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kenny6666 [7]
3 years ago
5

I need help please .!

Chemistry
1 answer:
Debora [2.8K]3 years ago
8 0

hydrogen (H2),and oxygen (O2),, that is the answer





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How does the polarity of water give it this solvent property
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being polar, it can easily dissolve other polar substances or substances with ionic bonds like nacl

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What is the name for potassium oxide's structure? Give two properties of potassium oxide.
Anna71 [15]

Answer:

- The name for the potassium oxide's structure is ionic.

Properties:

- High melting point.

- Soluble in water.

Explanation:

- The ionic structure it is formed by a cation (atom with positive charge) and an anion (atom with negative charge). In this case, potassium is the cation and the oxigen is the anion.

- Since potassium oxide is an ionic compound, it has a high melting point, because of the strong bonds. Also, it is soluble in polar solvents, like water, because its ions generate polarity in the molecule.

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In the Table of Atomic Particles, various symbols have two numbers. The top number represents the atomic mass of the particle, w
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Atomic number and number of electrons.
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Potassium and fluorine are both halogens?​
andreev551 [17]

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false, Potassium and fluorine are not halogens.

only fluorine here is halogen.

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7 0
3 years ago
The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th
MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

\frac{K_1}{K_2}=?

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

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

The expression used with catalyst and without catalyst is,

\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}

\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}

where,

K_2 = rate constant reaction -1

K_1 = rate constant reaction -2

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = 25^oC=273+25=298 K

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

\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340

0.4284 is the ratio of the rate constants.

7 0
3 years ago
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