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Snezhnost [94]
3 years ago
8

What mass of oxygen would be required for the complete combustion of 10 Grams of glucose?

Chemistry
1 answer:
enot [183]3 years ago
3 0

Answer:

For the complete combustion of 10 g of glucose 11.52 g of oxygen is required.

Explanation:

Given data:

Mass of glucose = 10 g

Mass of oxygen required for complete combustion = ?

Solution:

Chemical equation:

C₆H₁₂O₆ + 6O₂    →  6CO₂ + 6H₂O

Now we will convert the given mass of glucose into number of moles.

Number of moles = mass/ molar mass

Number of moles = 10 g/ 180.156 g/mol

Number of moles = 0.06 mol

Now we will compare the moles of glucose with oxygen.

         C₆H₁₂O₆            :              O₂

            1                     :               6

          0.06                :            6×0.06 = 0.36 mol

Mass of oxygen required:

Mass = number of moles × molar mass

Number of moles  = 0.36 mol × 32 g/mol

Number of moles = 11.52 g

For the complete combustion of 10 g of glucose 11.52 g of oxygen is required.

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Which of the following changes would have no effect on the equilibrium position of the reaction below? 2 NOBR (g) 2 NO (g)+ Br2
nasty-shy [4]

Answer:

D) cutting the concentrations of both NOBr and NO in half

Explanation:

The equilibrium reaction given in the question is as follows -

2NOBr ( g ) ↔  2NO ( g ) + Br₂ ( g )

The equilibrium constant for the above reaction can be written as -

K = [ NO ]² [ Br₂ ] / [ NOBr ] ²

Therefore from the condition given in the question , the changes that will not affect the equilibrium will be , reducing the concentration of both NOBr and NO to half ,

Hence ,

the new concentrations are as follows -

[ NoBr ] ' = 1/2 [ NoBr ]

[ NO ] ' = 1/2  [ NO ]

Hence the new equilibrium constant equation can be written as -

K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ]  ' ²

Substituting the new concentration terms ,

K ' = 1/2  [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ]  ²

K ' = 1/4  [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ]  ²

The value of 1 / 4 in the numerator and the denominator is cancelled -

K ' =   [ NO ] ² [ Br₂ ] /  [ NoBr ]  ²

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5 0
3 years ago
To prepare an acetic acid/acetate buffer, a technician mixes 30.6 mL of 0.0880 acetic acid and 21.6 mL of 0.110 sodium acetate i
enyata [817]

Answer: There are 0.00269 moles of acetic acid in buffer.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution in ml}}     .....(1)

Molarity of acetic acid solution = 0.0880 M

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Putting values in equation 1, we get:

0.0880M=\frac{\text{Moles of acetic acid}\times 1000}{30.6ml}\\\\\text{Moles of acetic acid}=\frac{0.0880\times 30.6}{1000}=0.00269mol

Thus there are 0.00269 moles of acetic acid in buffer.

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nikdorinn [45]

Answer:

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Explanation:

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