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3 years ago

CAN SOMEONE HELP ME WITH THESE QUESTIONS PLEASE

Chemistry

2 answers:

Leto [7]3 years ago

6 0

Answer:what was it so I can help u do u even got the answers yet I’ll tell you I’m not going to guess

Explanation:

Nostrana [21]3 years ago

5 0

What are they? .............

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The number of oxygen atoms in 48.0g of oxygen gas

Doss [256]

Answer:

Now u have 48 g of O2. There. Fore mole=weight/M. W. Of oxygen. Therefor 3mole.

After that if we to multiply the avogadro number with it. So 3 *NA

Now u want only atom calculation then we have 2 molecule of oxygen then multiply it with 2 too.

So final claculation is =3*2*NA.

Explanation:

your welcome

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3 0

3 years ago

Sodium metal is sometimes used as a cooling agent in heat exchange units because of its relatively high molar heat capacity of 2

IgorC [24]

Answer:

The specific heat of sodium is 1,23J/g°C

Explanation:

Using the atomic weight of sodium (23g/mol) and the atomic weight definition, we have that each mole of the substance has 23 grams of sodium.

starting from this, we use the atomic weight of sodium to convert the units from J / mol ° C to J / g ° C

$28,2 \frac{J}{mol C} x \frac{1mol}{23g} = 1,23 J/g C$

4 0

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

2 years ago

Water can form a solution by mixing with:

MA_775_DIABLO [31]

Explanation:

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2 years ago

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lions [1.4K]

Doing the same final thanks

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3 years ago

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