we are given
a swimming pool contains 2,850 kiloliters (kL) of water
2,850 kiloliters (kL)=2850000L
we know that
Firstly , we will find for 1L
now, we can multiply both sides by 2850000
so,
a swimming pool contains 752972.25892gal of water...........Answer
Answer:
1.43 M
Explanation:
We'll begin by calculating the number of mole of the solid. This can be obtained as follow:
Mass of solid = 8.60 g
Molar mass of solid = 21.50 g/mol
Mole of solid =?
Mole = mass / molar mass
Mole of solid = 8.60 / 21.50
Mole of solid = 0.4 mole
Next, we shall convert 280 mL to litre (L). This can be obtained as follow:
1000 mL = 1 L
Therefore,
280 mL = 280 mL × 1 L / 1000 mL
280 mL = 0.28 L
Thus, 280 mL is equivalent to 0.28 L.
Finally, we shall determine the molarity of the solution. This can be obtained as illustrated below:
Mole of solid = 0.4 mole
Volume = 0.28 L
Molarity =?
Molarity = mole / Volume
Molarity = 0.4 / 0.28
Molarity = 1.43 M
Thus, the molarity of the solution is 1.43 M.
Answer: The concentration of excess ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
in solution is 0.017 M.
Explanation:
1. 
moles of 
1 mole of 
give = 1 mole of 
Thus 0.019 moles of give = 0.019 mole of
2. moles of 
According to stoichiometry:
1 mole of 
gives = 2 moles of 
Thus 0.012 moles of give = 
moles of
As 1 mole of neutralize 1 mole of
0.019 mole of will neutralize 0.019 mole of
Thus (0.024-0.019)= 0.005 moles of will be left.
![[OH^-]=\frac{\text {moles left}}{\text {Total volume in L}}=\frac{0.005}{0.3L}=0.017M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B%5Ctext%20%7Bmoles%20left%7D%7D%7B%5Ctext%20%7BTotal%20volume%20in%20L%7D%7D%3D%5Cfrac%7B0.005%7D%7B0.3L%7D%3D0.017M)
Thus molarity of in solution is 0.017 M.
Answer:
= 12 mL H202
Explanation:
Given that, the concentration of H2O2 is given antiseptic = 3.0 % v/v
It implies that, 3ml H2O2 is present in 100 ml of solution.
Therefore, to calculate the amount of H202 in 400.0 mL bottle of solution;
we have;
(3.0 mL/ 100 mL) × 400 mL
= 12 mL H202
Answer:
Neon
Explanation:
1s² 2s² 2p⁶ 3s¹ or [Ne] 3s¹
The outer most shell is the 3s¹.
For this atom to achieve stability, if it loses the electron in the 3s shell, it would resemble an inert element with a complete octet configuration. Therefore, the atom would be like:
1s² 2s² 2p⁶ which is the configuration of Ne
