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Kay [80]
3 years ago
7

Match each number to how it would appear in scientific notation.

Chemistry
2 answers:
Natali5045456 [20]3 years ago
7 0

Answer:

8,330-->8.33 × 10^3

83.3-->8.33 × 10^1

0.00833-->8.33 × 10^-3

0.0833-->8.33 × 10^-2

83,330-->8.33 × 104

Verdich [7]3 years ago
4 0
A. 8,330 --> 8.33 × 10^3
B. 83.3 --> 8.33 × 10^1
C. 0.00833 --> 8.33 × 10^-3
D. 0.0833 --> 8.33 × 10^-2
E. 83,330 --> 8.33 × 10^4

Hope this helps!
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The arsenic in a 1.223 g sample of a pesticide was converted to H3AsO4 by suitable treatment. The acid was then neutralized, and
Vladimir79 [104]

Answer:

5.471% As₂O₃ in the sample.

Explanation:

<em>...the reaction is: Ag+ + SCN- => AgSCN(s) Calculate the percent As2O3 in the sample. (F.W. As2O3 = 197.84 g/mol).</em>

<em />

First, with the amount of KSCN we can find the moles of Ag in the filtrates. As we know the amount of Ag added we can know the precipitate of Ag and the moles of AsO₄ = 1/2 moles of As₂O₃ in the sample:

<em>Moles KSCN = Moles Ag⁺ in the filtrate:</em>

0.01127L * (0.100mol / L)= 0.001127moles Ag⁺

<em>Total moles Ag⁺:</em>

0.0400L * (0.0781mol/L) = 0.0031564 moles Ag⁺

<em>Moles of Ag⁺ in the precipitate:</em>

0.0031564 - 0.001127 = 0.0020294 moles Ag⁺

<em>Moles AsO₄ = Moles As:</em>

0.0020294 moles Ag⁺ * (1mol As / 3 moles Ag⁺) = 6.765x10⁻⁴ moles AsO₄

<em>Moles As₂O₃:</em>

6.765x10⁻⁴ moles AsO₄ * (1 mol As₂O₃ / 2 mol AsO₄) =

3.382x10⁻⁴ moles As₂O₃

<em>Mass As₂O₃:</em>

3.382x10⁻⁴ moles As₂O₃ * (197.84g/mol) = 0.0669g As₂O₃

Percent is:

0.0669g As₂O₃ / 1.223g sample * 100 =

<h3>5.471% As₂O₃ in the sample</h3>

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3 years ago
As the pendulum moves from point 2 to point 3, what happens to its mechanical energy?potential energy is converted to kinetic en
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Answer: Potential energy is converted to kinetic energy and back again.

Explanation:At points 1 and 3, the pendulum stops moving, and its mechanical energy is purely potential. At point 2, the pendulum is moving the fastest, and its mechanical energy is purely kinetic. Therefore, as the pendulum moves from point 1 to point 3, its potential energy is first converted to kinetic energy, then back to potential.

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2 years ago
How the calculation of the [OH-], pH and % ionization for 0.619 M ammonia (NH3) NH3 + H2O (liq) rightwards harpoon over leftward
fomenos

Answer:

[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52

Explanation:

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NH3 + H2O(l) ⇄ NH4+ + OH-

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Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]

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<em>And as </em>[NH₃] = 0.619M

1.8x10⁻⁵ = [X] [X] / [0.619M]

1.11x10⁻⁵ = X²

3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]

<h3>[OH⁻] = 3.34x10⁻³M</h3><h3 />

% ionization:

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