For this problem, we use the Beer Lambert's Law. Its usual equation is:
A = ∈LC
where
A is the absorbance
∈ is the molar absorptivity
L is the path length
C is the concentration of the sample solution
As you notice, we only have to find the absorbance. But since we are not given with the molar absorptivity, we will have to use the modified equation that relates % transmittance to absorbance:
A = 2 - log(%T)
A = 2 - log(27.3)
A = 0.5638
Answer:
The answer is 0.75M HCl
Explanation:
To calculate the concentration of 10 ml of HCl that would be required to neutralize 50.0 mL of 0.150 M NaOH, we use the formula:
To calculate the concentration of 10 ml of HCl that would be required to neutralize 50.0 mL of 0.150 M NaOH, we use the formula:
C1V1 = C2V2
C1 = concentration of acid
C2 = concentration of base
V1 = volume of acid
V2 = volume of base
From the information supplied in the question:
concentration of acid (HCl) is the unknown
volume of acid (HCl) = 10ml
concentration of base (NaOH) = 0.15M
volume of base (NaOH) = 50ml
C1 x 10ml = 0.15M x 50ml
C1 x 10 = 7.5
divide both side by 10
C1 = 0.75M
concentration of acid (HCl) is 0.75M
Oregon trail would be the best answer
3 moles FeCl2
Explanation:
For any chemical reaction, the balanced chemical equation tells you the ratio that must always exist between the reactants.
In your case, you have
FeCl2(aq) + 2NaOH(aq) → Fe(OH)2(s) ↓ + 2 NaCl(aq)
The stoichiometric coefficients that belong to iron(II) chloride and to sodium hydroxide, respectively, tell you the mole ratio that must exist between the two reactants when this reaction takes place.
Notice that you have a 1:2 mole ratio between the two reactants, so you can say that the reaction will always consume twice as many moles of sodium hydroxide than moles of iron(II) chloride.
Now, you know that 6 moles of iron(II) chloride are added to 6 moles of sodium hydroxide.
Use the aforementioned mole ratio to determine how many moles of iron(II) chloride will react with the moles of sodium hydroxide
6 moles NaOH ⋅ 1 mole FeCl2
———————
2 moles NaOH
= 3 moles FeCl 2
This tells you that in order for all the moles of sodium hydroxide to react, you need 3 moles of iron(II) chloride. The other 3 moles will not take part in the reaction, i.e. they are in excess.
So, you can say that
3 moles of FeCl 2 → will react
3 moles of FeCl 2 → will not react
Notice that sodium hydroxide is completely consumed before all the moles of iron(II) chloride get the chance to take part in the reaction.
This tells you that sodium hydroxide acts as a limiting reagent, i.e. it limits the amount of iron(II) chloride that takes part in the reaction from 6 moles to 3 moles.
Answer:
the scattering and reflection of light by dust particles
