Initial Conditions:
Volume=v1= 3.5 L
Temperature= T1 =298K
Final Conditions:
Temperature= T2= 285K
Volume= v2= ?
Use the general gas equation;
P1*v1/T1 = P2*v2/T2
Note:
Pressure is constant
so the equation become,
v1/T1 = v2/T2
3.5/298 = v2/285
v2 = 3.347 L
The tool used to measure pressure is a Barometer and is measured in Milibars.
Answer:
0.178 M
Explanation:
First we <u>determine the molar mass of NaOH</u>, using the <em>given molar masses </em>of the elements:
- Molar Mass of NaOH = Molar Mass of Na + Molar Mass of O + Molar Mass of H
- Molar Mass of NaOH = 40 g/mol
Then we<u> convert 5.000 grams of NaOH into moles</u>, using its <em>molar mass</em>:
- 5.000 g ÷ 40 g/mol = 0.125 mol
Now we <u>convert 700.0 mL into L</u>:
- 700.0 mL / 1000 = 0.700 L
Finally we <u>calculate the molarity of the solution</u>:
- Molarity = moles / liters
- 0.125 mol / 0.700 L = 0.178 M
Is that what you have?
Which energy graph represents the nonspontaneous transition of graphite into diamond?
-A graph with Reaction progression on the horizontal axis and energy on the vertical axis. A line starts low on the vertical axis, runs briefly parallel to the horizontal axis, slopes up sharply, curves down a short distance, and levels out.
-A graph with Reaction progression on the horizontal axis and energy on the vertical axis. A line starts midway up the vertical axis, runs briefly parallel to the horizontal axis, slopes up sharply, curves down sharply, and levels out below the initial starting point.
-A graph with Reaction progression on the horizontal axis and energy on the vertical axis. A line starts midway up the vertical axis, runs briefly parallel to the horizontal axis, slopes up a small amount, then curves down to level out below the initial starting point.
-A graph with Reaction progression on the horizontal axis and energy on the vertical axis. A line starts midway up the vertical axis, runs briefly parallel to the horizontal axis, slopes up a moderate amount, then curves down sharply to level out below the initial starting point.
