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lorasvet [3.4K]
2 years ago
10

You are working on an experiment that requires Iodine (I) to have a full outer shell. What element could you combine it with to

get a full outer shell? A. Lithium (Li) B. Sulfur (S) C. Neon (Ne) D. Oxygen (O)
Chemistry
2 answers:
kow [346]2 years ago
8 0

Answer:

A. Lithium

Explanation:

Hello,

In this case, we can find out the required element by noticing that iodine has seven valence electrons as it is in group VIIA, therefore, it needs one spare electron to successfully attain the octet, that is completing eight electrons in total. For that reason, since lithium is group in IA, it will be able to provide the missing electron to iodine in order to get the full outer shell as required, thereby, lithium iodide will be formed.

Take into account that sulfur will share two electrons so two iodine atoms will be required, neon does not provide any electron as it is a noble gas and oxygen behave as well as sulfur.

Regards.

Cloud [144]2 years ago
5 0

Answer:

Lithium

Explanation:

Answer via Educere/ Founder's Education

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Compound A is an organic compound which contains Carbon, Hydrogen and Oxygen. When 0.240g of the vapour of A is slowly passed ov
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hope it helpsss

4 0
2 years ago
The mass percent of oxygen in pure glucose, c6h12o6 is 53.3 percent. a chemist analyzes a sample of glucose that contains impuri
Dafna1 [17]
We can calculate the mass percent of an element by dividing its atomic mass by the mass of the compound and then multiply by 100:
     % by mass of element = (mass of element/mass of compound) x100%
Impurities like n-eicosane with the molecular formula C20H42 could account for the low percent by mass of oxygen in the sample because it has a zero percent oxygen based on its compound formula which indicates that it does not have the element oxygen.
7 0
2 years ago
Caffeine, a molecule found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 10.0 g of caffeine pr
denis-greek [22]

Answer:

194 g/mol.

Explanation:

Hello,

In this case, one first must compute the mass of each element as shown below:

C=18.13gCO_2*\frac{12gC}{44gCO_2} =4.945gC\\H=4.639gH_2O*\frac{2.016gH}{18.0152gH_2O}=0.519gH\\N=2.885gN_2\\O=10.0g-4.945g-0.519g-2.885g=1.651gO

Next, the corresponding moles:

C=4.945gC*\frac{1molC}{12gC}=0.412mol\\H=0.519gH*\frac{1molH}{1gH}=0.519mol\\N=2.885gN*\frac{1molN}{14gN}=0.206molN\\O=1.648gO*\frac{1molO}{16gO} =0.103molO

Then, each element's subscripts is found to be:

C=\frac{0.412}{0.103}=4\\H=\frac{0.519}{0.103}=5\\N=\frac{0.206}{0.103} =2\\O=\frac{0.103}{0.103}=1

Therefore, the empirical formula is:

C_4H_5N_2O

Nonetheless, it has a molar mass of 97bg/mol, thereby, by multiplying such formula by 2 one gets:

C_8H_10N_4O_2

Which has a molar mass of 194 g/mol being correctly contained in the given interval.

Best regards.

8 0
2 years ago
Among the alkali metals, the tendency to react with other substances. A. does not vary among the members of the group. . B. incr
yKpoI14uk [10]
B is correct. As you move down group 1, the elements become more reactive with other elements because the electrons have a weaker attraction to their own atoms nucleus which means attraction with other elements is much stronger, making the atom more reactive.
8 0
1 year ago
2CH4(g)⟶C2H4(g)+2H2(g)
Rasek [7]

Answer : The enthalpy change for the reaction is, 201.9 kJ

Explanation :

According to 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.

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. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The balanced reaction of CH_4 will be,

2CH_4(g)\rightarrow C_2H_4(g)+2H_2(g)    \Delta H^o=?

The intermediate balanced chemical reaction will be,

(1) CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)     \Delta H_1=-890.3kJ

(2) C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)     \Delta H_2=-136.3kJ

(3) 2H_2(g)+O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=-571.6kJ

(4) 2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)     \Delta H_4=-3120.8kJ

Now we will multiply the reaction 1 by 2, revere the reaction 2, reverse and half the reaction 3 and 4 then adding all the equations, we get :

(1) 2CH_4(g)+4O_2(g)\rightarrow 2CO_2(g)+4H_2O(l)     \Delta H_1=2\times (-890.3kJ)=-1780.6kJ

(2) C_2H_6(g)\rightarrow C_2H_4(g)+H_2(g)    \Delta H_2=-(-136.3kJ)=136.3kJ

(3) H_2O(l)\rightarrow H_2(g)+\frac{1}{2}O_2(g)    \Delta H_3=-\frac{1}{2}\times (-571.6kJ)=285.8kJ

(4) 2CO_2(g)+3H_2O(l)\rightarrow C_2H_6(g)+\frac{7}{2}O_2(g)     \Delta H_4=-\frac{1}{2}\times (-3120.8kJ)=1560.4kJ

The expression for enthalpy of the reaction will be,

\Delta H^o=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4

\Delta H=(-1780.6kJ)+(136.3kJ)+(285.8kJ)+(1560.4kJ)

\Delta H=201.9kJ

Therefore, the enthalpy change for the reaction is, 201.9 kJ

5 0
1 year ago
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