how many moles of aluminum oxide AI2O3 can you produce if you have two moles of AI

After 100 years, a person who had 100 g of pure radioactive nuclei with a half-life of 100 years would have 50 g of radioactive nuclei left.

The half-life of a radioactive substance is the time it takes for half of the substance's original amount to decay. In this case, since the half-life is 100 years, after 100 years, half of the original amount of radioactive nuclei would have decayed.

After the first 100 years, 50 g of radioactive nuclei would remain, and the other 50 g would have decayed. If we wait for another 100 years, half of the remaining 50 g, which is 25 g, would decay, leaving only 25 g of the original amount. This process will continue until all the radioactive nuclei have decayed.

It's worth noting that the rate of decay is exponential, which means that the amount of radioactive substance remaining decreases at a constant rate over time. Knowing the half-life of a radioactive substance is important in determining the amount of time it takes for the substance to decay to a safe level.

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Chemical change results in a new substance and it cannot be reversed by physical means.

Chemical changes is said to occur when a substance combines with another to form a new substance, called chemical synthesis or, alternatively, chemical decomposition into two or more different substances  and are not reversible except by further chemical reactions.

Examples of chemical change would be:

It is also worthy to note that in a physical change, no new substance is formed and also a  chemical change is always accompanied by one or more new substance.

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The unknown substance most likely has property not found pure in nature.(D)

Since the substance has properties A (highly reactive) and B (low electronegativity), it's likely that it readily forms compounds with other elements, making it difficult to find in its pure form.

Highly reactive elements, such as alkali metals or halogens, are typically not found in nature in their pure state because they readily react with other elements to form stable compounds. Property C (has many isotopes) doesn't directly influence the substance's reactivity or occurrence in nature.(D)

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When a 3.950 L sample of gas is cooled from 91.50°C to a volume of 2.550 L with no change in pressure, the new temperature is approximately -36.61°C.

To find the new temperature when a 3.950 L sample of gas is cooled from 91.50°C to a volume of 2.550 L, we can use the Charles' Law formula. Charles' Law states that the volume of a gas is directly proportional to its temperature, assuming that pressure remains constant.

Mathematically, this can be represented as:

V1/T1 = V2/T2

Here, V1 is the initial volume (3.950 L), T1 is the initial temperature (91.50°C), V2 is the final volume (2.550 L), and T2 is the final temperature, which we need to find.

First, convert the initial temperature from Celsius to Kelvin by adding 273.15:

T1 = 91.50°C + 273.15 = 364.65 K

Now, plug the values into the Charles' Law formula:

(3.950 L) / (364.65 K) = (2.550 L) / T2

To find T2, we can cross-multiply and divide:

T2 = (2.550 L) * (364.65 K) / (3.950 L)
T2 ≈ 236.54 K

Finally, convert the temperature back to Celsius by subtracting 273.15:

New temperature = 236.54 K - 273.15 ≈ -36.61°C

In conclusion, when a 3.950 L sample of gas is cooled from 91.50°C to a volume of 2.550 L with no change in pressure, the new temperature is approximately -36.61°C.

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The number of grams of oxygen required is 94.9 g, under the condition that it is used to  burn 28. 8 g of ammonia (NH₃)

NH₃ + 7O₂ → 4NO₂ + 6H₂O,

then the correct answer for the required question is Option B.

Now, the balanced chemical equation for the reaction of ammonia (NH₃) and oxygen (O₂) to create nitrogen dioxide (NO₂) and water (H₂O) is

4NH₃ + 7O₂ → 4NO₂ + 6H₂O

The given molar mass of NH₃ is 17.0305 g/mol and that of O₂ is 31.998 g/mol.
In order to  find out how many grams of O₂ are required to burn 28.8 g of NH₃, we have to first balance the equation:

4 NH₃+ 7O₂ → 4NO₂ + 6H₂O
Then there are  4 moles of NH₃, we need 7 moles of O₂.
Hence, molar mass of NH₃ is 17.0305 g/mol, so we can change 28.8 g of NH₃ to moles

28.8 g NH₃ × (1 mol NH₃/17.0305 g NH₃)
= 1.69 mol NH₃

Now we have to apply  stoichiometry to evaluate  how many moles of O₂ are required

1.69 mol NH₃ × (7 mol O₂/4 mol NH₃)
= 2.95 mol O₂

Therefore, we can convert moles of O₂ to grams:

2.95 mol O₂ × (31.998 g O₂/1 mol O₂)
= 94.9 g
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The complete question is
How many grams of oxygen (O2) is required to burn 28. 8 g of ammonia (NH3)?4NH3 + 7O2 → 4NO2 + 6H2O
Molar Mass
NH3=17. 0305 g/mol
O2=31. 998 g/mol
NO2=46. 0055 g/mol
H2O=18. 0153 g/mol
a)15. 3 g
b)94. 9 g
c)54. 1 g
d)108 g

To find the molarity of a solution, we need to know the number of moles of solute (in this case, sugar) and the volume of the solution in liters. We can use the given mass of sugar and the molar mass of sugar to find the number of moles:

Mass of sugar = 194.55 g

Molar mass of sugar (C12H22O11) = 342.3 g/mol

Number of moles of sugar = Mass of sugar / Molar mass of sugar

Number of moles of sugar = 194.55 g / 342.3 g/mol

Number of moles of sugar = 0.568 mol

Now we need to convert the given volume of the solution (250 mL) to liters:

Volume of solution = 250 mL

Volume of solution = 250 mL / 1000 mL/L

Volume of solution = 0.250 L

Finally, we can use the number of moles of sugar and the volume of the solution to calculate the molarity:

Molarity = Number of moles of sugar / Volume of solution

Molarity = 0.568 mol / 0.250 L

Molarity = 2.27 M

Therefore, the molarity of the sugar solution is 2.27 M.

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(a). [tex]C_2H_5OH + 3O_2[/tex] → [tex]2CO_2 + 3H_2O[/tex]

(b). The volume of air required to burn 250 grams of ethanol at 35°C and 790 mmHg is approximately 6.63 liters.

a. The balanced equation for the combustion of ethanol ([tex]C_2H_5OH[/tex]) in air to generate carbon dioxide ([tex]CO_2[/tex]) and water ([tex]H_2O[/tex]) is:

[tex]C_2H_5OH + 3O_2[/tex] → [tex]2CO_2 + 3H_2O[/tex]

b. We first need to calculate the number of moles of ethanol used in the reaction. The molar mass of ethanol is:

46.07 g/mol

Therefore, the number of moles of ethanol used is:

[tex]n = m/M = 250 g / 46.07 g/mol = 5.42 mol[/tex]

Therefore, the number of moles of oxygen required to burn 5.42 moles of ethanol is:

[tex]3n = 3 * 5.42 mol = 16.26 mol[/tex]

The ideal gas law is:

PV = nRT

V = nRT/P

Substituting the values, we get:

[tex]V = (16.26 mol)(0.08206 L.atm/(mol.K))(308.15 K) / 790 mmHg[/tex]

Simplifying, we get:

V = 6.63 L

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2.4453  ×  10⁹ KJ energy was evolved  when the total volume of hydrogen gas needed to fill the hindenburg was 2.09 × 10⁸ l at 1.00 atm and 25.1°

According to the given data,  

Volume of the hydrogen gas = 2.09 × 10⁸ L

Pressure of the gas = P = 1 atm

Temperature of the gas =T = 25.1 °C =298.1 K

We know that, for an ideal gas equation

PV=nRT

1 atm ×2.09 × 10⁸ L = n × 0.0820 atmL/molK × 298.1 K

⇒n = 1 atm ×2.09 × 10⁸ L/  0.0820 atmL/molK × 298.1 K

⇒n = 0.0855 ×10⁸ mol

ΔH for hydrogen gas is =-286 kJ/mol

For  0.0855 ×10⁸ mol energy evolved when hydrogen gas is burned =

0.0855 ×10⁸ mol × (-286 KJ/mol) = -2.4453  ×  10⁹ KJ

Therefore, 2.4453  ×  10⁹ KJ energy was evolved when it was burned.

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The complete question is-

The total volume of hydrogen gas needed to fill the hindenburg was 2.09 × 108 l at 1.00 atm and 25.1°. how much energy was evolved when it burned?

3.266 × 10²⁴ compounds in 434g of ammonium nitrate

To determine how many compounds are in 434g of ammonium nitrate, we will follow these steps:
Step 1: Determine the molar mass of ammonium nitrate (NH₄NO₃).
Ammonium nitrate consists of one nitrogen (N) atom, four hydrogen (H) atoms, and three oxygen (O) atoms in its chemical formula. The molar masses of N, H, and O are approximately 14 g/mol, 1 g/mol, and 16 g/mol, respectively.

Molar mass of NH₄NO₃ = 1(N) + 4(H) + 1(N) + 3(O)
= 14 + (4 × 1) + 14 + (3 × 16)
= 14 + 4 + 14 + 48
= 80 g/mol

Step 2: Calculate the number of moles of ammonium nitrate.
To find the number of moles, divide the given mass (434g) by the molar mass (80 g/mol).

Number of moles = 434 g / 80 g/mol
= 5.425 moles

Step 3: Calculate the number of compounds (molecules) in ammonium nitrate.
Use Avogadro's number (6.022 × 10²³ molecules/mol) to find the total number of molecules in 5.425 moles of ammonium nitrate.

Number of compounds = 5.425 moles × (6.022 × 10²³ molecules/mol)
= 3.266 × 10²⁴ molecules

So, there are approximately 3.266 × 10²⁴ compounds in 434g of ammonium nitrate.

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