<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>
7,040 energy to freeze the silver :)
The theoretical yield of urea : = 227.4 kg
<h3>Further explanation</h3>
Given
Reaction
2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)
128.9 kg of ammonia
211.4 kg of carbon dioxide
166.3 kg of urea.
Required
The theoretical yield of urea
Solution
mol Ammonia (MW=17 g/mol)
=128.9 : 17
= 7.58 kmol
mol CO₂(MW=44 g/mol) :
= 211.4 : 44
= 4.805 kmol
Mol : coefficient of reactant , NH₃ : CO₂ :
= 7.58/2 : 4.805/1
=3.79 : 4.805
Ammonia as limiting reactant(smaller ratio)
Mol urea based on mol Ammonia :
=1/2 x 7.58
=3.79 kmol
Mass urea :
=3.79 kmol x 60 g/mol
= 227.4 kg
Answer:
C
Explanation:
There will be a two long hair and two short hair
Answer: Limitation: They may be more expensive and time consuming than lab experiments. Limitation: There is no control over extraneous variables that might bias the results. This makes it difficult for another researcher to replicate the study in exactly the same way.
Explanation:
Hope this helps!
