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Zinaida [17]
3 years ago
13

Logging trees is often controversial. Why would some people support logging and some people not support logging?

Chemistry
2 answers:
Harlamova29_29 [7]3 years ago
7 0
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.
miss Akunina [59]3 years ago
5 0

Answer:

Explanation:

Logging is a controversial practice. It is a practice of cutting the logs of the trees or tree trunks. It is a process of cutting the trees, processing the logs of the trees for making furniture and other articles.

Logging is a source of livelihood for many people living in the forest, village and suburban regions. Also it is required to be done so as to reduce the population of dead wood trees, and unwanted tree species by the process of selective cutting. Therefore, people support logging.

But some people do not support logging. This is because of the fact logging on large scale can reduce the tree cover in the forest or any other region.

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A metal sample has a mass of 64.2 g. What is the density of this metal?
brilliants [131]

density= mass/volume

we need the volume of the metal to find the density,  in which case the answer would be 64.2/volume=density

8 0
3 years ago
If 200. mL of 0.60 M MgCl2(aq) is added to 400 mL of distilled water, what is the concentration of Mg and Cl in the resulting so
sladkih [1.3K]

Answer:

C. 0.20 M Mg ion & 0.40 M Cl ion

Explanation:

MgCl₂ is a ionic salt which is dissociated as this

MgCl₂  →  Mg²⁺  +  2Cl⁻

First of all, we have a solution of 200 mL, with [MgCl₂] = 0.6M

Molarity . volume = moles.

0.6 mol/l . 0.2l = 0.12 mol

  MgCl₂  →  Mg²⁺  +  2Cl⁻

0.12mol      0.12         0.24

This moles are also in 400mL of water, so the new concentration is

[Mg²⁺] = 0.12 m/0.6L = 0.2M

[Cl⁻] = 0.24 m/0.6L = 0.4M

Remember we initially have 200mL and then, we add 400 mL, so we supose aditive volume. (600mL)

8 0
3 years ago
What is in an inorganic molecule?
valentinak56 [21]
Inorganic molecules are composed of other elements. They can contain hydrogen or carbon, but if they have both, they are organic.
6 0
3 years ago
Why does opening the air valve of a tire at a constant temperature decrease the pressure
Shtirlitz [24]
The number of molecules decrease
8 0
3 years ago
The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?
anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

K_a\times K_b=K_w

K_b for sodium formate is K_b=\frac {K_w}{K_a}

Given that:

K_a of formic acid = 1.8\times 10^{-4}

And, K_w=10^{-14}

So,

K_b=\frac {10^{-14}}{1.8\times 10^{-4}}

K_b=5.5556\times 10^{-11}

Concentration = 0.35 M

HCOONa    ⇒     Na⁺ +    HCOO⁻

Consider the ICE take for the formate  ion as:

                                   HCOO⁻ + H₂O   ⇄   HCOOH + OH⁻

At t=0                              0.35                            -              -

At t =equilibrium           (0.35-x)                          x           x            

The expression for dissociation constant of sodium formate is:

K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}

5.5556\times 10^{-11}=\frac {x^2}{0.35-x}

Solving for x, we get:

x = 0.44×10⁻⁵  M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

5 0
3 years ago
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