Why is silicon used to make PV cells

1 answer:

3 0

Silicon is used in producing Pv celss because cells made from the crystallized polycrystalline silicon have efficiencies in the range of 14-18% and a moderate price.

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What elements are found in potassium oxide

GaryK [48]

Elements that are found in potassium oxide is oxygen.

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What is the concentration of an NaOH solution if 80.0 mL of 0.950M HCl acid is required to neutralize 50.0 mL of the base soluti

iogann1982 [59]

The question requires us to calculate the concentration of a NaOH solution, given the amount and concentration of HCl required to neutralize 50.0 mL of this base solution.

The following information was provided by the question:

concentration of HCl solution = C(HCl) = 0.950 M = 0.950 mol/L

volume of HCl solution = V(HCl) = 80.0 mL

volume of NaOH solution = V(NaOH) = 50.0 mL

To solve this problem, we need to understand what happens when NaOH and HCl react. A neutralization reaction occurs between a strong acid, such as HCl, and a strong base, such as NaOH, where the amount of H+ and OH- ions in solution are equal. We can write their reaction as:

NaOH + HCl -> NaCl + H2O

When the base is completely neutralized by the acid, it means that:

number of moles of NaOH = number of moles of HCl

The equality above is what we'll use to start our calculations.

Another important information to this question is that, by definition, the molar concentration is the number of moles of a compound divided by the volume of the solution:

$\text{molarity = }\frac{number\text{ of moles (mol)}}{\text{volume (L)}}\to\text{ M=}\frac{n}{V}$

Since we have the equality between the number of moles of acid and base, we can rewrite the equation above as:

$n=M\times V$

and use this to calculate the molar concentration (M) of NaOH.

Thus, so far we have that:

$n_{NaOH}=n_{HCl}\to M_{NaOH}\times V_{NaOH}=M_{HCl}\times V_{HCl}$

Since the volume of NaOH, molarity of HCl and volume of HCl were provided by the question, we can rearrange the equation above to calculate the molarity of NaOH:

$M_{NaOH}=\frac{M_{HCl}\times V_{HCl}}{V_{NaOH}}$

And, at last, we can apply the values provided by the problem (note that the volume here is being used in mL instead of L; this is fine as long as the volume of both solutions, acid and base, are used with the same unit):