The experimental mass was 51.5 grams and the theoretical mass was 50 grams. what was the percent error? show your work below. round your answer to two digits.

3%

Experimental:

We are given that the experimental mass was 51.5 grams and the theoretical mass was 50 grams and we are to find the percent error.

We know the formula for the percent error:

Percentage error = (| experimental value - actual value | / actual value) * 100

So substituting the given values in the above formula:

Percent error = (51.5 - 50)/50 * 100 = 0.03 * 100 = 3%

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

i just learnt this recently but i need to look through my notes again