Agas cylinder contains 2.0 mol of gas x and 6.0 mol of gas y at a total pressute of 2.1 atm. what is he partial pressure of gas y?

The answer is 1.6 atm. Let's first calculate the mole fraction of gas Y.

The mole fraction (x) is: x = n1/n, where n1 is a number of moles of an

individual gas in a mixture and n is total moles of the gas mixture. We

know that n1 of gas Y is 6.0 mol (n1 = 6.0 mol) and that there are in

total 8. mol of the gas mixture (n = 2.0 + 6.0 = 8.0 mol). Now calculate

the mole fraction of gas Y. x = 6.0/8.0 = 0.75. Now, let's use the mole

fraction of gas Y (x) and the total pressure (P) to calculate the

partial pressure of gas Y (P1): x = P1/P. P1 = x * P. If x = 0.75 and P =

2.1 atm, then the partial pressure of gas Y is: P1 = 0.75 * 2.1 atm =

1.6 atm.

this would equal 1.6 atm

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

the objects have the same charge.