Answer:
Oxygen needed for Stannous Oxide: 1.350g
Oxygen needed for Stannic Oxide: 2.710g
Explanation:
You're working with 10.00 grams of Tin mass for both Stannous Oxide and Stannic Oxide.
- 10.00 grams of Tin for Stannous Oxide is already 88.10% of the mass needed. You need to find how much 11.90% of Oxygen mass is needed to create the compound. Find a factor that you can multiply 88.10% by to get 100%
- 88.10 * x = 100
- Solve for x and you get 1.135
- Multiply that number by the mass of Tin (10.00 grams) to get the complete compound (Mixture of Tin and Oxygen).
- 10.00g * 1.135 = 11.35g (Tin + Oxygen)
- Subtract (Compound - Tin) to find Oxygen
- 11.35g - 10.00g = 1.350g (Oxygen)
Repeat the process with Stannic Oxide
- Find the factor that gets 78.70% to 100%
- 100/78.70 = 1.271
- Multiply by Tin mass
- 10.00g * 1.271 = 12.71g (Compound)
- Subtract Compound by Tin
- 12.71g - 10.00g = 2.710g (Oxygen)
Answer:
2.3x10⁷ W/m²
Explanation:
The intensity (I) of a laser is its potency (P) divided by the area (A) that it is affected. The potency is the energy applied (or absorbed) in a period, thus id 91.0% of the energy is absorbed, so:
E = 0.91*540 = 491.4 J
And,
P = E/t, where t is the time in seconds
P = 491.4/4.00
P = 122.85 J/s
P = 122.85 W
The are of a circular spot is:
A = (π/4)*d²
Where d is the diameter. Thus, with d = 2.60 mm = 0.0026 m
A = (π/4)*(0.0026)²
A = 5.31x10⁻⁶ m²
I = P/A
I = 122.85/5.31x10⁻⁶
I = 2.3x10⁷ W/m²
To identify the amount of chloride ions present in 0.1 mol
of MgCl2, first know the amount of mols of Chloride. The answer is 0.2 mol
since you just need to multiply 0.1 to 2 because there are 2 chloride in the
compound. Then, multiply the mols to 6.02x10^23. The answer would be
1.204x10^23 chloride ions.
Tri-blade radiation symbol
