<span>ideal gas law: PV = nRT so .....</span><span> V = PV/(RT) </span>
<span>
Initial number of moles of Cl, n = 0.943*5.11/(0.08206 × 286) mol = 0.2053 moles.
</span><span>
We know the molar mass of K (potassium) = 39.0 g/mol </span>
<span>sooo....
The Initial number of moles of K = 29.0 g/(39.0 g/mol) = 0.7436 moles</span>
<span>Find the balanced equation for the reaction : </span><span>2K + Cl2 → 2KCl </span>
<span>Mole ratio of K:Cl = 2:1 </span>
<span>So after the reaction, the amount of K needed = (0.2053 mol) × 2 = 0.4106 mol which is less than 0.7436 mol </span>
<span>
This means that K is in excess but Cl completely reacts. </span>
<span> So we know the mole ratio is Cl:KCl = 1 : 2
</span>
<span>Number of moles of Cl (completely) reacted = 0.2053 mol which means the n</span><span>umber of moles of KCl formed = (0.2053 mol) × 2 = 0.4106 mol </span>
<span>Molar mass of KCl = (39.0 + 35.5) g/mol = 74.5 g/mol </span>
<span>Mass of KCl formed = 0.4106 mol * 74.5 g/mol = 30.6 g</span>
Answer:
The correct option is;
d. Explicit knowledge
Explanation:
Explicit knowledge is the knowledge that can be easily articulated documented stored in a retrieval system accesses, transmitted and shared with others
Tacit knowledge is the skill developed by an individual based on actual experience such that such knowledge comprise of both facts and perspectives
Hence explicit knowledge and tacit knowledge are complementary
The operations performed by Kaia include documentation, storing in a retrieval system (her project report) and accessing what she documented, this is an example of explicit knowledge.
Answer: noble gasses
Explanation: zoom on on the red section and look at the key for what red means.
Answer:
4.26 %
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10
⁻⁴.</em>
<em />
Step 1: Given data
Initial concentration of the acid (Ca): 0.249 M
Acid dissociation constant (Ka): 4.50 × 10
⁻⁴
Step 2: Write the ionization reaction for nitrous acid
HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)
Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])
We will use the following expression.
![[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%20%5D%20%3D%20%5Csqrt%7BCa%20%5Ctimes%20Ka%20%7D%20%3D%20%5Csqrt%7B0.249%20%5Ctimes%204.50%20%5Ctimes%2010%5E%7B-4%7D%20%20%7D%20%3D%200.0106%20M)
Step 4: Calculate the percent ionization of nitrous acid
We will use the following expression.
![\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5BA%5E%7B-%7D%20%5D%7D%7B%5BHA%5D%7D%20%5Ctimes%20100%5C%25%20%3D%20%5Cfrac%7B0.0106M%7D%7B0.249%7D%20%5Ctimes%20100%5C%25%20%3D%204.26%5C%25)
