Answer:
The concentration of helium in the water is 2.405×10^-4 M
Explanation:
Concentration = Henry's law constant × partial pressure of helium
Henry's law constant = 3.7×10^-4 M/atm
Partial pressure of helium = 0.65 atm
Concentration = 3.7×10^-4 × 0.65 = 2.405×10^-4 M
Answer: C.
Explanation: Alcohol floats on oil and water sinks in oil. Water, alcohol, and oil layer well because of their densities, but also because the oil layer does not dissolve in either liquid. The oil keeps the water and alcohol separated so that they do not dissolve in one another. ... Water sinks because it is more dense than oil.
Answer:
21.6 g
Explanation:
The reaction that takes place is:
First we<u> convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:
- 9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄
- 64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂
0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.
Now we <u>calculate how many moles of water are produced</u>, using the <em>number of moles of the limiting reactant</em>:
- 0.6 mol CH₄ *
= 1.2 mol H₂O
Finally we<u> convert 1.2 moles of water into grams</u>, using its <em>molar mass</em>:
- 1.2 mol * 18 g/mol = 21.6 g
Answer:
6
Explanation:
p orbital can hold up to six electron. Argon electron configuration will be 1s²2s²2p⁶3s²3p⁶
Answer:
See explanation below
Explanation:
First, we need to understand that the monochlorination of an alkane like this one, involves substitution of one of the atoms of hydrogen of the molecule for an atom of chlorine.
This reaction takes place when the alkane reacts with Cl₂ in presence of light or heat.
When this happens, the first step involves the breaking of the double bond of the chlorine to form the ion Cl⁻.
The next step involves the substraction of the hydrogen of the molecule by the Chlorine. This will leave the alkane with a lone pair available for reaction.
The third step, the alkane with the lone pair of electron substract a chlorine for the beggining and form the mono chlorinated product.
The final step involves forming the remaining products with the remaining reagents there.
In the picture attached you have the mechanism and product for this reaction: