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Dmitriy789 [7]
3 years ago
5

Without doing any calculations, arrange the elements in CF2Cl2 in order of decreasing mass percent composition. Rank from highes

t percent to lowest.
a. C > F > Cl
b. F < Cl > C
c. Cl > C > F
d. Cl > F > C
Chemistry
1 answer:
Kitty [74]3 years ago
3 0

Answer:

a. C > F > Cl

Explanation:

We know that atomic mass of Chlorine is greater than of  Florine than that of carbon. Moreover, in CF2Cl2, therefore, there are two atoms of Cl, F and one atom of C. Therefore, in CF2Cl2 in order of decreasing mass percent composition C > F > Cl. Therefore, the correct option is a.

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Given what we know about ratios, we can confirm that in comparison to the iron pan, the aluminum pan will receive twice as much heat.

<h3>Why would the aluminum pan receive double the heat?</h3>
  • This has to do with the ratio being presented.
  • A ratio of 2:1 means that for every unit of heat to the iron, the aluminum receives 2.
  • In other words, the aluminum pan receives twice as much heat when compared to the iron pan.

Therefore, we can confirm that because the ratio of heat to the aluminum in comparison to the iron pan is 2:1, this means that for every 1 unit of heat to the iron pan, the aluminum pan will receive 2. This results in double the total heat received by the aluminum pan.

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3 years ago
How many grams of K2O will be produced from 0.50 g of K<br> and 0.10 g of O2?
Rudik [331]

Answer:

0.6g

Explanation:

Given parameters:

Mass of K = 0.5g

Mass of O₂  = 0.10g

Unknown:

Mass of K₂O  = ?

Solution:

To solve this problem, let us write the reaction equation first;

                   4K   +   O₂     →     2K₂O

The reaction above delineates the balanced chemical reaction.

To solve this problem, we need to know the limiting reactant. This reactant is the one that determines the amount and extent of the reaction because it is given in short supply. The other reactant is the one in excess.

Start off by find the number of moles of the reactant;

     Number of moles =  \frac{mass}{molar mass}

         Molar mas of K  = 39g/mol

          Molar mass of O₂   = 2(16) = 32g/mol

 Number of moles of K  = \frac{0.5}{39}   = 0.013moles

 Number of moles of O₂    = \frac{0.1}{32}   = 0.031moles

From the balanced reaction;

          4 moles of K reacted with 1 mole of O₂

         0.013 moles of K will react with \frac{0.013}{4}   = 0.0078 moles of O₂

We see that oxygen gas is in excess. We were given 0.031moles of the gas but only require 0.0078moles of oxygen gas.

The limiting reactant is potassium.

    therefore;

              4 moles of K produced 2 moles of K₂O

             0.013 moles of K will produce \frac{0.013 x 2}{4}   = 0.0065‬moles of K₂O

to find the mass of K₂O;

   Mass of K₂O  = number of moles x molar mass

                Molar mass of K₂O  = 2(39) + 16  = 94g/mol

  Mass of K₂O = 0.0065 x 94  = 0.6g

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