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Bumek [7]
3 years ago
5

Phosgene (carbonyl chloride), COCl2, is an extremely toxic gas that is used in manufacturing certain dyes and plastics. Phosgene

can be produced by reacting carbon monoxide and chlorine gas at high temperatures: CO(g) Cl2(g)⇌COCl2(g) Carbon monoxide and chlorine gas are allowed to react in a sealed vessel at 477 ∘C . At equilibrium, the concentrations were measured and the following results obtained: Gas Partial Pressure (atm) CO 0.830 Cl2 1.30 COCl2 0.220 What is the equilibrium constant, Kp, of this reaction
Chemistry
1 answer:
evablogger [386]3 years ago
8 0

Answer:

0.204.

Explanation:

  • For the following reaction:

<em>CO(g) + Cl₂(g) ⇌ COCl₂(g), </em>

Kp = (P of COCl₂)/(P of CO)(P of Cl₂)

P of COCl₂ = 0.22 atm, P of CO = 0.83 atm, P of Cl₂ = 1.3 atm.

∴ Kp = (P of COCl₂)/(P of CO)(P of Cl₂) = (0.22 atm)/(0.83 atm)(1.3 atm) = 0.204.

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Define saturated and unsaturated fats​
Sphinxa [80]

Answer:

<h3><u>Saturated and unsaturated</u>:-</h3>

Are a form of fat in which all or most of the fatty acid chains are single bonds. Glycerol and fatty acids are the two types of smaller molecules that make up fat.

<h3><u>Saturated fat is found in:</u></h3>
  • cakes
  • sausages
  • cheese
  • butter
<h3><u>Examples of unsaturated fats:- </u></h3>
  • Olive
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hope it helps...

6 0
3 years ago
When do electrons release photons​
Inessa05 [86]

Answer:

Later when they are older

Explanation:

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3 0
2 years ago
What is the average atomic mass of the element?
morpeh [17]

The average atomic mass of element X is 14.007 u.

The average atomic mass of X is the <em>weighted average</em> of the atomic masses of its isotopes.

We multiply the atomic mass of each isotope by a number representing its <em>relative importance</em> (i.e., its % abundance).

Thus,  

0.996 36 × 14.003 u =13.952 03 u

0.003 64 × 15.000 u = <u>0.054 60 u</u>

__________TOTAL = 14.007     u

7 0
3 years ago
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
72.40% iron,27.60% Oxygen Empirical formula
VARVARA [1.3K]

Explanation:

Fe. O

72.40/ 56 27.60/16

____________________

1.29/1.29 1.725/1.29

_____________________

1. :. 1

<h3>Emperical formula = FeO</h3>
6 0
3 years ago
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