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kodGreya [7K]
2 years ago
6

Identify the MOs that react to form cyclohexene. HOMO of 1,3-butadiene and LUMO of ethylene LUMO of 1,3-butadiene and LUMO of et

hylene HOMO of 1,3-butadiene and HOMO of ethylene LUMO of 1,3-butadiene and LUMO of 1,3-butadiene HOMO of 1,3-butadiene and HOMO of 1,3-butadiene

Chemistry
1 answer:
BARSIC [14]2 years ago
6 0

Answer:

HOMO of 1,3-butadiene and LUMO of ethylene

HOMO of ethylene LUMO of 1,3-butadiene

Explanation:

1,3 - butadiene underogoes cycloaddition reaction with ethylene to give cyclohexene.

According to Frontier molecular orbital theory HOMO of 1,3 butadiene and LUMO of ethylene and HOMO of ethylene and LUMO of ethylene underoges (4 + 2) in thermal or photochemical condition.

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An unopened, cold 2.00 L bottle of soda contains 46.0 mL of gas confined at a pressure of 1.30 atm at a temperature of 5.0oC. If
Katena32 [7]

Answer:

20.8mL

Explanation:

7 0
2 years ago
If an equal number of moles of reactants are used, do the following equilibrium mixtures contain primarily reactants or products
puteri [66]

<u>Answer:</u>

<u>For a:</u> The equilibrium mixture contains primarily reactants.

<u>For b:</u> The equilibrium mixture contains primarily products.

<u>Explanation:</u>

There are 3 conditions:

  • When K_{eq}>1; the reaction is product favored.
  • When K_{eq}; the reaction is reactant favored.
  • When K_{e}=1; the reaction is in equilibrium.

For the given chemical reactions:

  • <u>For a:</u>

The chemical equation follows:

HCN(aq.)+H_2O(l)\rightleftharpoons CN^-(aq.)+H_3O^+(aq.);K_{eq}=6.2\times 10^{-10}

The expression of K_{eq} for above reaction follows:

K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}

As, K_{eq}, the reaction will be favored on the reactant side.

Hence, the equilibrium mixture contains primarily reactants.

  • <u>For b:</u>

The chemical equation follows:

H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g);K_{eq}=2.51\times 10^{4}

The expression of K_{eq} for above reaction follows:

K_{eq}=\frac{[HCl]^2}{[H_2][Cl_2]}=2.51\times 10^{4}

As, K_{eq}>1, the reaction will be favored on the product side.

Hence, the equilibrium mixture contains primarily products.

4 0
2 years ago
What pressure (in atm) would be exerted by 76 g of fluorine gas (f2) in a 1.50 liter vessel at -37oc? (a) 26 atm(b) 4.1 atm(c) 1
nirvana33 [79]

<span>Let's assume that the F</span>₂ gas has ideal gas behavior. 

<span> Then we can use ideal gas formula,
PV = nRT

Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span>⁻¹ K⁻<span>¹) and T is temperature in Kelvin.</span>


Moles = mass / molar mass


Molar mass of F₂ = 38 g/mol

Mass of F₂  = 76 g

Hence, moles of F₂ = 76 g / 38 g/mol = 2 mol

<span>
P = ?
V = 1.5 L = 1.5 x 10</span>⁻³ m³

n = 2 mol

R = 8.314 J mol⁻¹ K⁻<span>¹
T = -37 °C = 236 K

By substitution,
</span>

P x 1.5 x 10⁻³ m³ = 2 mol x 8.314 J mol⁻¹ K⁻¹ x 236 K

                         p = 2616138.67 Pa

                         p = 25.8 atm = 26 atm


Hence, the pressure of the gas is 26 atm.

Answer is "a".

<span>

</span>
5 0
2 years ago
a student adds 3.5 moles of solute to enough water to make a 1500mL solution. what is the concentration?
aksik [14]
<h2>Hello!</h2>

The answer is:

MolarConcentration=\frac{3.5moles}{volume(1.5L)}=2.33molar

<h2>Why?</h2>

Since there is not information about the solute but only its mass, we need to assume that we are calculating the molar concentration of a solution or molarity. So, need to use the following formula:

MolarConcentration=\frac{mass(solute)}{volume(solution)}

Now, we know that the mass of the solute is equal  3.5 moles and the volume is equal to 1500 mL or 1.5L

Then, substituting into the equation, we have:

MolarConcentration=\frac{3.5moles}{1.5L}=2.33molar

Have a nice day!

7 0
2 years ago
Read 2 more answers
When 100 ml of 1.0 M Na3PO4 is mixed with 100 ml of 1.0 M AgNO3,a yellow precipitate forms and Ag+ becomes negligibly small. Whi
jok3333 [9.3K]

Answer:

Option A

Explanation:

Number of millimoles of Na3PO4 = 1 × 100 = 100

Number of millimoles of AgNO3 = 1 × 100 = 100

When 1 mole of Na3PO4 is dissociated we get 3 moles of sodium ions and 1 mole of phosphate ion

When 1 mole of AgNO3 is dissociated, we get 1 mole of Ag+ and 1 mole of NO3-

As Ag+ concentration is negligible, the dissociated Ag+ ion must have form the precipitate with phosphate ion and as number of moles of Ag+ and phosphate ion are same, therefore the concentration of phosphate ion must be negligible

Here as 100 millimoles of Na3PO4 is there, we get 300 millimoles of Na+ and 100 millimoles of PO43-

And as 100 millimoles of AgNO3 is there, we get 100 millimoles of Ag+ and 100 millimoles of NO3-

∴ Increasing order of concentration will be  PO43- < NO3- < Na+

4 0
2 years ago
Read 2 more answers
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