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3 years ago

Am I cute lol here's a pic of meh

Chemistry

2 answers:

Licemer1 [7]3 years ago

5 0

Answer:

lol sooo cute

Explanation:

charle [14.2K]3 years ago

5 0

Answer:

yo what's your number ya lookin kinda thicc

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Which element is a gas at STP?<br> 1)sulfur<br> 2)xenon<br> 3)potassium<br> 4)phosphorus

natulia [17]

I believe the answer is xenon

4 0

2 years ago

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What Chemicals including in Cosmetic?

Roman55 [17]

The chemicals including in cosmetic are:-

1.Talcum Powder

2.Nanoparticles

3.Coal Tar

4.Triclosan

5.Toluene

Hope this helps you.

7 0

2 years ago

Accuracy vs. Precision

Oksi-84 [34.3K]

Answer:

Accuracy is the closeness to the specific target and precision is the closeness of the measurements to each other.

3 0

3 years ago

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What is the molarity of a 17.0% by mass solution of sodium acetate, NaC2H3O2 (82.0 g/mol), in water? The density of the solution

sattari [20]

Answer:

[NaCH₃COO] = 2.26M

Explanation:

17% by mass is a sort of concentration. Gives the information about grams of solute in 100 g of solution. (In this case, 17 g of NaCH₃COO)

Let's determine the volume of solution, by density

Mass of solution / Volume of solution = Solution density

100 g / Volume of solution = 1.09 g/mL

100 g / 1.09 g/mL = 91.7 mL

17 grams of solute is contained in 91.7 mL

Molarity (M) = Mol of solute /L of solution

91.7 mL / 1000 = 0.0917L

17 g / 82 g/m = 0.207 moles

Molariy = 0.207 moles / 0.0917L → 2.26M

4 0

2 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

2 years ago

Am I cute lol here's a pic of meh​

Am I cute lol here's a pic of meh