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>
The pH = -log of hydrogen ion concentration
The more acid a solution is the higher is its concentration of H+ ions and the lower is its pH. Acids have a pH lower that 7.
Explanation:
elctronic configuration of manganese
Mn=1s²2s²2p⁶3s²3p⁶4s²3d⁵
ground state
Mn=Ar3d⁵4s²
note that Ar is argon
Percent error (%)= 
Accepted value is true value.
Measured values is calculated value.
In the question given Accepted value (true value) = 63.2 cm
Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm
1) Percent error (%) for first measurement.
Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm
Percent error (%)= 



Percent error = 0.158 %
2) Percent error (%) for second measurement.
Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm
Percent error (%)= 



Percent error = 0.316 %
3) Percent error (%) for third measurement.
Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm
Percent error (%)= 




Percent error = 0.791 %
Percent error for each measurement is :
63.1 cm = 0.158%
63.0 cm = 0.316%
63.7 cm = 0.791%
Stoichiometry <span>of the reaction:
</span><span>2 KClO</span>₃<span> = 2 KCl + 3 O</span>₂
↓ ↓
2 mole KClO₃ ----------> 3 mole O₂
2 mole KClO₃ ----------> ?
KClO₃ = 2 * 3 / 2
KClO₃ = 6 / 2
= 3 moles de KClO₃
hope this helps!