The plants which have a fibrous root system are
<h3>What is a fibrous root?</h3>
A fibrous root system is a root system which consists of large mass of roots of nearly equal size
Some few examples of plants with fibrous roots system are as follows:
- Rice plants
- Sugarcane plants
- Onions
- Pines
- Asparagus
- Maize
In conclusion; the plants which have a fibrous root system are grasses and corns.
Learn more about fibrous root system:
brainly.com/question/1600214
Answer:
theory is diffrent from law
Explanation:
a Theory can never be proven to be true nd a law can usually be expressed
Answer: The right conjugate of is
Explanation:
According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.
For the given chemical equation:
Here, is gaining a proton, thus it is considered as a brønsted-lowry base and after gaining a proton, it forms which is a conjugate acid.
Thus the right conjugate of is
Answer:
If the rock can't scratch a knife, but glass, it's a 7-10 (Hard Rock)
It's between Quartz and Topaz.
So the answer is,
<h2>F.</h2>
Answer:
a) 40 %
b)
c)
Explanation:
For a) we will have to calculate the <u>molar mass</u> of , so the first step is to find the <u>atomic mass</u> of each atom and multiply by the <u>amount of atoms</u> in the molecule.
C => 12*(6) = 72
H => 1*(12) = 12
O => 6*(16) = 96
Molar mass = 180 g/mol
Then we can calculate the percentage by mass:
For b) we have to start with the <u>reaction of glucose</u>:
Then we have to convert the grams of glucose to moles, the moles of glucose to moles of carbon dioxide and finally the moles of carbon dioxide to grams. To do this we have to take into account the<u> following conversion ratios</u>:
-) 180 g of glucose = 1 mol glucose
-) 1 mol glucose = 6 mol carbon dioxide
-) 1 mol carbon dioxide = 44 g carbon dioxide
For C, we have to start with the conversion from grams of glucose to moles, the moles of glucose to moles of oxygen and finally the moles of oxygen to molecules. To do this we have to take into account the <u>following conversion ratios</u>:
-) 180 g of glucose = 1 mol glucose
-) 1 mol glucose = 6 mol oxygen
-) 1 mol oxygen = 6.023x10^23 molecules of O2