3 years ago

Can you help me please..

Chemistry

1 answer:

Murrr4er [49]3 years ago

7 0

Answer:

Using Charles law of Gas

Explanation:

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PI3Cl2 is a nonpolar molecule. Based on this information, determine the I−P−I bond angle, the Cl−P−Cl bond angle, and the I−P−Cl

alexdok [17]

Answer: Yes I-P-Cl = 90

Explanation:

This is because the angle formed between I-P-Cl is perpendicular hence the angle is 90°

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3 years ago

If cacl2 is dissolved in water, what can be said about the concentration of the ca2+ ion? view available hint(s) if is dissolved

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The concentration of Ca2+ ions is half that of the Cl- ions.

8 0

3 years ago

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Calculate the number of mol of S by 1.81

rodikova [14]

It’s 6.022x10^23. This is how many atoms of any element are in one mole of that element. No matter what it is, there are always 6.022x10^23 atoms per mole. So divide 1.81x10^24 by 6.022x10^23 to get approximately 3 moles of sulfur.

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3 years ago

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

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

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

= 27.38 min

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

6 0

3 years ago

The oceanic crust is mostly associated with the —————.

Lilit [14]

I think it would the hydrosphere

4 0

3 years ago