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Jlenok [28]
3 years ago
10

A balance was tested against a standard calibration mass with a certified value of 200.002 g and produced the following readings

in grams: 195.587, 195.586, 195.588, 195.587, 195.589. The performance of this balance is best described as:
Chemistry
1 answer:
wlad13 [49]3 years ago
7 0

Answer:

Precise but not accurate.

Explanation:

We can tell the performance of the balance is precise, because the repeated measurements give values close to one another.

However, the performance of the balance is not accurate, as the mean value of the repeated measurements (195.587) is not close to the value considered as true (in this case the standard calibration mass with a certified value of 200.002 g).

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3 years ago
At 400 k, the equilibrium constant for the reaction is kc = 7.0. br2 (g) + cl2 (g) 2brcl (g) a closed vessel at 400 k is charged
jeka57 [31]
a) when Kc = concentration of products / concentration of reactants
  So according  to the reaction equation:
Br2(g) + Cl2(g) → 2BrCl(g)

∴ Kc =[BrCl] ^2 / [Br2][Cl2]

b) when q = [BrCl]^2 / [Br2][Cl2]
and we have [BrCl] = 3 m 
[Br2] = 1 m 
[Cl2] = 1 m
So by substitution:
q= 3^2 / 1*1 = 9 

- and we can see that q > Kc 
the reaction is not at equilibrium that means there are more products and the reaction shifts to the left to increase the reactants and decrease the products to achieve equilibrium.

C) by using ICE table:

              Br2(g) + Cl2(g) → 2BrCl (g)
initial       1               1               3
change  -X              -X            +X
Equ       (1-X)          (1-X)         (3+X)

when Kc = [Brcl]^2/[Cl2][Br2]
by substitution:
7 = (3+X)^2 / (1+X) (1+X)  by solving this equation for X
∴X = 0.215
so at equilibrium:
∴ [Br2] = [Cl2] = 1-0.215 = 0.785 m 
    [BrCl] = 3+0.215 = 3.215 m
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Answer:

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Explanation:

Hope this helped, nya~ :3

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Answer:

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