Areaction mixture in a 3.67 l flask at a certain temperature initially contains a mixture of 0.763 grams of h2 and 96.9 grams of i2. at equilibrium, the flask contains 90.4 grams if hi. calculate the equilibrium constant for this reaction at this temperature.
show your work using an ice table.

H2 + I2 = 2HI

we express Keq as follows:

Keq = [HI]^2 / [H2] [I2]

Where the terms represents the concentrations of the substances involved. THe concentrations are as follows:

H2 = 0.763 ( 1 / 18.02) / 3.67 = 0.0115 M

I2 = 96.9 ( 1 / 253.8) / 3.67 = 0.1040 M

HI (at equilibrium) = 90.4 ( 1 / 127.91) / 3.67 = 0.1956 M 

By the ICE table, we can calculate the equilibrium concentrations,
        H2                       I2                           HI
I      0.0115              0.1040                        0
C      -x                        -x                         +2x
 
E  0.0115-0.0963   0.1040-0.0963       0.1926 ===> x = 0.0963

Keq = (0.1926)^2 / (0.0077) (0.0848)
Keq = 56.81

1. they vary with a higher number because boiling is way hotter than melting

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.