Yes the ANSWER is Correct-
On first half life the mass will be 10 gram
On second half life the mass will be 5 gram
On third half life the mass will be 2.5 gram
Answer:
The correct option is;
24 moles
Explanation:
Here, we have the reaction as follows;
Sn(s) + 2HF(g) → SnF₂ (s) + H₂ (g)
Therefore, one mole of Sn reacts with 2 moles HF to form one mole of SnF₂ and one mole of H₂
Molar mass of H₂ = 2.01588 g/mol
Therefore, the number of moles of H₂ in 48 grams of H₂ is given by the relation;

Since one mole each of SnF₂ and H₂ are produced, the number of moles of SnF₂ produced = 24 moles.
The number of moles of SnF₂ that will be produced is 24 moles.
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
A chemical change affects on the molecular level of matter, which makes it irreversible. Combustion is a pretty good exmple. Physical changes are reversible and dont alter the formula. Hope this helped!