Answer:
5 g
Explanation:
The heat required to vaporize ice is the sum of
i) Heat required to melt ice at 0°C
ii) Heat required to raise the temperature from 0°C to 100°C
iii) Heat required to vaporize water at 100°C
Thus;
H = nLfus + ncθ + nLvap
H= n(Lfus + cθ + Lvap)
Lfus = 6.01 kJ/mol
Lvap = 41 kJ/mol
c = 75.38
n =?
2100 = n(6.01 + 75.38(100) + 41)
n = 2100 KJ/7585.01 kJ/mol
n = 0.277 moles
Mass of water = number of moles * molar mass
Mass of water = 0.277 moles * 18 g/mol
Mass of water = 5 g
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:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄ 
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Answer:
0.46 grams (C₆H₅)₂CO
Explanation:
To find the mass of benzophenone ((C₆H₅)₂CO), you need to (1) convert mmoles to moles and then (2) convert moles to grams (via molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units. The final answer should have 2 sig figs to match the sig figs of the given value (2.5 mmoles).
Molar Mass ((C₆H₅)₂CO): 13(12.011 g/mol) + 10(1.008 g/mol) + 15.998 g/mol
Molar Mass ((C₆H₅)₂CO): 182.221 g/mol
2.5 mmoles (C₆H₅)₂CO 1 mole 182.221 g
----------------------------------- x ------------------------ x ------------------- =
1,000 mmoles 1 mole
= 0.46 grams (C₆H₅)₂CO
