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3 years ago

Calculate the amount of F.A.S required to prepare 1000 ml of 0.1 molar standard solution of F.A.S

Chemistry

1 answer:

RideAnS [48]3 years ago

8 0

Answer:

Preparation and Standardization of 0.1 M Ferrous Ammonium...

Dissolve 40 g of ferrous ammonium sulfate in a previously cooled mixture of 40 ml of sulphuric acid and 200 ml of water.

Dilute with sufficient freshly boiled and cooled water to produce 1000 ml.

Standardize the solution in the following manner.

Explanation:

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3 years ago

According to newtons third law forces always occur in equal but____ pairs?

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'Every action has an equal but opposite reaction' They are equal, but opposite.

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3 years ago

Which statement best relates the strength and concentration of a base? At high enough concentrations, a weak base becomes strong

Lorico [155]

Answer: Option (b) is the correct answer.

Explanation:

When there are more number of hydroxide ions in a solution then there will be high concentration of $OH^{-}$ or hydroxide ions. As a result, more will be the strength of base in that particular solution.

A base is strong when it readily dissociate into its ions in the solution. When a base is strong, then it does not matter at what concentration it is dissolved in the solution because despite of its low concentration it will remain a strong base.

Thus, we can conclude that out of the given options, the statement even at low concentrations, a strong base is strong best relates the strength and concentration of a base.

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3 years ago

Read 2 more answers

Will the calculated Molarity of NaOH be too high or too low or unaffected if the following happen: when you answer the question,

vodomira [7]

Answer:

Explanation:

The result will be affected.

The mass of KHP weighed out was used to calculate the moles of KHP weighed out (moles = mass/molar mass).

Not all the sample is actually KHP if the KHP is a little moist, so when mass was used to determine the moles of KHP, a higher number of moles than what is actually present would be obtained (because some of that mass was not KHP but it was assumed to be so. Therefore, there is actually a less present number of moles than the certain number that was thought of.

During the titration, NaOH reacts in a 1:1 ratio with KHP. So it was determined that there was the same number of moles of NaOH was the volume used as there were KHP in the mass that was weighed out. Since there was an overestimation in the moles of KHP, then there also would be an overestimation in the number of moles of NaOH.

Thus, NaOH will appear at a higher concentration than it actually is.

7 0

3 years ago

1 C8H10(l) +21/2O2(g) → 8CO2(g) + 5H2O(g), Hcomb= ? Hf for C8H10(l) = +49.0kJ/mol C8H10(l) Use the balanced combustion reaction

nikklg [1K]

Answer:

$H_{comb}=-4406kJ/mol$

Explanation:

Hello,

In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

$H_{comb}=8*\Delta _fH_{CO_2}+5\Delta _fH_{H_2O}-\Delta _fH_{C_8H_{10}}$

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

$H_{comb}=8*(-393.5kJ/mol)+5(-241.8kJ/mol)-49.0kJ/mol\\\\H_{comb}=-4406kJ/mol$

Best regards.

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3 years ago