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3 years ago

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Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid fuel ro

cket motors. The reaction is Fe2O3(s) + 2Al(s) → 2Fe(l) + Al2O3(s) Click here to see the reaction. What is the maximum mass of aluminum oxide that could be produced along with 18.65 g iron?

1 answer:

Diano4ka-milaya [45]3 years ago

3 0

Answer:

16.98 g

Explanation:

The balacend equation is

Fe2O3(s) + 2Al(s) → 2Fe(l) + Al2O3(s)

So, when 2 moles of iron (Fe) are produced, 1 mol of aluminum oxide is produced. The mass molar of the elements are:

Fe = 56 g/mol; O = 16 g/mol; Al = 27 g/mol

Then, the mass molar of Al2O3 is:

2x27 + 3x16 = 102 g/mol

The stoichiometry relation will be:

2 mol of Fe -------------------- 1 mol of Al2O3

2x56 g of Fe -------------------- 102 g of Al2O3

18.65 g of Fe -------------------- x g of Al2O3

By a direct simple three rule:

112x = 1902.3

x = 16.98 g

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Name the types of radiation known to be emitted by radioactive elements.

nikitadnepr [17]

Answer : The types of radiation known to be emitted by radioactive elements are, alpha particles, beta particles, or gamma rays.

Explanation :

Radioactive decay : It the process in which an unstable atomic nucleus loses energy by emitting the radiations like, alpha particles, beta particles, or gamma rays.

The naturally occurring radioactive elements are, radium, thorium, and uranium.

Alpha particle : It is also known as alpha radiation or alpha ray that consists of 2 protons and 2 neutrons that are bound together into a particle that is identical to the helium nucleus. It is produced in the process of alpha decay.

Beta particle : It is also known as beta radiation or beta ray. During the beta decay process, a high energy and speed electron or positron are emitted by the radioactive decay of atomic nucleus.

Gamma particle : It is also a gamma radiation or gamma ray that is arising from the radioactive decay of atomic nuclei. It has shortest wavelength waves and imparts high photon energy can pass through most forms of matters because they have no mass.

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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]

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

<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 $CO_2=0.2829g$

Mass of $H_2O=0.1159g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

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

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.

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

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:

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

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

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

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

<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 0.00066 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 = $\frac{0.00066}{0.00066}=1$

<u>Step 3:</u> 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 $C_{10}H_{20}O$

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3 years ago

Read 2 more answers

flammability is a material s ability to burn in the presence of a. hydrogen. b. nitrogen. c. oxygen. d. carbon dioxide.

WARRIOR [948]

I'm guessing it's C-Oxygen because in order to burn something, or create fire, oxygen HAS to be present.

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3 years ago

Which of the following would be most useful in trying to obtain procedural information to replicate an experiment previously pub

Jet001 [13]

Answer: Peer-reviewed journal article is the most useful because the information in them had been carefully scrutinized and aproved by people who are experts in that particular field.

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3 years ago

3. A 31.2-g piece of silver (s = 0.237 J/(g · °C)), initially at 277.2°C, is added to 185.8 g of a liquid, initially at 24.4°C,

VARVARA [1.3K]

Answer:

$Cp_{liquid}=2.54\frac{J}{g\°C}$

Explanation:

Hello,

In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

$Q_{Ag}=-Q_{liquid}$

That in terms of the heat capacities, masses and temperature changes turns out:

$m_{Ag}Cp_{Ag}(T_2-T_{Ag})=-m_{liquid}Cp_{liquid}(T_2-T_{liquid})$

Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

$Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}$

Best regards.

6 0

4 years ago