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3 years ago

SnCL2 is solid while SnCl4 is liquid ...?

1 answer:

8 0

<span>As the oxidation state of a metal increases, its electronegativity increases and the bonds become increasingly covalent. SnCl2 forms a network solid where the atoms are bonded to one another throughout a network of polar covalent bonds. SnCl4 is molecular, and held together by weaker intermolecular forces called London dispersion forces. </span>

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The number of electrons in the valence shell determine what types of bonds an atom can form.

Nataly_w [17]

This statement is right !

8 0

3 years ago

Since a solution of KBr would probably cause significant damage to the car radiator and engine, you decide to use 125 g of the n

Vilka [71]

Answer:

When octane is used, the solution will have less effect on the freezing point depression of the solution

Explanation:

The complete question is:

Calculate the freezing point of a solution of 125 g KBr in 450 g water.

Since a solution of KBr would probably cause significant damage to the car radiator and engine, you decide to use 125 g of the nonelectrolyte octane (molar mass 114 g/mole). Will this have a greater or less effect on freezing point depression of the solution?

Step 1: Data given

Molar mass of KBr = 119.0 g/mol

Molar mass of octane = 114 g/mol

Mass of KBr = 125 grams

Mass of octane = 125 grams

Mass of water = 450 grams

Step 2: Calculate moles KBr

Moles KBr = mass KBr / molar mass KBr

Moles KBr = 125 grams / 119.0 g/mol

Moles KBr = 1.05 moles

Step 3: Calculate moles octane

Moles octane = 125 grams / 114 g/mol

Moles octane = 1.10 moles

Step 4: Calculate molality

Molality = moles compound / mass water

Molality KBr = 1.05 moles / 0.450 kg

Molality KBr = 2.33 molal

Molality octane = 1.10 moles / 0.450 kg

Molality octane = 2.44 molal

Step 5: Calculate the freezing point depression when KBr is used

ΔT = i*Kf * m

⇒with ΔT = the freezing point depression = TO BE DETERMINED

⇒with i = the van't Hoff factor of KBr = 2

⇒with Kf = the freezing point depression constant of water = 1.86 °C/m

⇒with m= the molality = 2.33 molal

ΔT = 2*1.86 * 2.33

ΔT = 8.68 °C

This means the freezing point of this solution is -8.68 °C

Step 6: Calculate the freezing point depression when octane is used

ΔT = i*Kf * m

⇒with ΔT = the freezing point depression = TO BE DETERMINED

⇒with i = the van't Hoff factor of the nonelectrolyte octane = 1

⇒with Kf = the freezing point depression constant of water = 1.86 °C/m

⇒with m= the molality = 2.44 molal

ΔT = 1* 1.86 * 2.44

ΔT = 4.54 °C

This means the freezing point of this solutions is -4.54 °C

When octane is used, the solution will have less effect on the freezing point depression of the solution

7 0

3 years ago

Please help me on this.

Anestetic [448]

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3 0

3 years ago

In a data set, the number that appears the most is called the ( ).

scZoUnD [109]

Answer:

Mode

Explanation:

The mode is the number that appears most frequently in a data set. A set of numbers may have one mode, more than one mode, or no mode at all. Other popular measures of central tendency include the mean, or the average (mean) of a set, and the median, the middle value in a set.

4 0

4 years ago

The nitrogen atoms in a molecule of N2 share a total of

astraxan [27]

Answer: Option (3) is the correct answer.

Explanation:

Atomic number of nitrogen is 7 and its electronic configuration is $1s^{2}2s^{2}2p^{3}$ . Thus, we can see that valency of nitrogen is 5.

So, one nitrogen atom has one lone pair of electron.

Therefore, when two nitrogen atoms will bond together then there will be three pairs of electrons bonded together and one lone pair of electron on each nitrogen atom.

6 0

3 years ago

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