Answer:
480.40 g.
Explanation:
- According to the balanced equation:
<em>2 SO₂(g) + O₂(g) → 2 SO₃(g),</em>
it is clear that 2.0 moles of SO₂ react with 1.0 mole of oxygen to produce 2.0 moles of SO₃.
- We can get the no. of moles of SO₃ produced:
∵ 2.0 moles of SO₂ produce → 2.0 moles of SO₃, from the stichiometry.
∴ 6.0 moles of SO₂ produce → 6.0 moles of SO₃.
- Then, we can get the mass of the produced 6.0 moles of SO₃ using the relation:
<em>mass = no. of moles x molar mass of SO₃</em> = (6.0 moles)(80.066 g/mol) = <em>480.396 g ≅ 480.40 g.</em>
15.5% by mass is
equivalent 15.5 g urea in 100 g solution or 155 g urea in 1 kg solution. <span>
<span>we know that molality = moles solute / kg solvent
<span>moles solute = 155 g / 60 g/mol = 2.58 moles urea
</span></span></span>
Since there are 155 g
urea in 1000g solution, hence the solvent is 845 g or 0.845 kg
So:<span>
<span>molality = 2.58 / 0.845 = 3.06 m</span></span>
Arrhenius Bases, so that other compounds that have the hydroxyl group (OH⁻)
hope this helps!
Answer:
We produce 2.0 moles of KCl when we have 2.00 moles K and and excess Cl2
Explanation:
Step 1: Data given
Number of moles K = 2.00 moles
Cl2 is in excess, so K is the limiting reactant
Step 2: The balanced equation
2K + Cl2 → 2KCl
Step 3: Calculate moles KCl
For 2 moles K we need 1 mol Cl2 to produce 2 moles KCl
We produce 2.0 moles of KCl when we have 2.00 moles K and and excess Cl2