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3 years ago

A gas has a volume of 350 mL at 318.15 K. If the volume changes to 4.00 L, what is the new temperature?

1 answer:

7 0

28°C.
Explanation:
The correct equation to use with these kinds of questions is the Combined Gas Law.
P
1
⋅
V
1
T
1
=
P
2
⋅
V
2
T
2
This equation, though, is a little unnecessary at times because of the possibility of extra constants.
In this case you have a pressure constant which enables us to shorten your longer equation down to Charles' Law because pressure remains the same and doesn't affect the calculation.
V
1
T
1
=
V
2
T
2
The next step is to isolate the variable that you want, which is, in this case,
T
2
.
T
2
=
V
1
T
1
V
2
Now it is time to replace your known variables with numbers
T
2
=
350
m
L
⋅
45
°
C
400
m
L
And calculate...
T
2
=
28.125
°
C
Significant digits...
T
2
=
28
°
C

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Anvisha [2.4K]

1.5 grams of glucose is produced from 2.20 g of CO₂.

To find the mass of glucose produced, first you must know the balanced reaction. For this, the Law of Conservation of Matter is followed.

The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

So, in this case, the balanced reaction is:

6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the amounts of moles of each reactant and product participate in the reaction:

CO₂: 6 moles
H₂O: 6 moles
C₆H₁₂O₆: 1 mole
O₂: 6 moles

So, you know that 2.20 g of CO₂ react, whose molar weight is 44.01 g/mole. By definition of molar mass, 1 mole of CO₂ has 44.01 g. So, the number of moles that 2.20 grams of the compound represent is calculated as:

$moles of CO_{2} =2.20 grams*\frac{1 mole}{44.01 grams}$

moles of CO₂= 0.05 moles

Now you must follow the following rule of three: if by stoichiometry of the reaction 6 moles of CO₂ produce 1 mole of C₆H₁₂O₆, 0.05 moles of CO₂ produce how many moles of C₆H₁₂O₆?

$moles of C_{6} H_{12} O_{6} =\frac{0.05moles of CO_{2} *1 mole of C_{6} H_{12} O_{6}}{6moles of CO_{2}}$

moles of C₆H₁₂O₆= 8.33*10⁻³

Being the molar mass of glucose 180.18 g/mole, the mass that 8.33*10⁻³ moles of the compound represent is calculated as:

$mass of glucose =8.33*10^{-3} moles*\frac{180.18 grams}{1 mole}$

mass of glucose= 1.5 grams

Then, 1.5 grams of glucose is produced from 2.20 g of CO₂.

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