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.
the answer is heat. while a car is in idol, the tailpipe gets very hot, (motorcycle, car, etc.) this also produces h20 which you can see dripping out of the tailpipe.
Silver (Ag) is the number of atoms per unit cell for each metal. Silver has a face-centred cubic (FCC) unit cell structure, where there are 8 corner atoms and 6 atoms on the faces, so there are a total of 4 atoms per unit cell.
The identical unit cells are defined in such a way that they take up space without touching one another. A crystal's internal 3D arrangement of atoms, molecules, or ions is known as its lattice. It consists of a large number of unit cells. Every point of the lattice is occupied by one of the three component particles.
Primitive cubic, body-centred cubic (BCC), and face-centred cubic are the three types of unit cells (FCC). The three different sorts of unit cells will be thoroughly covered in this section.
To learn more about the unit cell refer here:
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we are given the the two reactants: AgNO3 and Na2CO3 and is asked to write a balanced equation and a net ionic equation for the reaction of the two. This is a double-replacement reaction:
2AgNO3 (aq)+ Na2CO3 (aq)= Ag2CO3 + 2NaNo3 (aq)
2 Ag + + 2 N03- + 2Na+ + CO32- = Ag2CO3 + 2 Na+ 2NO3-
cancelling the spectator ions, 2Ag + + CO32- = Ag2CO3