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anyanavicka [17]
1 year ago
7

Using a good compound light microscope with a resolving power of 0.3 µm, a 10x ocular lens, and a 100x oil immersion lens, would

you be able to discern two objects separated by 3 µm? 0.3 µm? 300 nm? why would, or wouldn't, you be able to do this with the presented microscope?
Chemistry
1 answer:
irina [24]1 year ago
4 0
Answer : One <span>would be able to easily discern between two objects as they would be separated by all 4 distances as given in the question; because each one will be greater than or equal to the the resolving power of the compound light microscope.

Resolution means the separation between the two close images so, if we use a microscope with </span><span>0.3 µm, a 10x ocular lens, and a 100x oil immersion lens it will be able to resolve images which are greater of the given values.</span>
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The gaseous product of a reaction is collected in a 25.0-l container at 27
nataly862011 [7]
To find the number of moles of gas we can use the ideal gas law equation, we dont need to use the mass of gas given as we only have to find the number of moles 
PV = nRT 
P - pressure - 300.0 kPa 
V - volume - 25.0 x 10⁻³ m³
n - number of moles 
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 27 °C + 273 = 300 K
substituting these values in the equation 
300.0 kPa x 25.0 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 300 K 
n = 3.01 mol 
number of mols of gas - 3.01 mol
4 0
2 years ago
From the Bohr equation in the introduction, the calculated energy of an electron in the sixth Bohr orbit of a hydrogen atom is
Natalka [10]

Answer:

<em><u>= - 0.38 eV</u></em>

Explanation:

Using Bohr's equation for the energy of an electron in the nth orbital,

E = -13.6 \frac{Z^{2} }{n^{2} }

Where E = energy level in electron volt (eV)

Z = atomic number of atom

n = principal state

Given that n = 6

⇒ E = -13.6 × \frac{1^{2} }{6^{2} }

<em><u>= - 0.38 eV</u></em>

<em><u></u></em>

<em>Hope this was helpful.</em>

<em><u></u></em>

4 0
2 years ago
Which of the compounds above are strong enough acids to react almost completely with a hydroxide ion (pka of h2o = 15.74) or wit
luda_lava [24]

The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Further explanation </em></h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.

From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

K~=~10^{-pK}

K value> 1 indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:

pyridinium pKa = 5.25

acetone pKa = 19.3

butan-2-one pKa = 19

Let's look at the K value of each possible reaction:

pka H₂O = 15.74, pka of H₂CO₃ = 6.37)

  • 1. C₅H₆N pyridinium

* with OH⁻

C₅H₆N + OH- ---> C₅H₅N- + H₂O

pK = pKa pyridinium - pKa H₂O

pK = 5.25 - 15.74

pK = -10.49

K~=~10^{4.9}

K values> 1 indicate the reaction can take place

* with HCO3⁻

C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃

pK = 5.25 - 6.37

pK = -1.12

K`=~10^{1.12]

Reaction can take place

  • 2. Acetone C₃H₆O

* with OH-

C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O

pK = 19.3 - 15.74

pK = 3.56

K~=~10^{ -3.56}

Reaction does not happen

* with HCO₃-

C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃

pK = 19.3 - 6.37

pK = 12.93

K`=~10 ^{-12.93}

Reaction does not happen

  • 3. butan-2-one C₄H₇O

* with OH-

C₄H₇O + OH- ---> C₄H₆O- + H₂O

pK = 19 - 15.74

pK = 3.26

K~=~10^{-3.26}

Reaction does not happen

* with HCO₃⁻

C₄H₇O + HCO₃⁻ ---> C₄H₆O⁻ + H₂CO₃

pK = 19 - 6.37

pK = 12.63

K~=~ 10^{-12.63}

Reaction does not happen

So that can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Learn more </em></h3>

the lowest ph

brainly.com/question/9875355

the concentrations at equilibrium.

brainly.com/question/8918040

the ph of a solution

brainly.com/question/9560687

Keywords : acid base reaction, the equilibrium constant

5 0
1 year ago
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AURORKA [14]
4) fission is a nuclear reaction where the nucleus of a reactant breaks apart
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Give the number of significant figures in this number: 40.00
Tatiana [17]

A significant figure is every symbol that made the number itself.

In this case, the number 40.00 has four figures but only two of them are significant 40, this is because you haven't got any more decimals than the first zero.

If you have a case with zeros in front, you take to the first non zero digits.

For example, 0.071004 you wold express as 0.071 and those 7, and 1 are the significant ones.

4 0
2 years ago
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