The temperature change is calculated using the combined gas law
that is P1V1/T1 =P2V2/T2
P1= 100KPa
P2=90kpa
v1= 2.50 L
v2= 3.75 L
T1= 303 K
T2=?
T2 is therefore = P2V2T1/P1V1
=( 90 x 3.75 x303)/ (100 x2.50) = 409.05 K
Arrhenius acids and Bronsted Lowrey acids are the exact same. HCl is one example
3. Cl₂ + 2KI --> 2KCl + I₂
Cl = 2 Cl = 2
K = 2 K = 2
I = 2 I = 2
4. 2NaCl --> 2Na + Cl₂
Na = 2 Na = 2
Cl = 2 Cl = 2
5. 4Na + O₂ --> 2Na₂O
Na = 4 Na = 4
O = 2 O = 2
6. 2Na + 2HCl --> H₂ + 2NaCl
Na = 2 Na = 2
H = 2 H = 2
Cl = 2 Cl = 2
7. 2K + Cl₂ --> 2KCl
K = 2 K = 2
Cl = 2 Cl = 2
CH4 + 2O2 -> 2H2O + CO2
You need to add a 2 in front of the O2 and H2O in order to balance the equation.
Answer:
a) 88.48%
b) 0.05625 mol
Explanation:
2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g) Reaction 1
CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g) Reaction 2
a) CH₃CH₂OH = 46.0684 g/mol
CH₃CH₂OCH₂CH₃ = 74.12 g/mol
1 mol CH₃CH₂OH ______ 46.0684 g
x ______ 50.0 g
x = 1.085 mol CH₃CH₂OH
1 mol CH₃CH₂OCH₂CH₃ ______ 74.12 g g
y ______ 35.9 g
y = 0.48 mol CH₃CH₂OCH₂CH₃
100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃
w _____ 0.48 mol CH₃CH₂OCH₂CH₃
w = 88.48%
b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.
