Answer:
C
Explanation:
If you move the decimal to the left it transforms the expression into scientific notation.
Answer:
Explanation:
Group 4A contains a total of 4 electrons for each atom in their valence shell. Filling the orbital diagram, let's say, for carbon, notice that when we start with period 2, we have two elements in the s-block, that is, lithium and beryllium. They correspond to the two s electrons that belong to the valence shell of carbon.
Moving on, we have boron and carbon, the remaining 2 electrons. Now, starting with boron, we're in the p-block.
That said, looking at the second period, the electron configuration for the valence shell of a group 4A element would be:
Answer: Atoms with 11 protons, 10 neutrons and 11 electrons belong to the same element with 11 protons, 12 neutrons and 11 electrons.
Explanation:
Elements that contain same number of valence electrons belong to the same group. This is because they will have same reactivity (or properties) due to which they lie in the same group.
For example, element with 11 protons, 10 neutrons and 11 electrons is same as the element with 11 protons, 12 neutrons and 11 electrons.
Hence, both these atoms belong to the same element.
Thus, we can conclude that atoms with 11 protons, 10 neutrons and 11 electrons belong to the same element with 11 protons, 12 neutrons and 11 electrons.
The question is incomplete, the complete question is;
1.000 atm of Oxygen gas, placed in a container having apinhole opening in its side. leaks from the container 2.14 timesfaster thatn 1.000 atm of an unknown gas placed in this sameapparatus. Which of the following species could be theunknown gas?
A. CL2
B. SF6
C. Kr
D. UF6
E. Xe
Answer:
SF6
Explanation:
From Graham's law;
Let the rate of diffusion of oxygen be R1
Let the rate of diffusion of unknown base be R2
Let the molar mass of oxygen by M1
Let the molar mass of unknown gas be M2
Hence;
R1/R2 = √M2/M1
So;
2.14/1 = √M2/32
(2.14/1)^2 = M/32
M= (2.14/1)^2 × 32
M= 146.6
This is the molar mass of SF6 hence the answer above.
