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Alex
3 years ago
6

If in the following diagram the substance is in solid form during stage 1, what is happening during stage 3?

Chemistry
2 answers:
yulyashka [42]3 years ago
6 0

Answer : The stage 3 shows the liquid state of a substance.

Explanation :

In the given phase diagram,

Stage 1 : It shows the solid state of a substance.

Stage 2 : It shows the melting and freezing process of a substance. It means that when the phase changes from solid state to liquid state at constant temperature is know as melting process and when the phase changes from liquid state to solid state at constant temperature is know as freezing process.

Stage 3 : It shows the liquid state of a substance.

Stage 4 : It shows the evaporation and condensation process of a substance. It means that when the phase changes from liquid state to gaseous state at constant temperature is know as evaporation process and when the phase changes from gaseous state to liquid state at constant temperature is know as condensation process.

Stage 5 : It shows the gaseous state of a substance.

Hence, from the above information we conclude that the stage 3 shows the liquid state of a substance.

Rufina [12.5K]3 years ago
5 0
Its converting from a solid to a liquid.
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Consider the reaction: CH4 + 2O2 = CO2 + 2H2O
crimeas [40]

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The answer to your question is:

a)  80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction                             CH4   +   2O2   ⇒   CO2   +   2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

MW CH4 = 16 g

MW O2 = 32 g

                               16 g of CH4 ----------------  2(32) g of O2

                               20 g              --------------    x

                               x = (20 x 64) / 16 = 80 g of O2

b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .  

                                CH4   +   2O2   ⇒   CO2   +   2H2O

                                15 g         22 g

                                16 g of CH4 ----------------  64 g of O2

                                15 g of CH4  ---------------   x

                               x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

                                 CH4   +   2O2   ⇒   CO2   +   2H2O

                        64 g of O2 ------------------  44 g of CO2

                        22 g of O2 ------------------   x

                        x = (22 x 44)/ 64 = 15. 13 g of CO2

c. For the reaction CH4 + 2O2 = CO2 + 2H2O, if you have 10.31 g. of CH4 and an unknown amount of oxygen, and form 20.00 g. of CO2, i. Identify if there is a limiting reactant ______________ ii. Calculate the number of grams of the limiting reactant present if there is one. ______________  

                           CH4   +   2O2   ⇒   CO2   +   2H2O                              

                           10.31 g                     20 g

We can identify the limiting reactant if we know the quantity of the reactants, if we only know the quantity of one it is not posible to which is the limiting reactant.

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