It would be 4.2, hope this helps.
Answer:

Explanation:
(a) Balanced equation
Cu + 2AgNO₃ ⟶ Cu(NO₃)₂+ 2Ag
(b) Calculation
You want to convert atoms of Cu to atoms of Ag.
The atomic ratio is ratio is 2 atoms Ag:1 atom Cu

Answer:
2. The metal would lose one electrons and the non metal would gain one electrons
Explanation:
An atom of a certain element reacts with the atoms of other elements in order to fullfill its outermost shell (called valence shell).
We notice the following:
- The elements in Group 1 (which are metals) have only 1 electron in their valence shell
- The elements in Group 17 (which are non-metals) have 1 vacancy (lack of electron) in their valence shell
This means that in order for both an atom of group 1 and an atom of group 17 to fullfill the valence shell, they have to:
- The atom in group 1 has to give away its only electron of the valence shell
- The atom in group 17 has to gain one electron in order to fullfill the shell
Therefore, the correct option is
2. The metal would lose one electrons and the non metal would gain one electrons
There's a lot of capillaries in the lungs because the blood needs to be transferred and the capillaries are the smallest vessels that can do this.
Answer:
Molecular formula
Explanation:
Molecular formula in the first place is required to understand which compound we have. We then should refer to the periodic table and find the molecular weight for each atom. Adding individual molecular weights together would yield the molar mass of a compound.
Then, dividing the total molar mass of a specific atom by the molar mass of a compound and converting into percentage will provide us with the percentage of that specific atom.
E. g., calculate the percent composition of water:
- molecular formula is
; - calculate its molar mass: [tex]M = 2M_H + M_O = 2\cdot 1.00784 g/mol + 16.00 g/mol = 18.016 g/mol;
- find the percentage of hydrogen: [tex]\omega_H = \frac{2\cdot 1.00784 g/mol}{18.016 g/mol}\cdot 100 \% = 11.19 %;
- find the percentage of oxygen: [tex]\omega_O = \frac{16.00 g/mol}{18.016 g/mol}\cdot 100 \% = 88.81 %.