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Aloiza [94]
3 years ago
12

Which process occurs during transpiration?

Chemistry
2 answers:
miskamm [114]3 years ago
8 0

Answer:

It should be B

Explanation:

because I just searched it up if I'm wrong I'm wrong

defon3 years ago
3 0

Answer:transpiration just took the test

Explanation:

100%

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Are there moles of stars in the universe?
Natalka [10]

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Yes, a mole is defined as 6x1023 of something, so a mole of stars is 6x1023 stars. If we compare this number to our estimate of the total number of stars, we find that there is about 1/10 of a mole of stars in the Universe.

Hope this helps!

7 0
3 years ago
1. A sample of a hydrocarbon is burned completely
amid [387]

Answer:

Governor Peter Stuyvesant

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If 21.4 grams of hydrogen reacts completely with 120.3 grams of oxyge to form hydrogen peroxide, what percent by mass of hydroge
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In its pure form, it is a very pale blue liquid, slightly more viscous than water. Hydrogen peroxide is the simplest peroxide (a compound with an oxygen–oxygen single bond).

5 0
3 years ago
I really need help on this one!!
Lady bird [3.3K]

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Due to the decrease in the gravitational force, it will take longer time for ball to reach the ground while dribbling. Moreover, the hang time of players will also be longer. This means that the players will stay on air longer and will be able to jump higher.

3 0
3 years ago
Enter your answer in the box provided. How many grams of helium must be added to a balloon containing 6.24 g helium gas to doubl
Archy [21]

Answer : The mass of helium gas added must be 12.48 grams.

Explanation : Given,

Mass of helium (He) gas = 6.24 g

Molar mass of helium = 4 g/mole

First we have to calculate the moles of helium gas.

\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=\frac{6.24g}{4g/mole}=1.56moles

Now we have to calculate the moles of helium gas at doubled volume.

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

V\propto n

or,

\frac{V_1}{V_2}=\frac{n_1}{n_2}

where,

V_1 = initial volume of gas  = V

V_2 = final volume of gas = 2V

n_1 = initial moles of gas  = 1.56 mole

n_2 = final moles of gas  = ?

Now we put all the given values in this formula, we get

\frac{V}{2V}=\frac{1.56mole}{n_2}

n_2=3.12mole

Now we have to calculate the mass of helium gas at doubled volume.

\text{Mass of }He=\text{Moles of }He\times \text{Molar mass of }He

\text{Mass of }He=3.12mole\times 4g/mole=12.48g

Therefore, the mass of helium gas added must be 12.48 grams.

4 0
3 years ago
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