How many grams of carbon are in 300 g of glucose
1 answer:
You might be interested in
From the information given:
- The volume of the graduated cylinder = 50.0 mL
- when a sterling silver pendant is added, the volume increases to = 61.3 mL
∴
The volume of the sterling silver pendant is:
= 61.3 mL - 50.0 mL
= 11.3 mL
Since, 1 mL = 1cm³
Then;
11.3 mL = 11.3 cm³
- the density of the sterling silver = 10.25 g/cm³
Using the relation for Density; i.e.
mass = 10.25 g/cm³× 11.3 cm³
mass of the sterling silver = 115.825 grams
Recall that sterling silver has:
- 92.5% silver and;
- 7.5% copper
∴
The mass of the copper contained in the sterling silver pendant can be calculated as:
= 8.687 grams
Therefore, we can conclude that the mass of the copper contained in the sterling silver pendant is 8.687 grams
Learn more about the relation between Density, Mass, and Volume here:
Answer:
The [SO₃²⁻]
Explanation:
From the first dissociation of sulfurous acid we have:
H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)
At equilibrium: 0.50M - x x x
The equilibrium constant (Ka₁) is:
With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:
Similarly, from the second dissociation of sulfurous acid we have:
HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)
At equilibrium: 7.94x10⁻²M - x x x
The equilibrium constant (Ka₂) is:
Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:
Therefore, the final concentrations are:
[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M
[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M
[SO₃²⁻] = 7.07x10⁻⁵M
[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M
So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.
I hope it helps you!
Step (1):
Generation of electrophile: by the action of Lewis acid FeCl₃ on Cl₂ to serve as a source of Cl⁺ (Electrophile)
Step (2):
Addition of electrophile to form carbocation:
addition of electrophile to form C-Cl bond and form carbocation which is stabilized by resonance.
Step (3):
Loss of proton to re-form the aromatic ring by the action of FeCl₄⁻ which removes proton from carbon containing Cl and forming the aromatic ring again
Generation of electrophile: by the action of Lewis acid FeCl₃ on Cl₂ to serve as a source of Cl⁺ (Electrophile)
Step (2):
Addition of electrophile to form carbocation:
addition of electrophile to form C-Cl bond and form carbocation which is stabilized by resonance.
Step (3):
Loss of proton to re-form the aromatic ring by the action of FeCl₄⁻ which removes proton from carbon containing Cl and forming the aromatic ring again