Answer:
108.9g of Silver can be produced from 125g of Ag2S
Explanation:
The compound Ag2S shows that two atoms of Silver Ag, combined with an atom of Sulphur S to form Ag2S. We can as well say the combination ration of Silver to Sulphur is 2:1
•Now we need to calculate the molecular weight of this compound by summing up the molar masses of each element in the compound.
•Molar mass of Silver Ag= 107.9g/mol
•Molar mass of Sulphur S= 32g/mol
•Molecular weight of Ag2S= (2×107.9g/mol) + 32g/mol
•Molecular weight of Ag2S= 215.8g/mol + 32g/mol= 247.8g/mol
•From our calculations, we know that 215.8g/mol of Ag is present in 247.8g/mol of Ag2S
If 247.8g Ag2S produced 215.8g Ag
125g Ag2S will produce xg Ag
cross multiplying we have
xg= 215.8g × 125g / 247.8g
xg= 26975g/247.8
xg= 108.85g
Therefore, 108.9g of Silver can be produced from 125g of Ag2S
Answer:
there are no valence electrons left over, so the molecule has four bond pairs and no lone pairs.
Explanation:
Freezing, condensation, Deposition.
Answer:
When the two atoms move towards each other a compound is formed by sharing electron pairs supplied by each of the atoms to enable them have the stable 8 (octet) valency electrons in their outermost shell
Explanation:
The electronic configuration of the given element can be written as follows;
1s²2s²2p⁴
The given electronic configuration is equivalent to that of oxygen, therefore, we have;
The number of electrons in the valence shell = 2 + 4 = 6 electrons
Therefore, each atom requires 2 electrons to complete its 8 (octet) electrons in the outermost shell
When the two atoms move towards each other, they react and combine to form a compound by sharing 4 electrons, 2 from each atom, such that each atom can have an extra 2 electrons in its outermost orbit in the newly formed compound and the stable octet configuration is attained by each of the atoms in the newly formed compound.
Answer:
1.2×10² mmole of Na₂S₂O₃
Explanation:
From the question given above, the following data were obtained:
Volume = 0.6 L
Molarity = 0.2 mol/L
Mole of Na₂S₂O₃ =?
Molarity is simply defined as the mole of solute per unit litre of water. Mathematically, it is expressed as:
Molarity = mole /Volume
With the above formula, we can obtain the number of mole of Na₂S₂O₃ in the solution as illustrated below:
Volume = 0.6 L
Molarity = 0.2 mol/L
Mole of Na₂S₂O₃ =?
Molarity = mole /Volume
0.2 = Mole of Na₂S₂O₃ / 0.6
Cross multiply
Mole of Na₂S₂O₃ = 0.2 × 0.6
Mole of Na₂S₂O₃ = 0.12 mole
Finally, we shall convert 0.12 mole to millimole (mmol). This can be obtained as follow:
1 mole = 1000 mmol
Therefore,
0.12 mole = 0.12 mole × 1000 mmol / 1 mole
0.12 mole = 120 = 1.2×10² mmole
Thus, the chemist added 1.2×10² mmole of Na₂S₂O₃
