Answer:
1.58×10E18
Explanation:
Since we have the reduction potentials we could make decisions regarding which one will be the anode or cathode. Evidently, bromine having the more positive reduction potential will be the cathode while the iodine will be the anode.
E°cell= 1.07- 0.53= 0.54 V
E°cell= 0.0592/n logK
0.54 = 0.0592/2 logK
logK= 0.54/0.0296
logK= 18.2
K= Antilog (18.2)
K= 1.58×10^18
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Answer:
6.23 KOH 90% son necesarios
Explanation:
Una solución 1N de KOH requiere 1equivalente (En KOH, 1eq = 1mol) por cada litro de solución.
Para responder esta pregunta se requiere hallar los equivalentes = Moles de KOH para preparar 100mL = 0.100L de una solución 1N. Haciendo uso de la masa molar de KOH y del porcentaje de pureza del KOH se pueden calcular los gramos requeridos para preparar la solución así:
<em>Equivalentes KOH:</em>
0.100L * (1eq / L) = 0.100eq = 0.100moles
<em>Gramos KOH -Masa molar: 56.1056g/mol-:</em>
0.100moles * (56.1056g/mol) = 5.61 KOH se requieren
<em>KOH 90%:</em>
5.61g KOH * (100g KOH 90% / 90g KOH) =
<h3>6.23 KOH 90% son necesarios</h3>
Answer:
All atoms of the same element have always have the same amount of protons.
Explanation:
Atoms of the same element have always have the same amount of protons but not always the same electrons and neutrons. If an atom gains or loses one of its valance electrons, the electrons on the outermost shell, then it becomes ionized. Also not all atoms of the same element have the same amount of neutron. This is called an isotope. A good example would be Carbon 13. Normally, Carbon atoms have an atomic mass of 12 AMU or 12 atomic mass units. However, Carbon atoms have an atomic mass of 13 AMU, consisting of 7 neutrons instead of 6 neutrons. So the only thing that all atoms of the same element have in common is the amount of protons.
