Answer:
1.2 x 10²⁴atoms
Explanation:
Given parameters:
Molar mass of Mg = 24.3g/mol
Mass of Mg = 48.6g
Number of atoms in 1 mole = 6.02 x 10²³ atoms
Solution
We know that
1 mole contains 6.02 x 10²³ atoms
Now we have to find the number of moles that are in the given mass of the magnesium atom.
Number of moles = ![\frac{mass}{molar mass}](https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7Bmolar%20mass%7D)
Number of moles =
= 2mole
Now 2 moles of Mg will contain 2x 6.02 x 10²³ atoms = 1.2 x 10²⁴atoms
I really dont know so yea
Answer:
i wanna say its the third one
Explanation:
but im not sure
<u>Answer:</u> The mass of manganese(III) oxide produced is 113.03 g
<u>Explanation:</u>
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
......(1)
Given mass of zinc = 46.8 g
Molar mass of zinc = 65.38 g/mol
Plugging values in equation 1:
![\text{Moles of zinc}=\frac{46.8g}{65.38g/mol}=0.716 mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20zinc%7D%3D%5Cfrac%7B46.8g%7D%7B65.38g%2Fmol%7D%3D0.716%20mol)
The given chemical equation follows:
![Zn+2MnO_2+H_2O\rightarrow Zn(OH)_2+Mn_2O_3](https://tex.z-dn.net/?f=Zn%2B2MnO_2%2BH_2O%5Crightarrow%20Zn%28OH%29_2%2BMn_2O_3)
By the stoichiometry of the reaction:
If 1 mole of zinc produces 1 mole of manganese(III) oxide
So, 0.716 moles of zinc will produce =
of manganese(III) oxide
Molar mass of manganese(III) oxide = 157.87 g/mol
Plugging values in equation 1:
![\text{Mass of manganese(III) oxide}=(0.716mol\times 157.87g/mol)=113.03g](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20manganese%28III%29%20oxide%7D%3D%280.716mol%5Ctimes%20157.87g%2Fmol%29%3D113.03g)
Hence, the mass of manganese(III) oxide produced is 113.03 g