22.4 L<span>So, if 1 mole occupies 22.4 L, the imediate conclusion is that a bigger number of moles will occupy more than 22.4 L, and a smaller number of moles will occupy less than 22.4 L. In your case, 3 moles of gas will occupy 3 times more volume than 1 mole of gas.</span>
Answer:
It is 20. g HF
Explanation:
H2 + F2 ==> 2HF ... balanced equation
Since the question is asking us to find the mass of product formed, we will want to first convert the molecules of H2 into moles of H2 (we could do this at the end of the calculations, but it's just as easy to do it now).
moles of H2 present (using Avogadro's number):
3.0x1023 molecules H2 x 1 mole H2/6.02x1023 molecules = 0.498 moles H2
From the balanced equation, we see that 1 mole H2 produces 2 moles HF. Therefore, we can now find the theoretical mass of HF produced from 0.498 moles H2:
0.498 moles H2 x 2 moles HF/1 mol H2 = 0.996 moles HF formed.
The molar mass of HF = 20.01 g/mole, thus...
0.996 moles HF x 20.01 g/mole = 19.93 g HF = 20. g HF formed (to 2 significant figures)
Acid is 1. a substance that increases the hydrogen ion concentration of a solution
<span>2. removes hydroxide ions because of the tendency for H+ to combine with OH-</span>
Answer:
0.0295M
Explanation:
As you can see, in the mixture you have KSCN and other compounds. The KSCN in solution is dissolved in K⁺ ions and SCN⁻ ions. That means initial concentration of SCN⁻ ions is the same of KSCN, 0.0800M.
You are adding 35.0mL of this solution and the total volume of the mixture is 20.0mL + 35.0mL + 40.0mL = 95.0mL.
That means you are diluting your solution 95.0mL / 35.0mL = 2.714 times.
And the concentration of SCN⁻ is:
0.0800M / 2.714 =
<h3>0.0295M </h3>
Answer:
3.5 atm
Explanation:
As stated in the question pressure is required to counteract the natural tendency for water to dilute the more concentrated solution. The difference in concentrations will give us the answer using the osmotic pressure equation.
π = ( n/v) RT where n/v is the molarity (mol/L), R is the gas constant and T is the temperature.
The difference in osmotic pressure of the solutions is:
Δπ = Δ c RT where c is the difference in molar concentrations.
pressure required = Δπ = (0.190 - 0.048) M x 0.821 Latm/Kmol x 298 K
= 3.47 atm
