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WINSTONCH [101]
4 years ago
12

Which will conduct electricity: A solution of water and (NH₄)₂S or a solution of water and C₆H₁₂O₆?

Chemistry
1 answer:
TEA [102]4 years ago
8 0

Answer:

A solution of water and (NH₄)₂S

Explanation:

In a solution , the electricity can pass from it , if there is presence of free ions , as they are mobile and helps to conduct the electricity .

Therefore , the present of free ions is necessary for a solution to conduct electricity.

Hence, from the given options ,

  • Solution of water and (NH₄)₂S will conduct electricity as (NH₄)₂S will dissociate completely to the respective ions as they are dissolved in water , and the ions will help to conduct electricity.

Whereas ,

  • Solution of water and C₆H₁₂O₆ will not conduct electricity , as C₆H₁₂O₆ in water  will not dissociate completely , and does not produce ions , hence , won't conduct electricity.

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619°C

Explanation:

Given data:

Initial volume of gas = 736 mL

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Solution:

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Initial temperature = 15.0°C (15+273 = 288 K)

The given problem will be solve through the Charles Law.

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Mathematical expression:

V₁/T₁ = V₂/T₂

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T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = T₁V₂/V₁  

T₂ = 2.28 L × 288 K / 0.736 L

T₂ = 656.6 L.K / 0.736 L

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K to °C:

892.2 - 273.15 = 619°C

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Propane burns according to the following equation:
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Answer:

a. 324 mL is the volume of CO₂ measured

b. 0.223 moles of carbonate

Explanation:

a. We determine the moles of used O₂ by the Ideal Gases Law

STP are 1 atm of pressure and 273.15K of T°

We convert the volume from mL to L → 0.465 mL . 1L/ 1000 mL = 0.465 L

Now, we replace data: 1 atm . 0465L = n . 0.082 . 273.15K

 1 atm . 0465L / 0.082 . 273.15K  = n → 0.0207 moles

The balanced combustion is: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

Ratio is 5:3. 5 moles of oxygen are needed to produce 3 moles of CO₂

Then 0.0207 moles must produce (0.0207 . 3) / 5 = 0.0124 moles of CO₂

Let's apply again the Ideal Gases Law. Firstly we convert:

37°C + 273.15K = 310.15K

98.59kPa . 1atm / 101.3 kPa = 0.973 atm

0.973atm . V = 0.0124 mol . 0.082 . 310.15K

V = (0.0124 mol . 0.082 . 310.15K) / 0.973atm = 0.324 L → 324 mL

b. The reaction is: CaCO₃(s) → CaO(s) + CO₂(g)

We used the Ideal Gases Law to determine the moles of produced CO₂

P . V = n . R . T → P .V / R . T = n

We replace data: 1 atm . 5L / 0.082 . 273.15K = 0.223 moles

As ratio is 1:1, 0.223 moles of CO₂ were produced by the decomposition of 0.223 moles of carbonate

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3 years ago
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