Answer : If a substance is the limiting reactant, then it limits the formation of products because in the reaction it is present in limited amount.
Explanation :
While observing a chemical reaction, we can tell about whether a reactant is limiting or excess.
Step 1 : first write the chemical reaction and then balanced the chemical equation.
Step 2 : convert the given masses into the moles if mass of 
is 10.5 g and molar mass of is 28 g/mole and the mass of hydrogen is 0.40 g and molar mass of hydrogen is 2 g/mole.
Step 3 : Now we have to determine the limiting reagent and excess reagent.
Now we conclude that 
is the limiting reagent and hydrogen is an excess reagent.
Hypothesis :
Limiting reagent : It is the reagent in the chemical reaction that is totally consumed when the chemical reaction is complete. Limiting reagent limits the formation of products.
Answer: D
Explanation:
This is the answer because everyone knows he discovered gravity and he conducted scientific experiments to prove them which he also used math for
Hope this helps
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>
