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:
5.06atm
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (atm)
P2 = final pressure (atm)
V1 = initial volume (Litres)
V2 = final volume (Litres)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
P1 = 1.34 atm
P2 = ?
V1 = 5.48 L
V2 = 1.32 L
T1 = 61 °C = 61 + 273 = 334K
T2 = 31 °C = 31 + 273 = 304K
Using P1V1/T1 = P2V2/T2
1.34 × 5.48/334 = P2 × 1.32/304
7.34/334 = 1.32P2/304
Cross multiply
334 × 1.32P2 = 304 × 7.34
440.88P2 = 2231.36
P2 = 2231.36/440.88
P2 = 5.06
The final pressure is 5.06atm
B. The energy barrier between reactants and products
hope this helps!
(:
Answer:
1. the end result of meiosis is haploid daughter cells with chromosomal combinations different from those originally present in the parent.
2. Prophase, metaphase, and telophase