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pickupchik [31]
3 years ago
6

Which of the following happens during a lunar eclipse?

Chemistry
1 answer:
Svetlanka [38]3 years ago
5 0

Answer:

a. Earth casts a shadow on the moon.    

Explanation:

Lunar eclipse occurs when Earth comes in between sun and moon. The sunlight falling on moon is blocked by Earth and the shadow of earth falls on moon. This happens on full moon day when sun, earth and moon align in one line.The moon passes through the shadow of earth and appears dark.

Thus, the correct option is a.

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The type of charge an electron carries is B negative
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If further increases in substrate concentration do not result in further increases in reaction rate, then an enzyme is likely
igor_vitrenko [27]

Given what we know, we can confirm that if further increases in substrate concentration do not result in further increases in reaction rate, then an enzyme is likely saturated.

<h3>What does it mean for an enzyme to be saturated?</h3>

Enzymes work by binding to the substrate in specific zones of the enzyme. The zones are known as the active sites on enzymes. Since enzymes have a limited amount of these zones, once they are all bonded to a substrate, we can say that it is saturated.

Therefore, the saturation of enzymes allows us to explain how further increases in substrate concentration do not result in further increases in reaction rate.

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4 0
2 years ago
What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91
Vikentia [17]

Answer:  A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L

Temperature = 91^{o}C = (91 + 273) K = 364 K

As molar mass of O_2 is 32 g/mol. Hence, the number of moles of O_2 are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

PV = nRT\\P  \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

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