Question 1 answer: A
Question 2 answer: H
Question 3 answer: J
Question 4 answer: T
D all of the above answers are correct
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>
<h3><u>Answer;</u></h3>
When hydrogen is covalently bonded to an electronegative atom
<h3><u>Explanation;</u></h3>
- Hydrogen bonding is a special type of dipole-dipole attraction between molecules. It results from the attractive force between a hydrogen atom covalently bonded to a very electronegative atom such as a N, O, or F atom.
- Highly electronegative atoms attract shared electrons more strongly than hydrogen does, resulting in a slight positive charge on the hydrogen atom. The slightly positive hydrogen atom is then attracted to another electronegative atom, forming a hydrogen bond.
We can calculate the final temperature from this formula :
when Tf = (V1* T1) +(V2* T2) / (V1+ V2)
when V1 is the first volume of water = 5 L
and V2 is the second volume of water = 60 L
and T1 is the first temperature of water in Kelvin = 80 °C +273 = 353 K
and T2 is the second temperature of water in Kelvin = 30°C + 273= 303 K
and Tf is the final temperature of water in Kelvin
so, by substitution:
Tf = (5 L * 353 K ) + ( 60 L * 303 K) / ( 5 L + 60 L)
= 1765 + 18180 / 65 L
= 306 K
= 306 -273 = 33° C
