The reactant is Mercury (II) Oxide while the products are Mercury and Oxygen separately.
This is because the reactants are typically always on the left side of the yields symbol. In this decomposition reaction, it would still be the same as at the end of the reaction, there were to products produced: Mercury and Oxygen.
Products tend to always be on the right side of the yields symbol, they're what comes out of a reaction no matter what type.
Hope this helps!
Answer:
53.3 %.
Explanation: C2H4O2. = 2 * 12.011 + 4 * 1.008 + 2 * 15.999. = 60.052.
I could only find 7!
- independent variable
- dependent variable
- control group
- experimental group
- constant
- observation
- inference
Moles of solute for both a and b are the same = 1 mol
<h3>Further explanation</h3>
Given
a 500 cm³ of solution, of concentration 2 mol/dm³
b 2 litres of solution, of concentration 0.5 mol/dm³
Required
moles of solute
Solution
Molarity shows the number of moles of solute in every 1 liter of solution or mmol in each ml of solution
Can be formulated :
a.
V = 500 cm³ = 0.5 L
M = 2 mol/L
n=moles = M x V
n = 2 mol/L x 0.5 L
n = 1 mol
b.
V = 2 L
M = 0.5 mol/L
n=moles = M x V
n = 0.5 mol/L x 2 L
n = 1 mol
Answer:
Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.
2 S + 3 O₂ → 2 SO₃
The stoichiometric calculations is as follows:
6 g S * 1 mol/32.06 g S = 0.187 mol S
Moles O₂ needed = 0.187 mol S * 3 mol O₂/2 mol S = 0.2805 mol O₂
Since the molar mas of O₂ is 32 g/mol,
Mass of O₂ needed = 0.2805 mol O₂ * 32 g/mol = 8.976 g O₂
