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KIM [24]
3 years ago
6

If a long steady rain is expected which clouds should be expected

Chemistry
1 answer:
Inessa [10]3 years ago
4 0
Nimbostratus clouds bring long steady rains
You might be interested in
14. An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the Km
emmainna [20.7K]

Answer:

27 min

Explanation:

The kinetics of an enzyme-catalyzed reaction can be determined by the equation of Michaelis-Menten:

v = \frac{vmax[S]}{Km + [S]}

Where v is the velocity in the equilibrium, vmax is the maximum velocity of the reaction (which is directed proportionally of the amount of the enzyme), Km is the equilibrium constant and [S] is the concentration of the substrate.

So, initially, the velocity of the formation of the substrate is 12μmol/9min = 1.33 μmol/min

If Km is a thousand times smaller then [S], then

v = vmax[S]/[S]

v = vmax

vmax = 1.33 μmol/min

For the new experiment, with one-third of the enzyme, the maximum velocity must be one third too, so:

vmax = 1.33/3 = 0.443 μmol/min

Km will still be much smaller then [S], so

v = vmax

v = 0.443 μmol/min

For 12 μmol formed:

0.443 = 12/t

t = 12/0.443

t = 27 min

7 0
3 years ago
A sample of helium has a volume of 50.00 L at STP. How many helium atoms are in the sample? Show work
Setler79 [48]

Answer:

Explanation:

molar volume at STP=22.4 L

given volume=50.0 L

number of moles=given volume/molar volume

number of moles=50.0/22.4

number of moles=2.2

1 mole of helium =6.023*10^23 atoms

2.2 moles of helium =6.023*10^23*2.2=1.3*10^24

therefore 50.0 L of helium contain 1.33*10^24 atoms

4 0
3 years ago
7 organic compounds of plant origin​
Nat2105 [25]

Explanation:

Among the numerous types of organic compounds, four major categories are found in all living things carbohydrates, lipids, proteins and nucleic acids.

8 0
3 years ago
How many grams of oxygen are required to react with 13.0 grams of octane (C8H18) in the combustion of octane in gasoline?
egoroff_w [7]

Grams of oxygen are required to react with 13.0 grams of octane (C8H18) in the combustion of octane in gasoline is 45.5g

Octane is a hydrocarbon which burns in gasoline in presence of oxygen according to the given balanced chemical equation,

2C₈H₁₈ + 25O₂------> 16CO₂ + 18H₂0

Molar mass of octane = 114.23g/mol

Molar mass of Oxygen = 32g/mol

According to the stiochiometry of the balanced equation the mole ratio of Octane and Oxygen is 2:25

2 mole of octane needs 25 mole of oxygen

1 mole of octane needs 12.5 moleof oxygen

114.23g of octane needs 400g of oxygen

13g   of octane  needs 45.5g of oxygen

Mass of oxygen needed =45.5g

Hence, the Mass of oxygen needed is 45.5g for the combustion of octane in gasoline.

Learn more about Octane here, brainly.com/question/21268869

#SPJ4

5 0
2 years ago
In which case would recalibrating a thermometer be an important next step in an experiment dealing with boiling points?
Sonbull [250]

I believe the answer would be D

7 0
2 years ago
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