All categories

3 years ago

Assign an oxidation number to each atom in the products.

Chemistry

2 answers:

Rzqust [24]3 years ago

8 0

<h3>Answer:</h3>

Oxidation number refers to a number given to an atom which shows the number of number of electrons lost (or gained, if the number is negative), by an atom of that element in the compound.

Hence oxidation number of following atoms in given reaction are:

In NaOH:

Na = +1

O = -2

H = +1

In H2:

H = 0

Drupady [299]3 years ago

4 0

Answer : The oxidation number of Na, O and H in NaOH are, (+1), (-2) and (+1) respectively.

The oxidation number of H in $H_2$ is zero.

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.

The given chemical reaction is:

$2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq)+H_2(g)$

Now we have to determine the oxidation number of each atom in the products.

The given product is, $NaOH$

'Na' belongs to group 1 thus, the oxidation number is (+1)

The oxidation number of oxygen generally in all compound is (-2)

The oxidation number of hydrogen generally in all compound is (+1)

And as we know that, the sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

(+1)+(-2)+(+1)=0

Thus, the oxidation number of Na, O and H in NaOH are, (+1), (-2) and (+1) respectively.

The given product is, $H_2$

The oxidation number of hydrogen generally in all compound is (+1).

As we know that the oxidation number of atoms of elements at their standard states is always zero.

Thus, the oxidation number of H in $H_2$ is zero.

You might be interested in

What is the site of ATP synthesis

Fynjy0 [20]

ATP is synthesized in the mitochondria.
The mitochondria is known to be the powerhouse of a cell.

Hope it helped!

6 0

3 years ago

Read 2 more answers

Standard atmospheric pressure corresponds to the normal air pressure at sea level.

jekas [21]

Im not really sure what your asking.... Standard sea-level pressure, by definition, equals 760 mm (29.92 inches) of mercury, 14.70 pounds per square inch, 1,013.25 × 10 3 dynes per square centimetre, 1,013.25 millibars, one standard atmosphere, or 101.325 kilopascals.

""atmospheric pressure | Britannica.com""

6 0

3 years ago

Using wolfram alpha or some other reference, determine which of these elements would be liquid at 525 k (assume samples are prot

romanna [79]

The choices for this problem are bismuth, Bi; platinum, Pt; selenium, Se; calcium, Ca and copper, Cu. I think the correct answer would be selenium. The melting point of bismuth is at a temperature of 544.4 Kelvin. At a temperature of 525 K, it would exist as solid. Platinum melts at 2041.1 K. At 525 K, platinum would be in solid form. Selenium has a melting point at 494 K so that at a temperature of 525 K, it would exist in its liquid state. Calcium has a melting point of 1112 K so it would exist as solid at 525 K. Copper has a melting point at 1358 K, so it would still exist as solid at a temperature of 525 K. Therefore, the answer would only be selenium.

3 0

3 years ago

Which of the following instruments is a combination of a percussion and a stringed instrument?

murzikaleks [220]

Answer:B

Explanation:

3 0

3 years ago

Draw the product that valine forms when it reacts with excess CH3CH2OH and HCl followed by a wash with aqueous base.

-BARSIC- [3]

Answer:

Product: ethyl L-valinate

Explanation:

If we want to understand what it is the molecule produced we have to analyze the reagents. We have valine an amino acid, in this kind of compounds we have an amine group ( $NH_2$ ) and a carboxylic acid group ( $COOH$ ). Additionally, we have an alcohol ( $CH_3CH_2OH$ ) in the presence of HCl (a strong acid) in the first step, and a base ( $OH^-$ ).

When we have an acid and an alcohol in a vessel we will have an esterification reaction. In other words, an ester is produced. As the first step, the oxygen in the C=O (in the carboxylic acid group) would be protonated. In the second step, the ethanol attacks the carbon in the C=O of the carboxylic acid group producing a new bond between the oxygen in the ethanol and the carbon in the carboxylic acid. In step 3, a proton is transferred to produce a better leaving group ( $H_2O$ ). In step 4, a water molecule leaves the main structure to produce again the double bond C=O. Finally, a base ( $OH^-$ ) removes the hydrogen from the C=O bond to produce ethyl L-valinate

See figure 1

I hope it helps!

7 0

3 years ago