Answer:
0.0036 moles of NaOH are present
Explanation:
A solution of 0.1mol/dm³ contains 0.1 moles of solute (In this case, NaOH), per 1 dm³ of solution.
36cm³ are in dm³ (1000cm³ / 1dm³):
36cm³ * (1dm³/1000cm³) = 0.036dm³
The moles are:
0.036dm³ * (0.1mol/dm³) = 0.0036 moles of NaOH are present
Its D for plato.
by other people who asked
Metals in their pure form are highly flammable. In this item, we are given with the pure magnesium metal in ribbon form reacting with oxygen in the combustion process. The equation that would best illustrate this given is,
2Mg + O₂ --> 2MgO
In this item, we are to determine the amount of MgO that can will be formed.
Magnesium
(4.81 g)(1 mol Mg/24.31 g Mg)(2 mol MgO/2 mol Mg)(40.3 g MgO/1 mol MgO)
= 7.97 g MgO
Oxygen
(7.46 g)(1 mol O2/32 g O2)(2 mol MgO/1 mol O2)(40.3 g MgO/1 mol MgO)
= 18.79 g MgO
This means that the amount of MgO produced is 7.97 g.
<h3>
Answer:</h3>
H₂O
<h3>
Explanation:</h3>
The additional product that balances the reaction is water (H₂O).
- Therefore, the complete balanced equation for the reaction is;
H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + H₂O(l)
- Balancing a chemical equation ensures that the law of conservation of mass is obeyed by the reaction.
- The reaction is an example of a neutralization reaction where an acid, H₂SO₄, reacts with a base, NaOH to form a salt and water.
- In this case the salt formed is sodium sulfate (Na₂SO₄)
The number of mole of H in NH₃, given the data is 23.85 moles
<h3>Data obtained from the question</h3>
- Mole of NH₃ = 7.95 moles
- Mole of H in NH₃ =?
<h3>How to determine the mole of H in 7.95 moles of NH₃</h3>
1 mole of NH₃ contains 3 moles of H
Therefore,
7.95 moles of NH₃ will contain = 7.95 × 3 = 23.85 mole of H
Thus, 23.85 mole of H is present in 7.95 moles of NH₃
Learn more about mole:
brainly.com/question/13314627
#SPJ1