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3 years ago

Name three properties of solids that are different from those of liquids. Explain the differences for each.

1 answer:

4 0

1) Solids have a fixed shape
The particles of solids are held into their positions and are only able to vibrate about fixed points. The strength of the bonding means that the particles have their own shape, which they retain, and do not take the form of the container they are in.

2) Solids may not flow
The molecules of liquids are able to slide past one another due to weaker bonds between particles. This allows the liquid to flow; whereas, this movement is not possible in solids so they cannot flow.

3) Solids may fracture
The fact that solids possess a rigid structure means that they have the ability to fracture. Although some are able to withstand more stress than others, all solids may break. This breakage occurs when the intermolecular bonds are mechanically broken. This is not possible in the case of liquids because such rigid intermolecular bonds do not exist.

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Select all that apply.

kap26 [50]

I think it is A and C!

3 0

3 years ago

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=1.03g$

Mass of $H_2O=0.14g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago

Consider the chemical reaction in equilibrium.

Zarrin [17]

This is an application of Le Chatlier's principle: What happens when we add a reagent to one side of an equation? The reaction will shift to the other side. So heat is a reactant and we're adding more of it, the reaction must therefore, shift to the right ( or the products side).

4 0

3 years ago

Read 2 more answers

The bonding found in calcium chloride is

photoshop1234 [79]

The bonding found in calcium chloride is ionic bonds.

I hope this helps!

4 0

3 years ago

What is the chemical formula of sodium carbonate?

Gekata [30.6K]

Answer:

Na2CO3 is the formula for sodium carbornate

6 0

3 years ago