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DochEvi [55]
3 years ago
10

0.5 gm of mixture of NH4Cl and NaCl was boiled with 25 ml of 0.95 N NaOH in a vessel till all the ammonia is expelled.The residu

al solution is cooled and then made up to 100 ml.It was found that 10 ml of the diluted residual solution neutralised by 16 ml of 0.1 N (F=1.06)H2SO4. Calculate the % of pure NH4Cl in the mixture.
Ans:72.66%​
Chemistry
1 answer:
GalinKa [24]3 years ago
8 0

Answer:

72.66%

Explanation:

NH₄Cl reacts in presence of NaOH producing ammonia, NH₃, as follows:

NH₄Cl + NaOH → NaCl + NH₃ + H₂O

The residual NaOH reacts with H₂SO₄ as follows:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.

The equivalent-gram of H₂SO₄ are:

16mL * 0.1N * 1.06 = 1.696mEq.

As the complete residual solution is 100mL but the neutralization was made only with 10mL, the mEq you need to neutralize the residual NaOH is:

1.696mEq * (100mL / 10mL) = 16.96mEq.

The mEq of NaOH you add in the first are:

25mL * 0.95mEq = 23.75mEq

That means the NaOH that reacts = moles of NH₄Cl is:

23.75mEq - 16.96mEq = 6.79mEq = 6.79mmoles NH₄Cl =

6.79x10⁻³ moles NH₄Cl

In grams (Using molar mass NH₄Cl = 53.5g/mol):

6.79x10⁻³ moles NH₄Cl * (53.5g / mol) =

0.3633g of NH₄Cl are in the original mixture.

% is:

0.3633g/ 0.5g * 100 = 72.66%

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Which postulate of Dalton's atomic theory was later proven wrong?
ruslelena [56]

Answer:

Option c and d

Explanation:

John Dalton. In 1808, John Dalton proposed a theory known as Dalton’s Atomic Theory. The theory was published in a paper titled “A New Chemical Philosophy”. This theory was new to that era

The 5 postulates of Daltons' atomic theory are:

1. All the matters are made of atoms.

2. Atoms of different elements combine to form compounds

3. Compounds contain atoms in small whole-number ratios

4. Atoms can neither be created nor destroyed . (This was later proven wrong )

5. All atoms of an element are identical and have the same properties (This was later proven wrong as atoms of same element may be different in case of elements having isotopes )

Therefore, options c and d are the answer.

6 0
3 years ago
Can Someone help me pls?
Ilia_Sergeevich [38]

Answer:

For H3O concentration you do 10^-pH so if pH is 5 then H3O+ is 10^-5= 1*10^-5 H3O+ ions

For OH is one extra step. First find H3o+ ions using equation above then you have to use that to divide 1*10^-14

So if pH is 5....the H3O+ is 1*10^-5 then OH- = (1*10^-14)/(1*10^-5) =  1*10^-9 OH ions

as far as acid/base pH 0-6 is Acid 8-14 is Base. pH of 7 is neutral. Recheck your work *hint* *hint* water is neutral. Spit is above 7 so is base.

8 0
3 years ago
Here is my question..
OverLord2011 [107]
I would say 3.0 cause yeah yeah yeah yeah I’m iann Dior
7 0
3 years ago
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If 73.5 mL of 0.200 M KI(aq) was required to precipitate all of the lead(II) ion from an aqueous solution of lead(II) nitrate, h
frutty [35]

Answer:

There were 0.00735 moles Pb^2+ in the solution

Explanation:

Step 1: Data given

Volume of the KI solution = 73.5 mL = 0.0735 L

Molarity of the KI solution = 0.200 M

Step 2: The balanced equation

2KI + Pb2+ → PbI2 + 2K+

Step 3: Calculate moles KI

moles = Molarity * volume

moles KI = 0.200M * 0.0735L = 0.0147 moles KI

Ste p 4: Calculate moles Pb^2+

For 2 moles KI we need 1 mol Pb^2+ to produce 1 mol PbI2 and 2 moles K+

For 0.0147 moles KI we need 0.0147 / 2 = 0.00735 moles Pb^2+

There were 0.00735 moles Pb^2+ in the solution

3 0
3 years ago
What is the volume of 0.04 mol of a gas at 0.9 kPa and 20°C?
Annette [7]
A. 108 L is the winner!
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