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UNO [17]
4 years ago
5

Calculate the molality of the solution. (assume a density of 1.08 g/ml for the solution.)an aqueous nacl solution is made using

140 g of nacl diluted to a total solution volume of 1.20
Chemistry
1 answer:
Anna007 [38]4 years ago
6 0

molality is moles solute / kg of solvent

 

moles of NaCl = mass / molar mass

molar mass NaCl = 58.44 g/mol

moles NaCl = 140 g / 58.44 g/mol = 2.40 mol NaCl

 

The total volume of the solution is 1.2L or 1200 ml

The density of the solution is 1.08 g/ml

 

Therefore the total mass of the solution is 1.08 g/ml x 1200 ml

= 1296 g

 

But there are140 g of NaCl in this solution

 

So mass H2O = 1296 g - 140 g

= 1156 g

= 1.156 kg

 

molality = moles solute/ kg solvent

= 2.40 mol / 1.156 kg

= 2.08 molal

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rate of a certain reaction is given by the following rate law: rate Use this information to answer the questions below. What is
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Complete Question

The  rate of a certain reaction is given by the following rate law:

            rate =  k [H_2][I_2]

rate Use this information to answer the questions below.

What is the reaction order in H_2?

What is the reaction order in I_2?

What is overall reaction order?

At a certain concentration of H2 and I2, the initial rate of reaction is 2.0 x 104 M / s. What would the initial rate of the reaction be if the concentration of H2 were doubled? Round your answer to significant digits. The rate of the reaction is measured to be 52.0 M / s when [H2] = 1.8 M and [I2] = 0.82 M. Calculate the value of the rate constant. Round your answer to significant digits.

Answer:

The reaction order in H_2 is  n =  1

The reaction order in I_2 is  m = 1

The  overall reaction order z =  2

When the hydrogen is double the the initial rate is   rate_n  =  4.0*10^{-4} M/s

The rate constant is   k = 35.23 \  M^{-1} s^{-1}

Explanation:

From the question we are told that

   The rate law is  rate =  k [H_2][I_2]

   The rate of reaction is rate =  2.0 *10^{4} M /s

Let the reaction order for H_2 be  n and for I_2  be  m

From the given rate law the concentration of H_2 is raised to the power of 1 and this is same with I_2 so their reaction order is  n=m=1

   The overall reaction order is  

               z  = n +m

               z  =1 +1

               z  =2

At  rate =  2.0 *10^{4} M /s

        2.0*10^{4}  = k  [H_2] [I_2] ---(1)

= >    k  = \frac{2.0*10^{4}}{[H_2] [I_2]  }

given that the concentration of hydrogen is doubled we have that

            rate  = k [2H_2] [I_2] ----(2)

=>      k = \frac{rate_n  }{ [2H_2] [I_2]}

 So equating the two k

           \frac{2.0*10^{4}}{[H_2] [I_2]  } = \frac{rate_n  }{ [2H_2] [I_2]}

    =>    rate_n  =  4.0*10^{-4} M/s

So when

      rate_x =  52.0 M/s

        [H_2] = 1.8 M

         [I_2] =  0.82 \ M

We have

      52 .0 =  k(1.8)* (0.82)

     k = \frac{52 .0}{(1.8)* (0.82)}

      k = 35.23 M^{-2} s^{-1}

     

     

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∴ mass of CO₂ = 2.8125 × 10⁻⁴ × (12 + 16×2)

= 12.375 × 10⁻³ g

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