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2 years ago

Which of the following is NOT an example of a molecule A: H2O2 B:NCI3 C:F D:O3

Chemistry

2 answers:

timama [110]2 years ago

6 0

C. F because it's an atom not a molecule

romanna [79]2 years ago

6 0

Yes, the answer is C:F which has no atoms

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If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i

Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane ⇄ Isobutane

At equilibrium

1.0 M 2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M -

After equilibrium reestablishes:

(1.50-x)M (2.5+x)

The equilibrium expression will wriiten as:

$K_c=\frac{[Isobutane]}{[Butane]}$

$2.5=\frac{2.5+x}{(1.50-x)}$

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

3 0

2 years ago

Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a low

LenaWriter [7]

Answer:

Compound B has greater molar mass.

Explanation:

The depression in freezing point is given by ;

$\Delta T_f=i\times k_f\times m$ ..[1]

$m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}$

Where:

i = van't Hoff factor

$k_f$ = Molal depression constant

m = molality of the solution

According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.

The depression in freezing point of solution with A solute: $\Delta T_{f,A}$