The result is 4.16 L.
Based on the provided information, we calculated the following:
Molarity = 0.225 M,
Quantity of KI = 0.935 moles,
To find Volume: Molarity = moles/Volume, hence Volume = moles/Molarity.
Thus, Volume = 0.935/0.225, giving Volume = 4.16 L.
Consequently, 4.16 L of KI is required.
According to the periodic table:
the molar mass of barium is 137.2 grams
the molar mass of oxygen is 16 grams
the molar mass of hydrogen is 1 gram
The molar mass of Ba(OH)2 can be calculated as 137.2 + 2(16) + 2(1) = 171.2 grams.
The molar mass of 4H2O is computed as 4 [2(1) + 16] = 72 grams.
Consequently, the molar mass of Ba(OH)2·4H2O is 171.2 + 72 = 243.2 grams.
Therefore, a sample weighing 243.2 grams of <span>barium hydroxide tetrahydrate includes 72 grams of water, meaning that within 92.8 grams, the mass of water would be:
mass of water in 92.8 grams = (92.8 x 72) / 243.2 = 27.474 grams.
Thus, when heating a 92.8 gram sample of Ba(OH)2·4H2O (barium hydroxide tetrahydrate), 27.474 grams of water will be emitted.</span>
The element represented by the nlx notation of 5d2 is hafnium, designated by the symbol hf. It is categorized as a chemical element with the atomic number 72. Hafnium has a chemical resemblance to zirconium and occurs in various zirconium minerals, whereas it is distinct from zirconium which shares some chemical similarities.
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:
Explanation:
Charge of one electron = 
The formula for calculating charge is:
Given that: Charge = 
Total electrons, n = 
