All categories

3 years ago

Which element has five energy levels? radon iodine argon krypton

Chemistry

2 answers:

sergeinik [125]3 years ago

5 0

Answer:

iodine

Explanation:

As this chart shows:

Iodine:

Number of Energy Levels: 5

First Energy Level: 2

Second Energy Level: 8

Third Energy Level: 18

Fourth Energy Level: 18

Fifth Energy Level: 7

alexgriva [62]3 years ago

3 0

Answer:

Iodine

Explanation:

You might be interested in

IF YOU GIVE A LINK OR DON"T ACTUALLY ANSWER THIS QUESTION RIGHT JUST FOR POINTS I WILL REPORT YOUR ANSWER AND YOU

AVprozaik [17]

Answer:

No, the puddle was formed because of the sun, because if there was snow and it rained then it would have turned slippery or icy

8 0

3 years ago

Read 2 more answers

Rare earth elements plz

Slav-nsk [51]

Rare earth metals are a group of 17 elements - lanthanum, cerium, praseodymium, neodymium, promethium, samarium, europium, gadolinium, terbium, dysprosium, holmium, erbium, thulium, ytterbium, lutetium, scandium, yttrium - that appear in low concentrations in the ground

3 0

3 years ago

Read 2 more answers

A) Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of CO2 and 1.37 mg of H2O. If the compound contains o

Mumz [18]

Answer:

For a: The empirical formula for the given compound is $CH$

For b: The empirical and molecular formula for the given organic compound are $C_{10}H_{20}O$

Explanation:

For a:

The chemical equation for the combustion of hydrocarbon follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

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

We are given:

Conversion factor used: 1 g = 1000 mg

Mass of $CO_2=5.86mg=5.86\times 10^{-3}g$

Mass of $H_2O=1.37mg=1.37\times 10^{-3}g$

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 $5.86\times 10^{-3}g$ of carbon dioxide, $\frac{12}{44}\times 5.86\times 10^{-3}=1.60\times 10^{-3}g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in $1.37\times 10^{-3}g$ of water, $\frac{2}{18}\times 1.37\times 10^{-3}=0.152\times 10^{-3}g$ of hydrogen will be contained.

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.60\times 10^{-3}g}{12g/mole}=0.133\times 10^{-3}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.152\times 10^{-3}g}{1g/mole}=0.152\times 10^{-3}moles$

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 $0.133\times 10^{-3}$ moles.

For Carbon = $\frac{0.133\times 10^{-3}}{0.133\times 10^{-3}}=1$

For Hydrogen = $\frac{0.152\times 10^{-3}}{0.133\times 10^{-3}}=1.14\approx 1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

Hence, the empirical formula for the given compound is $CH$

For b:

The chemical equation for the combustion of menthol 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 $CO_2$ = 0.2829 g

Mass of $H_2O$ = 0.1159 g

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, $\frac{2}{18}\times 0.1159=0.013g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1005) - (0.077 + 0.013) = 0.105 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 = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.013g}{1g/mole}=0.013moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0105g}{16g/mole}=0.00065moles$

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 0.00065 moles.

For Carbon = $\frac{0.0064}{0.00065}=9.84\approx 10$

For Hydrogen = $\frac{0.013}{0.00065}=20$

For Oxygen = $\frac{0.00065}{0.00065}=1$

Step 3: Taking the mole ratio as their subscripts.

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

The empirical formula for the given compound is $C_{10}H_{20}O$

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 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

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

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

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

Hence, the empirical and molecular formula for the given organic compound are $C_{10}H_{20}O$

3 0

4 years ago

A sample of metal has a mass of 16.5916.59 g, and a volume of 5.815.81 mL. What is the density of this metal

Oxana [17]

Answer:

=> 2.8554 g/mL

Explanation:

To determine the formula to use in solving such a problem, you have to consider what you have been given.

We have;

mass (m) = 16.59 g

Volume (v) = 5.81 mL

From our question, we are to determine the density (rho) of the rock.

The formula:

$p = \frac{m}{v}$

Substitute the values into the formula:

$p = \frac{16.59 g}{5.81 mL} \\ = 2.8554 g/mL$

= 2.8554 g/mL

Therefore, the density (rho) of the rock is 2.8554 g/mL.

5 0

2 years ago

Lấy ví dụ về tính chất hoá học của chất

hodyreva [135]

Answer:

Ví dụ về tính chất hóa học bao gồm tính dễ cháy, tính độc, tính axit, khả năng phản ứng (nhiều loại) và nhiệt của quá trình cháy.

Explanation:

4 0

3 years ago