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maksim [4K]
3 years ago
12

Which is easier to observe, physical or chemical properties? why?

Chemistry
2 answers:
tiny-mole [99]3 years ago
8 0
<span>Physical properties are easier to observe than chemical properties as they include characteristics such as color, texture, etc. while to observe chemical properties, testing must be done of the substance such as heating the substance up to determine at what temperature it will combust</span>
atroni [7]3 years ago
7 0

Answer:  Physical Properties

Explanation: Physical Properties are the properties which can be observe through naked eyes.

Chemical properties are the properties which induces any chemical change in the identity of the molecule.

Physical properties include color, smell, texture etc which can be easily observed but chemical properties include mass, volume reactivity which can only be observed when any changes in the chemical identity of the substance takes place.

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The atomic radius increases down a column (group) and decreases along a row
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The reaction below shows how silver chloride can be synthesized.
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AgNO_{3} _{(aq)}  +  NaCl_{(aq)}  ----\ \textgreater \   NaNO_{3} _{(aq)}   +  AgCl _{(aq)}
mole ration of AgCl _{(aq)}  :  AgNO_{3} _{(aq)} =  1 : 1

∴ if moles of  AgNO_{3} _{(aq)} =  15.0 mol
then   "      "     AgCl _{(aq)}  =  15.0 mol

∴  the Option  2 is the answer.

5 0
3 years ago
Read 2 more answers
What is the best place to find detailed information about a hazardous chemical used in your work area
azamat

Answer:

MSDS  --  Material Safety Data sheet

Explanation:

The full form of MSDS is Material Safety Data sheet. A Material Safety Data Sheet is a document which contains the information related to the potential hazards (fire or reactivity and environmental and health ) and includes how to work or use the chemical product safely.

It is regarded as an essential starting point for development of the complete health as well as safety program. MSDS also contains other information such as the use of the chemical, its storage and also handling and the  emergency procedures that are related to the hazards of the material.

3 0
2 years ago
1. Energy changes occur as *
alexandr1967 [171]
Number 3 i think is <span>d.heat moves from an object of higher temperature to an object of lower temperature</span>
3 0
3 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
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