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3 years ago

Do you notice a pattern in the charge for elements in each group? Explain.

Chemistry

1 answer:

MrRissso [65]3 years ago

7 0

Answer:

Below are some patterns in the charge for elements in each group.

Group 1 elements : 1 valence electron and form ions with charge +1

Group 2 elements : 2 valence electrons and form ions with charge + 2

Group 3 elements : 3 valence electrons and form ions with charge + 3(there are some exceptions as well)

Elements in groups 4 and 5 are unpredictable also the D block elements consist of multiple oxidation states..

Group 6 elements : 6 valence electrons, form ions with charge -2

Group 7 elements: 7 valence electrons, form ions with charge -1

The octet rule is being followed, the elements form either ionic bond or covalent bond to fulfill it.

eg: when a K atom forms a K+ ion, the ion has the same electron configuration as the noble gas Ar (argon).

When an O atom gains 2 electrons to form the O²⁻ ion, the ion has the same electron configuration as the noble gas Ne (neon).

Note: Boron (B) is in Group 3 but doesn't form ions.

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Anvisha [2.4K]

Hey You!

The Correct Answer Is: True.

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3 years ago

Read 2 more answers

When CO2 decomposes into oxygen and carbon, it gives a gram ratio of 2.67:1 O2:C. When a 32.4g of CO2 decomposes, how many grams

Rufina [12.5K]

Answer : The mass of carbon and oxygen produced is 8.83 g and 23.6 g respectively.

Explanation :

Law of conservation of mass : It states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The balanced chemical reaction will be,

$CO_2\rightarrow O_2+C$

As we are given:

$\text{Mass of }O_2}:\text{Mass of }C=2.67:1{$

According to the law of conservation of mass,

Total mass of $CO_2$ = Mass of $O_2$ + Mass of C

Total mass of $CO_2$ = 2.67 + 1 = 3.67 g

Now we have to calculate the mass of $O_2$ and C.

$\text{Mass of }O_2=\frac{\text{Given mass of }CO_2}{\text{Total mass of }CO_2}\times \text{Given mass of }O_2=\frac{32.4g}{3.67g}\times 2.67=23.6g$

and,

$\text{Mass of }C=\frac{\text{Given mass of }CO_2}{\text{Total mass of }CO_2}\times \text{Given mass of }C=\frac{32.4g}{3.67g}\times 1=8.83g$

Therefore, the mass of carbon and oxygen produced is 8.83 g and 23.6 g respectively.

6 0

3 years ago

How can I find the answer for this ?

12345 [234]

For what? what is your question

7 0

3 years ago

Consider the reaction of NO and CO to form N2 and CO2, according to the balanced equation: 2 NO (g) + 2 CO (g) → N2 (g) + 2 CO2

Gekata [30.6K]

The image is not given in the question, it is attached below:

<u>Answer:</u> The excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.

<u>Explanation:</u>

In the given image:

Red spheres represent oxygen atoms, blue spheres represent nitrogen atoms and black spheres represent carbon atoms

The combination of 1 black and 2 red spheres will represent carbon dioxide $(CO_2)$ compound

The combination of 2 blue spheres will represent nitrogen molecule $(N_2)$

The combination of 1 blue and 1 red sphere will represent nitrogen monoxide $(NO)$ compound

The combination of 1 black and 1 red sphere will represent nitrogen monoxide $(NO)$ compound

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

We are given:

Given moles of NO = 6 moles

Given moles of CO = 4 moles

For the given chemical equation:

$2NO(g)+2CO(g)\rightarrow N_2(g)+2CO_2(g)$

By stoichiometry of the reaction:

If 2 moles of CO reacts with 2 moles of NO

So, 4 moles of CO will react with = $\frac{2}{2}\times 4=4mol$ of NO

As the given amount of NO is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, CO is considered a limiting reagent because it limits the formation of the product.

Hence, the excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.

3 0

3 years ago

Sea water contains roughly 158.0g of Nacl per liter. What is the molarity of sodium chloride in sea water?

IrinaK [193]

Molar mass NaCl = 58.5 g/mol

C = 158.0 g/L

Molarity = C / molar mass

M = 158.0 / 58.5

M = 2.7000 M

hope this helps!

6 0

3 years ago