Answer:
Proper waste management allows for separation at the source and subsequent recycling, which allows materials that required energy and raw materials to be reincorporated into the production cycle, as the finished material is incorporated back into the production cycle, the amount of energy required for the production of <u>the final product is less than if it were made from the start</u>.
Explanation:
A very useful example of energy conservation through waste management is the following: suppose that in your school they do proper waste management, therefore materials such as cardboard, paper and plastic are separated appropriately and are subsequently donate or sell to a company that is dedicated to recycling these materials, plastic is a material that requires a lot of energy, first for its elaboration and later for the creation of the final product such as the packaging of a shampoo or a chair, however, as the recycled plastic is already manufactured, <u>the recycling company saves said energy, which only requires crushing the obtained plastic and selling it to a company that produces the final product</u>.
Answer:
0.271 M NO₃⁻
Explanation:
To find the molarity of the nitrate ion (NO₃⁻), you need to (1) convert grams to moles (via molar mass), then (2) convert moles Al(NO₃)₃ to moles NO₃⁻, then (3) convert mL to L, and then (4) calculate the molarity. When (Al(NO₃)₃) dissolves in water, it dissociates into 3 nitrate ions. The final answer should have 3 sig figs.
(Steps 1 + 2)
Molar Mass (Al(NO₃)₃): 26.982 g/mol + 3(14.007 g/mol) + 9(15.998 g/mol)
Molar Mass (Al(NO₃)₃): 212.985 g/mol
1 Al(NO₃)₃ = 1 Al³⁺ and 3 NO₃⁻
6.25 g Al(NO₃)₃ 1 mole 3 moles NO₃⁻
------------------------- x ----------------- x ----------------------- = 0.0880 moles NO₃⁻
212.985 g 1 mole Al(NO₃)₃
(Steps 3 + 4)
325.0 mL / 1,000 = 0.3250 L
Molarity = moles / volume
Molarity = 0.0880 moles / 0.3250 L
Molarity = 0.271 M
Answer:
mass of the planet and its distance from other objects
Explanation:
Answer:
670.68°C
Explanation:
Given that:
volume of water = 50 ml but 1 g = 1 ml. Therefore the mass of water (m) = 50 ml × 1 g / ml = 50 g
specific heat (C) = 4.184 J/g˚C
Initial temperature = 20°C, final temperature = 22°C. Therefore the temperature change ΔT = final temperature - initial temperature = 22 - 20 = 2°C
The quantity of heat (Q) used to raise the temperature of a body is given by the equation:
Q = mCΔT
Substituting values:
Q = 50 g × 4.184 J/g˚C × 2°C = 418.4 J
Since the mass of lead = 5 g and specific heat = 0.129 J/g˚C. The heat used to raise the temperature of water is the same heat used to raise the temperature of lead.
-Q = mCΔT
-418.4 J = 5 g × 0.129 J/g˚C × ΔT
ΔT = -418.4 J / ( 5 g × 0.129 J/g˚C) = -648 .68°C
temperature change ΔT = final temperature - initial temperature
- 648 .68°C = 22°C - Initial Temperature
Initial Temperature = 22 + 648.68 = 670.68°C
