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:
Q = 10.8 KJ
Explanation:
Given data:
Mass of Al= 100g
Initial temperature = 30°C
Final temperature = 150°C
Heat required = ?
Solution:
Specific heat of Al = 0.90 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 150°C - 30°C
ΔT = 120°C
Q = 100g×0.90 J/g.°C× 120°C
Q = 10800 J (10800j×1KJ/1000 j)
Q = 10.8 KJ
Answer:
The answer is pyruvate → lactate
Explanation:
In the reaction of glycolysis, glucose breaks down to form pyruvate yielding ATP and NADH.
Under or during strenuous exercise, which is an anaerobic condition, lactate is formed by the reoxidization of NADH and the conversion of pyruvate to lactate.
Answer:
elements on the left-hand side of the periodic table such as sodium and magnesium prefer to lose electrons to form a cation because this requires less energy to obtain a stable octet, and vice-versa for the right-hand side of the periodic table e.g. fluorine. However, using this reasoning I am not sure why all transition metals tend to lose electrons rather than gain them.