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3 years ago

8

Each axis on a graph should be: equal O numbered Olabeled vertical

1 answer:

Anni [7]3 years ago

6 0

I think it’s Labeled or numbered....

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14.28 explain how you would distinguish between each pair of compounds using high- resolution mass spectrometry

vagabundo [1.1K]

The distinguish between each pair of compounds using high- resolution mass spectrometry by the exact mass rather than nominal mass are utilizes to measure the compound.

The mass spectrometry is involves the following steps :

The ionization
acceleration
deflection
detection

Mass spectrometry is the analytical method useful for the calculating the mass to charge ratio ( m / z ). the mass spectrometry is based on the newton's second law and the momentum.

Thus, the mass spectroscopy is method to measure the molecular mass of the compound and indirectly helps examine the isotopes and based on the newton's second law .

To learn more about mass spectroscopy here

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7 0

1 year ago

Please match the example with the classification of matter in which it belongs. Column A 1. soil: soil 2. chocolate milk: chocol

bekas [8.4K]

Answer:

34

Explanation:

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8 0

3 years ago

1. A smoothie contains 1 banana (B), 4 strawberries (St), 1 container of yogurt (Y), and 3 ice cubes (Ic). Write a balanced equa

kvv77 [185]

Answer:

1B +4St+1Y+3lc——-> BSt4Ylc3

Explanation:

I only know the answer for the first question.

4 0

3 years ago

. A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be r

Tom [10]

Answer : The energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

Explanation : Given,

Wavelength = $435.8nm=435.8\times 10^{-9}m$

conversion used : $1nm=10^{-9}m$

Formula used :

$E=h\times \nu$

As, $\nu=\frac{c}{\lambda}$

So, $E=h\times \frac{c}{\lambda}$

where,

$\nu$ = frequency

h = Planck's constant = $6.626\times 10^{-34}Js$

$\lambda$ = wavelength = $435.8\times 10^{-9}m$

c = speed of light = $3\times 10^8m/s$

Now put all the given values in the above formula, we get:

$E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}$

$E=4.56\times 10^{-19}J$

Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, $4.56\times 10^{-19}J$

3 0

3 years ago

For which solutes dissolved in a liquid is the overall enthalpy of solution exothermic? A-some gases and most solids B- some sol

lina2011 [118]

Answer:

B

Explanation:

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8 0

3 years ago

Read 2 more answers