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3 years ago

14

Cindy runs 2 kilometers every morning. She takes 2 minutes for the first 250 meters, 4 minutes for the next 1,000 meters, 1 minu

te for the next 350 meters, and 3 minutes for the rest.
Cindy's average speed for the entire run is?

2 answers:

beks73 [17]3 years ago

8 0

It's 200 on Edmentum

IceJOKER [234]3 years ago

5 0

She would be going 0.2 km/m I think. Let me know if it was correct please!

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As you move from bottom on the periodic table to the top shielding

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Answer:

AS we move from bottom to top on periodic table shielding decreased.

Explanation:

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction.

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

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3 years ago

Determine the value of the equilibrium constant, KgoalKgoalK_goal, for the reaction CO2(g)⇌C(s)+O2(g)CO2(g)⇌C(s)+O2(g), Kgoal=?

NISA [10]

Answer:

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$ .

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$K_{goal}=\frac{[C][O_2]}{[CO_2]}$

$2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2(g)$ ..[1]

$K_1=\frac{[CH_3COOH][O_2]^2}{[CO_2]^2[H_2O]^2}$

$2H_2(g)+O2(g)\rightleftharpoons 2H_2O(l)$ ..[2]

$K_2=\frac{[H_2O]^2}{[H_2]^2[O_2]}$

$CH_3COOH(l)\rightleftharpoons 2C(s)+2H_2(g)+O_2(g)$ ..[3]

$K_3=\frac{[C]^2[H_2]^2[O_2]}{[CH_3COOH]}$

[1] + [2] + [3]

$2CO_2(g)\rightleftharpoons 2C(s)+2O_2(g)$

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$K=K_1\times K_2\times K_3$

$K=5.40\times 10^{-16}\times 1.06\times 10^{10}\times 2.68\times 10^{-9}$

$K=1.53\times 10^{-14}$

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On comparing the K and $K_{goal}$ :

$K^2=K_{goal}$

$K_{goal}=\sqrt{K}=\sqrt{1.53\times 10^{-14}}=1.24\times 10^{-7}$

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$ .

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