Answer:
The correct answer is: 1.035 x 10⁻³ M
Explanation:
The dissociation equilibrium for acetic acid (CH₃COOH) is the following:
CH₃COOH(aq) ↔ CH₃COO⁻(aq) + H⁺(aq) Kc = 1.8 x 10⁻⁵
The expression for the equilibrium constant (Kc) is the ratio of concentrations of products over reactants. The products are acetate ion (CH₃COO⁻) and hydrogen ion (H⁺) while the reactant is acetic acid (CH₃COOH):
![Kc=\frac{[CH_{3} COO^{-} ][H^{+} ]}{[CH_{3} COOH]}= 1.8 x 10^{-5}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BCH_%7B3%7D%20COO%5E%7B-%7D%20%5D%5BH%5E%7B%2B%7D%20%5D%7D%7B%5BCH_%7B3%7D%20COOH%5D%7D%3D%201.8%20x%2010%5E%7B-5%7D)
Given: [CH₃COOH]= 0.016 M and [CH₃COO⁻]= 0.92 M, we replace the concentrations in the equilibrium expression and we calculate [H⁺]:
![\frac{(0.016 M)[H^{+} ]}{(0.92M)}= 1.8 x 10^{-5}](https://tex.z-dn.net/?f=%5Cfrac%7B%280.016%20M%29%5BH%5E%7B%2B%7D%20%5D%7D%7B%280.92M%29%7D%3D%201.8%20x%2010%5E%7B-5%7D)
⇒[H⁺]= (1.8 x 10⁻⁵)(0.92 M)/(0.016 M)= 1.035 x 10⁻³ M
Answer: Lithium fluoride is an inorganic compound with the chemical formula LiF.
Answer:
25% thorium will left after 50 days.
Explanation:
Half life:
A nuclear half is the time period of radioactive material in which its amount remain halved.
In given question it is stated that the half life thorium-234 is 25 days. Which means after passing the 25 days the amount of thorium must be halved of original amount.
For example,
If the original concentration was 100%, than after 25 days it will be 50%.
After 50 days amount of thorium left:
Number of half life = T (elapsed) / T half life
Number of half life = 50/25
Number of half life = 2
At first half life amount of thorium left = 100/2 = 50
After second half life amount of thorium left = 50/2 = 25
Total amount decayed = 50+25 = 75
Amount left after 50 days = 100-75 = 25
25% thorium will left after 50 days.
Answer is: <span>a. c6h14 and c10h20.
This pair will </span>most likely form a homogeneous solution because they are both nonpolar substances and "li<span>ke dissolves like".
Other pairs will not form homogeneous solution because nonpolar substances have low solubility in polar or ionic substances (for example LiBr is ionic and C</span>₅H₁₂ is nonpolar).
Answer:
The molecular formule for this unknow molecule is C2H4O2
Explanation:
The empirical formula is CH2O ( or better said CnH2nOn)
This means there are 3 elements in the formula of this molecule
⇒ Carbon (C) with a Molar mass of 12 g/mole
⇒ Hydrogen (H) with a Molar mass of 1 g/mole
⇒ Oxygen (O) with a Molar mass of 16 g/mole
We can also notice that the amount of hydrogen should 2x the amount of carbon ( also 2x the amount of oxygen).
The mass of the empirical formule = 12g/ mole + 2* 1 g/mole + 16 g/mole = 30 g/mole
To know what number is n in CnH2nOn we should divide the molecular mass by the empirical mass:
60 g/mole / 30g/mole = 2
this means n = 2
and this will give a molecular formule of C2H4O2
We can control this to calculate the molecular mass:
2*12 + 4* 1 + 2*16 = 24 + 4 + 32 = 60 g/mole
The molecular formule for this unknow molecule is C2H4O2
