If a person had 100 g of pure radioactive nuclei with a half-life of 100 years, then after 100 years he or she would have _____ of radioactive nuclei

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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We need 39.9 grams of sodium sulfate to prepare 750 mL of a 0.375 M solution.

Volume of the solution = 750 mL = 0.750 L
We know that, moles of solute = molarity × volume of solution (in L)

moles of sodium sulfate = 0.375 M × 0.750 L = 0.281 mol
Molar mass of sodium sulfate ([tex]Na_{2}SO_{4}[/tex])= (2 × 22.99 g/mol) + (4 × 16.00 g/mol) + (32.07 g/mol) = 142.04 g/mol
Therefore, grams of [tex]Na_{2}SO_{4}[/tex] = moles of [tex]Na_{2}SO_{4}[/tex] × molar mass of [tex]Na_{2}SO_{4}[/tex] = 0.281 mol × 142.04 g/mol = 39.9 g

We need 39.9 grams of sodium sulfate to prepare 750 mL of a 0.375 M solution.

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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?

A pH of 13 indicates a highly basic solution. To calculate the H3O+ concentration in household bleach, we can use the following formula:

pH = -log[H3O+]

Rearranging the formula, we get:

[H3O+] = 10^(-pH)

Substituting pH = 13 into the formula, we get:

[H3O+] = 10^(-13)

[H3O+] = 1 x 10^(-13) mol/L

Therefore, the H3O+ concentration in household bleach is approximately 1 x 10^(-13) mol/L.

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The oxidation states exhibited by C, Si, Ge, Sn, Pb are -4, +4, +2 or +4, +2 or +4, and +2 or +4, respectively.

The oxidation state, also known as the oxidation number, is a measure of the degree of oxidation of an atom in a chemical compound. The oxidation state can be determined by assigning electrons to each atom in a compound according to a set of rules.

In general, carbon (C) exhibits an oxidation state of -4 in compounds such as methane (CH₄), where it is bonded to four hydrogen atoms. Carbon can also exhibit positive oxidation states in compounds such as carbon dioxide (CO₂), where it is bonded to two oxygen atoms, and in carbonyl compounds, where it is bonded to a metal.

Silicon (Si), germanium (Ge), tin (Sn), and lead (Pb) all belong to the same group in the periodic table and therefore exhibit similar chemical properties. They can all exhibit positive oxidation states of +2 and +4. For example, silicon can exhibit an oxidation state of +4 in silicon dioxide (SiO₂) and +2 in silane (SiH₄). Germanium, tin, and lead also exhibit a similar range of oxidation states in their compounds.

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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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The voltage generated by a hydrogen-oxygen fuel cell at 73.5°C when the partial pressures of hydrogen and oxygen are 19.8 atm is 1.174 V.

The standard cell potential for the hydrogen-oxygen fuel cell is 1.23 V at 25°C. However, the Nernst equation takes into account the temperature and the partial pressures of the reactants. The Nernst equation is as follows:

Ecell = E°cell - (RT/nF)lnQ

where Ecell is the cell potential, E°cell is the standard cell potential, R is the gas constant (8.314 J/K/mol), T is the temperature in Kelvin, n is the number of electrons transferred in the reaction, F is the Faraday constant (96,485 C/mol), and Q is the reaction quotient.

To calculate Q, we need to know the concentrations of the reactants and products. In the case of a fuel cell, the reactants are the fuels, which are gases, and their concentrations are expressed as partial pressures. The reaction in a hydrogen-oxygen fuel cell is:

2H2 + O2 → 2H2O

The reaction quotient can be expressed as:

Q = (PH2)²(PO2)

where PH2 is the partial pressure of hydrogen and PO2 is the partial pressure of oxygen.

At 73.5°C, the temperature in Kelvin is 346.65 K. The partial pressures of hydrogen and oxygen are 19.8 atm. Substituting these values into the Nernst equation, we get:

Ecell = 1.23 V - (8.314 J/K/mol)(346.65 K/ (2*96,485 C/mol)) ln[(19.8 atm)²(19.8 atm)]

Ecell = 1.23 V - 0.056 V

Ecell = 1.174 V

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The thermodynamic equation for the reaction is:

[tex]HNO_3(g) + 4H_2(g)[/tex] → [tex]NH_3(g) + 3H_2O(g) \Delta H = -637 kJ[/tex]

This means that the reaction releases 637 kJ energy per mole ammonia produced. The amount of energy released when one mole of hydrogen gas reacts is 159.25 kJ,

However,  the amount of energy released when one mole of hydrogen gas reacts. From the balanced equation, we can see that one mole of ammonia is produced for every 4 moles of hydrogen gas that react. Therefore, the amount of energy released :

ΔH/4 = -637 kJ / 4 = -159.25 kJ

So, the amount of energy released when one mole hydrogen gas reacts is 159.25 kJ, and we consider this to be a positive value because the reaction is exothermic.

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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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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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If sodium increases in the extracellular fluid (ECF), water will move from the intracellular fluid (ICF) to the ECF through osmosis.

This is because sodium is an osmotically active particle, meaning that it affects the concentration of particles in a solution.

When the concentration of sodium in the ECF increases, it creates a hypertonic environment compared to the ICF, which is relatively hypotonic.

As a result, water will move from the hypotonic ICF to the hypertonic ECF in order to balance the concentration of particles between the two compartments.

This movement of water can lead to changes in cell volume and function, which is why maintaining proper electrolyte balance is important for normal cellular function.

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The mass of aluminum metal produced when 15.0 mL of a 0.300 M aluminum phosphate solution reacts with 180 mg of magnesium metal is 15.60 mg.


1. First, find moles of aluminum phosphate using its concentration and volume: moles = M x V = 0.300 mol/L x 0.015 L = 0.0045 mol.

2. Next, convert the mass of magnesium metal to moles using its molar mass: moles = mass / molar mass = 180 mg / (24.31 g/mol x 1000 mg/g) = 0.00741 mol.

3. Now, find the limiting reactant by comparing the mole ratios: (0.0045 mol AlPO₄) / (2) < (0.00741 mol Mg) / (3), so aluminum phosphate is the limiting reactant.

4. Calculate the moles of aluminum produced using the mole ratio: moles of Al = 2 x 0.0045 mol AlPO₄ = 0.009 mol.

5. Finally, convert the moles of aluminum to mass: mass = moles x molar mass = 0.009 mol x 26.98 g/mol x 1000 mg/g = 15.60 mg.

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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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The mass in grams of 3.45 x 10E24 atoms of carbon is 68.93 g.

To find the mass in grams of 3.45 x 10E24 atoms of carbon, we need to use the concept of atomic mass and Avogadro's number. The atomic mass of carbon is 12.01 g/mol, which means that one mole of carbon contains 6.022 x 10E23 atoms. This is known as Avogadro's number.

So, to find the mass of 3.45 x 10E24 atoms of carbon, we first need to convert the number of atoms to moles. We do this by dividing the given number of atoms by Avogadro's number:

3.45 x 10E24 atoms / 6.022 x 10E23 atoms/mol = 5.74 moles

Next, we can use the molar mass of carbon to find the mass of 5.74 moles of carbon:

5.74 moles x 12.01 g/mol = 68.93 g

Therefore, the mass in grams of 3.45 x 10E24 atoms of carbon is 68.93 g.

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The volume occupied by 3.67 moles of H₂ gas at STP is 82.19 L.

To calculate the volume, we use the equation V = n × Vm, where V is the volume, n is the number of moles, and Vm is the molar volume of a gas at STP (22.4 L/mol). At STP (standard temperature and pressure), one mole of any gas occupies 22.4 L. Given that we have 3.67 moles of H₂ gas, we can calculate the volume as follows:

1. Identify the number of moles (n): 3.67 moles of H₂
2. Find the molar volume of a gas at STP (Vm): 22.4 L/mol
3. Use the equation V = n × Vm
4. Substitute the values: V = 3.67 moles × 22.4 L/mol
5. Calculate the volume: V = 82.19 L

Therefore, 3.67 moles of H₂ gas occupy 82.19 L at STP.

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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 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