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steposvetlana [31]
3 years ago
12

Which two structures would provide a positive identification of a animal cell under a microscope

Chemistry
2 answers:
Ymorist [56]3 years ago
8 0
<span>The two structures are lysosomes and central vacuole

Hope it helps ;)

You look beautiful ;P (if thats you)</span>
babunello [35]3 years ago
4 0
Lack of cell walls no chloroplasts...hoped that helped
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Is this answer right ?
kumpel [21]

Answer:

<h2>YES</h2>

Explanation:

Explanation

<h2>#KEEPSTUDYING</h2><h2 />
5 0
3 years ago
Read 2 more answers
Can anyone give me the name of these? 40 points. Hope it's not too much
kolbaska11 [484]
1. 5 ethyl, 2 methyl octane
2. 1 ethyl, 2 methyl cyclopentane
3. 3,3,5,5- tetrafluoro heptane
4. 3,4-dimethyl hexene
5. 3,4-dimethyl cyclobutene
6. 3,5 diisopropyl cyclohexene
7. 3,3,4 trimethyl pentyne
8. 2,6 dibromo phenol

keep in mind that between 4-7, there could be #1 in front of the main name. for example with #4: 3,4-dimethyl-1- hexene. this honestly depends on the professor how he/she likes it. It is not necessary because if the number is not specified, it is assumed is #1
4 0
4 years ago
It takes 167 s for an unknown gas to effuse through a porous wall and 99 s for the same volume of n2 gas to effuse at the same t
irakobra [83]

Answer : The molar mass of the unknown gas will be 79.7 g/mol


Explanation : To solve this question we can use graham's law;


Now we can use nitrogen as the gas number 2, which travels faster than gas 1;

So, 167 / 99 = 1.687 So the nitrogen gas is 1.687 times faster that the unknown gas 1

We can compare the rates of both the gases;


So here, Rate of gas 2 / Rate of gas 1 = \sqrt{(molar mass 1 / molar mass 2)}

Now, 1.687 = square root [\sqrt{(molar mass 1) / (28.01 g/mol N_{2})} ]


When we square both the sides we get;


2.845 = (molar mass 1) / (28.01 g/mol N2)


On rearranging, we get,


2.845 X (28.01 g/mol N2) = Molar mass 1

So the molar mass of unknown gas will be = 79.7 g/mol

3 0
3 years ago
A scientific ________ must have a control, so that the variables that could affect the out come is reduced. A:experiment B:concl
Lostsunrise [7]

Answer:

A.experiment

Explanation:

hope this helps

4 0
3 years ago
If the lungs of a child hold 0.11 mol of air in a volume of 2.8 L, then the lungs of an average female adult, with a volume is 4
Aleks [24]

Answer:

Explanation:

A childs lung can hold .11mols/ per 2.8 L so that gives us a molarity of .039M

A adults lungs can hold .18 mols /per 4.6 so that gives us .039M aswell meaining that the lung capacity between the two is not different.

4 0
3 years ago
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