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pochemuha
3 years ago
15

Free Points to my friends!!!!! :D:D:D:D

Chemistry
2 answers:
solong [7]3 years ago
6 0

Answer:

Danke! (≧∇≦)

Explanation:

beks73 [17]3 years ago
4 0

Answer:

thanks!!!

Explanation:

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Which of the following events could increase Earth's temperature?
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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
Sunny_sXe [5.5K]

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}

     

     

3 0
3 years ago
2H2 + O2 + 2H20
kondaur [170]

Answer:

Hello

The answer is 25gr H2O

Explanation:

50grH2×1 molH2/2 gr H2×2mol H2O/2molH2=25gr H2O

8 0
3 years ago
How many grams of solute are dissolved in 125.0 mL of 5.00 M NaCl (MM = 58.45)?
Nookie1986 [14]

Answer: 36.53g

Explanation:

First we need to find the amount of NaCl that dissolves in 1L of the solution that produced 5M of NaCl

Molarity = 5M

MM of NaCl = 58.45

Molarity = Mass conc (g/L) / MM

Mass conc. (g/L) of NaCl = Molarity x MM

= 5 x 58.45 = 292.25g

Next, we need to find the amount that will dissolve in 125mL(i.e 0.125L)

From the calculations above,

292.25g of NaCl dissolved in 1L

Therefore Xg of NaCl will dissolve in 0.125L of the solution i.e

Xg of NaCl = 292.25 x 0.125 = 36.53g.

Therefore 36.53g of NaCl will dissolve in 125mL of the solution

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