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3 years ago

One liter of a solution ph 9 has how many more hydroxide ions (oh-) than 1 l of a solution of ph 4?

1 answer:

4 0

100,000 times more. The rule is that the product of the concentration of (H+) and (OH-) in molars is always 10^-14 (if it is more, they combine and produce simple water which brings the product back down to this number again. If it is less, water molecules will split to bring the product back up to equilibrium).
pH 9 means (H+) is 10^-9 molar. But that means (OH-) is 10^-5 at the same time.
pH 4 means (H+) is 10^-4 molar. But that means (OH-) is 10^-10 at the same time.
10^-5 / 10^-10 = 10^5 or 100,000 times more.

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State the oxidation number of S in <img src="https://tex.z-dn.net/?f=H_%7B2%7DSO_%7B3%7D" id="TexFormula1" title="H_{2}SO_{3

Nezavi [6.7K]

Taking into account the definition of oxidation number, the oxidation numbers of S in H₂SO₃ is 4.

<h3>Definition of oxidation number</h3>

The oxidation number is the charge that an atom has; is an integer that represents the number of electrons an atom puts into play when it forms a given compound.

In other words, the oxidation number of an element is a value that indicates the number of electrons that element gains or loses when it combines with another.

<h3>Oxidation number determination</h3>

To determine the oxidation state of different elements it is necessary to know that:

The oxidation number of hydrogen in a compound is +1, except in metal hydrides, where is –1.
The oxidation number of oxygen in a compound is –2, except in peroxides, where it is –1.

On the other side, the sum of the oxidation numbers of the existing elements in a chemical formula must add up to zero.

Then, considering the oxidation numbers of each element, multiplying it by the number of existing elements in the chemical formula and adding it and equaling it to zero, the value of the missing oxidation number can be obtained.

<h3>Oxidation numbers of S</h3>

In this case, the oxidation numbers of S in H₂SO₃ is calculated as:

2× (+1) + oxidation numbers of S + 3×(-2)= 0

2 + oxidation numbers of S -6= 0

oxidation numbers of S -4= 0

oxidation numbers of S= 4

Finally, the oxidation numbers of S in H₂SO₃ is 4.

Learn more about the oxidation number:

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5 0

2 years ago

Read 2 more answers

Hello,

nikitadnepr [17]

Answer:

$a)\ 2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3\\b)\ Decomposition\ Reaction$

Explanation:

Ferrous Sulphate $(FeSO4)$ is generally found as Lime-Green Crystals. On heating, these crystals almost immediately turn white-yellow. They then, break down to produce an anhydrous mixture of Sulphur Trioxide $(SO_3)$ , Sulphur Dioxide $(SO_2)$ as well as Ferric Oxide $(Fe_2O_3)$ .

We can hence, frame a skeletal equation of this reaction and try to balance it.

Hence,

$FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3$

Now,

a)In order to balance it through the 'Hit &Trial Method', we'll follow a series of steps:

1. First, lets compare the number of Fe (Iron) atoms on the RHS and LHS. We find that, the no. of Fe Atoms on the RHS is twice the number of Fe Atoms on the LHS. We hence, add a co-effecient 2 beside $FeSO_4$ .

2. Now, Iron atoms, Sulphur Atoms and Oxygen atoms occur 2, 2, 8 respectively on both the sides:

Hence, As all the other elements as well as iron, balance, we've arrived upon our Balanced Equation :

$2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3$

b) We know that, decomposition reactions are [generally] endothermic reactions in which Large Compounds decompose into smaller elements and compounds. Here, as Ferrous Sulphate decomposes into Sulphur Dioxide, Sulphur Trioxide and Ferric Oxide, the reaction that occurs here is Decomposition Reaction.

7 0

2 years ago

Binary compounds are formed by ............... ............... elements.

goblinko [34]

Answer: i think its A diatomic compound..

Explanation: hope i helped! sorry if im wrong!

6 0

2 years ago

Determine the value of the equilibrium constant, KgoalKgoalK_goal, for the reaction CO2(g)⇌C(s)+O2(g)CO2(g)⇌C(s)+O2(g), Kgoal=?

NISA [10]

Answer:

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$ .