Answer:
The answer to this question is 33.8
The correct answer is 12.044 × 10²³ molecules.
The molecular mass of H₂S is 34 gram per mole.
Number of moles is determined by using the formula,
Number of moles = mass/molecular mass
Given mass is 68 grams, so no of moles will be,
68/34 = 2 moles
1 mole comprises 6.022 × 10²³ molecules, therefore, 2 moles will comprise = 6.022 × 10²³ × 2
= 12.044 × 10²³ molecules.
Answer:
Part 1: - 1.091 x 10⁴ J/mol.
Part 2: - 1.137 x 10⁴ J/mol.
Explanation:
Part 1: At standard conditions:
At standard conditions Kp= 81.9.
∵ ΔGrxn = -RTlnKp
∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(81.9)) = - 1.091 x 10⁴ J/mol.
Part 2: PICl = 2.63 atm; PI₂ = 0.324 atm; PCl₂ = 0.217 atm.
For the reaction:
I₂(g) + Cl₂(g) ⇌ 2ICl(g).
Kp = (PICl)²/(PI₂)(PCl₂) = (2.63 atm)²/(0.324 atm)(0.217 atm) = 98.38.
∵ ΔGrxn = -RTlnKp
∴ ΔGrxn = - (8.314 J/mol.K)(298.0 K)(ln(98.38)) = - 1.137 x 10⁴ J/mol.
Deposition. Particles settle to the bottom of still water after being eroded.
Answer:
![V=0.0310L=3.10mL](https://tex.z-dn.net/?f=V%3D0.0310L%3D3.10mL)
Explanation:
Hello.
In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:
![n_{acid}=n_{KOH}](https://tex.z-dn.net/?f=n_%7Bacid%7D%3Dn_%7BKOH%7D)
Thus, since we are given 1.70 g of the acid, we compute the moles of acid that were titrated:
![n_{acid}=1.70g*\frac{1mol}{84.48g}=0.0201mol](https://tex.z-dn.net/?f=n_%7Bacid%7D%3D1.70g%2A%5Cfrac%7B1mol%7D%7B84.48g%7D%3D0.0201mol)
Which equal the moles of KOH. In such a way, since the molarity is defined as moles over liters (M=n/V), the liters are moles over molarity (V=n/M), thus, the resulting volume is:
![V=\frac{0.0201mol}{0.650mol/L}\\\\V=0.0310L=3.10mL](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B0.0201mol%7D%7B0.650mol%2FL%7D%5C%5C%5C%5CV%3D0.0310L%3D3.10mL)
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