Answer:
Your strategy here will be to use the molar mass of potassium bromide,
KBr
, as a conversion factor to help you find the mass of three moles of this compound.
So, a compound's molar mass essentially tells you the mass of one mole of said compound. Now, let's assume that you only have a periodic table to work with here.
Potassium bromide is an ionic compound that is made up of potassium cations,
K
+
, and bromide anions,
Br
−
. Essentially, one formula unit of potassium bromide contains a potassium atom and a bromine atom.
Use the periodic table to find the molar masses of these two elements. You will find
For K:
M
M
=
39.0963 g mol
−
1
For Br:
M
M
=
79.904 g mol
−
1
To get the molar mass of one formula unit of potassium bromide, add the molar masses of the two elements
M
M KBr
=
39.0963 g mol
−
1
+
79.904 g mol
−
1
≈
119 g mol
−
So, if one mole of potassium bromide has a mas of
119 g
m it follows that three moles will have a mass of
3
moles KBr
⋅
molar mass of KBr
119 g
1
mole KBr
=
357 g
You should round this off to one sig fig, since that is how many sig figs you have for the number of moles of potassium bromide, but I'll leave it rounded to two sig figs
mass of 3 moles of KBr
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
360 g
a
a
∣
∣
−−−−−−−−−
Explanation:
<em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>:</em><em> </em><em>3</em><em>6</em><em>0</em><em> </em><em>g</em><em> </em>
If 2.34 moles of Mg react with 3.56 moles of l2 and 1.76 moles of Mgl2 form, what is the percent yield?
Answer:
7.462
Explanation:
Well, every time that the tempurature is increased, the atmspheric pressure is increased by 0.574%. This would then mean that you would have 0.574 times
13. That would then equal 7.462. I hope this helps.
false, the rusting of iron can be prevented by painting, oiling, greasing or varnishing its surface.
Answer:
The final product of the reaction is (<em>2S,3S</em>)-2-ethoxy-3-methylpentane.
Explanation:
The given reaction undergoes 
mechanism in which the nucleophile attacks the backside and it is substituted by the elimination of bromine.
Due to the backside attack of nucleophile , the inverse in stereo-chemistry is observed.
After the substitution of ethoxy group, the configuration is assigned according to the priority it shows clock wise direction(R) - configuration.
When hydrogen faces the front side , it results shows inverse configuration i.e, S- configuration.
The chemical reaction is as follows.
