Since the container of the gas is rigid, the volume of the gas will remain constant. Therefore, when the number of particles were decreased in half then the pressure will also be half of the original given they both are subjected to the same temperature.
PV = nRT
V, T and R are constants so they can be lumped together to a constant k.
P/n = k
P1/n1 = P2/n2
since n2 = n1/2
P1/n1 = P2/<span>n1/2</span>
P2 = P1/2
Given buffer:
potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )
[KHC4H4O6] = 0.0451 M
[K2C4H4O6] = 0.028 M
Ka1 = 9.2 *10^-4
Ka2 = 4.31*10^-5
Based on Henderson-Hasselbalch equation;
pH = pKa + log [conjugate base]/[acid]
where pka = -logKa
In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2
pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]
= -log (4.31*10^-5) + log [0.0451]/[0.028]
pH = 4.15
Answer:
2.9 g
Explanation:
There is some info missing. I think this is the original question.
<em>Ammonium perchlorate is the solid rocket fuel used by the U.S. Space Shuttle. It reacts with itself to produce nitrogen gas, chlorine gas, oxygen gas, water, and a great deal of energy.
</em>
<em>What mass of water is produced by the reaction of </em><em>9.6 g</em><em> of ammonium perchlorate?</em>
<em />
Step 1: Given data
Mass of ammonium perchlorate: 9.6 g
Step 2: Write the balanced equation
2 NH₄ClO₄(s) → N₂(g) + Cl₂(g) + 2 O₂(g) + 4 H₂O(l)
Step 3: Calculate the moles corresponding to 9.6 g of ammonium perchlorate
The molar mass of ammonium perchlorate is 117.49 g/mol.

Step 4: Calculate the moles of water formed from 0.082 moles of ammonium perchlorate
The molar ratio of NH₄ClO₄ to H₂O is 2:4. The moles of water formed are 4/2 × 0.082 mol = 0.16 mol
Step 5: Calculate the mass corresponding to 0.16 moles of water
The molar mass of water is 18.02 g/mol.
