Bonds between carbon and oxygen are more polar than bonds between sulfur and oxygen. nevertheless, sulfur dioxide (SO₂) exhibits a dipole moment while carbon dioxide (CO₂) does not because of the difference in their shape, CO₂ is having linear geometry thus exhibit zero dipole moment while SO₂ is having bent shape thus exhibit dipole moment. So, despite the fact that bonds between carbon and oxygen are more polar than bonds between sulfur and oxygen. nevertheless, sulfur dioxide (SO₂) exhibits a dipole moment while carbon dioxide (CO₂) does not.
The balanced equation that illustrates the reaction is:
2C4H6 + 11O2 ......> 8CO2 + 6H2O
number of moles = mass / molar mass
number of moles of oxygen = 2.1 / 32 = 0.065625 moles
Now, from the balanced equation, we can note that:
11 moles of oxygen are required to produce 6 moles of water.
Therefore:
0.065625 moles of oxygen will produce:
(0.065625*6) / 11 = 0.03579 moles of water
number of moles = mass / molar mass
mass = number of moles * molar mass
mass of water = 0.03579 * 18 = 0.644 grams
Answer:
Yes
Explanation:
They are a unique type of eukaryote because they lack an important organelle: mitochondria. Mitochondria are essential for producing cellular energy in most eukaryotic cells. However, due to its habitat, it is able to acquire energy from a process called sulfur mobilization.
They are significant because they challenge the idea that eukaryotes need mitochondria to be classified as eukaryotic. However, they have other membrane-bound organelles such as a nucleus and Golgi apparatus, meaning they remain eukaryotic.
Research suggest they lost their mitochondria over time, rather than never having had them throughout their ancestry.
Because of all these reasons, they still meet the definition of a eukaryote.
Answer:
Concentration of nitrate in the new solution = 0.007 M
Explanation:
Given:
Concentration nitrate solution = 0.070 m
Volume of aliquote of the nitrate solution is add = 10.0 ml
Total volume = 100 ml
Find:
Concentration of nitrate in the new solution
Computation:
Number of M. mole = 0.070 m x 10.0 ml
Number of M. mole = 0.7 m-moles
Concentration of nitrate in the new solution = 0.7 m-moles / 100 ml
Concentration of nitrate in the new solution = 0.007 M