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3 years ago

A. Identify the structure drawn below.

Chemistry

1 answer:

nadya68 [22]3 years ago

7 0

Answer:

Hexane

Explanation:

You have a carbon structure with only single bonds. This means that the name will end in -ane.

There are 6 carbon atoms. This means that the name will begin with hex-.

The structure is hexane.

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Which group tends not to form ions or react?

Anastaziya [24]

Group 8 elements. They are unreactive and stable

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2 years ago

Give the n and l values and the number of orbitals for sublevel 5g.

Pepsi [2]

The n and l values and the number of orbitals for sublevel 5g is :

5g shell , n= 5 subshell g , l = 4, Number of orbitals for sublevel = 9.

There are total four quantum numbers:

1) Principal quantum number , n

2) Angular quantum number , l

3) Magnetic quantum number , ml

4) spin quantum number , ms

For 5g shell, n = 5

subshell g , l = 4 ....0 - s , 1 - p , 2 - d, 3 - f, 4 -g

number of orbitals in subshell = (2l + 1) ( 2×4 + 1) = 9

Thus, The n and l values and the number of orbitals for sublevel 5g is :

5g shell , n= 5 subshell g , l = 4, Number of orbitals for sublevel = 9.

To learn more about quantum numbers here

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9 months ago

How many grams of h20 are produced if 75g of CO2 are produced. (Work Shown with the answer would be preferred.)

Agata [3.3K]

2:7 ratio of ethane to O2 = 15:x 9solve for x)
x=52.5 mol O2

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2 years ago

Calculate the number of hydrogen atoms present in 40g of urea, (NH2)2CO

tekilochka [14]

Answer: There are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

Explanation:

Given: Mass of urea = 40 g

Number of moles is the mass of substance divided by its molar mass.

First, moles of urea (molar mass = 60 g/mol) are calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{40 g}{60 g/mol}\\= 0.67 mol$

According to the mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms.

So, the number of atoms present in 0.67 moles are as follows.

$0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms$

In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.

$4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms$

Thus, we can conclude that there are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

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