What is correct name of the compound whose formula is bf3?

Why is a 13 placed in front of the oxygen on the perodic table

xxMikexx [17]

Answer:

Oxygen comes 8th on the periodic table which is the atomic number.

Explanation:

but Oxygen 13 (Isotopes of Oxygen) is when oxygen has 8 protons and electrons, and 5 neutrons (8+5=13)

5 0

3 years ago

Several constructive processes helped build mountains on Earth. All BUT one of these is a way mountains can be built. That is A)

Naddik [55]

It is landslides i got it right on the test

7 0

3 years ago

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r

weqwewe [10]

Answer:

Therefore $\bigtriangledown H^\circ_{rxn}$ = -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by $\bigtriangledown H^\circ_{rxn}$

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s) $\bigtriangledown H^\circ_{rxn}$ =?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1) $\bigtriangledown H^\circ_{rxn}$ = +157KJ

P₄(g)+6Cl₂(g)→ 4PCl₃(g).............(2) $\bigtriangledown H^\circ_{rxn}$ = -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3) $\bigtriangledown H^\circ_{rxn}$ = -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4) $\bigtriangledown H^\circ_{rxn}$ =4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s) $\bigtriangledown H^\circ_{rxn}$ =( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s) $\bigtriangledown H^\circ_{rxn}$ = - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s) $\bigtriangledown H^\circ_{rxn}$ = -1835 KJ

Therefore $\bigtriangledown H^\circ_{rxn}$ = -1835 KJ

5 0

3 years ago

Which of these electron transitions correspond to absorption of energy and which to emission?

Keith_Richards [23]

The electron transitions representing absorption of energy are:

(a) n = 2 to n = 4

(d) n = 3 to n = 4

and those which represents emissions of energy are:

(b) n = 3 to n = 1

(c) n = 5 to n = 2

An electron emits energy when it jumps from a higher energy level to a lower energy level transitions.

An electron absorbs energy to jump from a lower energy level to a higher energy level transitions.

(a) n = 2 to n = 4

Here, the initial level (2) is lower than the final level (4).

Hence energy is absorbed.

(b) n = 3 to n = 1

Here, the initial level (3) is higher than the final level (1).

Hence energy is released.

(c) n = 5 to n = 2

Here, the initial level (5) is higher than the final level (2).

Hence energy is released.

(d) n = 3 to n = 4

Here, the initial level (3) is lower than the final level (4).

Hence energy is absorbed.

The question is incomplete. Find the complete question here:

Which of these electron transitions correspond to absorption of energy and which to emission?

(a) n = 2 to n = 4

(b) n = 3 to n = 1

(c) n = 5 to n = 2

(d) n = 3 to n = 4

Learn more about electron transitions at brainly.com/question/2079874

#SPJ4

4 0

2 years ago

What must be true of the net charge for any sample of sodium chloride

Ede4ka [16]

Answer:

Zero (0)

Explanation:

Since Na^+ has a charge of +1 and Cl^- has a charge of —1 in NaCl

Then the net = +1 —1 = 0

4 0

3 years ago

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