Answer is: the mass of the oxygen is 2.37 grams.
Balanced chemical reaction: 2Mg + O₂ → 2MgO.
m(Mg) = 3.6 g; mass of magnesium.
n(Mg) = m(Mg) ÷ M(Mg).
n(Mg) = 3.6 g ÷ 24.3 g/mol
n(Mg) = 0.149 mol.; amount of the magnesium.
n(O₂) = ?
From chemical reaction: n(Mg) : n(O₂) = 2 : 1.
n(O₂) = n(Mg) ÷ 2.
n(O₂) = 0.149mol ÷ 2.
n(O₂) = 0.075 mol; amount of the oxygen.
m(O₂) = m(O₂) · M(O₂).
m(O₂) = 0.075 mol · 32 g/mol.
m(O₂) = 2.37 g; mass of the oxygen.
The best answer among the choices given that describes compression would be "<span>region of high pressure in a medium caused by a passing wave." The concepts of air compression is widely used in the industries such as vehicle tires, balloons, and other inflatable materials requiring high air pressure.</span>
Answer:
70%
Explanation:
To find the percent yield, you need to (1) convert grams HCl to moles (via molar mass from periodic table), then (2) convert moles HCl to moles H₂ (via mole-to-mole ratio from equation), then (3) calculate the percent yield H₂ (via percent yield equation).
Zn (s) + 2 HCl (aq) --> ZnCl₂ (aq) + 1 H₂ (g)
Molar Mass (HCl): 1.008 g/mole + 35.45 g/mole
Molar Mass (HCl) = 36.458 g/mole
125 g HCl 1 mole HCl 1 mole H₂
--------------- x ------------------ x ------------------ = 1.7 moles H₂
36.458 g 2 moles HCl
(actual yield / theoretical yield) x 100% = percent yield
theoretical/calculated yield = 1.7 moles H₂
actual yield = 1.2 moles H₂
(1.2 moles H₂ / 1.7 moles H₂) x 100% = 71%
Therefore, the best percent yield of hydrogen produced is 70%.
For getting the result of this problem, the knowledge of periodic table is very important. From the periodic table we come to know that. The knowledge of atomic mass of magnesium is also required to solve the problem.
1 mole of magnesium = 24.3 gm
88.1 moles of magnesium = (24.3 * 88.1) gms
= 2140.83 gms
So 2140.83 grams are there in 88.1 moles of magnesium.
Answer:
Determining how many ozone molecules are lost in the atmosphere.
Explanation:
