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vagabundo [1.1K]

2 years ago

14

Which of the following is a possible set of quantum numbers for an electron n, l, m subscript l, m subscript s? (1, 1, 0, +one o

ver two ) (2, 1, 2, +one over two ) (3, 2, 0, -one over two ) (3, -2, 1, -one over two )

2 answers:

Dovator [93]2 years ago

8 0

The <span>possible set of quantum numbers for an electron n, l, m subscript l, m subscript s is D. 3, -2, 1, -one over two.</span>

Crank2 years ago

7 0

That's incorrect. l can only be between 0 and n-1. m can only be between -l and +l. Your answer is (3,2,0,-1/2)

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In a heating curve, when is the temperature constant? need answers!

Varvara68 [4.7K]

During a phase change

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3 years ago

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Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

kirza4 [7]

<u>Answer:</u> The molecular formula for the given organic compound is $C_{18}H_{20}O_2$

<u>Explanation:</u>

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 $CO_2=39.61g$

Mass of $H_2O=9.01g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

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

So, in 39.61 g of carbon dioxide, $\frac{12}{44}\times 39.61=10.80g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

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

So, in 9.01 g of water, $\frac{2}{18}\times 9.01=1.00g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

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

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles$

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

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles$

<u>Step 2:</u> 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 0.10 moles.

For Carbon = $\frac{0.9}{0.10}=9$

For Hydrogen = $\frac{1}{0.10}=10$

For Oxygen = $\frac{0.10}{0.10}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is $C_9H_{10}O$

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 = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

$n=\frac{268.34g/mol}{134g/mol}=2$

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

$C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2$

Thus, the molecular formula for the given organic compound is $C_{18}H_{20}O_2$ .

3 0

3 years ago

The molecular weight of table salt, NaCl, is 58.5 g/mol. A tablespoon of salt weighs 6.37 grams. Calculate the number of moles o

LiRa [457]

There are 0.109 moles of NaCl in one tablespoon of salt. This is found by dividing the number of grams you have by the molecular weight.

3 0

3 years ago

Read 2 more answers

3. The same well water contains dissolved calcium (Ca), magnesium (Mg), sodium (Na) and potassium (K) at 5.0, 1.0, 136 and 1.2 m

Verdich [7]

<u>Answer:</u> The concentration of calcium, magnesium, sodium and potassium are $1.25\times 10^{-4}mol/L$ , $4.20\times 10^{-5}mol/L$ , $5.91\times 10^{-3}mol/L$ and $3.08\times 10^{-5}mol/L$ respectively

<u>Explanation:</u>

To convert the mass from milligrams to grams, we use the conversion factor:

1 g = 1000 mg

To convert the given concentration fro grams to moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For Calcium:</u>

Concentration given = 5.0 mg/L = $5.0\times 10^{-3}g/L$

We know that:

Molar mass of calcium element = 40 g/mol

Putting values in equation 1, we get:

$\text{Concentration in mol/L}=\frac{5.0\times 10^{-3}g/L}{40g/mol}\\\\\text{Concentration in mol/L}=1.25\times 10^{-4}mol/L$

<u>For Magnesium:</u>

Concentration given = 1.0 mg/L = $1.0\times 10^{-3}g/L$

We know that:

Molar mass of magnesium element = 24 g/mol

Putting values in equation 1, we get:

$\text{Concentration in mol/L}=\frac{1.0\times 10^{-3}g/L}{24g/mol}\\\\\text{Concentration in mol/L}=4.20\times 10^{-5}mol/L$

<u>For Sodium:</u>

Concentration given = 136 mg/L = $136\times 10^{-3}g/L$

We know that:

Molar mass of sodium element = 23 g/mol

Putting values in equation 1, we get:

$\text{Concentration in mol/L}=\frac{136\times 10^{-3}g/L}{23g/mol}\\\\\text{Concentration in mol/L}=5.91\times 10^{-3}mol/L$

<u>For Potassium:</u>

Concentration given = 1.2 mg/L = $1.2\times 10^{-3}g/L$

We know that:

Molar mass of potassium element = 39 g/mol

Putting values in equation 1, we get:

$\text{Concentration in mol/L}=\frac{1.2\times 10^{-3}g/L}{39g/mol}\\\\\text{Concentration in mol/L}=3.08\times 10^{-5}mol/L$

Hence, the concentration of calcium, magnesium, sodium and potassium are $1.25\times 10^{-4}mol/L$ , $4.20\times 10^{-5}mol/L$ , $5.91\times 10^{-3}mol/L$ and $3.08\times 10^{-5}mol/L$ respectively

3 0

3 years ago

A 50 gram sample of a radioisotope undergoes 2 half-lives. How many grams

Veseljchak [2.6K]

Answer:

At the end of second half life 12.5 g will left

Explanation:

Given data:

Total Mass = 50 g

Half lives = 2

Mass remain at the end = ?

Solution:

At time zero = 50 g

At 1st half life = 50 g /2 = 25 g

At second half life = 25 g/2 = 12.5 g

So at the end of second half life 12.5 g will left.

3 0

3 years ago