Witch question could be tested in a scientific manner?

Explain how particles move before during and after melting

Answer:

Particles would move more freely, while still staying close together depending on the shape of the liquid

Explanation:

Melting is the process of going from a solid to a liquid due to the increase in heat/energy. This increase in heat/energy increases the speed at which the atoms within the object moves. Lets say we had an ice cube. While it is a cube, the particles inside the cube are slow and compact, staying close together.

When enough energy is gained, this makes the particles begin to move faster, gaining heat and energy which results in the ice cube melting and moving more freely than normal.

3 0

2 years ago

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What volume would a 0. 250 mole sample of h2 gas occupy if it had a which a a pressure of 1. 70 atm and a temperature of 35 C?

frosja888 [35]

The volume of 0. 250 mole sample of $H_{2}$ gas occupy if it had a pressure of 1. 70 atm and a temperature of 35 °C is 3.71 L.

Calculation,

According to ideal gas equation which is known as ideal gas law,

PV =n RT

P is the pressure of the hydrogen gas = 1.7 atm
Vis the volume of the hydrogen gas = ?
n is the number of the hydrogen gas = 0.25 mole
R is the universal gas constant = 0.082 atm L/mole K
T is the temperature of the sample = 35°C = 35 + 273 = 308 K

By putting all the values of the given data like pressure temperature universal gas constant and number of moles in equation (i) we get ,

1.7 atm×V = 0.25 mole ×0.082 × 208 K

V = 0.25 mole ×0.082atm L /mole K × 308 K /1.7 atm

V = 3.71 L

So, volume of the sample of the hydrogen gas occupy is 3.71 L.

learn more about ideal gas equation

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4 0

2 years ago

What is the molality of a solution of water and KCl if the freezing point of the solution is –3°C? (Kf = 1.86°C/m; molar mass of

Elina [12.6K]

Some of the solutions exhibit colligative properties. These properties depend on the amount of solute dissolved in a solvent. These properties include freezing point depression, boiling point elevation, osmotic pressure and vapor pressure lowering. Calculations are as follows:

<span> ΔT(freezing point) = (Kf)mi
3 = 1.86 °C kg / mol (m)(2)
3 =3.72m
m = 0.81 mol/kg</span>

4 0

4 years ago

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A molecule of acetone and a molecule of propyl aldehyde are both made from 3 carbon atoms, 6 hydrogen atoms, and 1 oxygen atom.

TEA [102]

Explanation:

The structures of both acetone and propanal are shown below:

In the formula of propanal there is -CHO functional group at the end.

In acetone -CO- group is present in the middle that is on the second carbon.

The molecular formula is C3H6O.

Both have same molecular formula but different structural formulas.

7 0

3 years ago

Given the value of the equilibrium constant (Kc) for the equation (a), calculate the equilibrium constant for equation (b)

matrenka [14]

Answer: The value of equilibrium constant for new reaction is $1.92\times 10^{-25}$

Explanation:

The given chemical equation follows:

$O_2(g)\rightarrow \frac{2}{3}O_3(g)+\frac{1}{2}O_2(g)$

The equilibrium constant for the above equation is $5.77\times 10^{-9}$

We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:

$3O_2(g)\rightarrow 2O_3(g)$

The equilibrium constant for this reaction will be the cube of the initial reaction.

If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{eq}'=(5.77\times 10^{-9})^3=1.92\times 10^{-25}$

Hence, the value of equilibrium constant for new reaction is $1.92\times 10^{-25}$

5 0

3 years ago