Answer:
312 g of O₂
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2KClO₃ —> 2KCl + 3O₂
From the balanced equation above,
2 mole of KClO₃ decomposed to 3 moles of O₂.
Next, we shall determine the number of mole of O₂ produced by the reaction of 6.5 moles of KClO₃. This can be obtained as follow:
From the balanced equation above,
2 mole of KClO₃ decomposed to 3 moles of O₂.
Therefore, 6.5 moles of KClO₃ will decompose to produce = (6.5 × 3)/2 = 9.75 moles of O₂.
Finally, we shall determine the mass of 9.75 moles of O₂. This can be obtained as follow:
Mole of O₂ = 9.75 moles
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mass of O₂ =?
Mole = mass / Molar mass
9.75 = Mass of O₂ / 32
Cross multiply
Mass of O₂ = 9.75 × 32
Mass of O₂ = 312 g
Thus, 312 g of O₂ were obtained from the reaction.
Answer:
That depends on what species it is
Explanation:
Like reptiles it is rattlesnakes
Spiders would be black widow
So it depends on the what species you what.
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.
Goods are objects that fulfill peoples needs and wants.
