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>
Um im pretty sure there are only about 100 different atoms...
Answer:
0.30M HOAc
Explanation:
Given 250.0ml (0.96M HOAc) => 800ml(??M HOAc)
Use the dilution equation...
(Molarity x Volume)concentrated soln = (Molarity x Volume)dilute soln
(0.96M)(250.0ml) = (Molarity diluted soln)(800ml)
Molarity diluted soln = (0.96M)(250.0ml)/(800ml) = 0.30M HOAc
Answer:
Approximately
under standard conditions.
Explanation:
Equation for the overall reaction:
.
Write down the ionic equation for this reaction:
.
The net ionic equation for this reaction would be:
.
In this reaction:
- Zinc loses electrons and was oxidized (at the anode):
. - Copper gains electrons and was reduced (at the cathode):
.
Look up the standard potentials for each half-reaction on a table of standard reduction potentials.
Notice that
is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction,
, is reduction and is likely on the table.
The reduction potential of
would be
, the opposite of the reverse reaction
.
The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:
.
1. For the first question, we must find the mass of the anhydrous salt, MgSO₄. The molar mass for MgSO₄·7H₂O is 246.47 g/mol, while that of MgSO₄ is 120.37 g/mol.
7.834 g MgSO₄·7H₂O * 1 mol MgSO₄·7H₂O/246.47 g * 1 mol MgSO₄/1 mol MgSO₄·7H₂O * 20.37 g MgSO₄/mol = <em>0.647 g MgSO₄</em>
2. Mass of Water = Mass of sample - Mass of anhydrous sale
Mass of water = 7.834 - 0.647 = 7.19 g
Percent Water in Hydrate = 7.19/7.834 * 100 = <em>91.74%</em>