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2 years ago

3. Do distilled water conduct electricity? What will happen if we add sugar to it and

Chemistry

2 answers:

nikitadnepr [17]2 years ago

8 0

Answer:

It cannot conduct electricity, however adding salt or sugar will make the water have impurities/other substance making it easier to conduct electricity

Explanation:

Distilled water by itself does not contain impurities, thus, it cannot conduct electricity.

When you put salt in water, the water molecules pull the sodium and chlorine ions apart so they are floating freely, increasing the conductivity.

For more information, please refer to the internet :D

Have fun studying, and goodluck!

If you are satisfied with this answer, please rate it or give brainliest.

Elden [556K]2 years ago

8 0

Answer:

• Distillate does not conduct electricity. This is because it lacks ions which may allow flow of electricity.

• If sugar is added to the distillate, it becomes a solution and contains ions which become mobile when electricity is passed through. Due to this mobility, the electrons are moved by these ions hence flow of current.

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Answer:

Kinetic Molecular Theory states that gas particles are in constant motion and exhibit perfectly elastic collisions. Kinetic Molecular Theory can be used to explain both Charles' and Boyle's Laws. The average kinetic energy of a collection of gas particles is directly proportional to absolute temperature only. Hope this helps!!

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What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2

atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

$Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}$

$Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}$

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

$Mole of AlCl_3 = Mole of Al * Mole ratio$

$Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}$

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = $133.34 \frac{g}{mol}$

Mass of AlCl3 can be calculated using mole formula as:

$1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}$

Multiplying both side by $133.34 \frac{g}{mol}$

$1.32 mole * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34 \frac{g}{mol}$

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

$Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}$

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

$Mole of AlCl_3 = mole of Cl_2 * mole ratio$

$Mole of AlCl_3 = 0.55 mole * \frac{2}{3}$

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

$0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}$

Multiplying both side by

$133.34 \frac{g}{mol}$

$0.367 mol of AlCl_3 * 133.34 \frac{g}{mol} = \frac{mass(g)}{133.34\frac{g}{mol}} * 133.34 \frac{g}{mol}$

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

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