Answer:
The transition elements or transition metals occupy the short columns in the center of the periodic table, between Group 2A and Group 3A.Explanation:
Answer:
Cells are extremely small.
Explanation:
As Mendel describes in this story, cells are so small they cannot normally be seen with the naked eye. ... The total organism remains the same throughout this process, and (usually) has a longer time on earth than any one of its cells.
Answer:
255.51cm3
Explanation:
Data obtained from the question include:
V1 (initial volume) =?
T1 (initial temperature) = 50°C = 50 + 273 = 323K
T2 (final temperature) = - 5°C = - 5 + 237 = 268K
V2 (final volume) = 212cm3
Using the Charles' law equation V1/T1 = V2/T2, the initial volume of the gas can be obtained as follow:
V1/T1 = V2/T2
V1/323 = 212/268
Cross multiply to express in linear form
V1 x 268 = 323 x 212
Divide both side by 268
V1 = (323 x 212)/268
V1 = 255.51cm3
Therefore, the initial volume of the gas is 255.51cm3
Answer:
0.78 atm
Explanation:
Step 1:
Data obtained from the question. This includes:
Mass of CO2 = 5.6g
Volume (V) = 4L
Temperature (T) =300K
Pressure (P) =?
Step 2:
Determination of the number of mole of CO2.
This is illustrated below:
Mass of CO2 = 5.6g
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Number of mole CO2 =?
Number of mole = Mass/Molar Mass
Number of mole of CO2 = 5.6/44
Number of mole of CO2 = 0.127 mole
Step 3:
Determination of the pressure in the container.
The pressure in the container can be obtained by applying the ideal gas equation as follow:
PV = nRT
The gas constant (R) = 0.082atm.L/Kmol
The number of mole (n) = 0.127 mole
P x 4 = 0.127 x 0.082 x 300
Divide both side by 4
P = (0.127 x 0.082 x 300) /4
P = 0.78 atm
Therefore, the pressure in the container is
The average atomic weight is, from the name itself, the average weight of all its naturally occurring isotopes. All you have to do is multiple the abundance of each isotope with its individual mass, then add them altogether.
Mass = (0.10*55)+(0.15*56)+(.75*57)
<em>Mass = 56.65 amu</em>