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Oksanka [162]
3 years ago
11

Consider the following reaction: Mn(s) + CuSO4(aq) → MnSO4(aq) + Cu(s) Which of the following statements regarding this reaction

is correct?
The sulfate ion is oxidized.

Each copper gains two electrons.

Manganese is the oxidizing agent.

Manganese is neither oxidized nor reduced.

Copper is the reducing agent.
Please answer quickly
Chemistry
1 answer:
Rudiy273 years ago
4 0

Answer:

Each copper gains two electrons

Explanation:

Oxidizing agents:

Oxidizing agents oxidize the other elements and itself gets reduced.

Reducing agents:

Reducing agents reduced the other element are it self gets oxidized.

Chemical equation:

CuSO₄ + Mn   →  MnSO₄  + Cu

This is single replacement reaction. Manganese replace the copper and react with sulfate to form copper sulfate.

Copper gains two electrons and gets reduced. The oxidation state of copper on left side is +2 while on right side its 0 so it gain two electrons and gets reduced.

while oxidation state of manganese on left side is 0 while on right side its +2. It means Mn lose to electrons and gets oxidized.

So  copper is oxidizing agent while manganese is reducing agent.

while the oxidation state of sulfate is not changed.

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Fluorine (F) and bromine (Br) are in the same group on the periodic table. How do atoms of these elements compare when they form
stiks02 [169]

Answer:

they both gain one electron

Explanation:

they are both halogens and they each have one unpaired electron so that unpaired electron join together to make them form a bond

5 0
3 years ago
The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make
natta225 [31]

Answer:

the energy of the third excited rotational state \mathbf{E_3 = 16.041 \ meV}

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = \dfrac{m_1 \times m_2}{m_1 + m_2}

μ = \dfrac{1 \times 35}{1 + 35}

μ = \dfrac{35}{36}

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ  = \\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \  kg

μ  = 1.6139 × 10⁻²⁷ kg

r_o = 127 \ pm = 127*10^{-12} \ m

The rotational level Energy can be expressed by the equation:

E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)

where ;

J = 3 ( i.e third excited state)  &

I = \mu r^2_o

E_J= \dfrac{h^2}{8  \pi  \mu r^ 2 \mur_o } \times J ( J +1)

E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8  \times  \pi ^2  \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2  } \times 3 ( 3 +1)

E_3= 2.5665 \times 10^{-21} \ J

We know that :

1 J = \dfrac{1}{1.6 \times 10^{-19}}eV

E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV

E_3 = 16.041  \times 10 ^{-3} \ eV

\mathbf{E_3 = 16.041 \ meV}

8 0
2 years ago
!!29 POINTS!! PLS HELP ASAP
ICE Princess25 [194]

Answer:

The person who just wanted points is annoying

Explanation:

WHY WOULD YOU DO THAT

8 0
3 years ago
Calculate the pH of the following simple solutions:
IceJOKER [234]

Answer :

(1) pH = 1.27

(2) pH = 13.35

(3) The given solution is not a buffer.

Explanation :

<u>(1) 53.1 mM HCl</u>

Concentration of HCl = 53.1mM=53.1\times 10^{-3}M

As HCl is a strong acid. So, it dissociates completely to give hydrogen ion and chloride ion.

So, Concentration of hydrogen ion= 53.1\times 10^{-3}M

pH : It is defined as the negative logarithm of hydrogen ion concentration.

pH=-\log [H^+]

pH=-\log (53.1\times 10^{-3})

pH=1.27

<u>(2) 0.223 M KOH</u>

Concentration of KOH = 0.223 M

As KOH is a strong base. So, it dissociates completely to give hydroxide ion and potassium ion.

So, Concentration of hydroxide ion= 0.223 M

Now we have to calculate the pOH.

pOH=-\log [OH^-]

pOH=-\log (0.223)

pOH=0.65

Now we have to calculate the pH.

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-0.65=13.35

<u>(3) 53.1 mM HCl + 0.223 M KOH</u>

Buffer : It is defined as a solution that maintain the pH of the solution by adding the small amount of acid or a base.

It is not a buffer because HCl is a strong acid and KOH is a strong base. Both dissociates completely.

As we know that the pH of strong acid and strong base solution is always 7.

So, the given solution is not a buffer.

5 0
3 years ago
What is true about energy in an ordinary chemical reaction? Energy is
Lisa [10]
The answer is statement #3.
4 0
3 years ago
Read 2 more answers
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