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3 years ago

5

Where can you find the atomic mass?

2 answers:

Paladinen [302]3 years ago

8 0

Answer:

hope you like it

Explanation:

Together, the number of protons and the number of neutrons determine an element's mass number: mass number = protons + neutrons. If you want to calculate how many neutrons an atom has, you can simply subtract the number of protons, or atomic number, from the mass number.

To calculate the atomic mass of a single atom of an element, add up the mass of protons and neutrons. Example: Find the atomic mass of an isotope of carbon that has 7 neutrons. You can see from the periodic table that carbon has an atomic number of 6, which is its number of protons.

artcher [175]3 years ago

7 0

Answer:

To calculate the atomic mass of a single atom of an element, add up the mass of protons and neutrons. Example: Find the atomic mass of an isotope of carbon that has 7 neutrons. You can see from the periodic table that carbon has an atomic number of 6, which is its number of protons.

Explanation:

To calculate the atomic mass of a single atom of an element, add up the mass of protons and neutrons. Example: Find the atomic mass of an isotope of carbon that has 7 neutrons. You can see from the periodic table that carbon has an atomic number of 6, which is its number of protons.

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Answer:

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3 years ago

Th e molar absorption coeffi cient of a substance dissolved in water is known to be 855 dm3 mol−1 cm−1 at 270 nm. To determine t

Olegator [25]

Answer : The percentage reduction in intensity is 79.80 %

Explanation :

Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

$A=\log \frac{I_o}{I}$

$\log \frac{I_o}{I}=\epsilon \times C\times l$

where,

A = absorbance of solution

C = concentration of solution = $3.25mmol.dm^{3-}=3.25\times 10^{-3}mol.dm^{-3}$

l = path length = 2.5 mm = 0.25 cm

$I_o$ = incident light

$I$ = transmitted light

$\epsilon$ = molar absorptivity coefficient = $855dm^3mol^{-1}cm^{-1}$

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

$\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)$

$\log \frac{I_o}{I}=0.6947$

$\frac{I_o}{I}=10^{0.6947}=4.951$

If we consider $I_o$ = 100

then, $I=\frac{100}{4.951}=20.198$

Here 'I' intensity of transmitted light = 20.198

Thus, the intensity of absorbed light $I_A$ = 100 - 20.198 = 79.80

Now we have to calculate the percentage reduction in intensity.

$\% \text{reduction in intensity}=\frac{I_A}{I_o}\times 100$

$\% \text{reduction in intensity}=\frac{79.80}{100}\times 100=79.80\%$

Therefore, the percentage reduction in intensity is 79.80 %

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Svetradugi [14.3K]

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