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charle [14.2K]
3 years ago
14

A good team member will actively listen, which means they will ______.

Chemistry
2 answers:
Wittaler [7]3 years ago
8 0
B is the answer I think better sure
Sergeeva-Olga [200]3 years ago
8 0

Answer:

C. use body language to understand how a speaker feels

Explanation:

let me know if i'm right

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Which pair of elements has reactivity that is similar to chlorine, Cl? F. Fluorine and Argon G. Fluorine and Iodine H. Sulfur an
kaheart [24]

halogens

Group 7A (or VIIA) of the periodic table are the halogens: fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At).

7 0
2 years ago
If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?
Semenov [28]

Answer:

see explanation below

Explanation:

The question is incomplete. Here's the complete question:

<em>Consider the following reaction: </em>

<em>SO2Cl2 -----> SO2(g) + Cl2(g) </em>

<em>Kc= 2.99 x 10^-7 at 227 degrees celcius </em>

<em>If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?</em>

This is a problem of equilibrium, therefore, we need to solve this using the expression of equilibrium constant. To do that, we need to wirte an ICE chart and solve from there:

       SOCl2 ---------> SO2 + Cl2     Kc = 2.99x10⁻⁷

i)       0.168                  0        0

c)         -x                    +x       +x

e)     0.168-x                x         x

Writting the Kc expression:

Kc = [SO2] [Cl2] / [SOCl2]

Replacing the values from the chart:

2.99x10⁻⁷ = x² / 0.168 - x

However, Kc is a very very small value, therefore, we can assume that the value of "x" would be very small too, and we can neglect the 0.168-x and just round it to 0.168:

2.99x10⁻⁷ = x²/0.168

2.99x10⁻⁷ * 0.168 = x²

√5.02x10⁻⁸ = x

x = 2.24x10⁻⁴ M

This means then, that the concentration of Cl2 in equilibrium would be:

<em>[Cl₂] = 2.24x10⁻⁴ M</em>

5 0
3 years ago
In the equilibrium constant expression for the reaction below what is the correct exponent for N2O4?
irga5000 [103]
As we have the balanced reaction equation is:

N2O4 (g) ↔ 2NO2(g)

from this balanced equation, we can get the equilibrium constant expression

KC = [NO2]^2[N2O4]^1

from this expression, we can see that [NO2 ] is with 2 exponent of  the stoichiometric and we can see that from the balanced equation as NO2
is 2NO2 in the balanced equation.

and [N2O4] is with 1 exponent of the stoichiometric and we can see that from the balanced equation as N2O4 is 1 N2O4 in the balanced equation. 

∴ the correct exponent for N2O4 in the equilibrium constant expression is 1 
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3 years ago
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The action was him hitting the ball the reaction was the ball moving after being hit
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