Chemistry, 19.07.2019 02:10, Brightcord9679

Determine the oxidation states of the elements in the compounds listed. none of the oxygen-containing compounds are peroxides or superoxides. (a) h2so4 (b) ca(oh)2 (c) broh (d) clno2 (e) ticl4 (f) nah'

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answered: ummammaar08

Answer :

Oxidation number or oxidation state : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

Rules for Oxidation Numbers are :

The oxidation number of a free element is always zero.The oxidation number of a monatomic ion equals the charge of the ion.The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.The oxidation number of oxygen (O) in compounds is usually -2.The oxidation number of a Group 17 element in a binary compound is -1.The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

Now we have to determine the oxidation state of the elements in the compound.

(a) H_2SO_4

Let the oxidation state of 'S' be, 'x'

$2(+1)+x+4(-2)=0\\\\x=+6$

Hence, the oxidation state of 'S' is, (+6)

(b) Ca(OH)_2

Let the oxidation state of 'Ca' be, 'x'

$x+2(-2+1)=0\\\\x=+2$

Hence, the oxidation state of 'Ca' is, (+2)

Let the oxidation state of 'Br' be, 'x'

$x+(-2)+1=0\\\\x=+1$

Hence, the oxidation state of 'Br' is, (+1)

(d) ClNO_2

Let the oxidation state of 'N' be, 'x'

$-1+x+2(-2)=0\\\\x=+5$

Hence, the oxidation state of 'N' is, (+5)

(e) TiCl_4

Let the oxidation state of 'Ti' be, 'x'

$x+4(-1)=0\\\\x=+4$

Hence, the oxidation state of 'Ti' is, (+4)

(f) NaH

Let the oxidation state of 'Na' be, 'x'

$x+(-1)=0\\\\x=+1$

Hence, the oxidation state of 'Na' is, (+1)

answered: Guest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

answered: Guest

What do you need with it’s not a question your just stating what it says

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Determine the oxidation states of the elements in the compounds listed. none of the oxygen-containin...