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3 years ago

12. List examples of physical and chemical properties.

Chemistry

2 answers:

alexandr1967 [171]3 years ago

7 0

Answer:its below

Explanation:

chemical

flammability

toxicity

enthalpy of formation

heat of combustion

oxidation states

half-life

coordination number

surface tension

reactivity

hygroscopy

physical

mass

volume

density

color

temperature

melting point

boiling point

reflectivity

elasticity

luster

permeability

ductility

pressure

viscosity

strength

solubility

electrical charge

opacity

hardness

gladu [14]3 years ago

3 0

Answer:Phase Transformation Temperatures

Density

Specific Gravity

Thermal Conductivity

Linear Coefficient of Thermal Expansion

Electrical Conductivity and Resistivity

Magnetic Permeability

Corrosion Resistance

Explanation:

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Ethanol (C2H5OH) melts a - 144 oC and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vapo

hammer [34]

Answer:

For a: The total heat required is 36621.5 J

For b: The total heat required is 58944.5 J

Explanation:

For a:

To calculate the heat required at different temperature, we use the equation:

$q=mc\Delta T$ .........(1)

where,

q = heat absorbed

m = mass of substance

$c$ = specific heat capacity of substance

$\Delta T$ = change in temperature

To calculate the amount of heat required at same temperature, we use the equation:

$q=m\times \Delta H$ ........(2)

where,

q = heat absorbed

m = mass of substance

$\Delta H$ = enthalpy of the reaction

The processes involved in the given problem are:

$1.)C_2H_5OH(l)(35^oC)\rightarrow C_2H_5OH(l)(78^oC)\\2.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

For process 1:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=35^oC\\\Delta T=[T_2-T_1]=[78-35]^oC=43^oC=43K$

Putting values in equation 1, we get:

$q_1=42.0g\times 2.3J/g.K\times 43K\\\\q_1=4153.8J$

For process 2:

We are given:

Conversion factor: 1 kJ = 1000 J

Molar mass of ethanol = 46 g/mol

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{35.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 773.04J/g\\\\q_2=32467.7J$

Total heat required = $[q_1+q_2]$

Total heat required = $[4153.8J+32467.7J]=36621.5J$

Hence, the total heat required is 36621.5 J

For b:

The processes involved in the given problem are:

$1.)C_2H_5OH(s)(-155^oC)\rightarrow C_2H_5OH(s)(-144^oC)\\2.)C_2H_5OH(s)(-144^oC)\rightarrow C_2H_5OH(l)(-144^oC)\\3.)C_2H_5OH(l)(-144^oC)\rightarrow C_2H_5OH(l)(78^oC)\\4.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

For process 1:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_s=0.97J/g.K\\T_2=-144^oC\\T_1=-155^oC\\\Delta T=[T_2-T_1]=[-144-(-155)]^oC=11^oC=11K$

Putting values in equation 1, we get:

$q_1=42.0g\times 0.97J/g.K\times 11K\\\\q_1=448.14J$

For process 2:

We are given:

$m=42.0g\\\Delta H_{fusion}=5.02kJ/mol=\frac{5.02kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=109.13J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 109.13J/g\\\\q_2=4583.5J$

For process 3:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=-144^oC\\\Delta T=[T_2-T_1]=[78-(-144)]^oC=222^oC=222K$

Putting values in equation 1, we get:

$q_3=42.0g\times 2.3J/g.K\times 222K\\\\q_3=21445.2J$

For process 4:

We are given:

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{38.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_4=42.0g\times 773.04J/g\\\\q_4=32467.7J$

Total heat required = $[q_1+q_2+q_3+q_4]$

Total heat required = $[448.14+4583.5+21445.2+32467.7]J=58944.5J$

Hence, the total heat required is 58944.5 J

8 0

3 years ago

What is the oxidation number of chromium in dichromate ion, cr2o72-?

castortr0y [4]

Oxidation number is charge of element in compound. Can be neutral, positive or negative.
Oxygen in dichromate has oxidation number -2, becauce there are seven oxygens, net oxidation number of oxygen is -14.
2·x(oxidation number of Cr) + 7· (-2) = -2 2x= +12
x= +6, oxidation number of one chromium is +6.

8 0

3 years ago

A sample of gas consists of 4.0 moles, and exists at stp. What is the volume of the gas?

faust18 [17]

Answer:

The volume of the gas is 89.60

Explanation:

When converting from moles to volume, you would multiply by 22.4 because 1 mole equals 22.4 liters at STP.

So...

4.0 moles * 22.4 liters =

89.60 liters

6 0

2 years ago

Read 2 more answers

I’m just doing this because it told me to in order to get the answer for my homework?

andrew-mc [135]

Ok good luck on your homework then

7 0

2 years ago

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What can the presence of suffixes such as -ite,-ate,or -de in a chemical name indicate?

laila [671]

Answer: Most polyatomic ions are -ate, related ions that have 1 fewer oxygen atom are -ite (must have at least 1 oxygen tho) and nonmetals are ide.

Explanation:

3 0

1 year ago