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polet [3.4K]
3 years ago
13

PLEASEEEE HELP WITH THIS ID REALLY REALLY APPRECIATE IT

Chemistry
1 answer:
dsp733 years ago
6 0

i can hardly see it sry...

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Artyom0805 [142]

Answer:

use a concaved miror

Explanation:

better and more clear

7 0
3 years ago
The specific heats at constant pressure of some common gases are provided as a thirdorder polynomial: �;<<< = � + �� +
3241004551 [841]

Answer:

1.991 kJ

Explanation:

Calculate the amount of heat ( J )

CH4 ;

coefficients are :  a = 19.89 , b = 5.02 * 10^-2 , c = 1.269 * 10^-5 , d = -11.01 * 10^-9

attached below is the detailed solution

3 0
3 years ago
At 25°C, the equilibrium constant Kc for the reaction in thesolvent CCl4 2BrCl <----> Br2 + Cl2 is 0.141. If the initial c
ivolga24 [154]

<u>Answer:</u> The equilibrium concentration of bromine gas is 0.00135 M

<u>Explanation:</u>

We are given:

Initial concentration of chlorine gas = 0.0300 M

Initial concentration of bromine monochlorine = 0.0200 M

For the given chemical equation:

                   2BrCl\rightleftharpoons Br_2+Cl_2

<u>Initial:</u>          0.02               0.03

<u>At eqllm:</u>    0.02-2x     x     0.03+x

The expression of K_c for above equation follows:

K_c=\frac{[Br_2]\times [Cl_2]}{[BrCl]^2}

We are given:

K_c=0.141

Putting values in above equation, we get:

0.141=\frac{x\times (0.03+x)}{(0.02-2x)^2}\\\\x=-0.96,+0.00135

Neglecting the value of x = -0.96 because, concentration cannot be negative

So, equilibrium concentration of bromine gas = x = 0.00135 M

Hence, the equilibrium concentration of bromine gas is 0.00135 M

8 0
3 years ago
: Citric acid, H3C6H5O7, is a triprotic acid. Consider a buffer system comprising H2C6H5O7 - and HC6H5O7 2- ions. What is the ne
olchik [2.2K]

Answer:

The Net reaction is    

  1.    H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction
  2.    H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH
  3.    HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH  

Explanation:

From the Question we are told that the buffers are

                H_2C_6H_5O_7^ - and  HC_6H_5O_7 ^{ 2-}

When NaOH is added the Net ionic reaction would be

  1.    H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction
  2.    H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH
  3.    HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH  

             

8 0
3 years ago
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Step I of the proposed mechanism involves the collision between NO2 and F2 molecules. This step is slow even though such collisi
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Answer:

bob

Explanation:

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7 0
3 years ago
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