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>
Solid Magnesium is considered as active metal so it reacts with strong acids like HCl and H₂SO₄ liberating Hydrogen gas according to the following equations:
Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g)
Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g)
so the amount of solid magnesium decrease by addition of strong acid to it.
Answer:
the surface tension of H20 is 72 dynes/cm at 25°C
<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.0275 M
<u>Explanation:</u>
Molarity is calculated by using the equation:
Moles of HI = 0.550 moles
Volume of container = 2.00 L
For the given chemical equation:
<u>Initial:</u> 0.275
<u>At eqllm:</u> 0.275-2x x x
The expression of 
for above equation follows:
![K_c=\frac{[H_2][I_2]}{[HI]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5D%5BI_2%5D%7D%7B%5BHI%5D%5E2%7D)
We are given:
Putting values in above expression, we get:
Neglecting the negative value of 'x' because concentration cannot be negative
So, equilibrium concentration of hydrogen gas = x = 0.0275 M
Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M
Answer:
add 7.5L of water
Explanation:
M1×V1=M2×V2
M is molarity, V is volume
0.7 × 10 = 0.4 × V2
V2= 17.5L
vol. of water to add= 17.5 - 10 = 7.5L
