Answer:
8.66 g of Al₂O₃ will be produced
Explanation:
4Al (s) + 3O₂ (g) → 2Al₂O₃ (s)
This is the reaction.
Problem statement says, that the O₂ is in excess, so the limiting reactant is the Al. Let's determine the moles we used.
4.6 g / 26.98 g/mol = 0.170 moles
Ratio is 4:2.
4 moles of aluminum can produce 2 moles of Al₂O₃
0.170 moles of Al, may produce (0.170 .2)/ 4 = 0.085 moles
Let's convert the moles of Al₂O₃ to mass.
0.085 mol . 101.96 g/mol = 8.66 g
Answer is: molarity of solution is 0,5 mol/dm³.
m(NaOH) = 10,0 g.
V(NaOH) = 500 ml = 0,5 dm³.
c(NaOH) = ?
n(NaOH) = m(NaOH) ÷ M(NaOH).
n(NaOH) = 10,0 g ÷ 40 g/mol.
n(NaOH) = 0,25 mol.
c(NaOH) = n(NaOH) ÷ V(NaOH).
c(NaOH) = 0,25 mol ÷ 0,5 dm³.
c(NaOH) = 0,5 mol/dm³.
False, I’m going from what I know but if it’s correct pls heart this
Answer:
<h2>Heterogeneous</h2>
Explanation:
<h3><em>Milk </em><em>seems</em><em> to</em><em> be</em><em> </em><em>homogeneous</em><em> mixture</em><em> </em><em>but </em><em>actually</em><em> </em><em>milk </em><em>is </em><em>a </em><em>heterogeneous</em><em> </em><em>mixture</em><em> </em><em>and </em><em>a </em><em>colloid</em><em> </em><em>solution</em><em>.</em></h3>
