Answer:
Explanation:
average speed=distance travelled/time taken
distance=speed*time
15*10=150 m
1200 m/min=20 m/s
0.17 minute=10.2 seconds
20*10.2=204 m
2500 cm/s=25 m/s
5000 milliseconds=5 seconds
25*5=125 m
125+150+204=479 m
10.2+5+10=25.2
479/25.2=<u>19.0 </u>
The procedure, which can be used to determine more accurately the concentration of the unknown acid is TO BACK-TITRATE WITH ADDITIONAL HYDROCHLORIC ACID TO NEUTRALIZE THE ADDITIONAL SODIUM HYDROXIDE THAT WAS ADDED.
Monoprotic acids are acids that can donate only one proton per each molecule and they have only one equivalence point. Examples of monoprotic acids are HCI, HNO3 and CH3COOH.
The back titration method is typically used when one needs to determine the concentration of an analyte provided there is a known molar concentration of excess reactants.
From the information given in the question above, we are told that excess NaOH was added. To correct this mistake, the right thing to do is to use additional HCl to carry out back titration, taking note of the quantity of acid that will be needed to neutralize the excess NaOH.
Answer:
The concentration of the most dilute solution is 0.016M.
Explanation:
First, a solution is prepared and then it undergoes two subsequent dilutions. Let us calculate initial concentration:
![[Na_{2}SO_{4}]=\frac{moles(Na_{2}SO_{4})}{liters(solution)} =\frac{mass((Na_{2}SO_{4}))}{molarmass(moles(Na_{2}SO_{4}) \times 0.100L)} =\frac{2.5316g}{142g/mol\times 0.100L } =0.178M](https://tex.z-dn.net/?f=%5BNa_%7B2%7DSO_%7B4%7D%5D%3D%5Cfrac%7Bmoles%28Na_%7B2%7DSO_%7B4%7D%29%7D%7Bliters%28solution%29%7D%20%3D%5Cfrac%7Bmass%28%28Na_%7B2%7DSO_%7B4%7D%29%29%7D%7Bmolarmass%28moles%28Na_%7B2%7DSO_%7B4%7D%29%20%5Ctimes%200.100L%29%7D%20%3D%5Cfrac%7B2.5316g%7D%7B142g%2Fmol%5Ctimes%200.100L%20%7D%20%3D0.178M)
<u>First dilution</u>
We can use the dilution rule:
C₁ x V₁ = C₂ x V₂
where
Ci are the concentrations
Vi are the volumes
1 and 2 refer to initial and final state, respectively.
In the first dilution,
C₁ = 0.178 M
V₁ = 15 mL
C₂ = unknown
V₂ = 50 mL
Then,
<u>Second dilution</u>
C₁ = 0.053 M
V₁ = 15 mL
C₂ = unknown
V₂ = 50 mL
Then,
Answer:
a. increase
Explanation:
Based on the kinetic molecular theory of gases, the average kinetic energy of the system will increase.
- The average kinetic energy is heat
- If temperature increases, heat of a system will also rise.
- According the kinetic molecular theory "the temperature of the gas is a measure of the average kinetic energy of the molecules"
Therefore, due to the increase in temperature, the average kinetic energy of the system increases.
Answer:
![[Fe^{+3}]=0.700 M](https://tex.z-dn.net/?f=%5BFe%5E%7B%2B3%7D%5D%3D0.700%20M)
![[NO_{3}^{-}]=2.10 M](https://tex.z-dn.net/?f=%5BNO_%7B3%7D%5E%7B-%7D%5D%3D2.10%20M)
Explanation:
Here, a solution of Fe(NO₃)₃ is diluted, as the total volume of the solution has increased. The formula for dilution of the compound is mathematically expressed as:
Here, C and V are the concentration and volume respectively. The numbers at the subscript denote the initial and final values. The concentration of Fe(NO₃)₃ is 1.75 M. As ferric nitrate dissociates completely in water, the initial concentration of ferric is also 1.75 M.
Solving for [Fe],
![[Fe^{+3}]=\frac{C_{1}.V_{1}}{V_{2} }](https://tex.z-dn.net/?f=%5BFe%5E%7B%2B3%7D%5D%3D%5Cfrac%7BC_%7B1%7D.V_%7B1%7D%7D%7BV_%7B2%7D%20%7D)
![[Fe^{+3}]=\frac{(1.75).(30.0)}{45.0+30.0 }](https://tex.z-dn.net/?f=%5BFe%5E%7B%2B3%7D%5D%3D%5Cfrac%7B%281.75%29.%2830.0%29%7D%7B45.0%2B30.0%20%7D)
For [NO₃⁻],
There are three moles of nitrate is 1 mole of Fe(NO₃)₃. This means that the initial concentration of nitrate ions will be three times the concentration of ferric nitrate i.e., it will be 5.25 M.
![[NO_{3}^{-}]=\frac{C_{1}.V_{1}}{V_{2} }](https://tex.z-dn.net/?f=%5BNO_%7B3%7D%5E%7B-%7D%5D%3D%5Cfrac%7BC_%7B1%7D.V_%7B1%7D%7D%7BV_%7B2%7D%20%7D)
![[NO_{3}^{-}]=\frac{(5.25)(30.0)}{30.0+45.0 }](https://tex.z-dn.net/?f=%5BNO_%7B3%7D%5E%7B-%7D%5D%3D%5Cfrac%7B%285.25%29%2830.0%29%7D%7B30.0%2B45.0%20%7D)
