True
After the process of Ore processing, Enrichment, Fuel production and being passed through the reactor core the last remaining step is spent fuel disposal.
Answer:
An emulsion is formed.
Explanation:
An association (emulsion) of two liquids is formed, in this case oil and vinegar, which when stirred, said mixture will separate.
Answer:
21 g/mL
Explanation:
To solve this problem, first look at the density equation, which is D=M/V, which D stands for density, M stands for mass, and V stands for volume. When you substitute in the variables, you get D=17.5/.82, which is equivalent to 21.34. However, since we need to pay attention to the sig fig rules for multiplying, we need to have the same amount of sig figs as the value with the least amount of sig figs, which is the number .82. .82 has two sig figs, so you round down. Your answer will be 21 g/mL.
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.
