The temperature of a liquid can exceed its boiling point. An example is water. Although at ordinary pressure of 1 atm, the boiling point is 100 degrees, water can still exist in higher temperatures but this time in another state. Superheated steam is the term used for water whose temperature has higher than the boiling point
As per the question, the mass of the nitrogen gas m = 22.25 gram.
The latent heat of vaporization of nitrogen = 199.0 j/g
As per the question, the nitrogen gas will condense. During condensation, the nitrogen gas will lose or release heat equal to its latent heat.
Hence, the heat released by nitrogen gas Q = ml = 22.25 × 199.0 J = 4427.75 J.
Hence, the amount of heat released will be 4427.75 J.
<h3>How can you figure out how much heat is in each gram?</h3>
The formula: can be utilized to determine energy. Q = mc ∆T. In the equation, Q stands for energy expressed in joules or calories, m for mass expressed in grams, c for specific heat, and T for temperature change, which is the difference between the final temperature and the initial temperature. Water has a specific heat of 1 calorie/gram °C.
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Li + H2O →<span> LiOH + H2
The equation is currently unbalanced, so to balance it out, you have to have the same number of each molecule on each side. It'll look like this:
</span>2Li + 2H2O → <span>2LiOH + H2
</span>
Also, in case you want to identify the phases as well, it'll be like this:
2Li (s) + 2H2O (l) → 2LiOH (aq) + H2 (g)
"s" is solid.
"l" is liquid.
"aq" is aquas.
"g" is gas.
Explanation:
At 365 K temperature sulfur tetrafluoride have a density of 0.260 g/L at 0.0721 atm.
What is an ideal gas equation?
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
First, calculate the moles of the gas using the gas law,
PV=nRT, where n is the moles and R is the gas constant. Then divide
the given mass by the number of moles to get molar mass.
Given data:
P= 0.0721 atm
n=\frac{mass}{molar \;mass}n=
molarmass
mass
R= 0.082057338 \;L \;atm \;K^{-1}mol^{-1}R=0.082057338LatmK
−1
mol
−1
T=?
Putting value in the given equation:
\frac{PV}{RT}=n
RT
PV
=n
density = \frac{2 \;atm\; X molar\; mass}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X T}density=
0.082057338LatmK
−1
mol
−1
XT
2atmXmolarmass
0.260 g/L = \frac{0.0721 \;atm\; X 108.07 g/mol}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X T}0.260g/L=
0.082057338LatmK
−1
mol
−1
XT
0.0721atmX108.07g/mol
T = 365.2158727 K= 365 K
Hence , at 365 K temperature sulfur tetrafluoride have a density of 0.260 g/L at 0.0721 atm.
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