Chemistry, 06.05.2020 08:28, LuluMathLover101

What element is in group 17 and period 2 of the periodic table?

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The periodic table is a tabular classification and arrangement of chemical elements in order of increasing atomic number into groups and periods.

The groups are vertical columns that span from group 1 - group 18The period is horizontal rows that span from period 1 - 7.

Chemical elements in the same groups tend to share the same properties.

The group 1 elements are called alkali metals. They are highly reactive metals. They form an electrovalent bond with group 17 elements (halogens).

The group 2 elements are alkaline earth metals. They are the most abundant metals on earth.

The group 3 - 12 elements are Transition Metals.Group 17 is the halogen family.Group 18 is the Noble gas.

On the periodic table, the element in group 17 is a member of the halogen family.

Since it is in period 2, which is the second row on the periodic table;

Then, the name of the element is Fluorine.

Therefore, we can conclude that the atomic element that is in group 17 and period 2 on the periodic table is Fluorine.

Learn more about the periodic table here:

link

answered: IsPink

Group 17 elements:

Flourine (F), Chlorine (Cl), Bromine (Br), Iodine (I), and Astatine (As).

Period 2 elements:

Lithium (Li), Beryllium (Be), Boron (B), carbon (C), Nitrogen (N), Oxygen (O), flourine (F), And Neon.

What element is in group 17 and period 2 of the periodic table?

answered: Guest

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh $pcl_{3}$ +δh $cl_{2}$ ] - [δh $pcl_{5}$ ]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

answered: Guest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.