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2 years ago

What is the difference between an acid and a base? Give an example of each.

Chemistry

1 answer:

xz_007 [3.2K]2 years ago

3 0

Acid is sour and base is bitter

Acid example= lemon

Base example= bitter melon

Hope this helps

You might be interested in

Which of the following is the best explanation for a covalent bond?

Scrat [10]

Answer:

A. electrons simultaneously attracted by more than one nucleus

Explanation:

Covalent bond is the bond which is formed with the sharing of the electrons between the two atoms which are taking part in the bond. It is generally formed between the atoms with similar electronegativity values.
It is the bond which is generally occurs within non metals as they share electrons to complete their octet.
The difference in the electronegativity values of the atoms involving in a covalent bond must not exceed the value of 1.7 .

Thus, the electrons are attracted by the two different nucleus of the atoms that are taking part in the bonding.

So, the correct answer is:- A. electrons simultaneously attracted by more than one nucleus

6 0

3 years ago

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

Romashka [77]

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

7 0

3 years ago

The energy that moving electric charges carry is called

katovenus [111]

The answer is electrical energy. This is the energy freshly resulting from electric potential energy or kinetic energy. When lightly used to label energy engaged or perhaps brought by an electrical circuit (for instance, one providing by an electric power usefulness) "electrical energy" talks about energy which has been transformed from electric potential energy.

8 0

3 years ago

Describe how modeling is useful in studying the structure of the atom.

pishuonlain [190]

Answer:

Scientists use models to study atoms because atoms are extremely small and can't even be seen. If atoms can't be seen, it's easier to study and.

Explanation:

hope it helps you

3 0

2 years ago

For a 0.50 molal solution of sucrose in water, calculate the freezing point and the boiling point of the solution.

jok3333 [9.3K]

Answer:

-0.93 °C; 100.26 °C

Step-by-step explanation:

(a) Freezing point depression

The formula for the freezing point depression ΔT_f is

ΔT_f = iKf·b

i is the van’t Hoff factor: the number of moles of particles you get from a solute.

For sucrose,

Sucrose (s) ⟶ sucrose (aq)

1 mole sucrose ⟶ 1 mol particles i = 1

ΔT_f = 1 × 1.86 × 0.50

ΔT_f = 0.93 °C

T_f = T_f° - ΔT_f

T_f = 0.00 – 0.93

T_f = -0.93 °C

(b) Boiling point elevation

The formula for the boiling point elevation ΔTb is

ΔTb = iKb·b

ΔTb = 1 × 0.512 × 0.50

ΔTb = 0.256 °C

Tb = Tb° + ΔTb

Tb = 100.00 + 0.256

Tb = 100.26 °C

6 0

3 years ago