What is the first step in creating a unit conversion factor?

In Universe , recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quant

lawyer [7]

Answer:

According to Hund's rule and the Aufbau principle in which the orbitals must be filled with electrons, they are not strictly applied in the real universe, because the intermediate and electron-filled atomic orbitals are very stable . Because there are four d-orbitals in universe L, a typical half-full configuration will be xd4 and its full configuration will be xd8, where x is the primary orbital for any specific element. Here is an example:

Vahadium ₂₃V

in real universe: [Ar]₈ 3d³4s²

in universe L: [Ar]₁₈ 3d⁴4s¹

Chromium

in real universe: [Ar]₈ 3d⁵4s¹

in universe L: [Ar]₁₈ 3d⁴4s²

Explanation:

8 0

3 years ago

1. Which phase change is accompanied by the release of heat? A) H20(s)--> H2O(g) B) H20(l) ->H2O(s) C) H20(l)→ H2O(g) D) H

docker41 [41]

Answer:

B, liquid to solid.

Explanation: Since heat is being released, the particles for H2O would clump up. Heat is basically being taken out.

5 0

3 years ago

Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is comb

Bogdan [553]

Answer: The molecular formula for the menthol is $C_{10}H_{20}O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=0.2829g$

Mass of $H_2O=0.1159g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.2829 g of carbon dioxide, $\frac{12}{44}\times 0.2829=0.077g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.1159 g of water, $\frac{2}{18}\times 0.1159=0.0129g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1005) - (0.077 + 0.0129) = 0.0106 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0129g}{1g/mole}=0.0129moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0106g}{16g/mole}=0.00066moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00066 moles.

For Carbon = $\frac{0.0064}{0.00066}=9.69\approx 10$

For Hydrogen = $\frac{0.0129}{0.00064}=19.54\approx 20$

For Oxygen = $\frac{0.00066}{0.00066}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 10 : 20 : 1

Hence, the empirical formula for the given compound is $C_{10}H_{20}O_1=C_{10}H_{20}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 156.27 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

$n=\frac{156.27g/mol}{156g/mol}=1$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O$

Thus, the molecular formula for the menthol is $C_{10}H_{20}O$

4 0

3 years ago

Read 2 more answers

Write your answer with correct significant figures (*no scientific notation): 847.3219 x 34.6 x 1.4560 =

Lunna [17]

Answer:

42686.04375

Explanation:

847.3219*34.6=2317.33774 multiply this by the next number 1.4560 to get 42686.04375

3 0

3 years ago

PLEASE ANSWER + BRAINLIEST !!!

Sunny_sXe [5.5K]

Hope this would help you

3 0

3 years ago