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Dimas [21]
3 years ago
7

Casey remembers that molecular motion increases as temperature increases. Identify what effect increased molecular motion.

Chemistry
2 answers:
Zanzabum3 years ago
8 0

Answer: Your answer would be D

Explanation: This is because whenever the collision increases the molecular motion increases with it.

Anastaziya [24]3 years ago
3 0

D

Molecular motion increases as temperature increases and therefore increasing the molecular motion increases collision frequency.

Explanation:

When a substance is heated, its molecules gain increased kinetic energy even as the substance increases in temperatures. With higher kinetic energy, this means the molecules are moving more rapidly and hence collision between the molecules will be more frequent and of higher energy. It is these high energy collisions that cause the molecules to move further away from each other with increased temperatures (why substances expand).  

Learn More:

For more on relationship between temperatures and molecular motion check out;

brainly.com/question/11659700

brainly.com/question/874953

brainly.com/question/2731193

brainly.com/question/2731193

brainly.com/question/2257506

#LearnWithBrainly

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If someone is whispering, but you can't hear them, how do you fix that problem?
Maurinko [17]
I'D SAY D IS THE ANSWER
5 0
3 years ago
Read 2 more answers
What is the rate law for the reaction 2A + 2B + 2C --> products
-Dominant- [34]

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
27.8 mL solution of 0.797 M HCHO2 with 0.928 M NaOH. What is the pH for the solution at the equivalence point in the titration?
kati45 [8]

Answer:

8.69 is the pH at the equivalence point

Explanation:

Formic acid, HCHO₂, reacts with NaOH as follows:

HCHO₂ + NaOH → NaCHO₂ + H₂O.

At the equivalence point you will have in the reaction just NaCHO₂ and H₂O. The concentration of NaCHO₂ will be:

<em />

<em>Moles: </em>0.0278L * 0.797mol/L = 0.02216moles

To reach the equivalence point it is necessary to add:

0.02216mol * (1L / 0.928mol) = 0.0239L

Total volume in the equivalence point:

0.0278L + 0.0239L = 0.0517L

Concentration: 0.02216moles / 0.0517L = 0.429M

The equilibrium of NaCHO₂, CHO₂⁻, in water is:

CHO₂⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + HCHO₂(aq)

Where Kb, 5.56x10⁻¹¹ is defined as:

5.56x10⁻¹¹ = [OH⁻] [HCHO₂] / [CHO₂⁻]

In the equilibrium, it is produced X OH⁻ and HCHO₂, and as concentration of NaCHO₂ is 0.429M:

5.56x10⁻¹¹ = [X] [X] / [0.429M]

2.383x10⁻¹¹ = X²

4.88x10⁻⁶ = X = [OH⁻]

As pOH = -log [OH⁻]

pOH = 5.31

And pH = 14 - pH

pH = 8.69 is the pH at the equivalence point

3 0
4 years ago
When 20 ml of 0.1 M HCl is mixed with 20
Bas_tet [7]

Answer:

it is b i took the test

Explanation:

4 0
2 years ago
If excess caso4(s) is mixed with water at 25 ∘c to produce a saturated solution of caso4, what is the equilibrium concentration
Ray Of Light [21]

<em>Answer:</em>

The equlibrium concentration sof Ca+2 ion willl be 4.9×10∧-3 M

<em>Data Given:</em>

              Ksp of CaSO4 = 2.4 × 10∧-5

              CaSO4 ⇔ Ca+2   +  SO4∧-2

<em>Solution:</em>

                Ksp = [Ca+2].[ SO4∧-2]

                 2.4 × 10∧-5 = [x].[x]= x²

                 x =  4.9×10∧-3 M

<em>Result:</em>

  • The conc. of Ca+2 ion is 4.9×10∧-3 M
3 0
3 years ago
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