Answer:
172 g Al
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s <em>gather all the information</em> in one place.
M_r: 26.98 101.96
4Al + 3O₂ ⟶ 2Al₂O₃
m/g: 325
(a) Calculate the <em>moles of Al₂O₃
</em>
n = 325 g Al₂O₃ × 1 mol Al₂O₃ /39.10 g Al₂O₃
n = 3.188 mol Al₂O₃
(b) Calculate the <em>moles of Al
</em>
The molar ratio is (4 mol Al/2 mol Al₂O₃)
n = 3.188 mol Al₂O₃ × (4 mol Al/2 mol Al₂O₃)
n = 6.375 mol Al
(c) Calculate the <em>mass of Al</em>
m = 6.375 mol Al × (26.98 g Al/1 mol Al)
m = 172 g Al
Note: The answer can have only <em>three</em> significant figures because that is all you gave for the mass of Al₂O₃.
Answer:
The answer is option b.
Explanation:
Amplitude is the distance apart each wave is.
Answer:
I don't fully understand what this is about...
Explanation:
sorry :(
<span>The particles are far apart from each other.</span>
Answer:-
The reaction of 2-bromopropane reacts with sodium iodide in acetone is an example of Sn2 reaction.
The I - attacks from backside to give the transition state for both.
If we compare the transition state for cyclobromopropane 2-bromopropane then we see in case of cyclobromopropane transition state, one of the H is very close to the incoming I -.
This results in steric strain and less stability of the transition state. Hence 2-bromopropane reacts with sodium iodide in acetone over 104 times faster than bromocyclopropane.
