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:
20.2 amu.
Explanation:
Let A represent isotope ²⁰X
Let B represent isotope ²²X
From the question given above, the following data were obtained:
For Isotope A (²⁰X):
Mass of A = 20
Abundance (A%) = 90%
For Isotope B (²²X):
Mass of B = 22
Abundance (A%) = 10%
Relative atomic mass (RAM) =?
The relative atomic mass (RAM) of the element can be obtained as follow:
RAM = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]
RAM = [(20 × 90)/100] + [(22 × 10)/100]
RAM = 18 + 2.2
RAM = 20.2 amu
Thus, relative atomic mass (RAM) of the element is 20.2 amu
It is 1-3-1-3
so 1Al2O3 + 3H2So4 ---> 1Al2(So4)3 + 3H2O
Answer: The pressure in atmospheres is 0.674 in the container if the temperature remains constant.
Explanation:
Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
(At constant temperature and number of moles)
where,
= initial pressure of gas = 205 kPa
= final pressure of gas = ?
= initial volume of gas = 4.0 L
= final volume of gas = 12000 ml = 12 L (1L=1000ml)
(1kPa=0.0098atm)
Therefore, the pressure in atmospheres is 0.674 in the container if the temperature remains constant.
Answer is: <span>concentration of fluoride in the water in parts-per-million is 1 ppm.
</span>Parts-per-million (10⁻⁶) is<span> present at one-millionth of a </span>gram per gram of sample solution, f<span>or example mg/kg.
</span>m(fluoride) = 500 g · 1000 mg/g = 500000 mg.
m(water) = d(water) · V(water).
m(water) = 1 kg/L · 500000 L.
m(water) = 500000 kg.
arts-per-million = 500000 mg ÷ 500000 kg = 1 mg/kg = 1 ppm.
