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lana66690 [7]
3 years ago
7

What mass of octane must be burned in order to liberate 5270 kj of heat? δhcomb = -5471 kj/mol?

Chemistry
1 answer:
katrin2010 [14]3 years ago
6 0
As,
                           5471 kJ heat is given by  =  1 mole of Octane
Then,
                    5310 kJ heat will be given by  = X moles of Octane

Solving for X,
                                  X  =  (5310 kJ × 1 mol) ÷ 5471 kJ

                                  X  =  0.970 moles of Ocatne

So, 0.970 moles of Octane will liberate 5310 kJ energy. Now changing moles to mass,
As,
                                  Moles  =  mass / M.mass
Or,
                                  Mass  =  Moles × M.mass
Putting values,
                                  Mass  =  0.970 mol × 114.23 g/mol

                                  Mass  =  110.83 g of Octane
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Answer:

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Explanation:

Given data includes:

Tris= 10mM

pH = 8.0

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In order to make 0.25 L solution buffer ; i.e (250 mL); we have the following component.

Stock Concentration             Volume to be             Final Concentration

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1 M Tris                                     2.5 mL                         10 mM

5 M NaCl                                  7.5 mL                        150 mM

1 M Imidazole                           75 mL                         300 mM    

M_1*V_1 = M_2*V_2. is the formula that is used to determine the corresponding volume that is added for each stock concentration

The stock concentration of Tris ( 1 M ) is as follows:

M_1*V_1 = M_2*V_2.

1*V_1= 0.01 M *250mL\\V_1 = 2.5mL

The stock concentration of NaCl (5 M ) is as follows:

M_1*V_1 = M_2*V_2.

1*V_1= 0.15 M *250mL\\V_1 = 7.5mL

The stock concentration of Imidazole (1 M ) is as follows:

M_1*V_1 = M_2*V_2.

1*V_1= 0.03 M *250mL\\V_1 = 75mL

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