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skelet666 [1.2K]
3 years ago
10

What is the percent yield for the reaction of nitrogen and hydrogen to produce ammonia, if the theoretical yield was 12.5 g but

the actual yield was only 10.2 g?
A. 123 %

B. 12.5%

C. 81.6 %

D. 10.2%

Chemistry
1 answer:
koban [17]3 years ago
4 0

Answer:

81.6%

Explanation:

10.2/12.5= 0.816

0.816=81.6%

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How many atoms are in a 1.8 mol sample of Magnesium (Mg)?
Gnom [1K]

Answer:

1.1 × 10²⁴ atoms Mg

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Moles
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

<em>Identify</em>

[Given] 1.8 mol Mg

[Solve] atoms Mg

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 1.8 \ mol \ Mg(\frac{6.022 \cdot 10^{23} \ atoms \ Mg}{1 \ mol \ Mg})
  2. [DA] Multiply [Cancel out units]:                                                                      \displaystyle 1.08396 \cdot 10^{24} \ atoms \ Mg

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.08396 × 10²⁴ atoms Mg ≈ 1.1 × 10²⁴ atoms Mg

5 0
3 years ago
Read 2 more answers
What amount of energy is required to change a spherical drop of water with a diameter of 1.80 mm to three smaller spherical drop
Gekata [30.6K]
This is a straightforward question related to the surface energy of the droplet. 

<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>

<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>

<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>

<span>The five smaller droplets need to have the same volume as the original. Therefore </span>

<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>

<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>

<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>

<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>

<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>
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