1. a 2.b 3 a 4. c. 4.a 5. d 6.a. 2Na + Cl2 2NaC 7.c. 3 8.3 9.b
Answer:
The volume of water to be added is 0.175 liters of water
Explanation:
The given concentration of the nitric acid = 55% (M/M)
The mass of the nitric acid solution = 100 gm
The concentration solution is to diluted to = 20% (M/M)
The 100 g 55%(M/M) nitric acid solution gives 55g nitric acid in 100 g of solution
Therefore, to have 20% (M/M) nitric acid solution with the 55 g nitric acid, we get
Let "x" represent the volume of the resulting solution, we have;
20% of x = 55 g of nitric acid
∴ 20/100 × x = 55 g
x = 55 g × 100/20 = 275 g
The mass of extra water to be added = The mass of the 20%(M/M) solution solution of nitric acid - The current mass of the 55%(M/M) solution of nitric acid
The mass of extra water to be added = 275 g - 100 g = 175 g
Volume = Mass/Density
The density of water ≈ 1 g/ml
∴ The volume of water to be added that gives 175 g of water = 175 g/(1 g/ml) = 175 ml. = 0.175 l
The volume of water to be added = 0.175 liters of water.
Answer:
46.40 g.
Explanation:
- It is a stichiometric problem.
- The balanced equation of the reaction: 4K + O₂ → 2K₂O.
- It is clear that 4.0 moles of K reacts with 1.0 mole of oxygen produces 2.0 moles of K₂O.
- We should convert the mass of K (38.5 g) into moles using the relation:
<em>n = mass / molar mass,</em>
n = (38.5 g) / (39.098 g/mol) = 0.985 mole.
<em>Using cross multiplication:</em>
4.0 moles of K produces → 2.0 moles of K₂O, from the stichiometry.
0.985 mole of K produces → ??? moles of K₂O.
∴ The number of moles of K₂O produced = (0.985 mole) (2.0 mole) / (4.0 mole) = 0.4925 mole ≅ 0.5 mole.
- Now, we can get the mass of K₂O:
∴ mass = n x molar mass = (0.5 mole) (94.2 g/mol) = 46.40 g.
Answer is: (2) Chemical energy is converted to electrical energy.
An electrochemical cell (voltaic or galvanic cell) is generating electrical energy from chemical reactions.
In galvanic cell, specie (for example zinc and zinc cations) from one half-cell, lose electrons (oxidation) and species from the other half-cell (for example copper and copper cations) gain electrons (reduction).
Oxidation on the zinc anode: Zn(s) → Zn²⁺(aq) + 2e⁻.
Reduction on the copper cathode: Cu²⁺(aq) + 2e⁻ → Cu(s).
I believe it’s B I apologize if it’s wrong
