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3 years ago

What is necessary to produce visible light from chemical compounds?

Chemistry

1 answer:

erastova [34]3 years ago

5 0

Answer:

the molecule must contain either pi bonds or atoms with non-bonding orbitals.

Explanation:

A non-bonding orbital is a lone pair on, say, oxygen, nitrogen or a halogen.

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How much water must be added to make a 1.0 M solution from 250. mL of a 2.75 M solution?

Stels [109]

700 mL ....................

4 0

3 years ago

Question 4 (1 point)

Rainbow [258]

а о

Explanation:

The given cation:

(Rf₂Al₂F₃)³⁺

The oxidation number gives the extent to which a specie is oxidized in a reaction.

This number is assigned based on some rules:

Elements in combined state whose atoms combines with themselves have an oxidation number of zero.
The charge carried on simple ions gives their oxidation number.
Algebraic sum of all the oxidation numbers of atoms in neutral compound is zero. In an ion with more than one kind of atom, the charge on it is the oxidation number.

for the specie given;

Known:

oxidation number of Al = +3

F = -1

charge = +3

let the oxidation number of Rf = k

2k + 2(3) + 3(-1) = +3

2k + 6 - 3 = 3

2k = 0

k = 0

The oxidation state of rutherfodium is 0

learn more:

Oxidation state brainly.com/question/10017129

#learnwithBrainly

8 0

3 years ago

In a solution of pure water, the dissociation of water can be expressed by the following: H2O(l) + H2O(l) ⇌ H3O+(aq) + OH−(aq) T

kakasveta [241]

Answer:

[H₃O⁺] = 2.90 × 10⁻¹⁰ M

Explanation:

1) Ionization equilibrium equation: given

H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)

2) Ionization equilibrium constant, at 25°C, Kw: given

Kw = 1.0 × 10⁻¹⁴

3) Stoichiometric mole ratio:

As from the ionization equilibrium equation, as from the fact it is stated, the concentration of both ions, at 25°C, are equal:

[H₃O⁺(aq)] = [OH⁻(aq)] = 1.0 × 10⁻⁷ M

⇒ Kw = [H3O⁺] [OH⁻] = 1.0 × 10⁻⁷ × 1.0 × 10⁻⁷ = 1.0 × 10⁻¹⁴ M

4) A solution has a [OH⁻] = 3.4 × 10⁻⁵ M at 25 °C and you need to calculate what the [H₃O⁺(aq)] is.

Since the temperature is 25°, yet the value of Kw is the same, andy you can use these conditions:

Kw = 1.0 × 10⁻¹⁴ M², and

Kw = [H3O⁺] [OH⁻]

Then you can substitute the known values and solve for the unknown:

1.0 × 10⁻¹⁴ M² = [H₃O⁺] × 3.4 × 10⁻⁵ M

⇒ [H₃O⁺] = 1.0 × 10⁻¹⁴ M² / ( 3.4 × 10⁻⁵ M ) = 2.9⁻¹⁰ M

As you see, the increase in the molar concentration of the ion [OH⁻] has caused the decrease in the molar concentration of the ion [H₃O⁺], to keep the equilibrium law valid.

6 0

3 years ago

Read 2 more answers

A 73.6 g sample of aluminum is heated to 95.0°C and dropped into 100.0 g of water at 20.0°C. If the resulting temperature of the

Softa [21]

Answer:

The specific heat of aluminium is 0.875 J/g°C

Explanation:

Step 1: Data given

The mass of the aluminium sample = 73.6 grams

Initial temperature of the sample = 95.0 °C

Mass of water = 100.0 grams

Initial temperature of water = 20.0 °C

Final temperature of water and aluminium = 30.0 °C

The specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of aluminium

Q gained = Q lost

Qwater = -Qaluminium

Q = m*c*ΔT

m(aluminium) * c(aluminium) * ΔT(aluminium) = - m(water) * c(water) * ΔT(aluminium)

⇒ mass of aluminium = 73.6 grams

⇒ c(aluminium) = TO BE DETERMINED

⇒ ΔT(aluminium) = The change of temperature = T2 - T1 = 30 .0 °C - 95.0 °C = -65.0°C

⇒ mass of water = 100.0 grams

⇒ c(water ) = The specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature of water = T2 - T1 = 30.0 - 20.0 = 10.0 °C

73.6g * c(aluminium) * -65.0 °C = 100.0g * 4.184 J/g°C * 10.0°C

-4784 * c(aluminium) = -4184

c(aluminium) = 0.875 J /g°C

The specific heat of aluminium is 0.875 J/g°C

7 0

3 years ago

Due to the fact that ionic compounds have strong intermolecular forces they are __________ at room temperature.

MrMuchimi

The correct answer should be solids

3 0

3 years ago