Answer:
88,88 % de O y 11,11 % de H
Explanation:
La composición porcentual se define como la masa que hay de cada mol de átomo en 100g. Las moles de agua en 100g son:
<em>Masa molar agua:</em>
2H = 2*1g/mol = 2g/mol
1O = 1*16g/mol = 16g/mol
Masa molar = 2 + 16 = 18g/mol
100g H2O * (1mol / 18g) = 5.556 moles H2O.
Moles de hidrógeno:
5.556 moles H2O * (2mol H / 1mol H2O) = 11.11 moles H
Moles Oxígeno = Moles H2O = 5.556 moles
La masa de hidrógeno es:
11.11mol * (1g/mol) 11.11g H
La masa de oxígeno es:
5.556 mol * (16g / 1mol) = 88.89g O
Así, el porcentaje de O es 88.89% y el de H es 11.11%. La opción correcta es:
<h3>88,88 % de O y 11,11 % de H</h3>
Answer: Option (b) is the correct answer.
Explanation:
When useful energy is conserved then this conservation is known as energy conservation.
For example, when lights of a room are switched on where no one is sitting then on switching off the lights we are conserving wastage of electricity.
Thus, we can conclude that out of the given options, the statement energy resources can be conserved by turning off lights when they are not in use is true.
Answer:
10.81, Boron
Explanation:
I love the explanation Khan Academy gives.
<span>masss % of KBr = mass of KBr/total mass of solution X 100 = 50/100 X 100 = 50 %</span>
Answer:
Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.
2 S + 3 O₂ → 2 SO₃
The stoichiometric calculations is as follows:
6 g S * 1 mol/32.06 g S = 0.187 mol S
Moles O₂ needed = 0.187 mol S * 3 mol O₂/2 mol S = 0.2805 mol O₂
Since the molar mas of O₂ is 32 g/mol,
Mass of O₂ needed = 0.2805 mol O₂ * 32 g/mol = 8.976 g O₂
