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son4ous [18]
3 years ago
5

Which property of ionic bonds is illustrated on the right? What other physical properties does this structure affect?

Chemistry
1 answer:
aniked [119]3 years ago
4 0

Answer:

Crystal structure; brittleness  

Step-by-step explanation:

The image illustrates the crystal structure of an ionic solid.

This structure affects the brittleness of the solid.

If you strike a sharp blow to the crystal with a hammer, the crystal layers will slide past each other.

The ions with the same charge will repel each other, and the crystal will fly apart.

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Answer:

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Explanation:

Did you just come out of the oven? Because you're hot

3 0
3 years ago
Calculate the new pressure of a gas if the gas at 50 ˚C and 81.0 kPa is heated to 100 ˚C at a constant volume.
Debora [2.8K]

Answer:

93.5 kPa

Explanation:

Step 1: Given data

  • Initial pressure (P₁): 81.0 kPa
  • Initial temperature (T₁): 50 °C
  • Final pressure (P₂): ?
  • Final volume (T₂): 100 °C

Step 2: Convert the temperatures to the Kelvin scale

When working with gases, we need to consider the absolute temperature. We will convert from Celsius to Kelvin using the following expression.

K = °C + 273.15

T₁: K = 50°C + 273.15 = 323 K

T₂: K = 100°C + 275.15 = 373 K

Step 3: Calculate the final pressure of the gas

At a constant volume, we can calculate the final pressure of the gas using Gay-Lussac's law.

P₁/T₁ = P₂/T₂

P₂ = P₁ × T₂/T₁

P₂ = 81.0 kPa × 373 K/323 K

P₂ = 93.5 kPa

7 0
3 years ago
Why is the Hubble telescope located in space? *
dolphi86 [110]

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Explanation:

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3 years ago
Naturally occurring boron has an atomic weight of 10.811. Its principal isotopes are 10B and 11B. Part A What is the abundance (
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Answer:

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Explanation:Please see attachment for explanation

4 0
4 years ago
glucose 6‑phosphate+H2O⟶glucose+Pi glucose 6‑phosphate+H2O⟶glucose+Pi K′eq1=270 K′eq1=270 ATP+glucose⟶ADP+glucose 6‑phosphate AT
ddd [48]

Answer:

-30.7 kj/mol

Explanation:

The standard free energy for the given reaction that is the hydrolysis of ATP is calculated using the formula:  ∆Go ’= -RTln K’eq

where,  

R = -8.315 J / mo

T = 298 K

For reaction,

1. K′eq1=270,

∆Go ’= -RTln K’eq

= - 8.315 x 298 x ln 270

=  - 8.315 x 298 x 5.59

= - 13,851.293 J / mo

= - 13.85 kj/mol

2. K′eq2=890

∆Go ’= -RTln K’eq

= - 8.315 x 298 x ln 890

=  - 8.315 x 298 x 6.79

=  - 16.82 kj/mol

therefore, total standard free energy

= - 13.85 + (-16.82)

=  -30.7 kj/mol

Thus, -30.7 kj/mol is the correct answer.

6 0
3 years ago
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