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2 years ago

A 52.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 26.0 mL of KOH at 25 ∘C.

1 answer:

7 0

The balanced equation for the above reaction is
HBr + KOH ---> KBr + H₂O
stoichiometry of HBr to KOH is 1:1
HBr is a strong acid and KOH is a strong base and they both completely dissociate.
The number of HBr moles present - 0.25 M / 1000 mL/L x 52.0 mL = 0.013 mol
The number of KOH moles added - 0.50 M / 1000 mL/L x 26.0 mL = 0.013 mol
the number of H⁺ ions = number of OH⁻ ions
therefore complete neutralisation occurs.
Therefore solution is neutral. At 25 °C, when the solution is neutral, pH = 7.
Then pH of solution is 7

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A vessel contained N2, Ar, He, and Ne. The total pressure in the vessel was 987 torr. The partial pressures of nitrogen, argon,

xz_007 [3.2K]

Answer:

The partial pressure of neon in the vessel was 239 torr.

Explanation:

In all cases involving gas mixtures, the total gas pressure is related to the partial pressures, that is, the pressures of the individual gaseous components of the mixture. Put simply, the partial pressure of a gas is the pressure it exerts on a mixture of gases.

Dalton's law states that the total pressure of a mixture of gases is equal to the sum of the pressures that each gas would exert if it were alone. Then:

PT= P1 + P2 + P3 + P4…+ Pn

where n is the amount of gases present in the mixture.

In this case:

PT=PN₂ + PAr + PHe + PNe

where:

PT= 987 torr
PN₂= 44 torr
PAr= 486 torr
PHe= 218 torr
PNe= ?

Replacing:

987 torr= 44 torr + 486 torr + 218 torr + PNe

Solving:

987 torr= 748 torr + PNe

PNe= 987 torr - 748 torr

PNe= 239 torr

<u><em>The partial pressure of neon in the vessel was 239 torr.</em></u>

4 0

2 years ago

If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is

mojhsa [17]

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a

compound.

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

Mn O

% composition 72.1 27.9

Divide by mass number 54.94 16

1.31 1.74

Divide by the smallest number 1.31 1.31

1 1.3

The resulting ratio is 1:1

Hence the Empirical formula is MnO, Manganese oxide

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2 years ago

A 8.5-liter sample of a gas at 2.0 atm and 300.0 K has 1.2 moles of the gas. If 0.65 mole of the gas is added to the sample at t

hodyreva [135]

Did you take the test? what was the answer Im stuck on this one too

7 0

1 year ago

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A white powder is added to a solution. The images show observations made before the powder is added, just after the powder has b

7nadin3 [17]

There was a change in its color from white to red which can only be changed by a chemical reaction

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2 years ago

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6. From the values of ΔH and ΔS, predict which of the following reactions would be spontaneous at 25ºC: Reaction A: ΔH = 10.5 kJ

qwelly [4]

Answer:

Both reaction A and reaction B are non spontaneous.

Explanation:

For a spontaneous reaction, change in gibbs free energy ( $\Delta G$ ) should be negative.

We know, $\Delta G=\Delta H-T\Delta S$ , where T is temperature in Kelvin scale.

Reaction A: $\Delta G=(10.5\times 10^{3})-(298\times 30)J/mol=1560J/mol$

As $\Delta G$ is positive therefore the reaction is non-spontaneous.

If at a temperature T K , the reaction is spontaneous then-

$\Delta H-T\Delta S< 0$

or, $T> \frac{\Delta H}{\Delta S}$

or, $T> \frac{10.5\times 10^{3}}{30}$

or, $T> 350$

So at a temperature greater than 350 K, the reaction is spontaneous.

Reaction B: $\Delta G=(1.8\times 10^{3})-(-113\times 298)J/mol=35474J/mol$

As $\Delta G$ is positive therefore the reaction is non-spontaneous.

If at a temperature T K , the reaction is spontaneous then-

$\Delta H-T\Delta S< 0$

or, $T> \frac{\Delta H}{\Delta S}$

or, $T> \frac{1.8\times 10^{3}}{-113}$

or, $T> -16$

So at a temperature greater than -16 K, the reaction is spontaneous.

3 0

2 years ago