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Vitek1552 [10]
3 years ago
7

Calculate the molecular weight of a dibasic acid.0.56gm of which is required 250ml of N/20 sodium hydroxide solution for neutral

ization.​
Chemistry
1 answer:
Alekssandra [29.7K]3 years ago
7 0

<u>Answer: </u>The molecular weight of the dibasic acid is 89.6 g/mol

<u>Explanation:</u>

Normality is defined as the amount of solute expressed in the number of gram equivalents present per liter of solution. The units of normality are eq/L. The formula used to calculate normality:

\text{Normality}=\frac{\text{Given mass of solute}\times 1000}{\text{Equivalent mass of solute}\times \text{Volume of solution (mL)}}      ....(1)

We are given:

Normality of solution = \frac{1}{20}=0.05N

Given mass of solute = 0.56 g

Volume of solution = 250 mL

Putting values in equation 1, we get:

0.05=\frac{0.56\times 1000}{\text{Equivalent mass of solute}\times 250}\\\\\text{Equivalent mass of solute}=\frac{0.56\times 1000}{0.05\times 250}=44.8g/eq

Equivalent weight of an acid is calculated by using the equation:

\text{Equivalent weight}=\frac{\text{Molar mass}}{\text{Basicity}}   .....(2)

Equivalent weight of acid = 44.8 g/eq

Basicity of an acid = 2 eq/mol

Putting values in equation 2, we get:

44.8g/eq=\frac{\text{Molar mass}}{2eq/mol}\\\\\text{Molar mass}=(44.8g/eq\times 2eq/mol)=89.6g/mol

Hence, the molecular weight of the dibasic acid is 89.6 g/mol

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7 0
3 years ago
How many moles of oxygen are formed when 58.6 g of KNO3 decomposes according to the following reaction? 4 KNO3(s) → 2 K2O(s) + 2
Leya [2.2K]

Answer:

0.725 mol

Explanation:

Moles are calculated as the given mass divided by the molecular mass.

i.e. ,

moles = ( mass / molecular mass )

since,

mass of KNO₃ = 58.6 g  ( given )

Molecular mass of KNO₃ = 101 g / mol

Therefore,

moles of KNO₃ = 58.6 g / 101 g / mol

moles of KNO₃ = 0.58 mol

From the balanced reaction ,

4 KNO₃ (s) ---> 2K₂O (s) + 2N₂ (g) + 5O₂ (g)

By the decomposition of 4 mol of KNO₃ , 5 mol of O₂ are formed ,

hence, unitary method is used as,

1  mol of KNO₃  gives 5 / 4 mol O₂

Therefore,

0.58 mol of KNO₃ , gives , 5 / 4  * 0.58 mol of O₂

Solving,

0.58 mol of KNO₃ , gives , 0.725 mol of O₂

Therefore,

58.6g of KNO₃ gives 0.725 mol of O₂.

3 0
3 years ago
explain using diagrams how potassium forms the compound potassium flouride when it reacts with flourine
Katarina [22]

Answer:

The answer to your question is given below.

Explanation:

Potassium (K) has 19 electrons with electronic configuration of 2, 8, 8, 1.

Fluorine (F) has 9 electrons with electronic configuration of 2, 7.

Fluorine needs 1 electron to complete it's octet configuration.

Hence, potassium (K), will lose 1 electron to fluorine (F) to form potassium ion (K+) with electronic configuration of 2, 8, 8. The fluorine atom (F) will receive the 1 electron from potassium to form the fluoride ion (F-) with electronic configuration of 2, 8.

**** Please see attached photo for further details.

7 0
3 years ago
an unknown sample required 21.05 mL of 0.02047 M KmnO4 to reach the end point. How many moles of KmnO4 reacted?
Angelina_Jolie [31]
.02047 = x / 0.02105 = 0.0004309mol
6 0
3 years ago
A 15.00 % by mass solution of lactose (C 12H 22O 11, 342.30 g/mol) in water has a density of 1.0602 g/mL at 20°C. What is the mo
Alik [6]

Answer:

22.82M

Explanation:

342.3g/mol is présent in 1000

what about in 15??

( 342.3g/mol × 1000 ) ÷ 15

3 0
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