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Pavel [41]
2 years ago
5

Which best describes an element

Chemistry
1 answer:
-Dominant- [34]2 years ago
8 0

Answer:

a pure substance

Explanation:

cant be broken down into any more  simpler substances

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Plz answer ASAP!
Alchen [17]
What types of atoms typically form covalent bonds?

The correct answer:

d) Nonmetals with nonmetals, because their difference in electronegativity is below 1.7.

En example to explain:
An example of a covalent bonding: HCl -> 2 nonmentals
-> a difference in electronegativity less than 1.7:

EN(Cl) = 3.0 and EN (H) = 2.1 (you can search these values in a periodic table)
/\EN = 3.0 - 2.1 = 0.9

0.9 < 1.7

I hope this helped you out!
4 0
3 years ago
Read 2 more answers
Forensic toxicology involves not only the area of toxicology, but also the use of areas such as analytic chemistry and pharmacol
grandymaker [24]

Answer:

The answer is True

Explanation:

Forensic toxicology involves or  employs or uses disciplines such as analytical chemistry, pharmacology and clinical chemistry to aid medical or legal investigation.

3 0
3 years ago
Read 2 more answers
What’s the answer to these 3 questions? thanks!
Citrus2011 [14]
Water has h bonding
H-H
Sodium fluoride
I think
4 0
2 years ago
A solid and a liquid are shaken together in a test tube to produce a clear blue liquid. Which of the following best describes th
harkovskaia [24]
<span>the behavior of the above pair of substances</span> is soluble

6 0
3 years ago
The reaction C 4 H 8 ( g ) ⟶ 2 C 2 H 4 ( g ) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ / mol. 262 kJ/mol. At 600.0 K,
ludmilkaskok [199]

Answer: 4.3\times 10^{-13}s^{-1}

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 775.0 = ?

Ea = activation energy for the reaction = 262 kJ/mol = 262000J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 775.0K

Now put all the given values in this formula, we get

\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]

\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150

(\frac{6.1\times 10^{-8}}{K_2})=141253.8

Therefore, the value of the rate constant at 775.0 K is 4.3\times 10^{-13}s^{-1}

5 0
2 years ago
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