B. Regulating Services
Explanation:
Pollination is an example of regulating ecosystem service.
- Ecosystem services are the direct and indirect contribution to human life.
- Humans relies on the ecosystem for a wide range of provisions.
- These services are categorized into different forms.
- Regulating services are the values derived from controlling the ecosystem.
- Pollination is one of such things.
- Plants are pollinated by insects. This helps to continue and sustain life.
- Pollination helps to fertilize plants and ensures reproduction.
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pH of buffer can be calculated as:
pH=pKa+log[salt]/[Acid]
As ka = 4.58 x 10-4
Concentration of [Salt] that is NO2(-1)=0.380M
Concentration of [Acid] that is HNO2=0.500M
So, pH= -log(4.58*10^-4)+log((0.380)/0.500))
=3.21
So pH of solution will be 3.21
From Avogadro we obtained a physical constant of matter which is Avogadro's number, and from both scientists we understand that elementary gases such as hydrogen, nitrogen, and oxygen were composed of two atoms.
<h3>What is Avogadro's number?</h3>
Avogadro's number, or Avogadro's constant, is the number of particles found in one mole of a substance.
The Avogadro's number is given as 6.02 x 10²³.
Summary of Josef Loschmidt and Amedeo Avogadro Contribution to chemistry.
- Equal volumes of gas contain equal numbers of molecules,
- Elementary gases such as hydrogen, nitrogen, and oxygen were composed of two atoms.
Thus, from Avogadro we obtained a physical constant of matter which is Avogadro's number, and from both scientists we understand that elementary gases such as hydrogen, nitrogen, and oxygen were composed of two atoms.
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Answer:
The correct answer is 574.59 grams.
Explanation:
Based on the given information, the number of moles of NH₃ will be,
= 2.50 L × 0.800 mol/L
= 2 mol
The given pH of a buffer is 8.53
pH + pOH = 14.00
pOH = 14.00 - pH
pOH = 14.00 - 8.53
pOH = 5.47
The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,
= -log (1.8 ×10⁻⁵)
= 5.00 - log 1.8
= 5.00 - 0.26
= 4.74
Based on Henderson equation:
pOH = pKb + log ([salt]/[base])
pOH = pKb + [NH₄⁺]/[NH₃]
5.47 = 4.74 + log ([NH₄⁺]/[NH₃])
log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73
[NH₄⁺]/[NH₃] = 10^0.73= 5.37
[NH₄⁺ = 5.37 × 2 mol = 10.74 mol
Now the mass of dry ammonium chloride required is,
mass of NH₄Cl = 10.74 mol × 53.5 g/mol
= 574.59 grams.