Answer:
C. H₂SO₄
Explanation:
Given expression:
H₂O + SO₃ → H₂SO₄
In the chemical reaction, the product and the reactant have the same species.
Chemical reactions obeys the law of conservation of matter.
This way, atoms combine with one another to give a desired product. The products will contain a mix of the reactants.
- Fluorine is not part of the reactants
- Carbon is not part of the reactants
- Carbon is not part of the reactants.
The percent yield of this reaction is calculated as follows
Mg3N2 + 3H2O =2NH3 + 3Mgo
calculate the theoretical yield,
moles=mass/molar mass
moles Mg3N2= 3.82 g/100g/mol= 0.0382 moles(limiting regent)
moles of H2o= 7.73g/18g/mol = 0.429 moles ( in excess_)
by use of mole ratio between Mg3N2 to MgO which is 1:3 the moles of MgO = 0.0382 x3 = 0.1146 moles
mass =moles x molar mass
the theoretical mass is therefore = 0.1146mole x 40 g/mol = 4.58 grams
The % yield = actual mass/theoretical mass x1000
= 3.60/4.584 x100= 78.5%
The number of moles of KF needed to prepare the solution is 3 moles
<h3>What is molarity?</h3>
Molarity is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:
Molarity = mole / Volume
<h3>How to determine the mole of KF </h3>
- Volume = 2500 mL = 2500 / 1000 = 2.5 L
- Molarity = 1.2 M
- Mole of KF =?
Molarity = mole / Volume
1.2 = mole of KF / 2.5
Cross multiply
Mole of KF = 1.2 × 2.5
Mole of KF = 3 moles
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Answer:
Uranium mines operate in many countries, but more than 85% of uranium is produced in six countries: Kazakhstan, Canada, Australia, Namibia, Niger, and Russia. Historically, conventional mines open pit or underground were the main source of uranium.
Explanation:
Hope this helps
<h2>
1.25 g of 
would be produced from the complete reaction of 25 mL of 0.833 mol/L 
with excess 
</h2>
Explanation:
To calculate the number of moles for given molarity, we use the equation:
According to stoichiometry:
1 mole of 
will give = 1 mole of
0.0208 moles of will give =
of
Mass of 
Thus 1.25 g of would be produced from the complete reaction of 25 mL of 0.833 mol/L with excess
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