Answer:
20.3 kJ of heat is absorbed when 9.00 g of steam condenses to liquid water.
Explanation:
Heat is being consumed during vaporization and heat is being released during condensation.
To vaporize 1 mol of water, 40.66 kJ of heat is being consumed.
Molar mass of water = 18.02 g/mol
Hence, to vaporize 18.02 g of water , 40.66 kJ of heat is being consumed.
So, to vaporize 9.00 g of water, 
of heat or 20.3 kJ of heat is being consumed
As condensation is a reverse process of vaporization therefore 20.3 kJ of heat is absorbed when 9.00 g of steam condenses to liquid water.
C. Temperature the average <span> kinetic energy of the particles in an object is directly proportional to its temperature </span>
Answer:
0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl
Explanation:
Strong acids are more acids than weak acids. In the same way, strong bases are more basic than weak bases that are in the same concentration.
Then, the more concentrated acid or base will be more acidic or basic.
CH3COOH. Weak acid
NaOH. Strong base
H2SO4. Strong acid
NH3. Weak base.
HCl. Strong acid
The less acid (More basic):
<h3>0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl</h3>
Strong base, weak base, weak acid, diluted strong acid, undiluted strong acid
Answer:
(a) 7.11x10⁻⁴ M/s
(b) 2.56 mol.L⁻¹.h⁻¹
Explanation:
(a) The reaction is:
O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)
The reaction rate of equation (1) is given by:
![rate = k*[O_{3}][NO]](https://tex.z-dn.net/?f=%20rate%20%3D%20k%2A%5BO_%7B3%7D%5D%5BNO%5D%20)
(2)
<u>We have:</u>
k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹
[O₃]₀ = 2.35x10⁻⁶ M
[NO]₀ = 7.74x10⁻⁵ M
Hence, to find the inital reacion rate we will use equation (2):
![rate = k*[O_{3}]_{0}[NO]_{0} = 3.91 \cdot 10^{6} M^{-1}s^{-1}*2.35\cdot 10^{-6} M*7.74 \cdot 10^{-5} M = 7.11 \cdot 10^{-4} M/s](https://tex.z-dn.net/?f=%20rate%20%3D%20k%2A%5BO_%7B3%7D%5D_%7B0%7D%5BNO%5D_%7B0%7D%20%3D%203.91%20%5Ccdot%2010%5E%7B6%7D%20M%5E%7B-1%7Ds%5E%7B-1%7D%2A2.35%5Ccdot%2010%5E%7B-6%7D%20M%2A7.74%20%5Ccdot%2010%5E%7B-5%7D%20M%20%3D%207.11%20%5Ccdot%2010%5E%7B-4%7D%20M%2Fs%20)
Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s
(b) The number of moles of NO₂(g) produced per hour per liter of air is:
t = 1 h
V = 1 L
![\frac{\Delta[NO_{2}]}{\Delta t} = rate](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%20rate)
![\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%207.11%20%5Ccdot%2010%5E%7B-4%7D%20M%2Fs%2A%5Cfrac%7B3600%20s%7D%7B1%20h%7D%20%3D%202.56%20mol.L%5E%7B-1%7D.h%7B-1%7D)
Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹
I hope it helps you!
