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

Neither model is able to show the reactivity of potassium.

answered: Guest

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied

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