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Verizon [17]
2 years ago
5

elements are arranged in groups by similar atomic structure on the periodic table. this allows for an elements properties to be

predicted based on general periodic trends. One of these trends, atomic radius , increases down a group and to the left along a period can be defined as
Chemistry
1 answer:
emmasim [6.3K]2 years ago
8 0

Answer:

SUB TO HAVIN RX

Explanation:

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What is the correct equilibrium constant
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Answer:[co2][CF4]

[COF2]^2

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3 years ago
The equations given in the problem introduction can be added together to give the following reaction: overall: N2O4→2NO2 However
Murljashka [212]

Answer:

Reverse the 2 NO_2 \longrightarrow 2 NO + O_2  reaction

Explanation:

Reactions:

2 NO_2 \longrightarrow 2 NO + O_2

N_2O_4 \longrightarrow 2 NO + O_2

Overall:

N_2O_4 \longrightarrow 2 NO_2

As can be seen, in the overall reaction we have N_2O_4 in the reactants like in the second reaction and NO_2 in the products. The NO_2 is in the first reaction but as a reactant so we need to reverse that reaction:

2 NO + O_2 \longrightarrow 2 NO_2

N_2O_4 \longrightarrow 2 NO + O_2

Combining:

N_2O_4 + 2 NO + O_2\longrightarrow 2 NO + O_2 + 2 NO_2

N_2O_4 \longrightarrow 2 NO_2

4 0
3 years ago
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>
7 0
3 years ago
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