Noble gases
Explanation:
Electronic configuration 1s² 2s² 2p⁶
The element belongs to the group of the noble gases.
- The noble gases have complete outer shell configuration of their atoms.
- we can infer that the configuration above is for an element in the p-block because the last sub-level filled is the p-orbital.
- The elements therefore belongs to the p-block
- The block is from group 111A to O
- Only the halogens and noble gases fits this picture from the option.
- The outer most p-subshell have three orbitals requiring 6 electrons to fill them up.
- This makes a complete and stable configuration.
- The highest energy level of 2 is also made up of 8 electrons, an octet.
- This is why we can conclude that they are noble gases.
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Answer:
2NaOH + (NH4) 2SO4 = Na2SO4(s) + 2NH3(g) + 2H2O(l)
Two moles of sodium hydroxide reacts with 1 Mike of ammonium sulphate to give 1 mole of Sodium sulphate, 2 moles of ammonia gas and 2 moles of water
Explanation:
Answer:
31.67 mph
Explanation:
To calculate the average speed of the truck, we must first obtain the total distance travelled by the truck followed by the total time taken for the truck to cover the distance travelled.
The following data were obtained from the question include:
Total distance) = 30 + 45 + 50 + 65 = 190 miles
Total time = 1 + 2 +1 +2 = 6 hours
Average speed =.?
Average speed = Total distance / Total time
Average speed = 190 /6
Average speed = 31.67 mph
Therefore, the average speed of the truck is 31.67 mph
The idea here is that you need to figure out how many moles of magnesium chloride,
MgCl
2
, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.
As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.
c
=
n
V
So, how many moles of magnesium chloride must be present in the target solution?
c
=
n
V
⇒
n
=
c
⋅
V
n
=
0.158 M
⋅
250.0
⋅
10
−
3
L
=
0.0395 moles MgCl
2
Now determine what volume of the target solution would contain this many moles of magnesium chloride
c
=
n
V
⇒
V
=
n
c
V
=
0.0395
moles
3.15
moles
L
=
0.01254 L
Rounded to three sig figs and expressed in mililiters, the volume will be
V
=
12.5 mL
So, to prepare your target solution, use a
12.5-mL
sample of the stock solution and add enough water to make the volume of the total solution equal to
250.0 mL
.
This is equivalent to diluting the
12.5-mL
sample of the stock solution by a dilution factor of
20
.
