Answer:
- 5.15×10²⁴ molecules of sulfur dioxide
- 3.63×10²³ molecules of carbon monoxide
- 6.02×10²³ molecules of ammonia
Explanation:
We begin from the relation that 1 mol of molecules contains NA of molecules
NA = 6.02×10²³
Now, we make rules of three:
1 mol has 6.02×10²³ molecules, therefore:
8.55 moles of SO₂ must have (8.55 . NA) / 1 = 5.15×10²⁴ molecules of dioxide
0.603 moles of CO must have (0.603 . NA) / 1 = 3.63×10²³ molecules of monoxide
Avogadro's Number of molecules of NH₃ are 6.02×10²³ molecules of ammonia
I’m pretty sure it would be B
Given:
A compound with:
Number of carbon atoms = 9
Number of double bonds = 1
A double bond between 5th and 6th carbon
A propyl group (CH2CH2CH3) branching off the 3rd carbon from the left
Try to illustrate the given and observe the formation of the atoms. Now, follow the correct IUPAC naming system. The name of the compound is
4-propyl-1-hexene
Count from the right to the left, the double bond is between the 1st and 2nd carbon, thus, 1-hexene. The propyl branches out the 4th carbon from the right, thus 4-propyl.
Answer: The reactivity of group 7 decreases as we move down the group because:
Explanation:
The elements of group 7 that is fluorine to iodine. The halogens are non metals and they react with metals to gain electrons. The metals loose electrons and the non metal gains it.
As we move down the group the atomic radius gets bigger( more electron and more proton) and as a result the outer shells move further away from the nucleus.
There is more distance between the negatively charged electrons and positively charged nucleus.
Therefore the force of attraction between the shells and nucleus is lesser or weaker.
This makes attracting an extra electron from metals very difficult which results in weaker reaction.
Consequently, the reactivity decreases as we move down the group 7
Answer:D the pressure increases
Explanation:
