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3 years ago

What are observations that indicate a chemical change has occurred ?

Chemistry

1 answer:

sukhopar [10]3 years ago

4 0

Answer:

Change of odor

Change of color

Change in temperature

Change in energy

Change of composition

Formation of gases

The decomposition of organic matter

The Change is irreversible

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How can we use rocks to determine how an area used to look in the past?

tatiyna

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8 0

3 years ago

Oxidation number of Al(OH)4

Kobotan [32]

Explanation:

The sum of all oxidation numbers in a neutral compound is zero. The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion. The oxidation number of oxygen in a compound is usually –2. The oxidation state of hydrogen in a compound is usually +1.

The oxidation state of Al in Al(OH)

−

x+4(+1−2)=−1

∴x=+3

The oxidation state of Mn in MnO

y+2(−2)=0

∴y=+4

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8 0

3 years ago

I2 is an example of a

Free_Kalibri [48]

Answer:

number

Explanation:

5 0

3 years ago

Read 2 more answers

The Michaelis constant for pancreatic lipase is 5 mM. At 60 C, lipase is subject to deactivation with a half-life of 8 min. Fat

Dmitriy789 [7]

Explanation:

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

Half life of lipase $t_{1/2}$ = 8 min x 60 s/min

= 480 s

Rate constant for first order reaction is as follows.

$k_{d} = \frac{0.6932}{480}$

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

Initial fat concentration $S_{o}$ = 45 $mol/m^{3}$

= 45 mmol/L

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

Conversion X = 0.80

Final concentration (S) = $S_{o} (1 - X)$

= 45 (1 - 0.80)

= 9 $mol/m^{3}$

or, = 9 mmol/L

It is given that $K_{m}$ = 5mmol/L

Therefore, time taken will be calculated as follows.

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

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

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

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