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3 years ago

Whats the mass number of the atom when you add six protons and 6 electrons?

Chemistry

1 answer:

Verdich [7]3 years ago

6 0

It is6 because the only the proton number and the neutron number affects the mass

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Which is the best pure substance gold or air?

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Gold is the best pure substance. The answer is gold

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How many L are in 1,500cm^3<br><img src="https://tex.z-dn.net/?f=1500%7Bcm%7D%5E%7B3%7D%20%3D%20%5C%3A%20...%20%20%5C%3A%20liter

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1.5 liters are in 1,500cm^3.

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2 years ago

a technician mixes 80 ml of a 5% solution with 10 ml of water. what is the final percentage strength of the solution prepared?

Vinvika [58]

A technician mixes 80 ml of a 5% solution with 10 ml of water. the final percentage strength of the solution prepared is 40 %.

given that :

8 ml of a 5 % solution mix with 10 ml . that means the 80 mL of 5 % solution is diluted with water of 10 mL

therefore, 80 × 5 = 10 × x %

x % = 40 %

Therefore, the final percentage strength of the solution is 40 %

Thus, A technician mixes 80 ml of a 5% solution with 10 ml of water. the final percentage strength of the solution prepared is 40 %.

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1 year 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.

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2 years ago

What two things must all matter have?

OleMash [197]

1.) Mass

2.) Can occupy space (Volume)

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