Explanation:
mega=10raised to the power 6
kilo=10 raised to the power 3
centi = 10 raised to the power negative 2
Milli = 10 raised to the power negative 3
nano = 10 raised to the power negative 9
pico= 10 raised to the power negative 12
micro = 10 raised to the power negative 6
Answer:
2
Step-by-step explanation:
A. Moles before mixing
<em>Beaker I:
</em>
Moles of H⁺ = 0.100 L × 0.03 mol/1 L
= 3 × 10⁻³ mol
<em>Beaker II:
</em>
Beaker II is basic, because [H⁺] < 10⁻⁷ mol·L⁻¹.
H⁺][OH⁻] = 1 × 10⁻¹⁴ Divide each side by [H⁺]
[OH⁻] = (1 × 10⁻¹⁴)/[H⁺]
[OH⁻] = (1 × 10⁻¹⁴)/(1 × 10⁻¹²)
[OH⁻] = 0.01 mol·L⁻¹
Moles of OH⁻ = 0.100 L × 0.01 mol/1 L
= 1 × 10⁻³ mol
B. Moles after mixing
H⁺ + OH⁻ ⟶ H₂O
I/mol: 3 × 10⁻³ 1 × 10⁻³
C/mol: -1 × 10⁻³ -1 × 10⁻³
E/mol: 2 × 10⁻³ 0
You have more moles of acid than base, so the base will be completely neutralized when you mix the solutions.
You will end up with 2 × 10⁻³ mol of H⁺ in 200 mL of solution.
C. pH
[H⁺] = (2 × 10⁻³ mol)/(0.200 L)
= 1 × 10⁻² mol·L⁻¹
pH = -log[H⁺
]
= -log(1 × 10⁻²)
= 2
Answer:
See explanation.
Explanation:
Hello!
In this case, we consider the questions:
a. Ideal gas at:
i. 273.15 K and 22.414 L.
ii. 500 K and 100 cm³.
b. Van der Waals gas at:
i. 273.15 K and 22.414 L.
ii. 500 K and 100 cm³.
Thus, we define the ideal gas equation and the van der Waals one as shown below:
Whereas b and a for hydrogen sulfide are 0.0434 L/mol and 4.484 L²*atm / mol² respectively, therefore, we proceed as follows:
a.
i. 273.15 K and 22.414 L.
ii. 500 K and 100 cm³ (0.1 L).
b.
i. 273.15 K and 22.414 L: in this case, v = 22.414 L / 1.00 mol = 22.414 L/mol
ii. 500 K and 100 cm³: in this case, v = 0.1 L / 1.00 mol = 0.100 L/mol
Whereas we can see a significant difference when the gas is at 500 K and occupy a volume of 0.100 L.
Best regards!
