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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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The hybridization associated with the central atom of a molecule in which all the bond angles are 180° is ____.
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3 years ago
The ratio of nitrogen to oxygen by mass in NO is 7.0:8.0. Identify the ratio of nitrogen to oxygen by mass in NO2 and N2O7 .
Rus_ich [418]

This problem is providing the ratio of nitrogen to oxygen by mass in nitrogen monoxide, NO, as 7.0:8.0 and asks for the same ratio but in NO₂ and N₂O₇. After doing the calculations, the results are 7.0:16.0 and 1.0:4.0 respectively.

<h3>Mass ratios:</h3>

In chemistry, one can calculate the mass ratios in chemical formulas according to the atomic mass of each atom. In such a way, one knows the mass ratio of nitrogen to oxygen in NO is 7.0:8.0 because we divide the atomic mass of nitrogen by oxygens:

\frac{14}{16}=\frac{7.0}{8.0}

Now, for chemical formulas with subscripts, one must multiply the atomic mass of the element by the subscript in the formula, which is the case of NO₂ and N₂O₇ as shown below:

NO_2:\frac{14}{16*2}=\frac{14}{32} =\frac{7.0}{16.0}  \\&#10;\\&#10;N_2O_7:\frac{14*2}{16*7}=\frac{28}{112} =\frac{1.0}{4.0}

Therefore, the results for NO₂ and N₂O₇ are 7.0:16.0 and 1.0:4.0 respectively

Learn more about atomic masses: brainly.com/question/5566317

3 0
3 years ago
f the Ksp for HgBr2 is 2.8×10−14, and the mercury ion concentration in solution is 0.085 M, what does the bromide concentration
goldfiish [28.3K]

Answer:

0.057 M

Explanation:

Step 1: Given data

Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴

Concentration of mercury (II) ion: 0.085 M

Step 2: Write the reaction for the solution of HgBr₂

HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻

Step 3: Calculate the bromide concentration needed for a precipitate to occur

The Ksp is:

Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²

[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M

7 0
3 years ago
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