Answer : The partial pressure of B in the container is, 25.94 atm
Explanation : Given,
Total pressure = 38.9 atm
The given chemical reaction is:
Initial pressure p 0 0
At eqm. 0 2p p
The total pressure expression will be:
Total pressure = 2p + p
38.9 atm = 2p + p
38.9 atm = 3p
p = 12.97 atm
Now we have to calculate the partial pressure of B in the container.
The partial pressure of B in the container = 2p = 2(12.97) = 25.94 atm
Thus, the partial pressure of B in the container is, 25.94 atm
Answer:
1.45 mol
Explanation:
Given data
- Volume of the gas (V): 8.77 L
- Temperature of the gas (T): 20 °C
- Pressure of the gas (P): 3.98 atm
Step 1: Calculate the absolute temperature (Kelvin)
We will use the following expression.
Step 2: Calculate the number of moles (n) of the gaseous sample
We will use the ideal gas equation.
Answer:
What will determine the number of moles of hydronium in an aqueous solution of a strong monoprotic acid? The amount of acid that was added.
Explanation:
<u>Given:</u>
Charge on potassium (K) = +1
Charge on selenium (Se) = -2
<u>To determine:</u>
The values of X and Y in K(X)Se(Y)
<u>Explanation:</u>
The molecule which is formed between K and Se is neutral.
The K+ and Se2- ions should combine such that the net charge on the molecule is zero.
X(K+) + Y(Se2-) = 0
X(+1) + Y(-2) = 0
The simplest solution to the above equation is obtained when X = 2 and Y = 1
Thus, the molecule would be: K₂Se₁
Ans: X = 2 and Y = 1; K₂Se₁
Answer:
The molar mass and atomic mass are essentially the same for an element
Explanation:
The molar mass of a substance can be obtained by dividing the mass of the substance by the no of moles of the substance present.
The atomic mass of an element is the number of protons and neutrons present in the substance.
These two measurements usually give the same values because they both make reference to the 1/12th the mass of carbon-12 for their measurement.
Because they both have the same reference point, though they have different calculating procedures, the results obtained will be similar.