Answer:
2.803013439419911 × 10⁻¹² J
Explanation:
Mass defect = mass of reactant - mass of product
(2.0140 + 3.01605) - (4.002603 + 1.008665)
5.03005 - 5.011268 = 0.018782 amu
mass in Kg = mass (amu) × 1.66053892173 × 10⁻²⁷ kg
mass in kg = 0.018782 × 1.66053892173 × 10⁻²⁷ = 3.1188242027932 × 10⁻²⁹kg
E = Δm c² where c is the speed of light = 2.9979 × 10⁸m/s
E = 3.1188242027932 × 10⁻²⁹kg × (2.9979 × 10⁸m/s)² = 2.803013439419911 × 10⁻¹² J
Answer:
ΔH = 180.6 kJ
Explanation:
Given that:
N2 (g) + 2O2(g) = 2NO2 (g) ΔH = 66.4 kJ
<u>2NO (g) + O2 (g) = 2NO2 (g) ΔH = -114.2 kJ </u>
N2 (g) + O2 (g) = 2NO (g) ΔH = ????
The subtraction of both equations would yield the unknown ΔH , therefore:
ΔH = 66.4 - ( - 114.2 kJ)
ΔH = 180.6 kJ
Well it depends on how long you let it boil, if it boils for a long time then the water levels will slowly decrease, if u let it sit for a while you can drink it.
<span>I believe the correct 2nd reaction is:</span>
cof2(g)⇌1/2 co2(g)+1/2 cf4(g)
where we can see that it is exactly one-half of the
original
Therefore the new Kp is:
new Kp = (old Kp)^(1/2)
new Kp = (2.2 x 10^6)^(1/2)
<span>new Kp = 1,483.24 </span>
Step 1: Write the unbalanced equation,
C₂H₆ + O₂ → CO₂ + H₂<span>O
There are 2 C at left hand side and 1 carbon at right hand side. So, multiply CO</span>₂ by 2 to balance C atoms at both side. So,
C₂H₆ + O₂ → 2 CO₂ + H₂O
Now, count number of H atoms at both sides. There are 6 H atoms at left hand side and 2 at right hand side. Multiply H₂O by 3 to balance H atoms.
C₂H₆ + O₂ → 2 CO₂ + 3 H₂O
At last, balance O atoms. There are 2 O atoms at left hand side and 3 O atoms at right hand side. Multiply O₂ with 1.5 (i.e. 3/2) to balance O atoms. i.e.
C₂H₆ + 3/2 O₂ → 2 CO₂ + 3 H₂O
Hence, the equation is balanced. If you want to make equation fraction free then multiply all equation with 2. i.e.
( C₂H₆ + 3/2 O₂ → 2 CO₂ + 3 H₂O ) × 2
2 C₂H₆ + 3 O₂ → 4 CO₂ + 6 H₂O
