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Igoryamba
3 years ago
8

Imprinting can be described as an example of a

Chemistry
1 answer:
yan [13]3 years ago
5 0

Answer:

We deduce that the correct option is option c: critical period

Explanation:

Hello!

Let's solve this!

The imprint is the learning that occurs in early ages, the example of the duckling is used, which follows anyone in early periods.

This is called the critical period.

We deduce that the correct option is option c: critical period

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Halons contain halogens, which are highly reactive with oxygen. true or false
Natalija [7]

Answer:

true

Explanation:

6 0
3 years ago
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How many moles of carbonate are there in sodium carbonate​
vaieri [72.5K]

There are 0.566 moles of carbonate in sodium carbonate.

<h3>CALCULATE MOLES:</h3>
  • The number of moles of carbonate (CO3) in sodium carbonate (Na2CO3) can be calculated by dividing the mass of carbonate in the compound by the molar mass of the compound.

  • no. of moles of CO3 = mass of CO3 ÷ molar mass of Na2CO3

  • Molar mass of Na2CO3 = 23(2) + 12 + 16(3)

  • = 46 + 12 + 48 = 106g/mol

  • mass of CO3 = 12 + 48 = 60g

  • no. of moles of CO3 = 60/106

  • no. of moles of CO3 = 0.566mol

  • Therefore, there are 0.566 moles of carbonate in sodium carbonate.

Learn more about number of moles at: brainly.com/question/1542846

5 0
2 years ago
Read 2 more answers
A 1.8 g sample of octane C8H18 was burned in a bomb calorimeter and the temperature of 100 g of water increased from 21.36 C to
melomori [17]

Answer:

HEAT OF COMBUSTION PER GRAM OF OCTANE IS 1723.08 J OR 1.72 KJ/G OF HEAT

HEAT OFF COMBUSTION PER MOLE OF OCTANE IS 196.4 KJ/ MOL OF HEAT

Explanation:

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

In other words, 3101.56 J of heat was evolved from the reaction of 1.8 g octane with water.

Heat of combustion of octane per gram:

1.8 g of octane produces 3101.56 J of heat

1 g of octane will produce ( 3101.56 * 1 / 1.8)

= 1723.08 J of heat

So, heat of combustion of octane per gram is 1723.08 J

Heat of combustion per mole:

1.8 g of octane produces 3101.56 J of heat

1 mole of octane will produce X J of heat

1 mole of octane = 114 g/ mol of octane

So we have:

1.8 g of octane = 3101.56 J

114 g of octane = (3101.56 * 114 / 1.8) J of heat

= 196 432.13 J

= 196. 4 kJ of heat

The heat of combustion of octane per mole is 196.4 kJ /mol.

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

8 0
3 years ago
What happens to an acid when it dissolves in water?
shusha [124]
The concentration of the acid decreases
5 0
3 years ago
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A balloon inflated in a room that is 34 C has a volume of 5.00 liters. The room is then heated to a temperature of 68 C. What is
Eduardwww [97]

Answer:-

5.554 litres

Explanation:-

Initial Volume V 1 = 5.00 litres

Initial temperature T 1 = 34 C + 273 = 307 K

Final temperature T 2 = 68 C + 273 = 341 K

Using the relation of Charles Law (V 1 / T 1) = (V 2 / T 2)

New Volume V 2 = V1 x T2 / T1

Plugging in the values

V2 = 5.00 litres x 341 K / 307 K

= 5.554 litres.

3 0
3 years ago
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