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2 years ago

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True or False? Nitrogen can get back into the atmosphere when dead plants and animals are decomposed by bacteria. The bacteria b

reak down the larger molecules into smaller ones including nitrates, ammonia, and ultimately N2. True or Fralse

2 answers:

Readme [11.4K]2 years ago

6 0

Answer: True

Explanation:

AleksandrR [38]2 years ago

5 0

Answer:

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Suppose that, from measurements in a microscope, you determine that a certain layer of graphene covers an area of 2.60μm2. conve

garri49 [273]

Given that 1 micrometer or micron (um) is equivalent by definition to 1 x 10^-6 m, this means that 1 square micron (um^2) is equivalent to (1 x 10^-6)^2 m^2, or 1 x 10^-12 m^2.

(2.60 um^2) * (1 x 10^-12 m^2 / 1 um^2) = 2.60 x 10^-12 m^2
Therefore the layer of graphene covers an area of 2.60 x 10^-12 m^2.
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Describe 5 importance matter to undergo changes

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Answer:

melting, freezing, sublimation, deposition, condensation, and vaporization. These changes

Explanation:

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The table below gives the numbers of protons, electrons, and neutrons in

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Answer:

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What is the pH of a 1.0 L buffer made with 0.300 mol of HF (Ka = 6.8 × 10⁻⁴) and 0.200 mol of NaF to which 0.150 mol of HCl were

irinina [24]

Answer:

pH = 2.21

Explanation:

Hello there!

In this case, according to the reaction between NaF and HCl as the latter is added to the buffer:

$NaF+HCl\rightarrow NaCl+HF$

It is possible for us to see how more HF is formed as HCl is added and therefore, the capacity of this HF/NaF-buffer is diminished as it turns acid. Therefore, it turns out feasible for us to calculate the consumed moles of NaF and the produced moles of HF due to the change in moles induced by HCl:

$n_{HF}^{new}=0.300mol+0.150mol=0.450mol\\\\n_{NaF}^{new}=0.200mol-0.150mol=0.050mol$

Next, we calculate the resulting concentrations to further apply the Henderson-Hasselbach equation:

$[HF]=\frac{0.450mol}{1.0L} =0.450M$

$[NaF]=\frac{0.050mol}{1.0L} =0.050M$

Now, calculated the pKa of HF:

$pKa=-log(6.8x10^{-4})=3.17$

We can proceed to the HH equation:

$pH=pKa+log(\frac{[NaF]}{[HF]} )\\\\pH=3.17+log(\frac{0.05M}{0.45M} )\\\\pH=2.21$

Best regards!

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2 years ago