Answer:
The final pressure of the gas is:- 21.3 kPa
Explanation:
Using Boyle's law
Given ,
V₁ = 10.0 L
V₂ = 45.0 L
P₁ = 96.0 kPa
P₂ = ?
Using above equation as:
The final pressure of the gas is:- 21.3 kPa
I am very confused, our teacher didn’t properly teach us this. How do you do it?
2 answers:
Answer:
You multiply the 3 numbers together to get your volume, in this case it would be 4058.488 cm^3 (cm cubed)
so
V: 4058.488cm^3 ( round up to 4058.5 for convenience)
M: 27579
D: ?
So we divide mass by the volume to get density, which is
27579 / 4058.5 = ~6.79536 (can round up to 7 or 6.8)
This graph can help a lot, so maybe try and memorize it, hope I helped.
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Answer:
a. Δ
b. 320.76° C
Explanation:
a.)
we can solve this type of question (i.e calculate Δ , for the gas-phase reaction ) using the Hess's Law.
Δ =
Given from the question, the table below shows the corresponding Δ for each compound.
Compound
Liquid EO -77.4
-74.9
-110.5
If we incorporate our data into the above previous equation; we have:
Δ = (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)
=
b.)
We are to find the final temperature if the average specific heat capacity of the products is 2.5 J/g°C
Given that:
the specific heat capacity (c) = 2.5 J/g°C
= 93.0°C &
the enthalpy of vaporization (Δ) = 569.4 J/g
If, we recall; we will remember that; Specific Heat Capacity is the amount of heat needed to raise the temperature of one gram of a substance by one kelvin.
∴ the specific heat capacity (c) is given as =
Let's not forget as well, that Δ =
If we substitute Δ for
in the above equation, we have;
specific heat capacity (c) =
Making () the subject of the formula; we have:
=
= 227.76°C +93.0°C
= 320.76°C
∴ we can thereby conclude that the final temperature = 320.76°C