Answer:
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)
Explanation:
Step 1: Data given
Volume = 500 mL = 0.500 L
The concentration sodium sulfate = 2.104 M
Step 2: The equation
Na2SO4 → 2Na+ + SO4^2-
For 1 mol Na2SO4 we have 2 moles sodium ion (Na+) and 1 mol sulfate ion (SO4^2-)
Step 3: Calculate the concentration of the ions
[Na+] = 2*2.104 M = 4.208 M
[SO4^2-] = 1*2.104 M = 2.104 M
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)
The answer is: 27 grams of aluminium.
Balanced chemical reaction: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂.
n(H₂) = 1.5 mol; amount of hydrogen.
Form chemical reaction: n(Al) : n(H₂) = 2 : 3.
n(Al) = 2 · 1.5 mol ÷ 3.
n(Al) = 1.0 mol; amount of aluminium.
m(Al) = n(Al) · M(Al).
m(Al) = 1 mol · 27 g/mol.
m(Al) = 27 g; mass of aluminium.
Answer:
this is a oxidation reaction
product is magnesium oxide
the balanced equation is
2Mg + O2 -> 2MgO
Answer:
<h2>2 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
From the question we have
We have the final answer as
<h3>2 g/mL</h3>
Hope this helps you
