The balloon's volume is 128 ml when the gas temperature rises to 320.0 K. Explanation: Given the following: T1 (initial temperature) = 300K, V1 (initial volume) = 120ml, T2 (final temperature) = 320 K, V2 (final volume) =?. Pressure is kept constant during this process. From the equation: Given that the pressure stays constant, we have: V2 = Putting the values into this formula yields: V2 = 128 ml, which indicates the volume of the gas when the temperature increases from 300 K to 320 K.
Answer:
0.1714 (w/w) %
Explanation:
Utilizando la ecuación:
16H+(aq) + 2Cr2O72−(aq) + C2H5OH(aq) → 4Cr3+(aq) + 2CO2(g) + 11H2O(l)
Se emplean 2 moles de ion dicromato (Cr₂O₇²⁻) para titular 1 mol de alcohol (C₂H₅OH)
35.46mL = 0.03546L de una solución de Cr₂O₇²⁻ a 0.05961M utilizada para alcanzar el punto de equivalencia en la titulación contiene:
0.03546L ₓ (0.05961 moles Cr₂O₇²⁻ / L) = 2.114x10⁻³ moles Cr₂O₇²⁻
Dado que 2 moles de dicromato reaccionan por cada mol de alcohol, los moles de alcohol en la muestra de plasma son:
2.114x10⁻³ moles Cr₂O₇²⁻ ₓ ( 1 mol C₂H₅OH / 2 moles Cr₂O₇²⁻) = 1.0569x10⁻³ moles de C₂H₅OH
Como la masa molar del alcohol es 46.07g/mol, la masa de alcohol es:
1.0569x10⁻³ moles de C₂H₅OH ₓ (46.07g / mol) = 0.04869g de C₂H₅OH
Por lo tanto, el porcentaje en masa de alcohol en sangre utilizando los 28.40g de plasma es:
(0.04869g de C₂H₅OH / 28.40g) × 100 = 0.1714 (w/w) %
Answer:
8 protons, 8 electrons, and 10 neutrons
Explanation:
Refer to the explanation and the attached image for further details. When AlCl3 reacts with (CH3)3CCH2Cl, a primary carbocation is produced. This primary carbocation then undergoes a 1,2-alkyl shift leading to the formation of a tertiary carbocation, which subsequently bonds with the benzene ring as depicted in the attached image.
