Answer:
B) irreversible process
Explanation:
The process given here is irreversible.
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
Hi!
The correct option would be 3.85x10^(24)
To find the number of atoms in 250g of potassium, we need to first calculate the number of atoms in
1 mole of Potassium = 39g which contains 6.022x10^(23) atoms of K
<em>(Avogadro's constant value for the amount of molecules/atoms in one mole of any substance)</em>
<em>Solution</em>
So as 39g of Potassium contains 6.022x10^(23) K atoms
1g of Potassium would contain 6.022x10^(23) / 39 = 1.544 x10^(22) atoms
So 250g of Potassium would contain 1.544x10^(22) x 250 = 3.86x10^(24) atoms
Answer:
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Explanation:
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Answer:
Percentage yield = 85.2%
Explanation:
Given data:
Mass of Mg = 21.3 g
Actual yield of MgO = 30.2 g
Percentage yield = ?
Solution:
Chemical equation:
2Mg + O₂ → 2MgO
Number of moles of Mg = mass/molar mass
Number of moles of Mg = 21.3 g / 24.3 g/mol
Number of moles of Mg = 0.88 mol
Now we will compare the moles of MgO with Mg.
Mg : MgO
2 : 2
0.88 : 0.88
Mass of MgO:
Mass of MgO= moles × molar mass
Mass of MgO= 0.88 mol × 40.3g/mol
Mass of MgO = 35.46 g
Actual yield of MgO = 30.2 g
Percentage yield:
Percentage yield = Actual yield/theoretical yield × 100
Percentage yield = 30.2 g/ 35.46 g × 100
Percentage yield = 85.2%
