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Dahasolnce [82]
2 years ago
11

According to the law of multiple proportions, if 12 g of carbon combine with 16 g of oxygen to form co, the number of grams of c

arbon that combine with 16 g of oxygen in the formation of co2 is ________.
Chemistry
1 answer:
scZoUnD [109]2 years ago
4 0

The amount of carbon combined with oxygen in order to form Carbon Dioxide(CO₂)  is 12 grams

The Law of Multiple proportions is one of the basic laws in chemistry along with the Law of Definite proportion. This law was proposed by John Dalton in 1803. The law states that if two elements react to form one or more compounds then the ratio of masses of both the elements is in the ratio of small whole numbers.

It is known that Carbon can form two different compounds while reacting with Oxygen that is: Carbon Monoxide, Carbon Dioxide. For Carbon Monoxide, 12g of Carbon needs to react with 16g of Oxygen. Whereas, In Carbon Dioxide, 12g of Carbon needs to react with 32g of Oxygen. So the mass of Carbon is equal in both cases is 12g.

                      C(12g)   +   O(16g)   ⇒  CO   (Carbon Monoxide)    

                      C(12g)   +  2O(32g) ⇒  CO₂  (Carbon Dioxide)

To know more about the Law of Multiple Proportion, Click here :

brainly.com/question/28458716

#SPJ4    

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The answer is Ga.

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4 0
4 years ago
Read 2 more answers
The yellow precipitate formed in the reaction between KI and Pb(NO3)2 is . b. The white precipitate formed in the reaction betwe
Thepotemich [5.8K]

Answer:

a. Here, the yellow precipitate is lead (II) iodide:

2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)

b. Here the white precipitate is barium sulfate:

BaCl(aq)+H_2SO_4(aq)\rightarrow BaSO_3(s)+2HCl(aq)

c. Here the brown precipitate is iron (III) hydroxide:

3NaOH(aq)+FeCl_3(aq)\rightarrow 3NaCl(aq)+Fe(OH)_3(s)

d. Here, the precipitate is copper (II) hydroxide:

CuSO_4(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+Na_2SO_4(aq)

Explanation:

Hello.

In this case, since the precipitation reactions are characterized by the formation of a product which is highly insoluble in water, specially very heavy products having heavy metals at the cations, for each case we proceed as follows:

a. Here, the yellow precipitate is lead (II) iodide:

2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)

b. Here the white precipitate is barium sulfate:

BaCl(aq)+H_2SO_4(aq)\rightarrow BaSO_3(s)+2HCl(aq)

c. Here the brown precipitate is iron (III) hydroxide:

3NaOH(aq)+FeCl_3(aq)\rightarrow 3NaCl(aq)+Fe(OH)_3(s)

d. Here, the precipitate is copper (II) hydroxide:

CuSO_4(aq)+2NaOH(aq)\rightarrow Cu(OH)_2(s)+Na_2SO_4(aq)

Best regards!

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Hello!

answer : c

eg = octahedral, mg = square planar, sp3d2


Hope that helps!

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