<span>Ammonia (NH3) is the combination of Nitrogen and Hydrogen
elements.
=> N2 + 3H2 => 2NH3
Ammonia is basically used as a fertilizer. It is a gas composed of nitrogen and
hydrogen. It is colorless with strong odor. Here are some other uses of Ammonia
aside from fertilizer:
=> used by manufacturer to produce synthetic fiber
=> Used in metallurgical process
Ammonia can be decomposed easily and it produce hydrogen that is very
convenient in welding.
Ammonia’s boiling point is -28.03 F and freezing point is -107.8F.
</span>
In this item, we are simply to find the ions that may bond and are able to form a formula unit. We are also instructed to give out their name. There are numerous possible combinations of ions to form a compound. Some answers are given in the list below.
1. Na⁺ , Cl⁻ , NaCl ---> sodium chloride (this is most commonly known as table salt)
2. C⁴⁺ , O²⁻ , CO₂ ---> carbon dioxide
3. Al³+ , Cl⁻ , AlCl₃ ----> aluminum chloride
4. Ca²⁺ , Cl⁻ , CaCl₂ ---> calcium chloride
5. Li⁺ , Br⁻ , LiBr ---> lithium bromide
6. Mg³⁺ , O²⁻ , Mg₂O₃ ----> magnesium oxide
7. K⁺ , I⁻ , KI ---> potassium iodide
8. H⁺ , Cl⁻ , HCl --> hydrogen chloride
9. H⁺ , Br⁻ , HBr ----> hydrogen bromide
10. Na⁺ , Br⁻ , NaBr ---> sodium bromide
Answer:
The correct answer is: pH= 4.70
Explanation:
We use the <em>Henderson-Hasselbach equation</em> in order to calculate the pH of a buffer solution:
![pH= pKa + log \frac{ [conjugate base]}{[acid]}](https://tex.z-dn.net/?f=pH%3D%20pKa%20%2B%20log%20%20%20%5Cfrac%7B%20%5Bconjugate%20base%5D%7D%7B%5Bacid%5D%7D)
Given:
pKa= 4.90
[conjugate base]= 4.75 mol
[acid]= 7.50 mol
We calculate pH as follows:
pH = 4.90 + log (4.75 mol/7.50 mol) = 4.90 + (-0.20) = 4.70
Answer:
81°C.
Explanation:
To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat released from water (Q = - 1200 J).
m is the mass of the water (m = 20.0 g).
c is the specific heat capacity of water (c of water = 4.186 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).
∵ Q = m.c.ΔT
∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).
(- 1200 J) = 83.72 final T - 7953.
∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.
<em>So, the right choice is: 81°C.</em>
