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3 years ago

What are the two main types of energy?

Chemistry

2 answers:

Sergio [31]3 years ago

7 0

kinetic and potential

Harlamova29_29 [7]3 years ago

3 0

Kinetic and potential energy, are the two main types of energy! Hope this helps:)

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Place whole number coefficients in the blanks to balance the chemical reaction.

mote1985 [20]

It is already balanced

4 0

3 years ago

Read 2 more answers

An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for

8090 [49]

Answer: The freezing point and boiling point of the solution are $-6.6^0C$ and $101.8^0C$ respectively.

Explanation:

Depression in freezing point:

$T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of water = $0^0C$

$k_f$ = freezing point constant of water = $1.86^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_f=-6.6^0C$

Therefore,the freezing point of the solution is $-6.6^0C$

Elevation in boiling point :

$T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^0C$

$k_b$ = boiling point constant of water = $0.52^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_b=101.8^0C$

Thus the boiling point of the solution is $101.8^0C$

4 0

3 years ago

Consider a gas in a container that can adjust its volume to maintain constant pressure. Suppose the gas is cooled. What happens

NNADVOKAT [17]

Answer:

The volume will also decrease.

Explanation:

This illustration clearly indicates Boyle's law.

Boyle's law states that the volume of a fixed mass of gas is directly proportional to the absolute temperature, provided the pressure remains constant. Mathematically, it is represented as:

V & T

V = KT

K = V/T

V1/T1 = V2/T2 =... = Vn/Tn

Where:

T1 and T2 are the initial and final temperature respectively, measured in Kelvin.

V1 and V2 are the initial and final volume of the gas respectively.

From the illustration above, the volume is directly proportional to the temperature. This implies that as the temperature increases, the volume will also increase and as the temperature decreases, the volume also will decrease.

4 0

2 years ago

Pleasee help <br> will name brainliest

zmey [24]

Answer:

0.125. work- divide the volume value by 1000

873. work- multiply the length value by 100

98100. work- Conversion factor: 1 kg = 100000 cg

1) Centigram = Kilogram * 100000

2) Centigram = 0.981 * 100000

3) Centigram = 98100

285.65. work- 12.5°C + 273.15 = 285.65K

446.85. work- 720K − 273.15 = 446.85°C

346.25. work- 73.1°C + 273.15 = 346.25K

Explanation:

i hope this helps:) brainliest plss??

8 0

2 years ago

Can anyone here help me with chemistry

Kay [80]

The answer is 20g N2

5 0

2 years ago