Ok i help you answer these question
I'm just taking a self educated guess of A sorry if I'm wrong :C
Answer:
16.2 J
Explanation:
Step 1: Given data
- Specific heat of liquid bromine (c): 0.226 J/g.K
- Volume of bromine (V): 10.0 mL
- Initial temperature: 25.00 °C
- Final temperature: 27.30 °C
- Density of bromine (ρ): 3.12 g/mL
Step 2: Calculate the mass of bromine
The density is equal to the mass divided by the volume.
ρ = m/V
m = ρ × V
m = 3.12 g/mL × 10.0 mL
m = 31.2 g
Step 3: Calculate the change in the temperature (ΔT)
ΔT = 27.30 °C - 25.00 °C = 2.30 °C
The change in the temperature on the Celsius scale is equal to the change in the temperature on the Kelvin scale. Then, 2.30 °C = 2.30 K.
Step 4: Calculate the heat required (Q) to raise the temperature of the liquid bromine
We will use the following expression.
Q = c × m × ΔT
Q = 0.226 J/g.K × 31.2 g × 2.30 K
Q = 16.2 J
Answer:
Explanation:
H₂SO₄ is a strong acid, which means that most of it ionizes in aqueous solution.
Since it is a diprotic acid (two hydrogen ions) its ionization occurs in two steps:
- H₂SO₄ (aq) → H⁺(aq) + HSO₄⁻(aq)
- HSO₄⁻ (aq) → H⁺(aq) + SO₄²⁻(aq)
Thus, almost all H₂SO₄ has ionized and its final concentration is almost nothing.
After the first ionization, the conentrations of H⁺(aq) and HSO₄⁻ are equal but by the second ionization more H⁺ ions are produced along with SO₄⁻.
You can show it as one step dissociation, assuming 100% dissociation (given this is a strong acid):
By the stequiometry you can build this table:
H₂SO₄ (aq) → 2H⁺(aq) + SO₄²⁻(aq)
Initial A 0 0
Change - x +2x +x
Equilibrium A - x 2x x
As explained, A - x is very low, and 2x is twice x. Thus,
The rank of the concentrations from highest to lowest is:
