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Alex17521 [72]
4 years ago
12

HELP: Nitrogen has two isotopes, N-14 and N-15, with atomic masses of 14.00031 amu and 15.001 amu, respectively. What is the per

cent abundance of N-15?
Chemistry
1 answer:
vitfil [10]4 years ago
6 0
The atomic mass<span> of an element is a weighted average of its </span>isotopes<span> in which the sum of the abundance of each isotope is equal to 1 or 100%.

</span><span>Let fraction of N-15 be x
then fraction of N-14 = 1-x 
</span>
<span>Formula for Calculating Isotope:
</span>
15.001x + 14.00031(1-x) = 14.0067

on calculation:

x = <span>0.006403
so, % = 0.006403 x 100  =  0.6403 %<span>
</span></span>
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An organic synthesis to make the pain reliever acetaminophen is supposed to produce 280 kg of product but instead produces 70 kg
Irina-Kira [14]

Answer: 75%

Explanation:

The following information can be gotten from the question:

Waste = 70kg

Theoretical yield = 280kg

Therefore, the actual yield will be the difference between the theoretical yield and the waste which will be:

= 280kg - 70kg = 210kg

The percent yield will now be:

= Actual yield / Theoretical yield × 100

= 210/280 × 100

= 3/4 × 100

= 75%

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Which two substances are covalent compounds?
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Answer:

Option B is correct: C6H1206(s) and HCl(g)

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Plant can accomplish something animal cells can’t. They can make their own food during
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Using stoichiometry, determine the mass of powdered drink mix needed to make a 1.0 M solution of 100 mL from C12H22O11
fgiga [73]

Answer:

34.23 g.

Explanation:

<em>Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.</em>

M = (no. of moles of solute)/(V of the solution (L)).

∴ M = (mass/molar mass)of C₁₂H₂₂O₁₁/(V of the solution (L)).

<em>∴ mass of C₁₂H₂₂O₁₁ = (M)(molar mass)(V of the solution (L)</em> = (1.0 M)((342.3 g/mol)/(0.10 L) = <em>34.23 g.</em>

5 0
4 years ago
If 5.0 g of each reactant were used for the the following process, the limiting reactant would be: 2KMnO4 + 5Hg2Cl2 + 16HCl -&gt
IceJOKER [234]

Answer:

The limiting reactant is Hg2Cl2.

Explanation:

Step 1: Data given

Mass of each reactant = 5.0 grams

KMnO4 MM=158 g/mol

Hg2Cl2 MM=472.1 g/mol

HCl MM=36.5 g/mol

HgCl2 MM=271.5 g/mol

MnCl2 MM=125.8 g/mol

KCl MM=74.6 g/mol

H2O MM=18 g/mol)

Step 2: The balanced equation

2KMnO4 + 5Hg2Cl2 + 16HCl -> 10HgCl2 + 2MnCl2 + 2KCl + 8H2O

Step 3: Calculate moles

KMnO4 = 5.00 grams / 158 g/mol = 0.0316 mol

Hg2CL2 = 5.00 grams / 472.1 g/mol = 0.0106 mol

HCl = 5.00 grams / 36.5 g/mol = 0.137 mol

Step 3: Calculate limiting reactant

For 2 moles of KMno4 we need 5 moles of Hg2Cl2 and 16 moles of HCl

Hg2Cl2 has the smallest amount of moles.

For 5 moles Hg2Cl2 ( 0.0106 mol) we need 0.0106 / (5/2) = 0.00424 mol KMnO4

For 5 moles Hg2Cl2 we need (16/5) *0.0106 = 0.03392 moles of HCl

So the limiting reactant is Hg2Cl2.

Step 4: Calculate moles of product produced:

2*0.0106 = 0.0212 moles of HgCl2

(2/5) * 0.0106 = 0.00424 moles of MnCl2 and 0.00424 moles of KCl

(8/5) * 0.0106 = 0.01696 moles H2O

7 0
3 years ago
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