The answer is the elements in a periodic table. <span>On the basis of the elements in the periodic table, they are divided into metals and non metals. </span>
Answer: 4.41 atm
Explanation:
Given that,
Original pressure of oxygen gas (P1) = 5.00 atm
Original temperature of oxygen gas (T1) = 25°C
[Convert 25°C to Kelvin by adding 273
25°C + 273 = 298K
New pressure of oxygen gas (P2) = ?
New temperature of oxygen gas (T2) = -10°C
[Convert -10°C to Kelvin by adding 273
-10°C + 273 = 263K
Since pressure and temperature are given while volume is held constant, apply the formula for Charle's law
P1/T1 = P2/T2
5.00 atm /298K = P2/263K
To get the value of P2, cross multiply
5.00 atm x 263K = 298K x V2
1315 atm•K = 298K•V2
V2 = 1315 atm•K / 298K
V2 = 4.41 atm
Thus, the new pressure inside the canister is 4.41 atmosphere
Hey I don’t know the answers but try to use the app Socratic or photomath
Answer:
0.0075 milliliters (7.5 microliters) will be taken from the stock solution and then diluted with 0.0249925 milliliters (24.9925 microliters) of water to make 0.025 milliliters (25 microliters) of NaOH solution.
Explanation:
This is a problem of dilution using the equation:
<em>initial concentration x initial volume = final concentration x final volume.</em>
The final volume to be prepared is 25 microliters.
The final concentration to be prepared is 3 M.
The initial volume to be taken is not known yet.
The initial concentration is 10 M.
Now, let's substitute these parameters into the the equation above.
10 x initial volume = 3 x 25
Initial volume = 3 x 25/10
= 7.5 microliters
Note that: 1 microliter = 0.001 milliliters
Hence,
7.5 microliters = 0.0075 milliliters
<u>This means that an initial volume of 0.0075 milliliters (7.5 microliters) will be taken from the stock solution. This amount will then be diluted with 0.0249925 milliliters (24.9925 microliters) of water to make 0.025 milliliters (25 microliters) of NaOH solution.</u>
Explanation:
In this experiment, carbon dioxide and water vapors combine to form H2CO3. After decomposition, the Na2CO3 had a mass of 2.86 grams. Determine ...
