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3 years ago

8

vanadium has an atomic mass of 50.9415 amu. it has two common isotopes.one isotopes has a mass of 50.9440 amu and a relative abu

ndance of 99.75% .

1 answer:

Kaylis [27]3 years ago

5 0

Explanation:

Average atomic mass of the vanadium = 50.9415 amu

Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975

Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu

Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025

Atomic mass of Isotope (II) of vanadium ,m' = ?

Average atomic mass of vanadium =

m × abundance of isotope(I) + m' × abundance of isotope (II)

50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025

m'= 49.944 amu

The atomic mass of isotope (II) of vanadium is 49.944 amu.

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A 360. g iron rod is placed into 750.0 g of water at 22.5°C. The water temperature rises to 46.7°C. What was the initial tempera

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Answer: The initial temperature of the iron was $515^0C$

Explanation:

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As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

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Now put all the given values in equation (1), we get

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3 years ago

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Answer:

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(3) [ASA]i = 3.79x10⁻³M

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Explanation:

<u>The Beer's Law is expressed by:</u>

$A = \epsilon \cdot l \cdot C$ (1)

<em>where A: is the absorbance of the species, ε: is the molar attenuation coefficient, l: is the pathlength and C: is the concentration of the species</em>

(1) <u>From </u><u>equation (1)</u><u>, the relation between the absorbance of the species and its concentration is directly proportional,</u> so if the aspirin concentration in solutions increases, the absorbance of the solutions will also increase.

(2) Starting in the given expression for the relationship between absorbance and concentration of ASA, we can calculate its concentration in the solution:

$A = 1061.5 \cdot [ASA]$

$[ASA] = \frac{A}{1061.5} = 3.79 \cdot 10^{-4}M$

Therefore, the aspirin concentration in the solution is 3.79x10⁻⁴ M

(3) To calculate the stock solution concentration, we can use the next equation:

$V_{i} [ASA]_{i} = V_{f} [ASA]_{f}$

<em>where Vi: is the stock solution volume=10mL, Vf: is the solution diluted volume=100mL, [ASA]i: is the aspirin concentration of the stock solution and [ASA]f: is the aspirin concentration of the diluted solution</em>

$[ASA]_{i} = \frac{V_{f} \cdot [ASA]_{f}}{V_{i}} = \frac {100mL \cdot 3.79\cdot 10^{-4} M}{10mL} = 3.79 \cdot 10^{-3} M$

Hence, the concentration of the stock solution is 3.79x10⁻³M

(4) To determine the aspirin mass in the tablet, we need to use the following equation:

$m_{ASA} = \eta_{ASA} \cdot M_{ASA} = [ASA]_{i} \cdot V_{0} \cdot M_{ASA}$

<em>where η: is the aspirin moles = [ASA]i V₀, M: is the molar mass of aspirin=180.158g/mol, V₀: is the volume of the volumetric flask=250mL and [ASA]i: is the aspirin concentration in the volumetric flask which is equal to the stock solution=3.79x10⁻³M</em>

$m_{ASA} = 3.79 \cdot 10^{-3} \frac{mol}{L} \cdot 0.250L \cdot 180.158 \frac{g}{mol} = 0.171 g$

Then, the aspirin mass in the tablet is 0.171 g.

I hope it helps you!

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Explanation:

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2 years ago