We read your photo as:

What are the answers to questions 11, 12, and 13 regarding chemical reactions, rusting, and gas laws?

Chemistry

ScanSolve AI Answer

Answer

d(ii)

4.84 atm\text{4.84 atm}

Step-by-step solution

4 steps

11.

Part a

The possible identity of Y is zinc nitrate (Zn(NO3)2\text{Zn(NO}_3)_2).

Part b

The name of gas X is **nitrogen dioxide

Part c

The equation for the formation of the white precipitate Q is

Na2[Zn(OH)4](aq)+2HCl(aq)→Zn(OH)2(s)+2NaCl(aq)+2H2O(l)\text{Na}_2[\text{Zn(OH)}_4](aq) + \text{2HCl}(aq) \rightarrow \text{Zn(OH)}_2(s) + \text{2NaCl}(aq) + \text{2H}_2\text{O}(l)

Part d(i)

Charles's Law states that for a fixed mass of gas at constant pressure, its volume is directly proportional to its absolute temperature.

One industrial use of gas X (nitrogen dioxide) is in the **manufacture of nitric acid

12.

a) These methods are similar because they all involve creating a physical barrier on the surface of the iron object, preventing it from coming into contact with oxygen and water, which are necessary for rusting.

b) Galvanized iron objects are better protected even when scratched due to sacrificial protection. Zinc is more reactive than iron, so when the protective zinc layer is scratched, the zinc preferentially corrodes (oxidizes) instead of the iron, thus protecting the iron from rusting.

13.

Part d(ii)

4.84 atm\text{4.84 atm}

Step 1: Convert temperatures from Celsius to Kelvin.

T1=37∘C+273=310 KT_1 = 37^\circ\text{C} + 273 = 310 \text{ K} T2=27∘C+273=300 KT_2 = 27^\circ\text{C} + 273 = 300 \text{ K}

  1. Use the Combined Gas Law formula.

    P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

  2. Rearrange the formula to solve for P2P_2.

    P2=P1V1T2V2T1P_2 = \frac{P_1 V_1 T_2}{V_2 T_1}

  3. Substitute the given values into the formula.

    P2=(1 atm)(2500 cm3)(300 K)(500 cm3)(310 K)P_2 = \frac{(1 \text{ atm})(2500 \text{ cm}^3)(300 \text{ K})}{(500 \text{ cm}^3)(310 \text{ K})}

  4. Calculate the final pressure.

    P2=750000155000 atmP_2 = \frac{750000}{155000} \text{ atm} P2≈4.8387 atmP_2 \approx 4.8387 \text{ atm} P2≈4.84 atmP_2 \approx \boxed{\text{4.84 atm}}
Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Chemistry Questions

QWhat are the answers to these chemistry questions?QIdentify and differentiate between physical and chemical changes, define atomicity, and explain…QAnswer the following questions: What is atomicity? What is rusting? Distinguish between physical and…QIdentify the substances and their uses, and describe the chemical reactions and properties for…QWhat are the answers to questions 8, 9, 10, and 11?
See all Chemistry questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp