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just olya [345]
4 years ago
6

Which of the following will stop gamma rays?

Chemistry
2 answers:
andrezito [222]4 years ago
7 0
Concrete will stop gamma rays. 

Good luck :)
pashok25 [27]4 years ago
4 0
Concrete. it cant be any other choice
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If smelly suzy has 109 scrunchies and 47 visco stickers for her hydroflasksksksksks hiw many visco items does she have in total
andreyandreev [35.5K]

Answer:

156

Explanation: SKSKSKSKSKSKKSKSKSKSKKS

6 0
3 years ago
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
How many grams of O₂ are required to react completely with 14.6 g of Na to form sodium oxide, Na₂O?
Bad White [126]

The balanced chemical reaction is :

O_2 + 4Na \ -> \ 2Na_2O

Number of moles of Na, n = \dfrac{14.6}{23} = 0.635 \  mol .

Now, from balance chemical reaction we can see that 1 mole of oxygen reacts with 4 moles of sodium.

So, number of moles of oxygen are :

n = \dfrac{0.635}{4}\  mole

So, amount of oxygen required is :

m = \dfrac{0.635 \times 32}{4}\  gm\\\\m = 5.08 \ gm

Therefore, 5.08 gram of oxygen will react with 14.6 gram of sodium.

7 0
3 years ago
Which best describes a hydrate?
Fofino [41]

Answer:

d

Explanation:

i did it

8 0
3 years ago
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Describe step by step how we can make a field of a variable quantity?
just olya [345]
Finding the variables in the experiment or study and measurinf
7 0
3 years ago
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