C. Fewer habitats for wild animals to live in
Explanation:
Biodiversity does not directly impact humans in that there are fewer habitats for wild animals to live in. This is a direct impact on animals and not humans.
Biodiversity is the variation of life forms in an area. It is measured by the ability of the ecosystem to sustain different life forms and keep their interaction with the environment.
It helps to maintain and sustain life.
To humans, reducing biodiversity;
- reduces the potential of medicine derived from other organisms. When their habitat is lost, we lose them.
- fewer ecosystems to appreciate for beauty and recreation. It leads to loss of the ecosystem.
- fewer wild food species for humans to eat.
Loss of habitat directly affects wild animals and not man.
learn more:
Biodiversity hotspots brainly.com/question/10723602
#learnwithBrainly
<u>Given information:</u>
Concentration of NaF = 0.10 M
Ka of HF = 6.8*10⁻⁴
<u>To determine:</u>
pH of 0.1 M NaF
<u>Explanation:</u>
NaF (aq) ↔ Na+ (aq) + F-(aq)
[Na+] = [F-] = 0.10 M
F- will then react with water in the solution as follows:
F- + H2O ↔ HF + OH-
Kb = [OH-][HF]/[F-]
Kw/Ka = [OH-][HF]/[F-]
At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x
10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x
x = [OH-] = 1.21*10⁻⁶ M
pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92
pH = 14 - pOH = 14-5.92 = 8.08
Ans: (b)
pH of 0.10 M NaF is 8.08
I would say B because c and d would decrease competition and a would do the same, or just kill the ecosystem.
Hope this helps and don't forget to hit that heart :)
Answer:
130ml of HCl(36%) in 4.90L solution => pH = 1.50
Explanation:
Need 4.90L of HCl(aq) solution with pH = 1.5.
Given pH = 1.5 => [H⁺] = 10⁻¹·⁵M = 0.032M in H⁺
[HCl(36%)] ≅ 12M in HCl
(M·V)concentrate = (M·V)diluted
12M·V(conc) = 0.032M·4.91L
=> V(conc) needed = [(0.032)(4.91)/12]Liters = 0.0130Liters or 130 ml.
Mixing Caution => Add 131 ml of HCl(36%) into a small quantity of water (~500ml) then dilute to the mark.
They both have 1 electron in their valence shell.....
