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3 years ago

How did I get the sand out the mixture ?

Chemistry

2 answers:

svetoff [14.1K]3 years ago

4 0

Answer:

They are called homogenous mixtures. If you put sand into a glass of water, it is considered to be a mixture. You can always tell a mixture, because each of the substances can be separated from the group in different physical ways. You can always get the sand out of the water by filtering the water away.

Oliga [24]3 years ago

3 0

Answer:

Sifter

Explanation: They are called homogenous mixtures. If you put sand into a glass of water, it is considered to be a mixture. You can always tell a mixture, because each of the substances can be separated from the group in different physical ways. You can always get the sand out of the water by filtering the water away.

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Answer:

Explanation:

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where

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$pH = \frac{1}{2} ( - log(1.75 \times {10 }^{ - 5} - log(1.00 \times {10}^{ - 3} ) ) \\ = \frac{1}{2} (4.757 + 3) \\ = \frac{1}{2} \times 7.757) \\ = 3.8785 \: \: \: \: \: \: \: \:$

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2 years ago

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Explanation:

According to the given data, we will calculate the following.

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= 480 s

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= $1.44 \times 10^{-3}s^{-1}
$

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= 45 mmol/L

Rate of hydrolysis $V_m_{o}$ = 0.07 mmol/L/s

Conversion X = 0.80

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= 45 (1 - 0.80)

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or, = 9 mmol/L

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Therefore, time taken will be calculated as follows.

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Now, putting the given values into the above formula as follows.

t = $-\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]$

= $-\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s
}{K_{M} ln (\frac{45 mmol/L
}{9 mmol/L
}) + (45 mmol/L - 9 mmol/L
)]$

= $1642.83 s \times \frac{1 min}{60 sec}$

= 27.38 min

Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.

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2 years ago

How did I get the sand out the mixture ￼?

How did I get the sand out the mixture ?