Answer:
441.28 g Oxygen
Explanation:
- The combustion of hydrogen gives water as the product.
- The equation for the reaction is;
2H₂(g) + O₂(g) → 2H₂O(l)
Mass of hydrogen = 55.6 g
Number of moles of hydrogen
Moles = Mass/Molar mass
= 55.6 g ÷ 2.016 g/mol
= 27.8 moles
The mole ratio of Hydrogen to Oxygen is 2:1
Therefore;
Number of moles of oxygen = 27.5794 moles ÷ 2
= 13.790 moles
Mass of oxygen gas will therefore be;
Mass = Number of moles × Molar mass
Molar mass of oxygen gas is 32 g/mol
Mass = 13.790 moles × 32 g/mol
<h3> = 441.28 g</h3><h3>Alternatively:</h3>
Mass of hydrogen + mass of oxygen = Mass of water
Therefore;
Mass of oxygen = Mass of water - mass of hydrogen
= 497 g - 55.6 g
<h3> = 441.4 g </h3>
Answer:
The periodic table of elements arranges all of the known chemical elements in an informative array. Elements are arranged from left to right and top to bottom in order of increasing atomic number. Order generally coincides with increasing atomic mass. The rows are called periods.
Explanation:
The acceleration of the ball was 0.6 m·s⁻².
<em>F = ma</em>
<em>a = F</em>/<em>m</em> = 5 N/9 kg × (1 kg·m·s⁻²/1 N) = 0.6 m·s⁻²
Answer:
The law of multiple proportions is the third postulate of Dalton's atomic theory. It states that the masses of one element which combine with a fixed mass of the second element are in a ratio of whole numbers.
Therefore, the masses of oxygen in the two compounds that combine with a fixed mass of carbon should be in a whole number ratio. In 100 grams of the first compound (100 is chosen to make calculations easier), there are 57.1 grams oxygen and 42.9 grams carbon. The mass of oxygen (O) per gram of carbon (C) is:
57.1 g O / 42.9 g C = 1.33 g O per g C
In the 100 grams of the second compound, there are 72.7 grams of oxygen (O) and 27.3 grams of carbon (C). The mass of oxygen per gram of carbon is:
72.7 g O / 27.3 g C = 2.66 g O per g C
Dividing the mass O per g C of the second (larger value) compound:
2.66 / 1.33 = 2
This means that the masses of oxygen that combine with carbon are in a 2:1 ratio. The whole-number ratio is consistent with the law of multiple proportions.
Explanation:
When carbon is burned in air carbon iv oxide gas is formed.
C (s) + O2 (g) = CO2(g) ΔH = - 393.5 kj/mol
The enthalpy change of the reaction is -393.5 j/mol which means that when one mole of carbon is completely burnt in air then 393.5 j of energy is evolved.
Thus, 1 mole = -393.5 j , then for 480 kj
= 480 × 1/393.5
= 1.2198 moles
1 mole of carbon iv oxide is equal to 44 g
thus, 1.2198 moles will be 1.2198 × 44 = 53.6712 g of CO2
