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Tpy6a [65]
3 years ago
6

In order to improve the condition of your muscular system, you could participate in aerobic activity at least __________. A. 3 m

inutes 5 times a week B. 10 minutes 4 times a week C. 30 minutes 3 times a week D. 50 minutes 3 times a week   Please select the best answer from the choices provided. A B C D
Chemistry
2 answers:
lara [203]3 years ago
8 0
The answer is C. 
30 mins 3 times a week. 
zlopas [31]3 years ago
5 0
I think it would be C
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The noble gases are grouped together in the periodic table. where are the noble gases in the table?
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Match the element with its description. Match Term Definition Potassium A) Has properties of both metals and nonmetals Barium B)
max2010maxim [7]

Answer:

A) Has properties of both metals and nonmetals - Barium.

B) Nonreactive gas - Neon.

C) Great conductor of heat and electricity - Boron.

D) Malleable and highly reactive - Potassium.

Explanation:

hope it helps .

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What is the relationship between the concentration of particles in a solution and the solution’s vapor pressure? As solute conce
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Answer:

As solute concentration increases, vapor pressure decreases.  

Step-by-step explanation:

As solute concentration increases, the number of solute particles at the surface of the solution increases, so the number of <em>solvent </em>particles at the surface <em>decreases</em>.

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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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