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jek_recluse [69]
4 years ago
11

Student follow up

Chemistry
2 answers:
sp2606 [1]4 years ago
8 0

Answer:

Hydrogen accept the two electron form zinc and get reduced. Zinc loses its to electron and get oxidized.

Explanation:

The oxidation reduction reactions are called redox reaction. These reactions are take place by gaining or losing the electrons and oxidation state of elements are changed.

Oxidation:

Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

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 reaction:

Zn(s) + 2H⁺(aq) →  Zn²⁺(aq) + H₂(g)

zinc is oxidized because its oxidation state increase from 0 to +2 and hydrogen is reduced because its oxidation state decreased from +2 to 0.

Hydrogen accept the two electron form zinc and get reduced. Zinc loses its to electron and get oxidized.

Maurinko [17]4 years ago
8 0
<h3>Answer:</h3>

Zinc (Zn) was oxidized

<h3>Explanation:</h3>

The ionic equation given is;

  • Zn(s) + 2H+ (aq) → Zn²⁺(aq) + H2(g)

To answer the question, we can first write the half equation for the above reaction;

The half equations are;

  • Zn(s) → Zn²⁺(aq) + 2e, and
  • 2H⁺(aq) + 2e → H₂(g)

From the half equations;

  • Zn lost two electrons to form Zn²⁺, while
  • 2H⁺ gained two electrons to form H₂

Therefore;

  • Zn underwent oxidation which is the loss of electrons while H+ underwent reduction which is the gain of electrons.

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A quantity of N2 gas originally held at 5.23 atm pressure in a 1.20 −L container at 26 ∘C is transferred to a 14.5 −L container
lara31 [8.8K]

Answer:

n (N₂) =  0.256 mol

n (O₂) = 1.0848 mol

n (Total) = 1.3408 mol

Pressure in new Container =  2.222 atm

Explanation:

Data Given:

For Nitrogen gas (N₂)

Pressure of N₂ gas =  5.23 atm

Volume of N₂ gas = 1.20 L

Temperature of N₂ gas = 26° C

Temperature of N₂ gas in Kelven (K) = 26° C +273

Temperature of N₂ gas in Kelven (K) = 299K

ideal gas Constant R = 0.08206 L atm K⁻¹ mol⁻¹

quantity of gas N₂ gas = ?

For Oxygen gas (O₂)

Pressure of O₂ gas =  5.21 atm

Volume of O₂ gas = 5.21 L

Temperature of O₂ gas = 26° C

Temperature of O₂ gas in Kelven (K) = 26° C +273

Temperature of O₂ gas in Kelven (K) = 299K

ideal gas Constant R = 0.08206 L atm K⁻¹ mol⁻¹

quantity of gas O₂ gas = ?

*we also have to find the total Pressure in the new container = ?

Formula Used

                         PV =nRT

                        n (N₂) = PV /RT . . . . . . . . . . . . . (1)

* Find the quantity of N₂

Put value in formula (1)

                n (N₂) = 5.23 atm x 1.20 L / 0.08206 L atm K⁻¹ mol⁻¹ x 299K

                n (N₂) =  6.276 atm .L /  24.52 L atm. mol⁻¹ x 299K

                n (N₂) =  0.256 mol

* Find the quantity of O₂

Put value in formula (1)

                n (O₂) = 5.21 atm x 5.10 L / 0.08206 L atm K⁻¹ mol⁻¹ x 299K

                n (O₂) =  26.6 atm .L /  24.52 L atm. mol⁻¹

                n (O₂) = 1.0848 mol

*Now to find the Total Quantity of both gases

                n(Total) =  n (N₂) + n (O₂)

                 n (Total) = 0.256 mol + 1.0848 mol

                 n (Total) = 1.3408 mol

**To find the Total Pressure in the new Container

Data to calculate Total Pressure in new container

Volume of gas = 14.5 L

Temperature of gases = 20° C

Temperature of gases in Kelven (K) = 20° C +273

Temperature of gases in Kelven (K) = 293K

ideal gas Constant R = 0.08206 L atm K⁻¹ mol⁻¹

Volume Pressure in new container = ?

Formula Used

                         PV =nRT

                        P = nRT / V . . . . . . . . . . . . . (2)

Put values in Equation (2)

           P =  1.3408 mol x 0.08206 L atm K⁻¹ mol⁻¹ x 293 K / 14.5 L

           P =  2.222 atm

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