The statement that best describes the effect of low ionization energies and low electronegativities on metallic bonding is the first one - the valence electrons are easily delocalized.
Due to these low energies and negativities, valence electrons can be moved around quite easily and their positions may be altered quite drastically.
Answer:
Molarity of NaOH = 1.8 M.
Explanation:
From the question given above, the following data were obtained:
Mass of NaOH = 36 g
Molar mass of NaOH = 40 g/mol
Volume = 500 mL
Molarity of NaOH =?
Next, we shall determine the number of mole in 36 g of NaOH. This can be obtained as follow:
Mass of NaOH = 36 g
Molar mass of NaOH = 40 g/mol
Mole of NaOH =?
Mole = mass / molar mass
Mole of NaOH = 36 / 40
Mole of NaOH = 0.9 mole
Next, we shall convert 500 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
500 mL = 500 mL × 1 L / 1000 mL
500 mL = 0.5 L
Finally, we shall determine the molarity of NaOH. This can be obtained as follow:
Mole of NaOH = 0.9 mole
Volume = 0.5 L
Molarity of NaOH =?
Molarity = mole / Volume
Molarity of NaOH = 0.9 / 0.5
Molarity of NaOH = 1.8 M
Endothermic reaction is the answer.
Chemical reaction: Ba(NO₃)₂ + H₂SO₄ → BaSO₄ + 2HNO₃.
V(H₂SO₄) = 250 mL ÷ 1000 mL/L = 0,25 L.
m(BaSO₄) = 0,55 g.
n(BaSO₄) = m(BaSO₄) ÷ M(BaSO₄).
n(BaSO₄) = 0,55 g ÷ 233,38 g/mol.
n(BaSO₄) = 0,00235 mol.
From chemical reaction: n(BaSO₄) : n(Ba(NO₃)₂) = 1 : 1.
n(Ba(NO₃)₂) = 0,00235 mol.
c(Ba(NO₃)₂) = n(Ba(NO₃)₂) ÷ V.
c(Ba(NO₃)₂) = 0,00235 mol ÷ 0,25 L.
c(Ba(NO₃)₂) = 0,0095 mol/L.
