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3 years ago

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All of the halogens, group 17, have seven valence electrons. which of these would represent the oxidation number of the halogens

such as chlorine and bromine?

1 answer:

fomenos3 years ago

6 0

It depends if it occurs naturally it has oxidation number of 0
but when it react with other element it has an oxidation number of -1

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In a mixture of 3 gases what is the total pressure of the gases if Gas A exerts a pressure

kirill [66]

Answer:

760 mm of Hg

Explanation:

If the gases A , B and C are non reacting , then according to <u>Dalton's </u><u>Law </u><u>of</u><u> </u><u>Partial </u><u>Pressure</u> the total pressure exerted is equal to sum of individual partial pressure of the gases .

If there are n , number of gases then ,

$P_{total}= P_1+P_2+P_3+\dots +P_n$

Here ,

Partial pressure of Gas A = 400mm of Hg
Partial pressure of Gas B = 220 mm of Hg
Partial pressure of Gas C = 140mm of Hg

Hence the total pressure exerted is ,

$P_{total}= P_{Gas\ A }+P_{Gas\ B }+P_{Gas\ C }$

Substitute ,

$P_{total}=( 400 + 220 + 140 )mm\ of \ Hg$

Add ,

$P_{total}= 760\ mm\ of \ Hg$

<u>Hence</u><u> the</u><u> </u><u>total</u><u> pressure</u><u> exerted</u><u> by</u><u> the</u><u> </u><u>gases </u><u>is </u><u>7</u><u>6</u><u>0</u><u> </u><u>mm </u><u>of </u><u>Hg</u><u>.</u>

<em>I </em><em>hope</em><em> this</em><em> helps</em><em>.</em>

4 0

2 years ago

Iridium has only two naturally occurring isotopes. Ir-191 with arelative mass of 190.96058 and Ir-193 with a relative mass of192

OLEGan [10]

<u>Answer:</u> The fractional abundance of Ir-191 is 0.372

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of Ir-191 be x and that of Ir-193 isotope be (1-x)

<u>For isotope 1 (Ir-191):</u>

Mass of isotope 1 = 190.96058 amu

Fractional abundance of Ir-191 = x

<u>For isotope 1 (Ir-193):</u>

Mass of isotope 1 = 192.96292 amu

Fractional abundance of Ir-193 = (1 - x)

Average atomic mass of iridium = 192.217 amu

Putting values in equation 1, we get:

$192.217=[(190.96058\times x)+(192.96292\times (1-x))]\\\\x=0.372$

Hence, the fractional abundance of Ir-191 is 0.372

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Lady_Fox [76]

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8 0

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Pls answer these and show work

Assoli18 [71]

There's no question??

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