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cluponka [151]
3 years ago
10

what will happen if potato strips are placed in a solution of the same concentration as the content of the potato

Chemistry
1 answer:
Snowcat [4.5K]3 years ago
6 0
If you mean the same glucose concentration as the potato then the potato and the solution will be in an Isotonic state where there is no net movement of water by osmosis so no water leaves the potato cells nor water enters the strips cells 

hope that helps
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Calculate the energy that is required to change 50.0 g ice at -30.0°C to a liquid at 73.0°C. The heat of fusion = 333 J/g, the h
OverLord2011 [107]

Answer:

There is 3.5*10^4 J of energy needed.

Explanation:

<u>Step 1:</u> Data given

Mass of ice at -30.0 °C = 50.0 grams

Final temperature = 73.0 °C

The heat of fusion = 333 J/g

the heat of vaporization = 2256 J/g

the specific heat capacity of ice = 2.06 J/gK

the specific heat capacity of liquid water = 4.184 J/gK

<u>Step 2:</u> Calculate the heat absorbed by ice

q = m*c*(T2-T1)

⇒ m = the mass of ice = 50.0 grams

⇒ c = the heat capacity of ice = 2.06 J/gK = 2.06 J/g°C

⇒ T2 = the fina ltemperature of ice = 0°C

⇒ T1 = the initial temperature of ice = -30.0°C

q = 50.0 * 2.06 J/g°C * 30 °C

q = 3090 J

<u>Step 3:</u> Calculate heat required to melt the ice at 0°C:

q = m*(heat of fusion)

q = 50.0* 333J/g

q =  16650 J

<u> </u>

<u>Step 4</u>: Calculate the heat required to raise the temperature of water from 0°C to 73.0°C

q = m*c*(T2-T1)

 ⇒ mass = 50.0 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = T2-T1 = 73.0 - 0  = 73 °C

q = 50.0 * 4.184 * 73.0 = 15271.6 J

<u>Step 5:</u> Calculate the total energy

qtotal = 3090 + 16650 + 15271.6 = 35011.6 J = 3.5 * 10^4 J

There is 3.5*10^4 J of energy needed.

8 0
3 years ago
2KCIO3 -2KCL + 302
Firlakuza [10]

Answer:

a. 6 moles of O₂

b. 4.06×10²⁴ molecules of O₂

c. 3.04 g of KCl

Explanation:

Reaction of decomposition is:

2KClO₃  →  2KCl  +  3O₂

a. See stoichiometry value

2 moles of potassium chlorate can decompose to 2 moles of potassium chloride and 3 moles of oxygen. Ratio is 2:3

If 2 moles of KClO₃ can decompose to 3 moles of O₂

Then 4 moles, may decompose to (4 . 3)/2 = 6 moles of O₂

b. In this case, the stoichiometry is the same.

Per 2 moles of KClO₃, I produce 2 moles of KCl

Then, 4.5 moles of KCl, were produced by 4.5 moles of KClO₃

We apply, the last relation:

(4.5 . 3) /2 = 6.75 moles of O₂ are also produced.

How many molecules are in 6.75 moles?

6.75 mol . 6.02×10²³ molecules/mol = 4.06×10²⁴ molecules of O₂

c. First of all, we convert the mass to moles:

5g . 1mol /122.55g = 0.0408 moles of salt

As ratio is 2:2, 0.0408 moles of salt, decompose to 0.0408 moles of KCl

We convert the moles to mass: 0.0408 mol . 74.55g /mol = 3.04g

5 0
2 years ago
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