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vampirchik [111]
3 years ago
8

What would be the advantage of using a molecular model.

Chemistry
1 answer:
ddd [48]3 years ago
7 0

Answer: Can pick it up and hold it, can rotate groups about bonds

and can look for symmetry

Explanation: A molecular model is a plastic model with interchangeable balls. Represents molecules and their processes.

The creation of mathematical models of molecular properties and behavior is molecular modelling, and their graphical depiction is molecular graphics, but these topics are closely linked and each uses techniques from the others.

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A sample of propane (C3H8) is placed in a closed vessel together with an amount of O2 that is 3 times the amount needed to compl
Aneli [31]

Answer:

final mole fraction of O₂ = 58.84% , CO₂ = 17.64% , H₂O = 23.52% .

final partial pressure of O₂ = 2.942 atm , CO₂ = 0.882 atm , H₂O = 1.176 atm .

Explanation:

Assuming that propane is present as a gas , and also that the combustion is complete:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

then taking as a reference propane sample= 1 mol , then

initial moles of O₂ = 3* 5 moles = 15 moles

final moles of O₂= 3* 5 moles - 5 moles = 10 moles

final moles of CO₂ = 3 moles

final moles of H₂O = 4 moles

total number of moles = 10 + 3 + 4 = 17 moles

final mole fraction of O₂ = 10/17 = 0.5884 = 58.84%

final mole fraction of  CO₂ = 3/17 = 0.1764 = 17.64%

final mole fraction of  H₂O = 4/17 = 0.2352 = 23.52%

From Dalton's law for ideal gases , the partial pressure p=P*x then

final partial pressure of O₂ = 5 atm * 10/17 = 2.942 atm

final partial pressure of CO₂ = 5 atm * 3/17 = 0.882 atm

final partial pressure of H₂O = 5 atm * 1/17 = 1.176 atm

4 0
3 years ago
Read 2 more answers
An ethylene gas torch requires 3200L of gas at 3.0 atm. What will be the pressure of the gas if the Ethylene is supplied by a 25
drek231 [11]

Answer:

38.4 atm.

Explanation:

  • We can use the general law of ideal gas:<em> PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

  • If n and T are constant, and have different values of P and V:

<em>(P₁V₁) = (P₂V₂).</em>

P₁ = 3.0 atm, V₁ = 3200 L,

P₂ = ??? atm, ​V₂ = 250.0 L.

∴ P₂ = (P₁V₁)/(V₂) = (3.0 atm)(3200 L)/(250.0 L) = 38.4 atm.

Read more on Brainly.com - brainly.com/question/12731388#readmore

4 0
3 years ago
Light used to produce food in plants is chemical or physical change?
Nitella [24]
This is Chemical change
4 0
3 years ago
What is the compound of HgS
qaws [65]

Answer:

Mercury sulfide or mercuric sulfide,

Explanation:

5 0
4 years ago
Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25∘C diamond changes to graphite so slowly that the entha
rjkz [21]

Answer:

-1.9 KJ/mol

Explanation:

In order to solve the problem, we have to rearrange the equations in a way in which all molecules of O₂ and CO₂ are eliminated:

2C(diamond) + 2O₂(g) → 2CO₂(g)     ΔH₁= 2 x (-395.4 KJ) ------> we multiply by 2 both reactants and products

2 CO₂(g) → 2CO(g) + O₂(g)         ΔH₂= 566.0 KJ

CO₂(g) → C(graphite) + O₂(g)     ΔH₃= -1 x (-393.5 KJ) ------> we use reverse rxn

2CO(g) → C(graphite) + CO₂(g)   ΔH₄= -172.5 KJ

When we cancel the molecules that appear both in reactants and products, the total reaction is the following:

2C(diamond) → 2C(graphite)

ΔHt= ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄ = 2 x (-395.4 KJ) + 566.0 KJ + (-1 x (-393.5 KJ)) - 172.5 KJ

ΔHt= 347.2 KJ

This is for 2 mol of C(diamond) which are converted in 2 mol of C(graphite). To obtain ΔH for the reaction of 1 mol C(diamond) to 1 mol (graphite) we have to divide into 2:

ΔH= -3.8 KJ/2mol= -1.9 KJ/mol

5 0
3 years ago
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