Answer:
kilogram is used to measure to mass
Answer:
After the transfer the pressure inside the 20 L vessel is 0.6 atm.
Explanation:
Considering O2 as an ideal gas, it is at an initial state (1) with V1 = 3L and P1 = 4 atm. And a final state (2) with V2 = 20L. The temperature remain constant at all the process, thus here applies the Boyle-Mariotte law. This law establishes that at a constant temperature an ideal gas the relationship between pressure and volume remain constant at all time:
Therefore, for this problem the step by step explanation is:
Clearing P2 and replacing
Answer:
Explanation:
The period law state that when elements are listed in order of their atomic numbers, the elements fall into recurring groups, so that there is a recurrence of similar properties at regular intervals.
Na and K in the periodic table fall into the same group, this is because they both have one electrons in their outermost shell.
Na 11 -1s2 2s2 2p6 3s1
K 19 - 1s2 2s2 2p6 3s2 3p6 4s1
They share similar chemical and physical properties. Na and K are very reactive metals, they can loose/donate their outermost electron to non metals in other to attain stable octet state.
The form ionic compound when they react with non metals.
Answer:
\left \{ {{y=206} \atop {x=82}}Pb \right.
Explanation:
isotopes are various forms of same elements with different atomic number but different mass number.
Radioactivity is the emission of rays or particles from an atom to produce a new nuclei. There are various forms of radioactive emissions which are
- Alpha particle emission \left \{ {{y=4} \atop {x=2}}He \right.
- Beta particle emission \left \{ {{y=0} \atop {x=-1}}e \right.
- gamma radiation \left \{ {{y=0} \atop {x=0}}γ \right.
in the problem the product formed after radiation was Pb-206. isotopes of lead include Pb-204, Pb-206, Pb-207, Pb-208. they all have atomic number 82. which means the radiation cannot be ∝ or β since both radiations will alter the atomic number of the parent nucleus.
Only gamma radiation with \left \{ {{y=0} \atop {x=0}}γ \right. will produce a Pb-206 of atomic number 82 and mass number 206 , since gamma ray have 0 mass and has 0 atomic number.equation is shown below
\left \{ {{y=206} \atop {x=82}}Pb\right ⇒ \left \{ {{y=206} \atop {x=82}}Pb\right + \left \{ {{y=0} \atop {x=0}}γ\right.
Thus the atomic symbol is \left \{ {{y=206} \atop {x=82}}Pb\right
