Answer:
1) Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make 0.157 mol ammonium sulfate when you neutralize 11.00 g ammonium hydroxide.
2) 2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O.
Explanation:
- Firstly, we should balance the equation of heptane combustion.
- We can balance the equation by applying the conservation of mass to the equation.
- The balanced equation is: <em>2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O.</em>
- This means that every 2.0 moles of ammonium hydroxide (NH₄OH) will produce 1.0 mole of ammonium sulfate (NH₄)₂SO₄ when it is neutralized by sulfuric acid.
- We need to calculate the no. of moles in 11.0 g of ammonium hydroxide that is neutralized using the relation: <em>n = mass/molar mass.
</em>
n of 11.0 g of ammonium hydroxide (NH₄OH) = mass/molar mass = (11.0 g)/(35.04 g/mol) = 0.314 mol.
<u><em>Using cross multiplication:
</em></u>
2.0 moles of NH₄OH make → 1.0 mole of (NH₄)₂SO₄.
0.314 mol of NH₄OH make → ??? moles of (NH₄)₂SO₄.
∴ The no. of moles of (NH₄)₂SO₄ that will be made from neutralizing (11.0 g) of NH₄OH = (0.314 mol)(1.0 mol)/(2.0 mol) = 0.157 mol.
<em>∴ Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make </em><em>0.157</em><em> mol ammonium sulfate when you neutralize 11.00 g ammonium hydroxide.</em>
½H2(g) + ½I2(g) → HI(g) ΔH = +6.2 kcal/mol
or...
½H2(g) + ½I2(g) + 6,2kcal/mole → HI(g)
________
21.0 kcal/mole + C(s) + 2S(s) → CS2(l)
or...
C(s) + 2S(s) → CS2(l) ΔH = +2,1 kcal/mole
_________
ΔH > 0 ----------->>> ENDOTHERMIC REACTIONS
Answer:
Explanation:
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Oxidizing agents:
Oxidizing agents oxidize the other elements and itself gets reduced.
Reducing agents:
Reducing agents reduced the other element are it self gets oxidized.
Consider the following reaction:
2AgCl + Zn → 2Ag + ZnCl₂
In this reaction oxidation state of Zn on left side is 0 while on right side +2 so it gets oxidized and oxidation state of Ag on left side is +1 and on right side 0 so it get reduced.
4NH₃ + 3O₂ → 2N₂ + 6H₂O
In this reaction oxidation state of nitrogen on left side is -3 while on right side 0 so it gets oxidized and oxidation state of oxygen on left side is 0 and on right side -2 so it get reduced.
Fe₂O₃ + 2Al → Al₂O₃ + 2Fe
In this reaction oxidation state of iron on left side is +3 while on right side 0 so it gets reduced and oxidation state of Al on left side is 0 and on right side +3 so it get oxidized.
The empirical formula of the compound is C. NiF₂.
<em>Step 1</em>. Calculate the <em>moles of each element</em>
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our job is to calculate the molar ratio of Ni to F.
Moles of Ni = 9.11 g Ni × (1 mol Ni /(58.69 g Ni) = 0.1552 mol Ni
Moles of F = 5.89 g F × (1 mol F/19.00 g F) = 0.3100 mol F
<em>Step 2</em>. Calculate the <em>molar ratio</em> of the elements
Divide each number by the smallest number of moles
Ni:F = 0.1552:0.3100 = 1:1.997 ≈ 1:2
<em>Step 3</em>: Write the <em>empirical formula</em>
EF = NiF₂
