Select ALL factors in conservation: *Social conditions such as need for electricity, famine, and war.
Scientific data related to the ecosystem and the effect of environmental changes.
Political action by governments and other organizations such as environmental
protection groups.
Economic issues such as cost of wood products, fuel for heat, price of electricity, and
income levels of local people.
Answers
Answer:
ALL of these are factors in conservation.
Explanation:
Related Questions
Find the density of an object that has a volume of 2.3 x 10^2 and a mass of3.5 x 10^3. *
Draw a well-labelled diagram showing how your body digests food
Which compound is classified as an electrolyte? a) C6H12O6 b)Ca(OH)2 c)C12H22O11 d)CH3OH
When a sodium chloride (NaCl) solution is given to some intravenously (as an IV directly into the blood), it is usually a 0.9% NaCl solution. The reason is that a 0.9% NaCl is isotonic to your red blood cells. Use that information to answer the following questions: In the solution above, which molecule is acting as the solute and which molecule is acting as the solvent? If you replaced the 0.9% NaCl solution with pure water instead, would the water be hypertonic, hypotonic, or isotonic to your red blood cells? If you replaced the 0.9% NaCl solution with pure water instead, would the net (overall) movement of water be into the cell, out of the cell, or would there be no net movement of water? If you replaced the 0.9% NaCl solution with pure water instead and waited, what specifically would happen to your red blood cells?
Answers
Answer: 1.037M
Explanation:
Since the rate constant unit is per seconds, therefore it is a first order reaction.
First order reaction equation is given as
InA= -kt +InAo
Where,Ao is the initial concentration of reactant =0.600M
A is the concentration of reactant at a specifies time t=3×60=180s
and k is the rate constant
InA = -6.50×10^-3 ×180 +In(0.6)
InA = -1.17 + 0.5108
InA= -1.680
A = e-1.680
A= 1.037M
Therefore the concentration after 3minutes is 1.037M
B. NH3 < PH3< CH4
C. CH4 < PH3 < NH3
D. NH3 < CH4< PH3
E. PH3< NH3 < CH4
Answers
Answer:
B. NH3 < PH3< CH4
Explanation:
Hello,
In this case, taking into account that the boiling point of ammonia, methane and phosphorous trihydrate are -33.34 °C, -161.5 °C and -87.7 °C, clearly, methane has the lowest boiling point (most negative) and ammonia the greatest boiling point (least negative), therefore, ranking is:
B. NH3 < PH3< CH4
Best regards.
Answers
Leo ger is the nuemonic
- loss electrons = oxidized
- gain electrons = reduced
You can see K gains a positive charge so it is safe to assume it lost some (negatively charged) electrons
Answers
Answer:
The daughter nuclides of these two decay processes are and
.
Explanation:
The beta emission is represented by:
A = (Z + 1) + (n - 1) = is invariant
n: neutron
p: proton
Hence, the daughter nuclide of the beta emission of Ir-192 is:
Now, electron capture is represented by:
A = (Z - 1) + (n + 1) = is invariant
Then, the daughter nuclide of the electron capture of Ir-192 is:
Therefore, the daughter nuclides of these two decay processes are and
.
I hope it helps you!
Answers
Answer:
hydrogen have electronic configuration of 1s¹ to acquire a stable state it can either lose electron and form H+cation or gain an electron( to compete its 1s subshell ) to form H- anion. As it has ±1 valency it is placed neither in group 1(alkaline metals ) nor in group 17 (halogens) .
Answers
The conclusions drawn by Ruthford after the experiment are that most of the atom is empty and that the nucleus of the atom is positively charged.
What is the atom?´
- It is the particular mentor of the matter.
- It is the element that makes up matter.
After the discovery of the atom, many scientists sought to understand this element more specifically, especially in relation to its composition. These discoveries were strengthened over time, and Rutherford was the one who established how the atom really is, with a positive charge in the nucleus and an electrosphere around it.
Complete question:
Ernest Rutherford completed his famous gold foil experiment in 1911. In this experiment, alpha particles were fired at a thin sheet of gold foil. He observed that most of the alpha particles passed straight through the gold foil unimpeded, but a small number of alpha particles were deflected. Which of the following conclusions about atomic structure were made from Rutherford’s gold foil experiment?
Most of the atom is empty.
The nucleus is positively charged.
The atom is a massive sphere.
The atom is indivisible.
Learn more about atoms:
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Answer: gold foil. It's in the experiment's name