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Arte-miy333 [17]
3 years ago
7

Number 14 please fast answer

Chemistry
2 answers:
frutty [35]3 years ago
8 0

Answer: A. Oxygen

Explanation:

Oxygen - 46.6%

Oxygen is the most abundant element in the Earth's crust. Oxygen makes up 467,100 ppm (parts per million) of the Earth's crust, or 46.6%. It exists as a major compound of the silicate minerals where it combines with other elements. It also exists as a compound in carbonates and phosphates. Oxygen has industrial, medical, and commercial purposes. It is used with acetylene to cut and weld metals. It is used in hospitals to ease respiratory disease and can also be used to manufacture explosives among other numerous uses.

Silicon - 27.7%

Silicon is the second most common element present in the crust with an abundance of 276,900 ppm. It exists as a compound in the mantle and the crust. In the crust, it exists combined with oxygen to form silicate minerals. It is found in the sand which is an abundant and easily accessible resource on Earth. Silicon is also recovered from quartzite, mica, and talc. From silicon, we get silicones used in hydraulic fluids, electric insulators, and lubricants among others. Solid silicon is majorly used as a semiconductor especially in computer hardware. It is used to make transistors in the electronic industry. Silicon is used in the aluminum industry in the making of aluminum alloys. It is used in the production of ceramics, glass, cosmetics, insecticides, some types of steel, and pharmaceutical products.

Aluminum - 8.1%

At 80,700 ppm, aluminum is the third most abundant element in Earth's crust. Aluminum does not exist as a lone element, and it is found as a compound. Abundant compounds of Aluminum include aluminum oxide, aluminum hydroxide, and potassium aluminum sulfate. Aluminum is extracted from its compounds largely through the Bayer and Hall-Heroult processes. The element ideal because of its light weight and Aluminum alloys are extensively used to make utensils foils, packaging materials. It is used in the manufacture of parts of automobiles, rockets, and machinery.

Iron - 5%

Iron exists in the earth’s crust, and its composition is 50,500 ppm. Iron extracted in the form of iron ores existing in iron oxide forms such as hematite and magnetite. Blast furnaces are used to recover iron from the iron ores. Iron has many extensive applications such as the making of steel. Iron is also used to make utensils and kitchen appliances. It is also used to make cast iron and wrought iron which is extensively utilized in many industries. Iron element will always react with water and oxygen and the surface of iron is ordinarily lustrous and silvery gray, but it has a tendency of oxidizing in the open air to produce hydrated iron oxide commonly known as rust. In its purest form iron is relatively soft and is hardened and strengthened in the smelting process by adding a small percentage of carbon. The addition of carbon of between 0.002% and 2.1% will result in steel that could be 1000 times harder compared to pure iron.

Hope this helps!!! Good luck!!! ;)

aleksley [76]3 years ago
5 0

Answer:

Go to google, I think you would find information about your question

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Why was Rutherford's Gold Foil Experiment so important?
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When nahco3 completely decomposes, it can follow this balanced chemical equation: 2nahco3 → na2co3 h2co3 determine the theoretic
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Theoretical yield = 2.397

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percent yield = 98.456%

When nahco3 completely decomposes, it can follow this balanced chemical equation:

2nahco3 → na2co3 h2co3

If the mass of the NaHCO3 sample is 3.80 g, we must use stoichiometry to calculate the theoretical yields of each of the products.

mass of NaHCO₃ = 3.80 g

molar mass of NaHCO₃ = 84 g/mol

so the no of moles of NaHCO₃ = 3.80/84 =  0.0452 mol

You see, one mole of sodium carbonate and one mole of hydrogen carbonate are produced from two moles of sodium bicarbonate.

so, the no of moles of sodium carbonate = 0.0452/2 = 0.0226 mol

∴ mass of sodium carbonate ( Na₂CO₃) = no of moles of Na₂CO₃ × molar mass of Na₂CO₃

=  0.0226 × 106 ≈ 2.397 g

no of moles of hydrogen carbonate = 0.0452/2 = 0.0226 mol

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percentage yield = experimental yield/theoretical yield × 100

here experimental yield of Na₂CO₃ = 2.36 g

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mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of
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<u>Answer:</u> The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

<u>For nitrogen gas:</u>

Molar mass of nitrogen gas = 28 g/mol

\text{Moles of nitrogen gas}=\frac{x}{28}mol

<u>For hydrogen gas:</u>

Molar mass of hydrogen gas = 2 g/mol

\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol

Equating the moles of the individual gases to the moles of mixture:

0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g

To calculate the mass percentage of substance in mixture we use the equation:

\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100

Mass of the mixture = 13.22 g

  • <u>For nitrogen gas:</u>

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%

  • <u>For hydrogen gas:</u>

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

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