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:
646.8 g
Explanation:
It really depends what salt silver cation makes in the context of this problem. In this problem, we have a single displacement reaction in which copper metal displaces silver cation from a salt to produce silver metal. Based on the charge of the anion in the salt, we may have differently balanced chemical equations.
Since silver cation mostly makes insoluble salts and we need a soluble one for a single displacement reaction, we'll use silver nitrate as our example, as it's the most common soluble silver salt. We have the following reaction:
Notice that 1 mole of copper reacts to produce 2 moles of silver. This may be mathematically written as:
Express moles as a ratio between mass and molar mass:
Rearrange the equation for the mass of silver metal:
Substitute the given data:
The products for the complete combustion of a hydrocarbon in excess air is carbon dioxide and water. Any hydrocarbon when reacted with oxygen will always yield the said products. Incomplete combustion, on the other hand, yields carbon monoxide and water.
The easiest way to approximate ph level is through the ph paper measurement system because using pH paper would be best for doing a quick measurement in the field. It does not essentially need to be calibrated or standardized and the papers make available an instant estimate.
