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3 years ago

What is the relationship between kinetic energy and velocity ?

1 answer:

5 0

Kinetic energy is directly proportional to the mass of the object and to the square of its velocity: K.E. = 1/2 m v2. If the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

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Which of the following is a nonrenewable energy source? A. coal B. solar C. wind D. hydropower E. burning biomass

VARVARA [1.3K]

The best answer among the following choices would be A) since the other options are NOT nonrenewable energy sources.

5 0

3 years ago

How is solid different from gas ?

Orlov [11]

Answer:

gas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.

Explanation:

5 0

3 years ago

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The number of moles of solute in 200.ml of a 0.500 m solution is

IrinaK [193]

Answer:

I believe the answer is 0.100.

Explanation:

Hope my answer has helped you!

8 0

3 years ago

The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be

Mrac [35]

Answer: The freezing point of solution is 5.35°C

Explanation:

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 4.90°C/m

$m_{solute}$ = Given mass of solute (naphthalene) = 2.60 g

$M_{solute}$ = Molar mass of solute (naphthalene) = 128.2 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC$

Hence, the freezing point of solution is 5.35°C

3 0

3 years ago

Nicotine is 74.1% carbon, 8.6% hydrogen, and 17.3% nitrogen by mass. What is its molecular formula if its molar mass is 162.26 g

WINSTONCH [101]

first we need to find the empirical formula of nicotine

empirical formula is the simplest ratio of whole numbers of elements making up a compound

the percentage compositions for each element has been given. So we can calculate for 100 g of the compound.

masses of elements and the number of moles

C - 74.1 g - 74.1 g/12 g/mol = 6.17 mol

H - 8.6 g - 8.6 g / 1 g/mol = 8.6 mol

N - 17.3 g - 17.3 g / 14 g/mol = 1.23

divide all by the least number of moles

C - 6.17 / 1.23 = 5.01

H - 8.6 / 1.23 = 6.99

N - 1.23 / 1.23 = 1.00

when the atoms are rounded off to the nearest whole numbers

C - 5

H - 7

N - 1

empirical formula is C₅H₇N

we have to find what the mass of 1 empirical unit is

mass - 5 x 12 g/mol + 7 x 1 g/mol + 14 g/mol = 81 g

molecular mass is 162.26 g/mol

we have to find how many empirical units make up 1 molecule

number of empirical units = molecular mass / mass of 1 empirical unit

= 162.26 g/mol / 81 g = 2.00

there are 2 empirical units

molecular formula is - 2 (C₅H₇N)

molecular formula - C₁₀H₁₄N₂

4 0

3 years ago

Read 2 more answers