Answer:
0.0821 L•atm/mol•K
Explanation:
The value of the ideal gas can be obtained as follow:
NOTE: We shall use the standard value for each variable. Based on the options given above, we shall use the equivalent units as well.
Pressure (P) = 1 atm
Volume (V) = 22.4 L
Number of mole = 1 mole
Temperature (T) = 273 K
Gas constant (R) =?
PV = nRT
Divide both side by nT
R = PV / nT
R = (1 × 22.4) / (1 × 273)
R = 22.4 / 273
R = 0.0821 L•atm/mol•K
Thus, the ideal gas constant is 0.0821 L•atm/mol•K
Answer:
250cm³ de disolución se deben preparar
Explanation:
El porcentaje volumen volumen (%v/v) se define como el volumen de soluto -En este caso, H2O2- en 100cm³ de disolución.
Se desea preparar una solución que contenga 20cm³ de H2O2 por cada 100cm³
Como solamente se cuenta con 50cm³ de H2O2, el volumen que se debe preparar es:
50cm³ H2O2 * (100cm³ Disolución / 20cm³ H2O2) =
<h3>250cm³ de disolución se deben preparar</h3>
Answer: yellow
When you mix solutions of lead (II) nitrate and potassium iodide. The precipitate is yellow in colour and the compound is lead (II) Iodide. Pb (NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq). Yes, it is a double displacement reaction.
Explanation: The lead nitrate solution contains particles (ions) of lead, and the potassium iodide solution contains particles of iodide. When the solutions mix, the lead particles and iodide particles combine and create two new compounds, a yellow solid called lead iodide and a white solid called potassium nitrate. When a solution of potassium iodide is added to a solution of lead nitrate taken in a test tube, the precipitation of a yellowish solid is observed. This yellowish solid is lead iodide. Potassium nitrate is formed along with lead iodide. When lead nitrate is mix with potassium iodine, a precipitation reaction occurs and yellow precipitate of lead iodide is formed.