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lions [1.4K]
4 years ago
9

If a molecule demonstrates paramagnetism, then : I. The substance can have both paired and unpaired electrons. II. The bond orde

r is not a whole number. III. It can be determined by drawing a Lewis structure. IV. It must be an ion.
Chemistry
1 answer:
jolli1 [7]4 years ago
8 0

Answer:The bond order is not a whole number.

Explanation:

If a molecule is paramagnetic, then there are unpaired electrons in some of its molecular orbitals. This will cause the bond order not to be a whole number when treated by applying the ordinary molecular orbital theory to the molecule.

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Which general equation represents an endothermic reaction?
igomit [66]

Answer:

D

Explanation:

Reactants+Energy lead to Products

3 0
3 years ago
Read 2 more answers
In the reaction CuO(s) + CO2(g) → CuCO3(s), a. CO2 is the Lewis acid and CuCO3 is its conjugate base. b. O2– acts as a Lewis bas
adoni [48]

<u>Answer:</u> The correct answer is Option d.

<u>Explanation:</u>

According to Lewis acid-base concept:

The substance which is donating electron pair is considered as Lewis base and the substance which is accepting electron pair is considered as Lewis acid.

For the given chemical reaction:

CuO(s)+CO_2(g)\rightarrow CuCO_3(s)

CO_2 is accepting electron pair and is getting converted to CO_3^{2-}. Thus, it is considered as Lewis acid.

O^{2-} present in CuO is a Lewis base because it is donating electron pair.

Thus, the correct answer is Option d.

4 0
3 years ago
PLEASE HELP ME I HAVE NO IDEA “Heat (thermal) energy is being applied to the substance whenever the Bunsen Burner is on. Before
svetlana [45]

TLDR: The energy was being used simply to heat the substance up.

Whenever something melts, it performs what is called a "phase transition", where the state of matter moves from one thing to something else. You can see this in your iced drink at lunch; as the ice in the cup of liquid heats up, it reaches a point where it will eventually "change phase", or melt. The same can be achieved if you heat up that water enough, like if you're cooking; when you boil eggs, the water has so much thermal energy it can "change phase" and become a gas!

However, water doesn't randomly become a boiling gas, it has to heat up for a while before it reaches that temperature. For a real-life example, the next time you cook something, hold you hand above the water before it starts boiling. You'll see that that water has quite a high temperature despite not boiling.

There's a lot of more complex chemistry to describe this phenomena, such as the relationship between the temperature, pressure, and what is called the "vapor pressure" of a liquid when describing phase changes, but for now just focus on the heating effect. When ice melts, it doesn't seem like its heating up, but it is. The ice absorbs energy from its surroundings (the warmer water), thus heating up the ice and cooling down the water. Similarly, the bunsen burner serves to heat up things in the lab, so before the solid melts in this case it was simply heating up the solid to the point that it <u>could</u> melt.

Hope this helps!

5 0
4 years ago
Convert 296 degrees Celsius to Kelvin.<br> A. 23 K<br> B. 552 K<br> C. 569 K<br> D. 550 K
katovenus [111]

Answer:

option c) 552 K is correct

Explanation:

we know that

0° C +273= 273 K

so

296° C+273=569 K

7 0
2 years ago
Read 2 more answers
What is the rate law for the reaction 2A + 2B + 2C --&gt; 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
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