Answer:
a) 2 (H+) ions
b) 1 (SO3²-) ions
c) 1.36 × 10^-22 grams.
Explanation:
According to this question, sulfurous acid has a chemical formula; H2SO3. It is made up of hydrogen and sulfite ion. Hydrogen ion (H+) is the cation while sulfite ion (SO32-) is the anion.
Based on the chemical formula, there are 2 moles of hydrogen ions that reacts with 1 mole of sulfite ion as follows:
2H+ + SO3²- → H2SO3
Hence;
- there are 2 hydrogen ions (2H+) present in H2SO3.
- there is 1 sulfite ion (SO3²-) present in H2SO3.
c) The mass of one formula unit of H2SO3 is calculated thus:
= 1.008 (2) + 32.065 + 15.999(3)
= 2.016 + 32.065 + 47.997
= 82.08 a.m.u
Since, 1 gram is = 6.02 x 10^23 a.m.u
82.08 a.m.u = 82.08/6.02 × 10^-23
= 13.6 × 10^-23
= 1.36 × 10^-22 grams.
Yes, especially if its from a lake, pond, or ocean since water moves around. Dinasours could have spit out or maybe dripped some water, so basically yes.
Answer:
The pressure of CH3OH and HCl will decrease.
The final partial pressure of HCl is 0.350038 atm
Explanation:
Step 1: Data given
Kp = 4.7 x 10^3 at 400K
Pressure of CH3OH = 0.250 atm
Pressure of HCl = 0.600 atm
Volume = 10.00 L
Step 2: The balanced equation
CH3OH(g) + HCl(g) <=> CH3Cl(g) + H2O(g)
Step 3: The initial pressure
p(CH3OH) = 0.250atm
p(HCl) = 0.600 atm
p(CH3Cl)= 0 atm
p(H2O) = 0 atm
Step 3: Calculate the pressure at the equilibrium
p(CH3OH) = 0.250 - X atm
p(HCl) = 0.600 - X atm
p(CH3Cl)= X atm
p(H2O) = X atm
Step 4: Calculate Kp
Kp = (pHO * pCH3Cl) / (pCH3* pHCl)
4.7 * 10³ = X² /(0.250-X)(0.600-X)
X = 0.249962
p(CH3OH) = 0.250 - 0.249962 = 0.000038 atm
p(HCl) = 0.600 - 0.249962 = 0.350038 atm
p(CH3Cl)= 0.249962 atm
p(H2O) = 0.249962 atm
Kp = (0.249962 * 0.249962) / (0.000038 * 0.350038)
Kp = 4.7 *10³
The pressure of CH3OH and HCl will decrease.
The final partial pressure of HCl is 0.350038 atm
Letter C would be the correct answer
What the heck is this 9287262729272727272727171
