J.J. Thomson hypothesized and discovered that the atom was not the smallest unit of matter but that instead there were much smaller units. He discovered "sub-atomic particles" which make up atoms. The sub-atomic particle that Thomson discovered was the electron. He discovered this through a process of experiments testing cathode rays.
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:
There is 37.36 grams of K3PO4 produced
Explanation:
Step 1: Data given
Mass of H3PO4 = 29.6 grams
KOH is in excess
Molar mass of KOH = 56.11 g/mol
Molar mass of H3PO4 = 97.99 g/mol
Step 2: The balanced equation
3KOH(aq) + H3PO4(aq) ⇔ K3PO4(aq)+3H2O(l)
Step 3: Calculate mass of KOH
Mass KOH = mass KOH / molar mass KOH
Mass KOH = 29.6 grams / 56.11 g/mol
Mass KOH = 0.528 moles
Step 4: Calculate moles of K3PO4
Since KOH is the limiting reactant, We need 3 moles of KOH for each moles of H3PO4, to produce 1 mole of K3PO4 and 3 moles of H2O
For 0.528 moles of KOH we'll have 0.528/3 = 0.176 moles of K3PO4
Step 5: Calculate mass of K3PO4
Mass K3PO4 = moles K3PO4 * molar mass K3PO4
Mass K3PO4 = 0.176 moles * 212.27 g/mol
Mass K3PO4 = 37.36 grams
There is 37.36 grams of K3PO4 produced
A molecular size affects the rate of evaporation when the larger the intermolecular forces in a compound, the slower the evaporation rate and this correlates with temperature change.
Molecular size seems to have an effect on evaporation rates in that the larger a molecule gets or grows from a base chemical formula, its evaporation rate will get slower.
<h3>What is the molecular size?</h3>
This is a measure of the area a molecule occupies in three-dimensional space as this relates to the physical size of an individual molecule.
Hence, we can see that a molecular size affects the rate of evaporation the larger the forces, the lower the rate.
Read more about<em> molecular size</em> here:
brainly.com/question/16616599
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