Answer:
you havent posted an equation
Explanation:
Answer no 1)
The correct option is A) He did not try to learn it. It was natural.
As Sergei did not put any effort in learning the art of tongue rolling, hence we can assume that this characteristic was found to be naturally present in Sergei. This hence shows that Sergei has acquired the characteristic of tongue rolling through inheritance.
Answer No 2)
We can learn whether other persons in the class have the characteristic of tongue rolling or not by observing them. In the students who can roll their tongue we can conduct a survey asking them whether they tried to learn the trait or did not put any effort to learn tongue rolling. The students who learned the skill by practicing might have acquired the characteristic from learning.
Answer No 3)
We can survey the class for features like eye color, hair color, height, weight. We can ask certain questions in our survey like whether the characteristic can be seen in either the mother, father or other siblings.
Answer:
See Explanation
Explanation:
We can establish that there are chloride ions present in water if we add dilute trioxonitrate V acid to a sample of the water in drops followed by aqueous silver trioxonitrate V.
If a white precipitate is observed and the precipitate is found to be soluble in excess aqueous ammonia solution, then we can confirm the presence of chloride ion in the water.
its returned to the soil by rhizobium
66.2% yield
The balanced equation for the reaction is:
2C6H10 + 17O2 ==> 12CO2 + 10H2O
So for every 2 moles of C6H10 consumed, we should get 12 moles of CO2. Or to simplify, for each mole of C6H10, we should get 6 moles of CO2. Now let's calculate the molar mass of C6H10 and CO2 and then determine how many moles of each we really have.
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Atomic weight oxygen = 15.999
Molar mass C6H10 = 6*12.0107 + 10*1.00794 = 82.1436 g/mol
Molar mass CO2 = 12.0107 + 2*15.999 = 44.0087 g/mol
Moles C6H10 = 8.88 g / 82.1436 g/mol = 0.10810337 mol
Moles CO2 = 18.9 g / 44.0087 g/mol = 0.429460538 mol
Since we had 0.10810337 moles of C6H10, we should have gotten
6*0.10810337 = 0.648620221 moles of CO2, but only got 0.429460538 moles. So let's divide the actual yield by the theoretical yield to get the percentage yield.
0.429460538 / 0.648620221 = 0.662114014 = 66.2%