30% of 107.75 g is 32.325g so there are about 8.08125 mol of He which is
4.86 x 10^24 particles
70% of 107.75 g is 75.425g so there are about .9 mol of Kr which is
5.42 x 10^23 paticles
In the whole sample, there are about 5.41 x 10^24 particles
The question is incomplete, here is the complete question:
If you performed a combustion reaction of 2-ethyl-1-methylpropene, what products would you expect to be present?
Answer: The products of the combustion of 2-ethyl-1-methylpropene are carbon dioxide and water.
<u>Explanation:</u>
Combustion reaction is defined as the reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.

We are given a chemical compound, which is 2-ethyl-1-methylpropene. The chemical formula of this compound is 
The chemical equation for the combustion of 2-ethyl-1-methylpropene follows:

By Stoichiometry of the reaction:
4 moles of 2-ethyl-1-methylpropene reacts with 37 moles of oxygen gas to produce 24 moles of carbon dioxide and 26 moles of water
Hence, the products of the combustion of 2-ethyl-1-methylpropene are carbon dioxide and water.
Answer:
As we read from left to right across the periodic table atomic numbers are increased by one each of element.
Explanation:
As we move from left to right across the periodic table the atomic number is increased by one and the number of valance electron in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.
Answer:
88 grams CO2
Explanation:
Convert from grams to moles to mole ratio to grams again
32 g CH4 * 1 mol CH4/16 g CH4 * 1 mol CO2/1 mol CH4 * 44 grams CO2 = 88 grams CO2
Lemme know if u need more explanation!