Answer:
One mole of the hydrochloric acid requires one mole of hydroxide ions to be neutralized.
Explanation:
Hydrochloric acid is a very strong acid an it fully dissociates at all concentrations. The neutralization of hydrochloric acid is conventional and any inorganic base such as sodium hydroxide can be used. Having a one molar hydrochloric acid solution means that for each mole of hydrochloric acid produces one mole of protons requiring one mole of hydroxide hydroxide ions to neutralize.
This can be illustrated by the equation shown below.
I think it’s 5000 because it says that their is over 4,124 valid species of minerals.
Answer:
C.
Explanation:
The electronic configuration of N (7 electrons): 1s² 2s² 2p³.
The orbital 1s is filled with two electrons and their spinning direction is opposite and also electrons of 2s.
3p contains (3 electrons) should fill the 3 orbitals firstly. Every orbital contains 1 electron and be in the same spin direction.
So, the right choice is c.
A is wrong because 2 electrons of 3p are paired in the first orbital before filling every orbital.
B is wrong because the 2 electrons of 1s and 2s are in the same direction and also 2 electrons of 3p are paired in the first orbital before filling every orbital.
D is also wrong the 2 electrons of 1s and 2s are in the same direction and the electron in the second orbital of 3p are in opposite direction of the other 2 electrons.
We know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl
Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.
To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH
Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572. That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH
Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH
Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
the original mixture was 16% KOH
I hope this helps.
None of the alpha particles fired at the foil are being repelled back, like they were in the Rutherford atom simulation.I hope this correct.
