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yuradex [85]
3 years ago
7

Which one is it ASAP

Chemistry
1 answer:
Phantasy [73]3 years ago
8 0
They would attract to each other so the first one. 
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Which of the following is used for chemical symbols today? a. drawings c. letters b. icons d. numbers
devlian [24]

Answer:

Letters

Explanation:

For example, today we use the periodic table which is full of elements named with 1 or 2 letters. Like how Helium is He and Sodium is Na. Hope this helps!!!

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Consider the reaction
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Value of Kc is E= 2.5
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Why can a solution be classified as a mixture
Vanyuwa [196]
Because they’re both made up of two substances that are not chemically combined
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Describe the intermolecular forces that must be overcome to convert these substances from a liquid to a gas: (a) SO2, (b) CH3COO
Andre45 [30]
You need to find which intermolecular forces are between the molecules
dipole-dipole,h bonds, etc.
I'm not very good at explaining but this is what my prof said to help us

Identify the class of the molecule or molecules you are given. Are they nonpolar species, ions or
do they have permanent dipoles? Is there only one species or are there two?
In the case of ONE species (i.e., a pure substance), the intermolecular forces will be between
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ION or IONIC. If you are dealing with dipoles, then the intermolecular forces will be DIPOLE-
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or VAN DER WAALS or INDUCED DIPOLE-INDUCED DIPOLE (the last three are desciptions
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In the case of TWO species (i.e., a mixture), the intermolecular forces will be between molecules of
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7 0
3 years ago
A very simple assumption for the specific heat of a crystalline solid is that each vibrational mode of the solid acts independen
MatroZZZ [7]

Answer:

4.7 kJ/kmol-K

Explanation:

Using the Debye model the specific heat capacity in kJ/kmol-K

c = 12π⁴Nk(T/θ)³/5

where N = avogadro's number = 6.02 × 10²³ mol⁻¹, k = 1.38 × 10⁻²³ JK⁻¹, T = room temperature = 298 K and θ = Debye temperature = 2219 K  

Substituting these values into c we have

c = 12π⁴Nk(T/θ)³/5  

= 12π⁴(6.02 × 10²³ mol⁻¹)(1.38 × 10⁻²³ JK⁻¹)(298 K/2219 K)³/5

= 9710.83(298 K/2219 K)³/5

= 1942.17(0.1343)³

= 4.704 J/mol-K

= 4.704 × 10⁻³ kJ/10⁻³ kmol-K

= 4.704 kJ/kmol-K

≅ 4.7 kJ/kmol-K

So, the specific heat of diamond in kJ/kmol-K is 4.7 kJ/kmol-K

7 0
3 years ago
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