Answer:
Volume occupied by 10 grams of Lithium Oxide at STP is 7.52 L
Explanation:
STP (Standard Temperature and Pressure) : It is used while performing calculation on gases and provide standards of measurements while performing experiment.
According to IUPAC (International Union of Pure And applied Chemistry) , standard temperature is 273.15 K and Pressure is 100 kPa (0.9869 atm). At STP condition 1 mole of substance occupies 22.4 L volume .
Molar mass of Lithium Oxide = 29.8 g/mol
= 2(6.9) + 15.99 = 29.8
Mass of 1 mole = 29.8 g
1 mole occupies, Volume = 22.4 L
29.8 g occupies, V = 22.4 L
1 g occupies ,V = 
1 g occupies ,V = 0.7516 L
10 g tex]Li_{2}O[/tex] occupies ,V = 
L
V = 7.52 L
So, volume occupied by Lithium Oxide At STP is 7.52 L
Answer: 8.38 seconds
Explanation:
Integrated rate law for second order kinetics is given by:
= initial concentartion = 0.860 M
a= concentration left after time t = 0.230 M
k = rate constant =
Thus it will take 8.38 seconds for the concentration of A to decrease from 0.860 M to 0.230 M .
Answer:
![[A_t]=54.5\ g](https://tex.z-dn.net/?f=%5BA_t%5D%3D54.5%5C%20g)
Explanation:
Given that:
Half life = 14.0 days
Where, k is rate constant
So,
The rate constant, k = 0.04951 days⁻¹
Initial concentration [A₀] = 60.0 g
Time = 46.7 hrs
Considering, 1 hr = 0.041667 days
So, time = 1.9458 days
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
So,
![[A_t]=60.0\times e^{-0.04951\times 1.9458}\ g](https://tex.z-dn.net/?f=%5BA_t%5D%3D60.0%5Ctimes%20e%5E%7B-0.04951%5Ctimes%201.9458%7D%5C%20g)
Answer:
new volume is 1215mL
Explanation:
using Charles law , the volume of a fixed mass of gas is directly proportional to its temperature provide that pressure is kept constant.
V1=900mL ,
convert the temperatures from Celsius to kelvin temperature.
T1 =27°C =27+273 =300K
T2 =132°C = 132+273=405K
V2= 1215mL
