P = 9411kPa

A person makes a 400 dollar deposit in a bank account in January. By December, the value of that deposit willA. be three hundred dollars.
B. actually double.
C. remain the same, plus interest.
D. be two hundred dollars.

Answers

The amount in his bank account should grow because he will have accumulated some interest, even if the $400 stays the same, so this would be c.

The answer choice is B) It should actually double.

Considering it's been in the bank so long, it would have increased because of interest rates.

The following Lewis structures represent valid resonance forms. true or false

Answers

The given Lewis structure does not show valid resonance form. So,the given statement is false.

What is Lewis structure?

The Lewis structure is the representation of the of valence shell in the molecule.

What is valence electron?

The electron present in outermost shell and which involve in chemical bond formation is called valence electron.

What is resonance structure?

Resonance is way to explain the localisation of electron in molecule where Lewis formula fails to describe.

The given statement is false because both structure shows different number of valence electron.

To learn more about Lewis structure here.

brainly.com/question/27037762

#SPJ2

Answer:

true

Explanation:

because Lewis represent valid resonance

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