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Advocard [28]
2 years ago
12

Back talk it’s quiet ain’t no back talk

Chemistry
2 answers:
ale4655 [162]2 years ago
7 0

Answer:

?

Explanation:

????????????????????????????????????????????????????????

I hope you find the answer to your question?

kvv77 [185]2 years ago
7 0

Answer:

u r so real for this

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Which of the following does not serve as a way to neutralize the charge in a body?
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B. Adding more protons to a positively charged body until the number of protons matches the number of electrons.

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A ____________ star would have the hottest surface temperature.
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Blue

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If you look at a flame, blue is always at the bottom right? So that would be common sense that blue would be the hottest.

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Which of the following happens to a molecule of an object when the object is heated? (1 point)
vfiekz [6]

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They get more energy, so they vibrate!

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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
The density of liquid mercury is 13.69 g/cm^3. How many atoms of mercury are in a 15.0 cm^3 sample? Use "E" for "x10" and use si
balu736 [363]
The formula is  m = D x V
D = <span>13.69 g/cm^3.
</span>V = <span>15.0 cm^3 
the mass of the liquid mercury is m= </span>13.69 g/cm^3 x 15.0 cm^3 = 195g
the molar mass of Hg is 200,
1 mole of Hg = 200g Hg, so #mole of Hg= 195 / 200 = 0.97 mol
but we know that
 1 mole  = 6.022 E23 atoms
0.97 mole=?

6.022 E23 atoms x 0.97 / 1 mole = 5.84 E23 atoms
6 0
3 years ago
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