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3 years ago

5

Why do ionic borids form between metal and no metals

1 answer:

mamaluj [8]3 years ago

4 0

Nonmetals don’t have enough electronegativity to transfer electrons so the electrons are shared between them

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Based on the Lewis/electron dot representation

zaharov [31]

B

Explanation:

I aint sure tho i am sorry

3 0

2 years ago

Mi One should know the properties of the components of the mixture to separate it. Explain with an example.

Alekssandra [29.7K]

One should know that the properties of the components of the mixture.
Suppose a mixture contains Sodium or potassium .If you put them in open air it may catch fire.
Suppose a mixture has flammable components like kerosene,spirit .If you put them in fire it may harm you .

4 0

3 years ago

What is the chemical formula of the salt produced by the neutralization of potassium hydroxide with sulfuric acid?KSO3KSO4K(SO4)

kakasveta [241]

Answer: The salt produced will be $K_{2}SO_{4}$

Explanation:

During a neutralization reaction, an acid reacts with a base for producing the correspondent salt, and water.

The strong acids release all the protons avalaible when are dissolved, such as sulfuric acid. As you can see, sulfuric acid have 2 protons ready for being released ( $H_{2}SO_{4}$ ); and those places have to be occcupied for other ions equivalents to the H+: K+ from KOH in this case.

Therefore the answer will be $K_{2}SO_{4}$ .

6 0

3 years ago

Given that Δ H ∘ f [ Br ( g ) ] = 111.9 kJ ⋅ mol − 1 Δ H ∘ f [ C ( g ) ] = 716.7 kJ ⋅ mol − 1 Δ H ∘ f [ CBr 4 ( g ) ] = 29.4 kJ

JulsSmile [24]

Answer:

283.725 kJ ⋅ mol − 1

Explanation:

C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1

$\frac{1}{2}$ Br2(g) ⇒ Br(g) , Δ H ∘ = 111.9 kJ ⋅ mol − 1

C(s) ⇒ C(g) , Δ H ∘ = 716.7 kJ ⋅ mol − 1

4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1

eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1

so,

average bond enthalpy is $\frac{1134.9}{4}$ = 283.725 kJ ⋅ mol − 1

4 0

3 years ago

Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

Explanation:

According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

⇒ $-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}$ -----------------------------(1)

Given that the rate of disappearance of oxygen = $-\frac{d[O_{2} ]}{dt}$ = 3.64 x 10⁻³ M/s

So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = ?

from equation (1) we can write

$\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]$

⇒ $\frac{d[SO_{3}] }{dt}$ = 2 x 3.64 x 10⁻³ M/s

⇒ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

7 0

3 years ago