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konstantin123 [22]
3 years ago
15

What is the concentration of nitrate ions in a 0.225 msr(no3)2 solution?

Chemistry
2 answers:
12345 [234]3 years ago
7 0

Answer : The concentration of nitrate ions is 0.450 m.

Solution : Given,

Molarity of Sr(NO_3)_2 = 0.225 m

The balanced ionic equation is,

Sr(NO_3)_2\rightarrow Sr^{+2}+2NO^-_3

From the reaction, we conclude that the 1 mole of Sr(NO_3)_2 produces 2 moles of NO^-_3 ions. That means the concentration of nitrate ions is twice the value of Sr(NO_3)_2.

then, 0.225 m of Sr(NO_3)_2 gives 2 × 0.225 of NO^-_3 ions

Now the concentration of NO^-_3 ions is equal to 0.450 m.


BabaBlast [244]3 years ago
5 0

We are given that there are 2 moles of NO3 from the 0.225 M Sr(NO3)2 solution. Therefore we simply use stoichiometry to solve this problem.

 

Concentration NO3 = 0.225 M Sr(NO3)2 * (2 moles NO3 / 1 mole Sr(NO3)2)

<span>Concentration NO3 = 0.45 M</span>

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What is the rate law for the reaction 2A + 2B + 2C --&gt; products
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Answer:

R = 47.19 [A]*([B]^2)*[C]

Explanation:

The rate law for the reaction 2A + 2B + 2C --> products

Is being sought.

The reaction rate R could be expressed as  

R = k ([A]^m)*([B]^n)*([C]^p)                      (1)

where m, n, and p are the reaction orders with respect to (w.r.t.) components A, B and C respectively. This could be reduced to

R = ka ([A]^m)                   (2)

Where ka=(k[B]^n)*([C]^p);    

R = kb ([B]^n)                    (3)

Where kb=(k[A]^m)*([C]^p); and  

R = kc ([C]^p)                     (4)

Where kc=(k[A]^m)*([B]^n).

Equations (2), (3) and (4) are obtained for cases when the concentrations of two components are kept constant, while only one component’s concentration is varied. We can determine the reaction wrt each component by employing these equations.  

The readability is very much enhanced when the given data is presented in the following manner:

Initial [A]  0.273   0.819   0.273   0.273

Initial [B]  0.763   0.763   1.526   0.763

Initial [C]  0.400   0.400    0.400   0.800

Rate           3.0       9.0       12.0       6.0

Run#  1  2 3  4

Additional row is added to indicate the run # for each experiment for easy reference.

First, we use the initial rate method to evaluate the reaction order w.r.t. each component [A], [B] and [C] based on the equations (2), (3) and (4) above.

Let us start with the order wrt [A]. From the given data, for experimental runs 1 and 2, the concentrations of reactants B and C were kept constant.

Increasing [A] from 0.273 to 0.819 lead to the change of R from 3.0 to 9.0, hence we can apply the relation based on equation (2) between the final rate R2, the initial rate R1 and the final concentration [A2] and the initial concentration [A1] as follows:

R2/R1=ka[A2]^m/ka[A1]^m=([A2]/[A1])^m

9.0/3.0 = (0.819/0.273)^m

3 = (3)^m = 3^1  -> m = 1

Similarly, applying experimental runs 1 and 3 could be applied for the determination of n, by employing equation (3):  

R3/R1=kb[B3]^n/kb[B1]^n=([B3]/[B1])^n

12/3= (1.526/0.763)^n

4= 2^n, -> n = 2

And finally for the determination of p we have using runs 4 and 1:

R4/R1=kc[C4]^p/kc[C1]^p=([C4]/[C1])^p

6/3= (0.8/0.4)^p

2= 2^p , -> p = 1

Therefore, plugging in the values of m, n and p into equation (1), the rate law for the reaction will be:

R = k [A]*([B]^2)*[C]

The value of the rate constant k could be estimated by making it the subject of the formula, and inserting the given values, say in run 1:

k = R /( [A]*([B]^2)*[C]) = 3/0.273*(0.763^2)*0.4 =

47.19

Finally, the rate law is

R = 47.19 [A]*([B]^2)*[C]

7 0
4 years ago
When 15.0 mL of a 6.42×10-4 M sodium sulfide solution is combined with 25.0 mL of a 2.39×10-4 M manganese(II) acetate solution d
Artist 52 [7]

Answer:

Q = 3.59x10⁻⁸

Yes, precipitate is formed.

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The reaction of Na₂S with Mn(CH₃COO)₂ is:

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8 0
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