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d1i1m1o1n [39]
3 years ago
6

You have seven closed containers, each with equal masses of chlorine gas (cl2). you add 10.0 g of sodium to the first sample, 20

.0 g of sodium to the second sample, and so on (adding 70.0g of sodium to the seventh sample). sodium and chlorine react to form sodium chloride according to the equation: 2na(s) + cl2(g)  2nacl(s) after the reaction is complete, you collect and measure the amount of sodium chloride
Chemistry
1 answer:
Phoenix [80]3 years ago
7 0

The reaction between Na and Cl2 is

2Na + Cl2 ---> 2NaCl

So two moles of Na will react with one mole of Cl2

However here in the given question the moles of Cl2 is constant as mass is constant. Say we have 71 of Cl2 in each container, the moles = 71 / 35.5 = 2 moles of Cl2

Moles of Na = Mass / atomic mass

The container will have following moles of Na

1 ) Moles = 10 / 23 = 0.435  (these will react with 0.2175 moles of Cl2 to give 0.435 moles of NaCl]

2) moles = 20 / 23 = 0.87 ( (these will react with 0.435 moles of Cl2 to give 0.87 moles of NaCl]

3) moles = 30 / 23 = 1.3 ( (these will react with 0.65 moles of Cl2 to give 1.3 moles of NaCl]

4) moles = 40 / 23 = 1.74 (these will react with 0.87 moles of Cl2 to give 1.7 moles of NaCl]

5) moles = 50 / 23 = 2.18 (these will react with 1.09 moles of Cl2 to give 2.18 moles of NaCl]

6) moles = 60 / 23 = 2.61  (these will react with 1.3 moles of Cl2 to give 2.61 moles of NaCl]

7) moles = 70 / 23 =  3 moles  (these will react with 1.5 moles of Cl2 to give 3 moles of NaCl]

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What was the key design change for hfc-134a a/c systems versus CFC 12 a/c systems
BabaBlast [244]
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For the reaction shown, identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent. KNO3 →
Vinvika [58]

Answer : The oxidizing element is N and reducing element is O. 

KNO_{3} is act as an oxidizing agent as well as reducing agent.

Explanation :

An Oxidizing agent is the agent which has ability to oxidize other or a higher in oxidation number.

Reducing agent is the agent which has ability to reduce other or lower in oxidation number.

The given reaction is :

KNO_{3} \rightarrow KNO_{2} +O_{2}

KNO_{3}  act as an oxidizing agent.

The oxidation number of N in KNO_{3} is calculated as:

(+1)+(x)+3(-2) = 0

x = +5

And the oxidation number of N in KNO_{2}  is calculated as:

(+1)+(x)+2(-2) = 0

x = +3

From the oxidation number method, we conclude that the oxidation number  reduced this means KNO_{3} itself get reduced to KNO_{2} and it can act as an oxidizing agent.

KNO_{3}  act as a reducing agent.

KNO_{3} \rightarrow KNO_{2} +O_{2}

The oxidation number of O in KNO_{3} is calculated as:

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x = -2

The oxidation number of O in O_{2} is Zero (o).

Now, we conclude that the oxidation number increases this means KNO_{3} itself get oxidized to O_{2} and it can act as reducing agent.





                     

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