Answer:
0.92^n
Explanation:
Given that :
Initial amount of vinegar = 1 Litre
Number of litres removed repeatedly = 0.08 Litre
Since the amount removed each time is constant, then ;
Initial % = 100% = 100/100 = 1
. Using the relation :
Amount of vinegar in mixture :
Initial * (1 - amount removed / initial amount)^n
n = number of times repeated
1 * (1 - 0.08/1)^n
1 * (1 - 0.08)^n
1 * 0.92^n
Hence,
For nth removal,
Concentration will be :
0.92^n ; for n ≥ 1
Answer : The amount of oxygen gas collected are, 0.217 mol
Explanation :
Using ideal gas equation :
where,
P = pressure of gas = 
(1 atm = 760 torr)
V = volume of gas = 5 L
T = temperature of gas = 
n = number of moles of gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:
Thus, the amount of oxygen gas collected are, 0.217 mol
Explanation:
it helps to make juices.
It helps to make concentrated acid into dilute acid.
Answer:
Kc = 1.09x10⁻⁴
Explanation:
<em>HF = 1.62g</em>
<em>H₂O = 516g</em>
<em>F⁻ = 0.163g</em>
<em>H₃O⁺ = 0.110g</em>
<em />
To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:
Kc = [H₃O⁺] [F⁻] / [HF]
<em>Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:</em>
<em />
[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M
[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M
[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M
Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]
<h3>Kc = 1.09x10⁻⁴</h3>
