All categories

3 years ago

How mant atoms are in 3N2

Chemistry

2 answers:

geniusboy [140]3 years ago

8 0

Explanation:

you times 3 and 2 to get six.

Ainat [17]3 years ago

8 0

In 2 formula units of silver nitrate, there are 2 silver atoms, 2 nitrogen atoms, and 6 oxygen atoms. We can improve this answer by using the mole concept.

You might be interested in

What is the molarity of a solution made by dissolving 14.8 g of ammonium hydroxide NH4OH, in enough water to make 250.0 mL of so

kicyunya [14]

Answer:

Molarity= 1.69M

Explanation:

m= 14.8, Mm= 35, V= 0.25dm3, C= ?

Moles = m/M= C×V

Substitute and Simplify

m/M= C×V

14.8/35= C×0.25

C= 1.69M

6 0

3 years ago

7. The youngest part of the ocean floor is found ____.

aev [14]

Answer:

The answer is Near The Ocean Ridges.

5 0

3 years ago

What is a potential risk of genetically modified food?

Ivahew [28]

It would be D since it could be unhealthy for you.

6 0

3 years ago

Read 2 more answers

How many grams of ammonia (NH3) can be produced by the synthesis of excess hydrogen gas (H2) and 253.8 grams of nitrogen gas (N2

kogti [31]

Answer:

308.2 g of NH₃.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3H₂ + N₂ —> 2NH₃

Next, we shall determine the mass of N₂ that reacted and the mass of NH₃ produced from the balanced equation. This can be obtained as follow:

Molar mass of N₂ = 2 × 14 = 28 g/mol

Mass of N₂ from the balanced equation = 1 × 28 = 28 g

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3 = 17 g/mol

Mass of NH₃ from the balanced equation = 2 × 17 = 34 g

Summary:

From the balanced equation above,

28 g of N₂ reacted to produce 34 g of NH₃.

Finally, we shall determine the mass of NH₃ produced by the reaction of 253.8 g of N₂. This can be obtained as illustrated below:

From the balanced equation above,

28 g of N₂ reacted to produce 34 g of NH₃.

Therefore, 253.8 g of N₂ will react to produce = (253.8 × 34)/28 = 308.2 g of NH₃.

Thus, 308.2 g of NH₃ were obtained from the reaction.

8 0

3 years ago

What orbitals are used to form the 10 sigma bonds in propane (ch3ch2ch3)? Label each atom with the appropriate hybridization. Dr

SVETLANKA909090 [29]

Mixing of pure orbitals having nearly equal energy to form equal number of completely new orbitals is said to be hybridization.

For the compound, $CH_3CH_2CH_3$ the electronic configuration of the atoms, carbon and hydrogen are:

Carbon (atomic number=6): In ground state= $1s^{2}2s^{2}2p^{2}$

In excited state: $1s^{2}2s^{1}2p^{3}$

Hydrogen (atomic number=1): $1s^{1}$

All the bonds in the compound is single bond( $\sigma$ -bond) that is they are formed by head on collision of the orbitals.

The structure of the compound is shown in the image.

The Carbon-Hydrogen bond is formed by overlapping of s-orbital of hydrogen to p-orbital of carbon.

In order to complete the octet the required number of electrons for carbon is 4 and for hydrogen is 1. So, the electron in $1s^{1}$ of hydrogen will overlap to the 2p^{3}-orbital of carbon.

Thus, the hybridization of Hydrogen is $s$ -hybridization and the hybridization of Carbon is $sp^{3}$ -hybridization.

The hybridization of each atom is shown in the image.

3 0

3 years ago