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masya89 [10]
3 years ago
6

Why do the Noble gases have the highest ionization energies across a period?

Chemistry
1 answer:
Kitty [74]3 years ago
3 0

Answer:

ionization energy decreases from top to bottom in groups, and increases from left to right across a period. So technically, the noble games have the largest ionization energy.

Explanation:

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Molecules can exist as: <br><br> A)solids <br> B) liquids <br> C) gases <br> D) all of the above
DerKrebs [107]

I believe it is D. Hope this helps you!

6 0
3 years ago
The half life of radon-222 is 3.8 days. How Much of a 100g sample is left after 15.2 days
professor190 [17]

Answer:  

6.2 g  

Explanation:  

In a first-order decay, the formula for the amount remaining after <em>n</em> half-lives is  

N = \frac{N_{0}}{2^{n}}  

where  

<em>N</em>₀ and <em>N</em> are the initial and final amounts of the substance  

1. Calculate the <em>number of half-lives</em>.  

If t_{\frac{1}{2}} = \text{3.8 da}  

n = \frac{t}{t_{\frac{1}{2}}} = \frac{\text{15.2 da}}{\text{3.8 da}}= \text{4.0}

2. Calculate the <em>final mass</em> of the substance.  

\text{N} = \frac{\text{100 g}}{2^{4.0}} = \frac{\text{100 g}}{16} = \text{6.2 g}

4 0
3 years ago
Write a balanced equation for the reaction of ca with hcl
Aleksandr-060686 [28]
Ca(s)+2Hcl(aq) ------>CaCl2(s)+H2(g)
3 0
3 years ago
Calculate Ho298 for the process
Inga [223]

Explanation:

As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

Sb + \frac{3}{2}Cl_2 \rightarrow SbCl_{3}    \Delta H^0_1 =  -314 kJ  ..........(1)

SbCl_{3} + Cl_2 \rightarrow SbCl_{5}    \Delta H^0_2 = -80kJ   ..............(2)

The final reaction is as follows:  

Sb + \frac{5}{2}Cl_{2} \rightarrow SbCl_{5}  \Delta H^0_3 = ?  .............(3)

Therefore, adding (1) and (2) we get the final equation (3) and value of \Delta H^{0}_{3} at 298 K will be as follows.

             \Delta H^{0}_{3} = \Delta H^{0}_{1} + \Delta H^{0}_{2}    

                       = -314 kJ + (-80) kJ

                       = -394 kJ

Thus, we can conclude that H^{o} at 298 K for the given process is -394 kJ.

4 0
3 years ago
Identify each of the following as a representative element or a transition element:
charle [14.2K]
The correct answer is B. Platinum is the transition element among the choices. The elements belonging to this group are those having a partially filled d or f subshell. It usually refers to the d-block transition elements of the periodic table.
3 0
3 years ago
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