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3 years ago

Determine the ratio of the diatomic element with equal percentage abundance

Chemistry

1 answer:

Irina-Kira [14]3 years ago

7 0

Answer:

I ratio is

47 : 43 ratio

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A sample of barium nitrate is placed into a jar containing water. The mass of the barium nitrate sample is 27g. Assume the water

34kurt

So we have Barium nitrate with a solubility of 8.7g in 100g water at 20°C.

using that relation
i.e.
8.7g (barium nitrate) =100g (water)
1g barium nitrate = 100/8.7 g water

27g barium nitrate = (100/ 8.7 ) × 27
= 310.34 g

therefore,
you need 310.34g of water is in the jar.

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4 years ago

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What is the oxidation number of NH4OH?

BlackZzzverrR [31]

NH4 itself is called ammonium ion with positive charge of 1+. hence the oxidation state will be +1

OH itself is called hydroxide ion with negative charge of 1-. hence the oxidation state will be -1

there's actually no such compound as stated in your chemical formula. although it's simply called ammonium hydroxide by looking at the chemical formula, in fact the compound should rightly be called as aqueous ammonia with chemical formula of NH3 (aq). this is because the ammonia molecule will ionise due to the presence of water molecule to be protonated, hence forming ammonium ion and hydroxide ion in the solution.

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3 years ago

Consider the reaction for the dissolution of solid magnesium hydroxide.

sp2606 [1]

Answer:

Molar solubility is 1.12x10⁻⁴M

Explanation:

The dissolution of magnesium hydroxide is:

Mg(OH)₂(s) ⇄ Mg²⁺ + 2OH⁻

The molar solubility represents the moles of the solid that the solution can dissolve, that could be written as:

Mg(OH)₂(s) ⇄ X + 2X

Where X is solubility.

If you obtained a [OH⁻] = 2.24x10⁻⁴M and you know [OH⁻] = 2X:

2X = 2.24x10⁻⁴M

X = 2.24x10⁻⁴M/2

X =1.12x10⁻⁴M

<h3>Molar solubility is 1.12x10⁻⁴M</h3>

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3 years ago

Which of the following is not true of alloys?

Scilla [17]

Option d: copper.

Because copper is an element, not a mixture.

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4 years ago

Please help!!!!! I need the correct answer quickly!!!

xxMikexx [17]

Answer :

The oxidation state of oxygen (O) in $OF_2$ is, (+2)

The oxidation state of carbon (C) in $CO$ is, (+2)

The oxidation state of nitrogen (N) in $K_3N$ is, (-3)

Explanation :

Oxidation number : 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.

When the atoms are present in their elemental state then the oxidation number will be zero.

Rules for Oxidation Numbers :

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, but it is -1 in peroxides.

The oxidation number of a Group 1 element in a compound is +1.

The oxidation number of a Group 2 element in a compound is +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.

(a) The given compound is, $OF_2$

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

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

The oxidation state of oxygen (O) in $OF_2$ is, (+2)

(b) The given compound is, $CO$

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

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

The oxidation state of carbon (C) in $CO$ is, (+2)

(c) The given compound is, $K_3N$

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

$3(+1)+x=0\\\\3+x=0\\\\x=-3$

The oxidation state of nitrogen (N) in $K_3N$ is, (-3)

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3 years ago

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Determine the ratio of the diatomic element with equal percentage abundance​

Determine the ratio of the diatomic element with equal percentage abundance