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3 years ago

A phase change from a solid to a gas is called: deposition freezing boiling sublimation

1 answer:

3 0

I’M SO HAPPY!
I’M LEARNING THIS, TOOOOO!

So, you know
Solid - Liquid = Melting
Liquid - Gas = Vaporization (Boiling)
Gas - Liquid = Condesation
Liquid - Solid = Freezing

What you need to know is
Solid - Gas
Gas - Solid

Sublimation is the process of solid to a gas
Deposition is the process of a gas to a solid

Think of Sublimation as a promotion directly from a solid to a gas
Think of Deposition as going down from the high rank of Gas to a Solid

Both Sublimation and Deposition can be used in an example with Nitrogen

The answer is Sublimation
P.S. Sorry I typed a lot to explain

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4 years ago

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m

Bond [772]

Answer:

$m_{H_2O}=4.86gH_2O$

Explanation:

Hello,

In this case, the described chemical reaction is:

$C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O$

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

$n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6$

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

$n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6$

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

$m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O$

Best regards.

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3 years ago

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