Chemistry, 14.07.2020 21:01, zhvianna

sodium azide nan3 is dissolved in water. acidified silver nitrate is added to the solution and a white precipitate forms. aqueous ammonia is then added to the solution and a white precipitate forms. aqueous ammonia is then added to the white precipitate. the azide ion n3- has similar chemical properties to the cl- ion. what is the formula of the white ppt and what is observed on adding aqueous ammonia?

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answered: Guest

1. and 2. question: in which equation? or i don't understand you, 3. a), 4. c), 5. c), 6. b), 7. d), 8. d) if small number is underlined(behind symbol) , b) if big number is underlined(in front of symbol)9. what is underlined there? 10. c)

answered: Guest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

answered: Guest

Because it was a dramatically big guy

answered: Guest

when the volume of gas is changed from 717.73 ml to 852 ml the temperature will change from 315 c to 452 c

calculation

this is calculated using the charles law formula

that is, v1/t1 = v2/t2

v1=?

t1 = 315 c into kelvin = 315 +273 = 588 k

v2 = 852 ml

t2 = 425c + 273 =698 k

make v1 the subject of the formula ( by multiplying both side by t1)

v1=( v2 xt1) / t2

v1= (852 ml x 588 k)/ 698 k= 717.73 ml

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sodium azide nan3 is dissolved in water. acidified silver nitrate is added to the solution and a whi...