Answer:
THE MASS OF NITROGEN GAS IN THIS CONDITIONS IS 0.0589 g
Explanation:
In an ideal condition
PV = nRT or PV = MRT/ MM where:
M = mass = unknown
MM =molar mass = 28 g/mol
P = pressure = 2 atm
V = volume = 25 mL = 0.025 L
R = gas constant = 0.082 L atm/mol K
T = temperature = 290 K
n = number of moles
The gas in the question is nitrogen gas
Molar mass of nitrogen gas = 14 * 2 = 28 g/mol
Then equating the variables and solving for M, we have
M = PV MM/ RT
M = 2 * 0.025 * 28 / 0.082 * 290
M = 1.4 / 23.78
M = 0.0589 g
The mass of the nitrogen gas at ideal conditions of 2 atm, 25 mL volume and 290 K temperature is 0.0589 g
Answer:
146.85 g/mol
Explanation:
PV=nRT
n=mass/molar mass
covert from mmhg to atm = 0.184 atm
convert from ml to L= 0.108 L
convert from degree C to K= 456.15 K
convert from mg to g= 0.07796g
then rearrange the formula:
n=PV/RT
=(0.184)(0.108)/(0.08206)(456.15)
n= 5.308*10^(-4)
rearrange the n formula interms of molar mass:
Molar mass= mass/n
=0.07796/(5.308*10^-4)
molar mass= 146.85g/mol
The decomposition time : 7.69 min ≈ 7.7 min
<h3>Further explanation</h3>
Given
rate constant : 0.029/min
a concentration of 0.050 mol L to a concentration of 0.040 mol L
Required
the decomposition time
Solution
The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time
For first-order reaction :
[A]=[Ao]e^(-kt)
or
ln[A]=-kt+ln(A0)
Input the value :
ln(0.040)=-(0.029)t+ln(0.050)
-3.219 = -0.029t -2.996
-0.223 =-0.029t
t=7.69 minutes
Answer:
₂He⁴
Explanation:
Alpha radiations are emitted as a result of radioactive decay. The atom emit the alpha particles consist of two proton and two neutrons. Which is also called helium nuclei. When atom undergoes the alpha emission the original atom convert into the atom having mass number less than 4 and atomic number less than 2 as compared to parent atom the starting atom.
Properties of alpha radiation:
Alpha radiations can travel in a short distance.
These radiations can not penetrate into the skin or clothes.
These radiations can be harmful for the human if these are inhaled.
These radiations can be stopped by a piece of paper.
₉₂U²³⁸ → ₉₀Th²³⁴ + ₂He⁴ + energy
