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k0ka [10]
3 years ago
14

Find the number of grams in 16.95 mol hydrogen peroxide (H2O2). Round your

Chemistry
1 answer:
Feliz [49]3 years ago
3 0

Answer: There are 576.46 number of grams present in 16.95 mol hydrogen peroxide (H_{2}O_{2}).

Explanation:

Number of moles is defined as the mass of substance divided by its molar mass.

The molar mass of H_{2}O_{2} is 34.01 g/mol. Hence, mass of hydrogen peroxide present in 16.95 moles is calculated as follows.

Moles = \frac{mass}{molarmass}\\16.95 mol = \frac{mass}{34.01 g/mol}\\mass = 576.46 g

Thus, we can conclude that there are 576.46 number of grams present in 16.95 mol hydrogen peroxide (H_{2}O_{2}).

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A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
Which isotope has the greatest number of protons?
tino4ka555 [31]

Answer:

The isotope with the greatest number of protons is:

  • <u>option D:  Pu-239, with 94 protons</u>

Explanation:

The number of <em>protons</em> is the atomic number and is a unique number for each type of element.

You can tell the number of protons searching the element in a periodic table and reading its atomic number.

Thus, this is how you tell the number of protons or each isotope

Sample       Chemical symbol  Element       atomic number   # of protons

A Pa-238       Pa                         protactinium         91                        91

B U-240         U                          uranium                 92                       92

C Np-238       Np                        neptunium            93                       93

D Pu-239        Pu                        plutonium              94                       94

8 0
3 years ago
Find density of nitrogen dioxide at 75*C and 0.805 atm.
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Answer:

1 (348) (D2) = 273 (2.05) (0.805) D2= 1.29 g/L

Explanation:

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3 years ago
Which of these is true for natural selection but not artificial selection?
ElenaW [278]
Can you show a picture so I can help??? :)
8 0
2 years ago
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