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satela [25.4K]
3 years ago
13

Element R and Element Q have the same number of valence electrons. These elements have similar chemical behavior, but element R

has fewer energy levels than element Q
Describe what their positions on the Periodic Table have in common.
Chemistry
1 answer:
I am Lyosha [343]3 years ago
7 0

Answer:

I dont now because I am in 6 class

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Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine. CCl4(g) + (1/2)O
madam [21]

The question is incomplete, here is the complete question:

Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine.

CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c=4.4\times 10^9 at 1,000 K

Calculate Kc for the reaction 2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)

<u>Answer:</u> The value of K_c' for the final reaction is 1.936\times 10^{19}

<u>Explanation:</u>

The given chemical equations follows:

CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c

We need to calculate the equilibrium constant for the equation, which is:

2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)

As, the final reaction is the twice of the initial equation. So, the equilibrium constant for the final reaction will be the square of the initial equilibrium constant.

The value of equilibrium constant for net reaction is:

K_c'=(K_c)^2

We are given:

K_c=4.4\times 10^9

Putting values in above equation, we get:

K_c'=(4.4\times 10^9)^2=1.936\times 10^{19}

Hence, the value of K_c' for the final reaction is 1.936\times 10^{19}

3 0
3 years ago
Watch the animation and identify the correct conditions for forming a hydrogen bond. check all that apply. check all that apply.
OLga [1]
The ch4 molecule exhibits hydrogen bonding.
This statement is false. A CH4 molecule do not have a hydrogen bonding instead it has dipole dipole attraction.

Hydrogen bonding occurs when a hydrogen atom is covalently bonded to an N, O, or F atom.
This would be a true statement. A hydrogen bond is present when an atom of hydrogen shares electrons with O, N or F atom.

A hydrogen bond is equivalent to a covalent bond.
This is a false statement. A hydrogen bond is an intermolecular force of attraction while covalent bond is a intramolecular force. So, they would mean different things.

a hydrogen bond is possible with only certain hydrogen-containing compounds.
This would be true. Without the presence of an hydrogen atom definitely there would be no hydrogen bond.

a hydrogen atom acquires a partial positive charge when it is covalently bonded to an f atom.
This would be true since a HF is a polar molecule.
6 0
3 years ago
Read 2 more answers
One brand of vinegar has a pH of 4.5. Another brand has a pH of 5.0. The equation for the pH of a substance is pH = –log[H+], wh
grigory [225]
[H+] in first brand:
4.5 = -log([H+])
[H+] = 10^(-4.5)

[H+] in second brand:
5 = -log[H+]
[H+] = 10^(-5)

Difference = 10^(-4.5) - 10^(-5)
= 2.2 x 10⁻⁵

The answer is A.
7 0
3 years ago
Read 2 more answers
A gas occupies a volume of 4.50 L at 44 kPa. What would be the new volume at 50 kPa?
oksano4ka [1.4K]

Explanation:

P1= 44 kpa

P2= 50 kpa

V1= 4.50 L

V2= ?

P1 V1= P2 V2

44 × 4.50 = 50 × V2

198= 50 × V2

V2 = 198/ 50

V2= 3.96 L "the new volume"

5 0
3 years ago
CS2 (s) + 3 O2 (g) → CO2 (g) + 2 SO2 (g)
Natasha_Volkova [10]

Answer:

2.067 L ≅ 2.07 L.

Explanation:

  • The balanced equation for the mentioned reaction is:

<em>CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g),</em>

It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.

  • At STP, 3.6 L of H₂ reacts with (?? L) of oxygen gas:

It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.

<u><em>using cross multiplication:</em></u>

1.0 mol of O₂ represents → 22.4 L.

??? mol of O₂ represents → 3.1 L.

∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.

  • To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:

<u><em>Using cross multiplication:</em></u>

3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.

0.1384 mol of O₂ produce → ??? mol of SO₂.

∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.

  • Again, using cross multiplication:

1.0 mol of SO₂ represents → 22.4 L, at STP.

0.09227 mol of SO₂ represents → ??? L.

∴ The no. of liters of SO₂ will be produced = (0.09227 mol)(22.4 L)/(1.0 mol) = 2.067 L ≅ 2.07 L.

8 0
3 years ago
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