It seems more and more there are fewer conservation organizations who speak for the forest, and more that speak for the timber industry. Witness several recent commentaries in Oregon papers that are by no means unique. I’ve seen similar themes from other conservation groups across the West in recent years.
Many conservation groups have uncritically adopted views that support more logging of our public lands based upon increasingly disputed ideas about forest health and fire ecology, as well as the age-old bias against natural processes like wildfire and beetles.
For instance, an article in the Portland Oregonian quotes Oregon Wild’s executive director Sean Stevens bemoaning the closure of a timber mill in John Day Oregon. Stevens said: “Loss of the 29-year-old Malheur Lumber Co. mill would be ‘a sad turn of events’” Surprisingly, Oregon Wild is readily supporting federal subsidies to promote more logging on the Malheur National Forest to sustain the mill.
Answer:
It will take 28.5 minutes
Explanation:
<u>Step 1: </u>Data given
Mass of Cu = 4.50 grams
8.00 A of current are used
Molar mass of Cu = 63.5 g/mol
Step 2: Calculate time needed
Cu2+ →Electricity → Cu
we notice a flow of 2 electrons ⇒ This means the Faraday constant = 2F
Since Molar mass of Cu is 63.5 g/mol
63.5 grams of Cu is deposited by 2*96500 C
4.50 grams of Cu ((2*96500)/63.5) * 4.50 = 13677.17 C
Q = It
13677.17 = 8t*60 seconds
t = 28.5 minutes
The answer is D, all of them have very different masses
Answer:
The order of solubility is AgBr < Ag₂CO₃ < AgCl
Explanation:
The solubility constant give us the molar solubilty of ionic compounds. In general for a compound AB the ksp will be given by:
Ksp = (A) (B) where A and B are the molar solubilities = s² (for compounds with 1:1 ratio).
It follows then that the higher the value of Ksp the greater solubilty of the compound if we are comparing compounds with the same ionic ratios:
Comparing AgBr: Ksp = 5.4 x 10⁻¹³ with AgCl: Ksp = 1.8 x 10⁻¹⁰, AgCl will be more soluble.
Comparing Ag2CO3: Ksp = 8.0 x 10⁻¹² with AgCl Ksp = AgCl: Ksp = 1.8 x 10⁻¹⁰ we have the complication of the ratio of ions 2:1 in Ag2CO3, so the answer is not obvious. But since we know that
Ag2CO3 ⇄ 2 Ag⁺ + CO₃²₋
Ksp Ag2CO3 = 2s x s = 2 s² = 8.0 x 10-12
s = 4 x 10⁻12 ∴ s= 2 x 10⁻⁶
And for AgCl
AgCl ⇄ Ag⁺ + Cl⁻
Ksp = s² = 1.8 x 10⁻¹⁰ ∴ s = √ 1.8 x 10⁻¹⁰ = 1.3 x 10⁻⁵
Therefore, AgCl is more soluble than Ag₂CO₃
The order of solubility is AgBr < Ag₂CO₃ < AgCl
