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Kitty [74]
3 years ago
11

A nucleus with four protons has total positive charge

Chemistry
2 answers:
Sladkaya [172]3 years ago
7 0
It would have a charge of 4+
Jlenok [28]3 years ago
5 0
It would be 4 Bc Protons are positive charges
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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]

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4 years ago
A flask at room temperature contains exactly equal amounts (in moles) of nitrogen and xenon.
zalisa [80]

Answer:

a) Same

b) Nitrogen

c) Same

d) Nitrogen

Explanation:

a)

The formula for partial pressure of a gas is equal to

p_B = n_B \frac{RT}{V}

Here nB is the number of moles .

The number of moles for both the gases are same and hence the partial pressure for the two gases will also be same.

b) The greater average velocity is calculated by using following formula

v_{RMS} = \sqrt{3RTM}

Here M is the molar mass.

Molar mass of nitrogen is greater than the molar mass of xenon and hence nitrogen will have higher greater average velocity

c) As we know, the average kinetic energy of gas particles is dependent on the absolute temperature of gas and if all the gases are at same temperature, their kinetic energy will also be same. Since nitrogen and xenon are at same temperature, their kinetic energy will be same

d) Effusivity is depended directly on  the thermal conductivity,  density and  and the specific heat capacity.

All these three parameters are higher in case of nitrogen. Thus, it will effuse first

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4 years ago
What is the purpose of control set up
Archy [21]
The purpose<span> of preparing </span>Set-up<span> _ is to act as a </span>control set-up<span> to ensure that the (measured variable) is only caused by the(changed variable) and not any other variables in the experiment.</span>
3 0
4 years ago
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