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MissTica
3 years ago
10

How many significant digits are there in the measurement 50,600 mg A)4 B)5 C)6 D)7

Chemistry
2 answers:
Arte-miy333 [17]3 years ago
7 0

Five. The measurement 50,600 mg has five significant digits.

I presume that you are using the comma as a decimal separator.

The <em>rules for significant figures</em> are
1. Nonzero digits are always significant.
2. Any zeros between two significant digits are significant.
3. Final or trailing zeros are significant only if they are to the right of a decimal point.

• According to Rule 1, the <em>5 and 6</em> are significant.
• According to Rule 2, the <em>0 between the 5 and 6</em> is significant
• According to Rule 3, the <em>final two zeros</em> are significant.

Thus, there are five significant digits in the measurement 50,600 mg.

Note: If the comma is a thousands separator, the number has only three significant digits.

stellarik [79]3 years ago
3 0

Answer:

5.

Explanation:

Significant figures may be defined as the numbers of the digit that can carry to contribute the meaning of the exact measurement resolution. Significant figures follows some specific scientific rules.

All non zero digits are significant figures. The zeros that are present between the non zero digits are significant figures. The zeros that are present after the significant figures are considered as significant. Here, 5 significant figures are 5,0,6,0 and 0. Hence, the total significant figures are 5.

Thus, the correct answer is option (B).

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In their notebook you see that it takes 9 hours for a sixth of a 0.5M solution of BC2 to react. Unfortunately, you have somewher
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of BC_2 that should used originally is C_Z_o = 0.4492M

Explanation:

     From the question we are told that

         The necessary elementary step is  

                  2BC_2 ----->4C + B_2

          The  time taken for sixth of 0.5 M of reactant to react t = 9 hr

           The time available is t_a = 3.5 hr

             The desired concentration to  remain C  = 0.42M

Let Z be the reactant ,   Y be the first product and X the second product

Generally the elementary rate  law is mathematically as

                    -r_Z = kC_Z^2 = - \frac{d C_Z}{dt}

Where k is the rate constant ,  C_Z is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

 For second  order reaction

            \frac{1}{C_Z}  - \frac{1}{C_Z_o}  = kt

Where C_Z_o is the initial concentration of Z which a value of   C_Z_o = 0.5M

       From the question we are told that it take  9 hours  for the concentration of  the reactant to become

                 C_Z =  C_Z_o - \frac{1}{6}  C_Z_o

                  C_Z = 0.5  - \frac{0.5}{6}

                       = 0.4167 M

So      

                     \frac{1}{0.4167}  - \frac{1}{0.50}  =  9 k

                          0.400 = 9 k

                =>    k = 0.044\  L/ mol \cdot hr^{-1}

  For   C_Z = 0.42M

                \frac{1}{0.42} - \frac{1}{C_Z_o}  = 3.5 * 0.044

                2.38 -  0.154  =    \frac{1}{C_Z_o}

                           2.226  =    \frac{1}{C_Z_o}

                            C_Z_o = \frac{1}{2.226}

                             C_Z_o = 0.4492M

                       

           

             

         

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