To solve the question we will assume that the gas behaves like an ideal gas, that is to say, that there is no interaction between the molecules. Assuming ideal gas we can apply the following equation:
Where,
P is the pressure of the gas
V is the volume of the gas
n is the number of moles
R is a constant
T is the temperature
Now, we have two states, an initial state, and a final state. The conditions for each state will be.
Initial state (1)
P1=975Torr=1.28atm
V1=3.8L
T1=-18°C=255.15K
Final state(2), STP conditions
P2=1atm
T2=273.15K
V2=?
We will assume that the number of moles remains constant, so the nR term of the first equation will be constant. For each state, we will have:
Since nR is the same for both states, we can equate the equations and solve for V2:
We replace the known values:
At STP conditions the gas would occupy 5.2L. First option
Why not search it up and figure it out and then write it how you would?
The correct answer is d. <span>6H20 + 6CO2.
The reactant in the chemical reaction </span>6H2O + 6CO2 ---> C6H12O6 + 6O2 is 6H20 + 6CO2. Remember that the reactant is always at the left side of the equation. So the correct answer is <span>6H20 + 6CO2 since it's in the left of the equation. I hope this answer helped you. </span>
Answer:
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Answer:
pH = 9,32
Explanation:
The compound with 2 ionizable groups has the following equilibriums:
H₂M ⇄ HM⁻ + H⁺ pka = 6,2
HM⁻ ⇄ M²⁻ + H⁺ pka = 9,5
The reaction of M²⁻ with HCl is:
M²⁻ + HCl → HM⁻ + Cl⁻
The moles of M²⁻ are:
0,100L×1,0M = 0,1moles
And moles of HCl are:
0,060L×1,0M = 0,06moles
That means that moles of M²⁻ will be 0,1-0,06 = 0,04mol and moles of HM⁻ will be the same than HCl, 0,06mol
Using Henderson-Hasselbalch formula:
pH = pka + log₁₀ [M²⁻] / [HM⁻]
Replacing:
pH = 9,5 + log₁₀ [0,04] / [0,06]
<em>pH = 9,32</em>
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I hope it helps!
