Noble gas notation write the electron configuration for the manganese atom
1 answer:
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Answer:
Oxide of M is and sulfate of
Explanation:
0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.
Let moles of hydrogen gas be n.
Temperature of the gas ,T= 17°C =290 K
Pressure of the gas ,P= 741 mmHg= 0.9633 atm
Volume occupied by gas , V = 0.303 L
Using an ideal gas equation:
Moles of hydrogen gas produced = 0.01225 mol
Moles of metal =
So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.
x = 2.9 ≈ 3
Formulas for the oxide and sulfate of M will be:
Oxide of M is and sulfate of
.
Answer:
We can not swim in 1.00 × 10²⁷ molecules of water
Explanation:
The given number of molecules of water = 1.00 × 10²⁷ molecules
The Avogadro's number, , gives the number of molecules in one mole of a substance
≈ 6.0221409 × 10²³ molecules/mol
Therefore
Therefore, we have;
The number of moles of water present in 1.00 × 10²⁷ molecules, n = (The number of molecules of water) ÷
∴ n = (1.00 × 10²⁷ molecules)/(6.0221409 × 10²³ molecules/mol) = 1,660.53902857 moles
The mass of one mole of water = The molar mass of water = 18.01528 g/mol
The mass, 'm', of water in 1,660.53902857 moles of water is given as follows;
Mass = (The number of moles of the substance) × (The molar mass of the substance)
∴ The mass of the water in the given quantity of water, m = 1,660.53902857 moles × 18.01528 g/mol ≈ 29.9150756 kg.
The density pf water, ρ = 997 kg/m³
Volume = Mass/Density
∴ The volume of the water present in the given quantity of water, v = 29.9150756 kg/(997 kg/m³) ≈ 30.0050909 liters
The volume of the water present in 1.00 × 10²⁷ molecules of water ≈ 30.0 liters
The average volume of a human body = 62 liters
Therefore, we can not swim in the given quantity of 1.00 × 10²⁷ molecules = 30.0 liters water