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:
I think it should be Carbon.
Chemical property can be referred to as a reaction into which a substance is changed
Answer:
The energy required to ionize the ground-state hydrogen atom is 2.18 x 10^-18 J or 13.6 eV.
Explanation:
To find the energy required to ionize ground-state hydrogen atom first we calculate the wavelength of photon required for this operation.
It is given by Bohr's Theory as:
1/λ = Rh (1/n1² - 1/n2²)
where,
λ = wavelength of photon
n1 = initial state = 1 (ground-state of hydrogen)
n2 = final state = ∞ (since, electron goes far away from atom after ionization)
Rh = Rhydberg's Constant = 1.097 x 10^7 /m
Therefore,
1/λ = (1.097 x 10^7 /m)(1/1² - 1/∞²)
λ = 9.115 x 10^-8 m = 91.15 nm
Now, for energy (E) we know that:
E = hc/λ
where,
h = Plank's Constant = 6.625 x 10^-34 J.s
c = speed of light = 3 x 10^8 m/s
Therefore,
E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s)/(9.115 x 10^-8 m)
<u>E = 2.18 x 10^-18 J</u>
E = (2.18 x 10^-18 J)(1 eV/1.6 x 10^-19 J)
<u>E = 13.6 eV</u>
