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Vedmedyk [2.9K]
2 years ago
12

समश्याजिक क्या है हाइड्रोजन का समस्थानियको समपस्य?​

Chemistry
1 answer:
LUCKY_DIMON [66]2 years ago
8 0

Answer:

GO

Explanation:

You might be interested in
Write three complete and balanced pairs of below mention electrochemical half reactions. For each pair of reactions, identify wh
kirza4 [7]

Answer: a) Cu\rightarrow Cu^{2+}+2e^-  : anode

b. 2H^++2e^-\rightarrow H_2 : cathode

c. O^{2-}\rightarrow \frac{1}{2}O_2+2e^-:  anode

Explanation:

Electrochemical cell is a device which converts chemical energy into electrical energy. It consist of two electrodes, anode and cathode.

Oxidation i.e. loss of electrons , which results in an increase in oxidation number occurs over anode.

Reduction i.e. gain of electrons, which results in decrease in oxidation number occurs over cathode.

a. Cu\rightarrow Cu^{2+}+2e^-  : oxidation :  anode

b. 2H^++2e^-\rightarrow H_2 : reduction : cathode

c. O^{2-}\rightarrow \frac{1}{2}O_2+2e^-: oxidation : anode

6 0
3 years ago
50.00 mL of 0.10 M HNO 2 (nitrous acid, K a = 4.5 × 10 −4) is titrated with a 0.10 M KOH solution. After 25.00 mL of the KOH sol
boyakko [2]

Answer:

b. 3.35

Explanation:

To calculate the pH of a solution containing both acid and its salt (produced as a result of titration) we need to use Henderson’s equation i.e.

pH = pKa + log ([salt]/[acid])     (Eq. 01)

Where  

pKa = -log(Ka)        (Eq. 02)

[salt] = Molar concentration of salt produced as a result of titration

[acid] = Molar concentration of acid left in the solution after titration

Let’s now calculate the molar concentration of HNO2 and KOH considering following chemical reaction:

HNO2 + KOH ⇆ H2O + KNO2    (Eq. 03)

This shows that 01 mole of HNO2 and 01 mole of KOH are required to produce 01 mole of KNO2 (salt). And if any one of them (HNO2 and KOH) is present in lower amount then that will be considered the limiting reactant and amount of salt produced will be in accordance to that reactant.

Moles of HNO2 in 50 mL of 0.01 M HNO2 solution = 50/1000x0.01 = 0.005 Moles

Moles of KOH in 25 mL of 0.01 M KOH solution = 25/1000x0.01 = 0.0025 Moles

As it can be seen that we have 0.0025 Moles of KOH therefore considering Eq. 03 we can see that 0.0025 Moles of KOH will react with only 0.0025 Moles of HNO2 and will produce 0.0025 Moles of KNO2.

Therefore

Amount of salt produced i.e [salt] = 0.0025 moles       (Eq. 04)

Amount of acid left in the solution [acid] = 0.005 - 0.0025 = 0.0025 moles (Eq.05)

Putting the values in (Eq. 01) from (Eq.02), (Eq. 04) and (Eq. 05) we will get the following expression:

pH= -log(4.5x10 -4) + log (0.0025/0.0025)

Solving above we get  

pH = 3.35

5 0
3 years ago
Can someone help me
Sphinxa [80]

Answer: 0.24

Explanation:

M=\frac{moles}{L}\\ M=\frac{0.6 moles}{2.5L} \\M=0.24

4 0
3 years ago
Find the number of Grams
zaharov [31]

Answer: 462 g

Explanation:molar mass is M= 63.55 +2·(12.01+14.01)= 115.59 g/mol.

Mass m= n·M = 4.0 mol·115.59 g/mol= 462.36 g

8 0
2 years ago
Acids naturally present in food are safe to eat because they usually are
tangare [24]
A.Weak

B.concentrated

C.dilute

D.strong



Answer: A - Weak

7 0
2 years ago
Read 2 more answers
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