Answer:
<em> = 0.2 mL.</em>
Explanation:
Given a 0.5 M solution of NaOH as stock solution, 10.0mL of 0.010M can be prepared via dilution with distilled water, by using the formula:
where C1 and V1 are initial concentration and volume respectively; same as C2 & V2 for fina.
Let C1 = 0.5M, V2 = ?
C2 = 0.010M; V2 = 10mL
⇒Volume of stock solution to be diluted, V2
= 
× 0.010
<em> = 0.2 mL.</em>
Glasswares used would be pipette (for smaller volume experiment) and measuring cylinder. 0.2mL would be measured and then made upto the 10mL mark of the measuring cylinder.
I hope this was a detailed explanation given the missing details of "Trial 1" in the question.
Answer:
=3 means is 3 or greater so that would be f and g subshells
=0 means is 0 or greater so that would be s, p, d, f and g subshells
=1 means is 1 or greater so that would be p, d, f, and g subshells
=4 means is 4 or greater so that would be g only
Answer:
See explanation.
Explanation:
Hello,
In this case, we could have two possible solutions:
A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:
That is the mass of copper (II) sulfate contained in 1 mol of substance.
B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:
So you can solve for the moles of the solute:
Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:
But this is just a supposition.
Regards.
Answer:
B. as a food preservative in the manufacture of detergents
.........
Calculate the ratio by using Henderson-Hasselbalch equation:
pH = pKa + log [neutral form] / Protonated form
3.05 = 2.21 + log [neutral form] / [Protonated form]
3.05 - 2.21 = log [neutral form] / [Protonated form]
0.84 = log [neutral form] / [Protonated form]
[neutral form] / [protonated form] = anti log 0.84 = 6.91
