35g of CaCl2 is dissolved in 2.75L of water to form solution (s). CaCl2 dissociates in water according to the given equation :
CaCl2 Ca2+ (aq)+2C1-(aq) (dissociation equation)
M(CaCl2)= 111 g/mol
a-Calculate the molar concentration of CaCl2 in solution (s).
ration of Ca2+ ion and clion

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

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

i believe the answer is 5,000