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3 years ago

Could someone explain how they got this answer, explain step by step plz

Chemistry

1 answer:

gulaghasi [49]3 years ago

7 0

Answer:

6.018 amu

Explanation:

Let 6–Li be isotope A.

Let 7–Li be isotope B.

Let the abundance of 6–Li be A%

Let the abundance of 7–Li be B%

The following data were obtained from the question:

Atomic mass of isotope A (6–Li) =.?

Atomic mass of isotope B (7–Li ) = 7.015 amu.

Abundance of 7–Li (B%) = 92.58%

Abundance of 6–Li (A%) = 100 – B% = 100 – 92.58 = 7.42%

Atomic mass of Lithium = 6.941amu

The atomic mass of isotope A (6–Li) can be obtained as follow:

Atomic mass = [(Mass of A x A%)/100] + [(Mass of B x B%)/100]

6.941 = [(mass of A x 7.42)/100] + [(7.015x92.58)/100]

6.941 = [(mass of A x 7.42)/100] + 6.494487

(mass of A x 7.42)/100 = 6.941 – 6.494487

(mass of A x 7.42)/100 = 0.446513

Mass of A x 7.42 = 100 x 0.446513

Mass of A x 7.42 = 44.6513

Divide both side by 7.42

Mass of A = 44.6513 / 7.42

Mass of A = 6.018 amu

Therefore, the mass of 6–Li is 6.018 amu

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3 years ago

It is desired to produce 2.25 grams of dichloromethane (CH2Cl2) by the following reaction. If the percent yield of dichlorometha

Allushta [10]

Answer:- 3.12 g carbon tetrachloride are needed.

Solution:- The balanced equation is:

$CH_4+CCl_4\rightarrow 2CH_2Cl_2$

From given actual yield and percent yield we will calculate the theoretical yield that would be further used to calculate the grams of carbon tetrachloride.

percent yield formula is:

percent yield = $(\frac{actual}{theoretical})100$

$65.5=(\frac{2.25}{theoretical})100$

$theoretical=(\frac{2.25(100)}{65.5})$

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$3.44gCH_2Cl_2(\frac{1molCH_2Cl_2}{84.93gCH_2Cl_2})(\frac{1molCCl_4}{2molCH_2Cl_2})(\frac{153.82gCCl_4}{1molCCl_4})$

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3 years ago

What happens when an atom of sulfur combines with two atoms of chlorine to produce SCI2?

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2 years ago

Suppose you have just added 200.0 ml of a solution containing 0.5000 moles of acetic acid per liter to 100.0 ml of 0.5000 M NaOH

uranmaximum [27]

Answer:

The final pH is 3.80

Explanation:

Step 1: Data given

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Number of moles acetic acid = 0.5000 moles

Volume of NaOH = 100.0 mL = 0.100 L

Molarity of NaOH = 0.500 M

Ka of acetic acid = 1.770 * 10^-5

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles

moles = molarity * volume

Moles NaOH = 0.500 M * 0.100 L

Moles NaOH = 0.0500 moles

Step 4: Calculate the limiting reactant

For 1 mol CH3COOH we need 1 mol NaOH to produce 1 mol CH3COONa and 2 moles H2O

NaOH is the limiting reactant. It will completely be consumed (0.0500 moles). CH3COOH is in excess. There will react 0.0500 moles . There will remain 0.500 - 0.0500 = 0.450 moles

There will be produced 0.0500 moles CH3COONa

Step 5: Calculate the total volume

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Total volume = 0.300 L

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Step 7: Calculate pH

pH = pKa + log[A-]/ [HA]

pH = -log(1.77*10^-5) + log (0.167/ 1.5)

pH = 4.75 + log (0.167/1.5)

pH = 3.80

The final pH is 3.80

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Answer:

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