Worked solutions please? (1 Viewer)

[tommoo]

Cheers.

 

[enigma_1]

question 1)
Delta H = -(mCAT)/n

I'm using A = Triangle btw

m = 1kg
C = 4.18 specific heat capacity of water
AT = (100 - 50) = 50

because of boiling point which is 100 degrees so that's why it's 100 - 50

1367kJ/mol = - 1kg x 4.18 x (100 - 50)
----------------------------------n

n = -1kg x 4.18 x 50
------------- 1367

:. n = 0.1528895391mol

using n = m/M

m = M x n
m = 0.1528895391 x 46
m = 7g so B



question 2)

n(Phosphorus pemtoxide) = 1.42/M = 1.42/(30.97(2)+16(5)) =0.01000422714mol

moles of H3PO4 in the first eqn (look ayt the molar ratio) = 0.01000422714 x 2 = 0.02000845428mol

in second equation, mol of NaOH is 3 times that of HSPO4 :. you do 0.02000845428mol x 3 = 0.06002536283mol of NaOh

:. Volume of NaOH to find use the formula C=n/V

C = 0.30mol/L
mol = 0.06002536283mol

:. V = n/c = 0.06002536283mol /0.30 = 0.2L which is c

 

[faisalabdul16]

Can confirm both of Enigma's answers are correct

 

[tommoo]

question 1)
Delta H = -(mCAT)/n

I'm using A = Triangle btw

m = 1kg
C = 4.18 specific heat capacity of water
AT = (100 - 50) = 50

because of boiling point which is 100 degrees so that's why it's 100 - 50

1367kJ/mol = - 1kg x 4.18 x (100 - 50)
----------------------------------n

n = -1kg x 4.18 x 50
------------- 1367

:. n = 0.1528895391mol

using n = m/M

m = M x n
m = 0.1528895391 x 46
m = 7g so B



question 2)

n(Phosphorus pemtoxide) = 1.42/M = 1.42/(30.97(2)+16(5)) =0.01000422714mol

moles of H3PO4 in the first eqn (look ayt the molar ratio) = 0.01000422714 x 2 = 0.02000845428mol

in second equation, mol of NaOH is 3 times that of HSPO4 :. you do 0.02000845428mol x 3 = 0.06002536283mol of NaOh

:. Volume of NaOH to find use the formula C=n/V

C = 0.30mol/L
mol = 0.06002536283mol

:. V = n/c = 0.06002536283mol /0.30 = 0.2L which is c

Thanks so much.

 

[enigma_1]

Thanks so much.

Nws

 

[Cleverusername5321]

http://i.imgur.com/5vA96qA.jpg
sdgfjnsdlkgfnsdjkfgdsnbfsd someone beat me to it.