Chemistry, 07.04.2020 20:33 jailinealvarado24

Arrange the elements in each set in order of decreasing electronegativity (enter with no spaces e. g. N>P>Si):

I. N>P>Si N, P, Si
II. As>Ga>Ca Ca, Ga, As
III. Cl>Br>P Br, Cl, P IV. F>O>I I ,F, O

Answers: 2

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Answer from: breannabryan1017

Explanation:

The trend of electronegativity in the periodic table is given as;

The electronegativity of atoms increases as you move from left to right across a period in the periodic table. The electronegativity of atoms decreases as you move from top to bottom down a group in the periodic table.

I. N, P, Si

N is in period 2 while P and Si are both in period 3. N has the greatest electronegativity aamong the three. Since electronegativity increases from left to right in a period, P has a greater electronegativity than Si. The trend is given as; N>P>Si

II. Ca, Ga, As

Since electronegativity increases from left to right in a period, As has the highest electronegativity because it is in group 5 period 4, followed by Ga which is in group 3 period 5, then the least which is Calcium present in group 2, period 5. The trend is given as; As>Ga>Ca

III. Br, Cl, P

Since electronegativity increases from left to right in a period, Cl has the highest value. Bromine has a higher value of electronegativity because it is a group 7 element. The least electronegative is P. The trend is given as; Cl>Br>P

IV. I ,F,O

Fluorine is the most electronegative element. Followed by oxygen. The electronegativity of atoms decreases as you move from top to bottom down a group in the periodic table. The least is Iodine, the trend is given as; F>O>I

Answer from: Quest

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.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

Ch4 is nonpolar and insoluble in water. since all of the hydrogen atoms are evenly spread around the carbon atom even through the c h bond is polar all the poles cancel each other out.

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