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GalinKa [24]
3 years ago
15

Air should be classified as:

Chemistry
1 answer:
Alex777 [14]3 years ago
6 0

Answer:

element

Explanation:

it is use full for you thanks

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Mass=1.01 kg; volume =1000cm3
guajiro [1.7K]

Density = 1.01 g/cm^3 or 1.01 kg/dm^3 or 1010 kg/m^3

Density = mass/volume = 1010 g/1000 cm^3 = 1.01 g/cm^3 = 1.01 kg/dm^3

= 1010 kg/m^3


7 0
2 years ago
Anybody good at chemistry?
Luba_88 [7]

Answer:

Explanation:

1)

Given data:

Mass of lead = 25 g

Initial temperature = 40°C

Final temperature = 95°C

Cp = 0.0308 j/g.°C

Heat required = ?

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 95°C -  40°C

ΔT = 55°C

Q = 25 g × 0.0308 j/g.°C  × 55°C

Q = 42.35 j

2)

Given data:

Mass  = 3.1 g

Initial temperature = 20°C

Final temperature = 100°C

Cp = 0.385 j/g.°C

Heat required = ?

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 100°C -  20°C

ΔT = 80°C

Q = 3.1 g × 0.385 j/g.°C  × 80°C

Q = 95.48 j

3)

Given data:

Mass of Al = ?

Initial temperature = 60°C

Final temperature = 30°C

Cp = 0.897 j/g.°C

Heat released = 120 j

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 30°C -  60°C

ΔT = -30°C

120 j = m × 0.897 j/g.°C  × -30°C

120 j = m × -26.91  j/g

m = 120 j / -26.91  j/g

m =  4.46 g

negative sign show heat is released.

4)

Given data:

Mass of ice = 1.5 g

Change in temperature  = ?

Cp = 0.502 j/g.°C

Heat added= 30.0 j

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

30.0 j = 1.5 g × 0.502 j/g.°C  × ΔT

30.0 j = 0.753 j/°C  × ΔT

30.0 j /0.753 j/°C  = ΔT

39.84 °C  =  ΔT

3 0
3 years ago
Help me please!!!!!!!!!!!!
Lisa [10]
PLEASDR PLEASE PLEASE PLEASE PLEASE PLEASE HELP PLEASE PLEASE MAKE PLEASE PLEASE HELP ME PLEASE PLEASE HELP POG PLEASE PLEASE HELP US PLEASE HELP ME HELP ME PLEASE HELP ME THANKS FOR PLE ME PLEASE
3 0
2 years ago
Can anyone help to complete this question?
makkiz [27]
1. Compound. 2. Dividing. 3. Relative Formula Mass. 4. Hundred.
5 0
3 years ago
A storage tank contains n, o, and co2. The partial pressure of n and o gas are 600 torr and 150 torr. The total pressure is 825
earnstyle [38]

The partial pressure of carbon dioxide is 75 torr.

<h3>What is Dalton's theory?</h3>

Dalton's theory of gas states that total pressure of the mixture of gas sample is equal to the sum of the partial pressure of all the gases present in that sample.

Given that,

  • Total pressure of mixture = 825 torr
  • Partial pressure of Nitrogen = 600 torr
  • Partial pressure of oxygen = 150 torr
  • Partial pressure of carbon dioxide = ?

On putting values, we get

Partial pressure of carbon dioxide = 825 - (600 + 150) = 75 torr

Hence Partial pressure of carbon dioxide is 75 torr.

To know more about Dalton law, visit the below link:

brainly.com/question/9211800

#SPJ4

5 0
1 year ago
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