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aniked [119]
3 years ago
11

MUSCLES IN A KANGAROO'S LEGS WORK BECAUSE OF THE CELLS THAT MAKE UP THE MUSCLE. WHICH COMPONENT OF CELL THEORY DOES THIS BEST IL

LUSTRATE? :)

Chemistry
2 answers:
valkas [14]3 years ago
7 0

Answer: OPTION B

Explanation:

Cells are generated from nonliving materials. Muscles in a kangaroo's legs work because of the cells that make up the muscle. The component of cell theory this best illustrate is: Cells are the basic unit of structure and function of all living things

Fed [463]3 years ago
4 0

Answer:

b

Explanation:

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How many moles are equal to 83.4 L of O2?
ladessa [460]

Answer:

3.72mol

Explanation:

Hello,

In this case, we consider that at STP conditions (273 K and 1 atm) we know that the volume of 1 mole of a gas is 22.4 L, thereby, for 83.4 L, the resulting moles are:

22.4L\rightarrow 1mol\\83.4L\rightarrow X\\X=\frac{83.4L*1mol}{22.4L}=3.72mol

This is a case in which we apply the Avogadro's law which relates the volume and the moles as a directly proportional relationship.

Best regards.

8 0
3 years ago
A gas mixture at room temperature contains 10.0 mol CO and 12.5 mol O2. (a) Compute the mole fraction of CO in the mixture. (b)
zlopas [31]

Answer:

a. the mole fraction of CO in the mixture of CO and O2.

mole fraction = moles of CO/ Total moles of the mixture

Mole fraction of CO = 10/(10+12.5)=0.444

b. Reaction - CO(g)+½O2(g)→CO2(g)

Stoichiometry: 1 mole of CO react with 0.5mole of O2 to give 1 mole of CO2

So given,

At a certain point in the heating, 3.0 mol CO2 is present. Determine the mole fraction of CO in the new mixture.

3mol of CO2 is produced from 3 mols of CO and 1.5mol of O2

This means that unused mols are : 7mols of CO and 11mols of O2

Total product mixture = 3 + 7 + 11 = 21mols

mole fraction of CO = 7/21 = 0.33

7 0
3 years ago
Which of the following statements about Ideal Gas Law is supported by Charles's Law?
grin007 [14]

Answer:

B

Explanation:

volume occupied by fixed amount of gas is directly proportional to its absolute temperature.

5 0
3 years ago
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What is the mass, in grams of a pure iron cube that has a volume of 4.20cm^3
Dmitriy789 [7]
d_{Fe}=7874\frac{kg}{m^{3}}=7,874\frac{g}{cm^{3}}\\
V=4,2cm^{3}\\\\
m=dV=7,784\frac{g}{cm^{3}}*4,2cm^{3}\approx 32,7g
6 0
3 years ago
Why is secondary growth important to plants
stepan [7]
Secondary growth is important to plants because it involves thickening of the plant axis.It also increased amounts of vascular tissue.

I tried sorry if it’s not worded perfect :)
5 0
3 years ago
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