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butalik [34]
3 years ago
8

In which of the following choices is the oxidation number incorrect?

Chemistry
1 answer:
8_murik_8 [283]3 years ago
4 0

Cl2(s); oxidation number 1 is the incorrect choices in oxidation number.

Explanation:

In the elemental form oxidation state is zero. Here chlorine is present in elemental form so oxidation state is zero.

Oxidation number depends on the number of electrons gained or lost by an atom of the element say in compound formation.

If electron is gained oxidation number becomes negative.

If electron is lost then oxidation number is positive.

If the octet rule is fulfilled that valence shell is filled them atomic number gets zero. Since Cl2 is in neutral state the oxidation number is 0.

Oxidation number in general can be made out by checking the valency of the element as oxidation number is also equal to the valency.

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3 years ago
W
DerKrebs [107]

Answer:

F = 30 N

Explanation:

Given data:

Mass of toy train = 1.5 kg

Acceleration of train = 20 m/s²

Amount of force acting on it = ?

Solution:

The net force on object is equal to the its mass multiply by its acceleration.

Formula:

F = ma

F = force

m = mass

a = acceleration

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7 0
3 years ago
Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin
STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of KI and AgNO_3.

\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}

Molar mass of KI = 166 g/mole

\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole

and,

\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

KI+AgNO_3\rightarrow KNO_3+AgI

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of AgNO_3

So, 0.0178 mole of KI react with 0.0178 mole of AgNO_3

From this we conclude that, AgNO_3 is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgI

From the reaction, we conclude that

As, 1 mole of KI react to give 1 mole of AgI

So, 0.0178 moles of KI react to give 0.0178 moles of AgI

Thus,

Moles of AgI = Moles of I^- anion = Moles of Ag^+ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}

\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

3 0
3 years ago
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