The rate constant for this first‑order reaction is
1 answer:
You might be interested in
Answer: The volume of 0.640 grams of gas at Standard Temperature and Pressure (STP) is 0.449 L.
Explanation:
Given: Mass of gas = 0.640 g
Pressure = 1.0 atm
Temperature = 273 K
As number of moles is the mass of substance divided by its molar mass.
So, moles of (molar mass = 32.0 g/mol) is as follows.
Now, ideal gas equation is used to calculate the volume as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
Thus, we can conclude that the volume of 0.640 grams of gas at Standard Temperature and Pressure (STP) is 0.449 L.
4.648 gm of solute is needed to make 37.5 mL of 0.750 M KI solution.
Solution:
We will start with the Molarity
Also we know 1000 ml = 1 L
Therefore 37.5 ml by 1000ml we obtained 0.0375L
Equation for solving mole of solute
Now, multiply 0.750M by 0.0375
Substitute the known values in the above equation we get
Also we know that Molar mass of KI is 166 g/mol
So divide the molar mass value to get the no of grams.
So 4.648 gm of Solute is required for make 37.5 mL of 0.750 M KI solution.