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>
Answer:
[Ar] 3d10 4s2 4p3 is shorthand, 1s22s22p63s23p63d104s24p3 long hang
Explanation:
The shorthand is made using the lowest & closest noble gas, and picking up where it leaves off as follows, and longhand is made from a followed pattern you can easily find
Ok then! So mitosis is when a cell splits and doesn't lose/gain any chromosomes. In meiosis the chromosomes join and split evenly at the cell's "poles". Chromosomes will be lost evenly through this process.
Answer:
C₂Cl₄
Explanation:
To know if free rotation around a bond in a compound is possible, we need to see the structure of the compound (picture in attachment).
In single bonds, which are formed by σ bonds, the atoms are not fixed in a single position, and free rotation is permitted.
Double and triple bonds are formed by a σ bond and one or two π bonds, respectively. These bonds do not allow rotation, since it is not possible to twist the ends without breaking the π bond.
The chloroethylene (C₂Cl₄) has two carbons with an sp2-sp2 hybridization, they are bonded together by a double bond. <u>Free rotation on this bond is not possible, because six atoms, including the carbon atoms, doubly bonded and the four chlorine atoms bonded to them, must be on the same plane. </u>
I think it would be minimize so u can have more friction