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2 years ago

Of elements N, O, Cl, Na, and Which two would likely have similar chemical properties and why

1 answer:

5 0

Oxygen and Nitrogen would be the most similar of the elements listed, because they are the closest in the periodic table. This isn't a very good reason for anything, but the two do have some similar properties. They are both non-metals, they are both highly electronegative, they are both diatmoic gasses in their natural states, they have a similar number of valence electrons, they are both generally oxidizing agents. Oxygen and Chlorine are also quite similar, but not quite as similar as Oxygen and Nitrogen.

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BRAINLIEST AND 20PTS!!!!!!

Elden [556K]

1) Answer is: 5.
Chemical dissociation of aluminium sulfide in water:
Al₂S₃(aq) → 2Al³⁺(aq) + 3S²⁻(aq).
There are five ions, two aluminium cations and three sulfide anions.

2) Answer is: 4.
Chemical dissociation of aluminium fluoride in water:
AlF₃(aq) → Al³⁺(aq) + 3F⁻(aq).
There are four ions, one aluminium cation and three fluoride anions.
Aluminium has oxidation +3, because it lost three electrons, to gain electron configuration as noble gas neon.

7 0

1 year ago

Draw the four structures of the compounds with molecular formula C5H10O that contain a carbon-carbon double bond, an unbranched

kondaur [170]

Explanation:

Molecular formula of the compound =

The carbon chain in the molecule will look like :

1. $H_2C=CH-CH_2-CH_2-CH_2-OH$

2. $H_3C-CH=CH-CH_2-CH_2-OH$

3. $H_3C-CH-CH=CH-CH_2-OH$

4. $H_3C-CH_2-CH_2-CH=CH-OH$

No branching is present, so that means the valency of the carbon will be fulfilled by 10 hydrogen atoms and 1 oxygen atom

3 0

2 years ago

At a fixed volume, a four-fold increase in the temperature of a gas will lead to _______ in pressure.

aliina [53]

Answer:

Increase in pressure

Explanation:

The rise in temperature will increase the pressure of the gas.

6 0

2 years ago

Read 2 more answers

Calculate the pH of each of following buffered solutions.?a. 0.10 M acetic acid/0.25 M sodium acetate b. 0.25 M acetic acid/0.10

Masteriza [31]

Answer:

a. 5.10.

b. 4.35.

c. 5.10.

d. 4.35.

Explanation:

a. 0.10 M acetic acid/0.25 M sodium acetate

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.25)/(0.10)

∴ pH = 4.744 + 0.34 = 5.084 ≅ 5.10.

b. 0.25 M acetic acid/0.10 M sodium acetate

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.10)/(0.25)

∴ pH = 4.744 - 0.34 = 4.346 ≅ 4.35.

c. 0.080 M acetic acid/0.20 M sodium acetate

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.20)/(0.08)

∴ pH = 4.744 + 0.34 = 5.084 ≅ 5.10.

d. 0.20 M acetic acid/0.080 M sodium acetate

For acidic buffer:

∵ pH = pKa + log [salt]/[Acid]

∴ pH = - log(Ka) + log [salt]/[Acid]

Ka for acetic acid = 1.8 x 10⁻⁵.

∴ pH = - log(1.8 x 10⁻⁵) + log(0.08)/(0.20)

∴ pH = 4.744 - 0.34 = 4.346 ≅ 4.35.

6 0

2 years ago

Rubbing alcohol contains 615g of isopropanol (C3H7OH) per liter (aqueous solution). Calculate the molality of this solution. Giv

faltersainse [42]

Answer:

Solution of isopropanol is 10.25 molal

Explanation:

615 g of isopropanol (C3H7OH) per liter

We gave the information that 615 g of solute (isopropanol) are contained in 1L of water. We need to find out the mass of solvent, so we use density.

Density of water 1g/mL → Density = Mass of water / 1000 mL of water

Notice we converted the L to mL

Mass of water = 1000 g (which is the same to say 1kg)

Molality are the moles of solute in 1kg of solvent, so let's convert the moles of isopropanol → 615 g . 1mol / 60g = 10.25 moles

Molality (mol/kg) = 10.25 moles / 1kg = 10.25 m

4 0

2 years ago