Answer:
Sodium Chloride has Ionic bond while Hydrogen Chloride has covalent bond.
Explanation:
Na has 11 electrons (2, 8, 1) and need to give away 1 electron to be stable
Cl has 17 electrons ( 2, 8, 7) and needs 1 electron to be stable.
Na transfers 1 electron to CL to form Ionic bond.
While
Hydrogen has 1 electron and shares with Chlorine to be stable.
Covalent bond involves sharing.
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Answer:
0.5 M
Explanation:
From the question given above, the following data were obtained:
Mass of NaOH = 80 g
Volume of solution = 4 L
Molarity =?
Next, we shall determine the number of mole in 80 g of NaOH. This can be obtained as follow:
Mass of NaOH = 80 g
Molar mass of NaOH = 23 + 16 + 1
= 40 g/mol
Mole of NaOH =?
Mole = mass / molar mass
Mole of NaOH = 80 / 40
Mole of NaOH = 2 moles
Finally, we shall determine the molarity of the solution. This can be obtained as follow:
Mole of NaOH = 2 moles
Volume of solution = 4 L
Molarity =?
Molarity = mole / Volume
Molarity = 2/4
Molarity = 0.5 M
Therefore, the molarity of the solution is 0.5 M.
Answer:
C. CH3COOH, Ka = 1.8 E-5
Explanation:
analyzing the pKa of the given acids:
∴ pKa = - Log Ka
A. pKa = - Log (1.0 E-3 ) = 3
B. pKa = - Log (2.9 E-4) = 3.54
C. pKa = - Log (1.8 E-5) = 4.745
D. pKa = - Log (4.0 E-6) = 5.397
E. pKa = - Log (2.3 E-9) = 8.638
We choose the (C) acid since its pKa close to the expected pH.
⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):
- pH = pKa + Log ([CH3COO-]/[CH3COOH])
∴ pH = 4.5
∴ pKa = 4.745
⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])
⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])
⇒ 0.5692 = [CH3COO-]/[CH3COOH]
∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-])/[CH3COOH]
⇒ 1.8 E-5 = [H3O+](0.5692)
⇒ [H3O+] = 3.1623 E-5 M
⇒ pH = - Log ( 3.1623 E-5 ) = 4.5
Balanced equation:
<span>CaO + 2 HCl --> CaCl2 + H2O </span>
<span>Calculate moles of each reactant: </span>
<span>60.4 g CaO / 56.08 g/mol = 1.08 mol CaO </span>
<span>69.0 g HCl / 36.46 g/mol = 1.89 mol HCl </span>
<span>Identify the limiting reactant: </span>
<span>Moles CaO needed to react with all HCl: </span>
<span>1.89 mol HCl X (1 mol CaO / 2 mol HCl) = 0.946 mol CaO </span>
<span>Because you have more CaO than that available, HCl is the limiting reactant. </span>
<span>Calculate moles and mass CaCl2: </span>
<span>1.89 mol HCl X (1 mol CaCl2 / 2mol HCl) X 111.0 g/mol = 105 g CaCl2</span>
