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4 years ago

When ice melts, what happens to the water molecules?

Chemistry

2 answers:

EastWind [94]4 years ago

5 0

Answer:

A) The attractions between the water molecules weaken.

Explanation:

During melting, The attractions between the water molecules weaken. The kinetic energy or "moving energy" increases as the particles are moving more. As the ice melts into liquid water, the molecules are still attracted to each other, but less than in the solid state.

Naddik [55]4 years ago

4 0

Not much.

Ice is only water, when the molecules are submitted to low temeperatures and have formed a crystalline arrangement of low density. In ice, the molecules arange themselves in hexagons with hydrogen bonds binfing each molecule.

When ice melts it turnes into water and the molecules leave the crystalline shape and come back to their normal density. In water, not all molecules are binded forming short and irregular lines.

→ In the image below, you can see the arrangement of ice and water.

Hope it helped,

BioTeacher101

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The equilibrium constant is equal to 5.00 at 1300 K for the reaction:2 SO2(g) + O2(g) ⇌ 2 SO3(g). If initial concentrations are

oee [108]

This is an incomplete question, here is a complete question.

The equilibrium constant is equal to 5.00 at 1300 K for the reaction:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

If initial concentrations are [SO₂] = 1.20 M, [O₂] = 0.45 M, and [SO₃] = 1.80 M, the system is

A) at equilibrium.

B) not at equilibrium and will remain in an unequilibrated state.

C) not at equilibrium and will shift to the left to achieve an equilibrium state.

D) not at equilibrium and will shift to the right to achieve an equilibrium state.

Answer : The correct option is, (A) at equilibrium.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

The expression for reaction quotient will be :

$Q=\frac{[SO_3]^2}{[SO_2]^2[O_2]}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$Q=\frac{(1.80)^2}{(1.20)^2\times (0.45)}=5.0$

The given equilibrium constant value is, $K_c=5.00$

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When $Q>K_c$ that means product > reactant. So, the reaction is reactant favored.

When $Q$ that means reactant > product. So, the reaction is product favored.

When $Q=K_c$ that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the $Q=K_c$ that means product = reactant. So, the reaction is in equilibrium.

Hence, the correct option is, (A) at equilibrium.

7 0

3 years ago

What would you expect to happen to the energy while a piece of paper is burning?

abruzzese [7]

D.

Energy cannot be created nor destroyed. It is just transferred through sound, heat and light.

6 0

3 years ago

Read 2 more answers

what would happen to the volume of 10L carbon dioxide, if the pressure is doubled and temperature is constant?

lara [203]

Answer:

The volume will goes to decrease.

Explanation:

The given problem will be solved through the Boyle's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

when pressure is going to increase the volume will goes to decrease and vice versa.

Consider the initial volume is 10 L at initial pressure of 15 atm when pressure is increased to 30 atm the volume will be,

Now we will put the values in formula,

P₁V₁ = P₂V₂

15 atm × 10 L = 30 atm × V₂

V₂ = 150 atm. L/ 30 atm

V₂ = 5 L

5 0

3 years ago

A sample of 2.45g aluminum oxide decomposes into 1.3g of aluminum and 1.15g of oxygen. What is the percentage composition of the

Vitek1552 [10]

Answer:

% $Al = 53.1%$ %

% $O = 46.9$ %

Explanation:

If we know the grams of a chemical compound in a specific reaction, it is possible to know the percentage of each atom that composes it.

For the Aluminum Oxide in this problem, we know its total weight and the grams of each component.

therefore we can determine the percentage ratio of its components through:

For Al

% $Al = \frac{mass of Al}{mass of Aluminium oxide} . 100$ %

% $Al = \frac{1,3 g}{2,45 g} . 100$ %

% $Al = 53.1%$ %

In the same way for oxygen

% $O = \frac{mass of O}{mass of Aluminium oxide} . 100$ %

% $O = \frac{1,5 g}{2,45 g} . 100$ %

% $O = 46.9$ %

5 0

3 years ago

You are carefully watching the temperature of your melting point apparatus as it is heating up. At 132 C it is still a white sol

Lisa [10]

Answer:

See the answer below

Explanation:

<em>Since the experiment is set out to determine the melting point of the white solid, after missing the melting point due to distraction, there are two possible solutions and both involves a repeat of the experiment.</em>

1. The first one is to allow the molten substance to solidify again and then repeat the experiment. This time around, a critical attention should be paid to be able to notice the melting point temperature once the temperature gets to 132 C.

2. The second solution would be discard the molten substance and repeat the experiment with the a new solid one. Similarly, critical attention should be paid once the temperature gets to 132 C since it is sure that the melting point lies within 132 and 138 C.

6 0

4 years ago