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jarptica [38.1K]
3 years ago
11

Suppose the isotopic ratio of the two boron isotopes 10B (10.013 amu) and 11B (11.009 amu) in a sample has been altered from the

ratio found in nature and now contains 33.36% 10B in the sample. Determine the atomic weight of this sample of this new boron element.
Chemistry
1 answer:
Black_prince [1.1K]3 years ago
6 0

Answer:

The correct answer is 10.676 amu.

Explanation:

Based on the given information, the concentration of 10B left in the sample is 33.36%. Therefore, the percentage of 11B present will be,

11B = 100% - 33.36% = 66.64%

Now the atomic weight of the new boron element can be determined by adding the atomic masses of both the isotopes multiplied by its percentage.

Therefore,

= (10.013 amu * 33.36%) + (11.009 amu * 66.64%) / 100

= 10.676 amu

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A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4
san4es73 [151]

Answer:

83.20 g of Na3PO4

Explanation:

1 mole of Na3PO4 contains 3 moles of Na+.

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                 = 0.700 x 725/1000

                     = 0.5075 mole

If 1 mole of Na3PO4 contains 3 moles of Na ion, then 0.5075 Na ion will be contained in:

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mole of Na3PO4 = mass/molar mass = 0.1692

Hence, mass of Na3PO4 = 0.1692 x molar mass

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4 years ago
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4 0
3 years ago
A chemical reaction is shown below:
BabaBlast [244]

Answer:

Mass = 8.46 g

Explanation:

Given data:

Mass of water produced = ?

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Mass of oxygen = 15 g

Solution:

Chemical equation:

C₆H₁₂O₆ + 6O₂     →   6H₂O + 6CO₂

Number of moles of glucose:

Number of moles = mass/molar mass

Number of moles = 20 g/ 180.16 g/mol

Number of moles = 0.11 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 15 g/ 32 g/mol

Number of moles = 0.47 mol

now we will compare the moles of water with oxygen and glucose.

               C₆H₁₂O₆           :            H₂O

                   1                   :              6

                 0.11                :           6/1×0.11 = 0.66

                   O₂               :            H₂O

                   6                   :              6

                 0.47                :           0.47

Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.

Mass of water produced:

Mass = number of moles × molar mass

Mass = 0.47 mol  ×18 g/mol

Mass = 8.46 g

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