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3 years ago

What is the average atomic mass of titanium on that planet

Chemistry

1 answer:

Licemer1 [7]3 years ago

6 0

Answer:
average atomic mass = 46.8509984 amu

Explanation:
To get the average atomic mass, we will follow these steps:
1- multiply the atomic mass of each isotope by its percentage of abundance
2- add the products

We are given that:
1- The first isotope has an atomic mass of 45.95293 and percentage of abundance equals 72%
2- The second isotope has an atomic mass of 47.94795 and percentage of abundance equals 11%
3- The third isotope has an atomic mass of 49.94479 and percentage of abundance equals 17%

Following the mentioned steps, we can get the average atomic mass as follows:
average atomic mass = (45.95293)(0.72)+(47.94795)(0.11)+(49.94479)(0.17)
= 46.8509984 amu

Hope this helps :)

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Methane is the chief constituent of petroleum fuel A:yes B:no

Ad libitum [116K]

Yes

Methane also is the chief constituent of natural gas, which contains from 50 to 90 percent methane (depending on the source), and occurs as a component of firedamp (flammable gas) along coal seams.

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3 years ago

12.A piece of magnesium is in the shape of a cylinder with a height of 5.62 cm

yaroslaw [1]

Answer:

Density, d = 1.779 g/cm³

Explanation:

The density of a material is given by its mass per unit volume.

Here, height of a piece of magnesium cylinder, h = 5.62 cm

Its diameter, d = 1.34 cm

Radius = 0.67 cm

Volume of he cylinder,

$V=\pi r^2 h\\\\\text{Putting the value of r and h, we get :}\\\\V=(\pi \times (0.67)^2\times 5.62)\ cm^3$

$d=\dfrac{m}{V}\\\\d=\dfrac{14.1\ g}{(\pi \times (0.67)^2\times 5.62)\ cm^3}\\\\d=1.779\ g/cm^3$

So, the density of the sample is 1.779 g/cm³.

4 0

3 years ago

How many electrons elements are present in period 2 ?

AVprozaik [17]

There are Eight elements

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3 years ago

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There are 7 named classes of hazardous materials. O True O False

Anika [276]

False

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2 years ago

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

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 $CO_2=1.03g$

Mass of $H_2O=0.14g$

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 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

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 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

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 = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

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

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago