Answer:
5.47 x 10^23 atoms in Ag (sliver)
Explanation:
0.909 mol of Ag x 6.02 x 10^23 atoms of Ag/ 1 mol of Ag
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
The maximum amount of hydrogen gas that can be prepared is if all the hydrogen from both compounds is released.
The hydrogen in 4.94 g of SrH2 is calculated from the mass ratios between Sr and H
1) H2 in SrH2
Sr atomic mass = 87.62 g/mol
H2 molar mass = 2.02 g/mol
Mass of 1 mol of SrH2 = 87.62 g / mol + 2.02 g/mol = 89.64 g/mol
Ratio of H2 to SrH2 = 2.02 g H2 / 89.64 g SrH2
Proportion: 2.02 g H2 / 89.64 gSrH2 = x / 4.93 g SrH2
=> x = 4.93 g SrH2 * 2.02 g H2 / 89.64 g SrH2 = 0.111 g H2
2) H2 in H2O
2.02 g H2 / 18.02 g H2O * 4.14 g H2O = 0.464 g H2
3) Total mass of hydrogen = 0.111 g + 0.464 g = 0.575 g
Answer: 0.575 g
<h3>
Answer:</h3><h3>a) 9.033 × 10²³ particles</h3><h3>b) 4.068 × 10²⁴ particles</h3><h3>c) 1.51 × 10²³ particles</h3>
Explanation:
For us to answer these questions, we have to know two formulas:
- Number of particles = moles × Avogadro's Number
- Moles = Mass ÷ Molar Mass
Therefore:
a) particles of Na = 1.50 mol × (6.022 × 10²³) particles/mol
= 9.033 × 10²³ particles
b) particles of Pb = 6.755 mol × (6.022 × 10²³) particles/mol
= 4.068 × 10²⁴ particles
c) particles of Si
= (7.02 g ÷ 28.085 g/mol) × (6.022 × 10²³) particles/mol
= 1.51 × 10²³ particles