A truck weighs 7280 pounds. If the pressure exerted by its tires on the ground is 87. 5 pounds per square centimeter,what is the area of one tire that in contact with the road

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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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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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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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 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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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 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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The evidence that best supports the student's claim that "life is possible because of the unique mixture of gases that cycle through the Earth's spheres" is the presence and balance of oxygen, nitrogen, and carbon dioxide in the atmosphere.

These gases play a crucial role in maintaining life on Earth by supporting respiration, regulating temperature, and enabling the carbon cycle, which allows organisms to exchange and utilize carbon for growth and energy production.

Oxygen: Oxygen is a vital gas for sustaining life on Earth. It is a key component of the atmosphere, making up about 21% of its composition. Oxygen is essential for respiration, the process by which organisms extract energy from food.

Through respiration, organisms break down glucose (derived from food) and use oxygen to produce energy-rich molecules called adenosine triphosphate (ATP).

This energy is necessary for cellular functions and metabolic activities. Many organisms, including humans, require oxygen to survive.

Nitrogen: Nitrogen is the most abundant gas in the Earth's atmosphere, accounting for approximately 78% of its composition. Although nitrogen is relatively inert and does not directly participate in biological processes, it is crucial for life.

Nitrogen is an essential component of amino acids, proteins, and nucleic acids (DNA and RNA), which are fundamental building blocks of life. Nitrogen fixation, a process carried out by certain bacteria, converts atmospheric nitrogen into forms that can be used by plants and other organisms.

This allows nitrogen to enter the food chain and support the growth and development of living organisms.

Carbon Dioxide: Carbon dioxide is a greenhouse gas and an integral part of the Earth's carbon cycle. It plays a significant role in regulating the planet's temperature through the greenhouse effect.

Carbon dioxide traps heat in the atmosphere, preventing excessive heat loss into space and maintaining a suitable temperature range for life. Additionally, carbon dioxide is essential for photosynthesis, a process carried out by plants and other autotrophic organisms.

During photosynthesis, carbon dioxide is absorbed, and with the help of sunlight, it is converted into glucose and oxygen. This process not only provides oxygen for respiration but also allows organisms to utilize carbon for growth, energy production, and the formation of organic compounds.

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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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By considering the concept of Faraday's constant and Avogadro's number we can say that one mole of hydrogen is ionized at 13.598V, removing around 6.022 × 10²³ electrons.

To determine the number of electrons removed when ionizing one mole of hydrogen using 13.598V, we can use the formula:

N = (1 mole) * (Avogadro's number)

where N represents the number of particles (in this case, electrons) in one mole of the substance.

Avogadro's number is approximately 6.022 × 10²³ particles/mol.

Therefore, the number of electrons removed can be calculated as:

N = (1 mole) * (6.022 × 10²³ particles/mol)

= 6.022 × 10²³ electrons

Thus, when ionizing one mole of hydrogen using 13.598V, approximately 6.022 × 10²³ electrons are removed.

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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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The complete combustion of one gallon of gasoline containing 2370.0 grams of octane produces 6888.2 grams of carbon dioxide.

In 2017, people in the US generated approximately 9.85 x 10¹⁴ grams of carbon dioxide by burning 143 billion gallons of gasoline.



1. Write the balanced chemical equation for the combustion of octane:
  2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

2. Determine the molecular weight of octane (C₈H₁₈) and carbon dioxide (CO₂):
  C₈H₁₈: (8 x 12.01) + (18 x 1.01) = 114.23 g/mol
 CO₂: (1 x 12.01) + (2 x 16.00) = 44.01 g/mol

3. Use stoichiometry to find the grams of CO₂ produced from the combustion of 2370.0 grams of octane:
  (2370.0 g octane) x (16 mol CO₂/ 2 mol octane) x (44.01 g CO₂ / mol CO₂) = 6888.2 g CO₂

4. Calculate the total grams of CO₂ generated by burning 143 billion gallons of gasoline in the US in 2017:
  (143 billion gallons) x (6888.2 g CO₂ / gallon) = 9.85 x 10¹⁴ grams of CO₂

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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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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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Since Q is less than K, the reaction will proceed towards products to reach equilibrium. So, the correct option is the reaction is not at equilibrium and it will proceed toward the products.

When the rates of forward and reverse reactions are equal, equilibrium is the condition where there is no overall change in the concentrations of reactants and products. When a system is in equilibrium, the concentrations of all reactants and products are constant over time, and the system appears to be in a state of rest. An equilibrium constant [tex](K_e_q)[/tex], which represents the ratio of the concentrations of products to reactants at equilibrium for a reaction, can be used to characterize the state of equilibrium.

Therefore, the correct option is B.

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The actual yield from the reaction was 92.4 grams. The answer is 2)

To find the actual yield of calcium chloride from the reaction, we can use the percent yield formula:

Percent Yield = (Actual Yield / Theoretical Yield) x 100%

We know that the theoretical yield of calcium chloride is 115 grams, and the percent yield is 80.34%. Rearranging the formula to solve for actual yield, we get:

Actual Yield = (Percent Yield / 100%) x Theoretical Yield

Plugging in the given values, we get:

Actual Yield = (80.34% / 100%) x 115 grams

Simplifying and solving for actual yield, we get:

Actual Yield = 92.4 grams

Therefore, the actual yield from the reaction was 92.4 grams, which is the second option in the given choices, i.e., option 2.

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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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(a) To determine if this can be done, we need to calculate the maximum amount of polysilicon that can be etched while keeping the SiO2 removal below 50 Å.

Let's assume the initial thickness of SiO2 is 1000 Å. Since the selectivity is 30, the maximum amount of polysilicon that can be etched is:

50 Å * (1/30) = 1.67 Å

Now, taking into account the overetch of 10%, the total amount of polysilicon that can be etched is:

1.67 Å / (1-0.1) = 1.85 Å

So, we need to etch a maximum of 1.85 Å of polysilicon.

The total thickness of the polysilicon and SiO2 layers is:

0.35 um + 1000 Å = 1350 Å

To find the required polysilicon uniformity, we can use the following equation:

(1 - uniformity) * 0.35 um = 1.85 Å

Solving for uniformity, we get:

uniformity = 1 - (1.85 Å / 0.35 um) = 0.9947 or 99.47%

So, the required polysilicon uniformity is 99.47%.

(b) To find the maximum polysilicon film thickness, we can use the same approach as above.

Let's assume the initial thickness of SiO2 is 1000 Å. The maximum amount of polysilicon that can be etched is:

50 Å * (1/30) = 1.67 Å

The total thickness of the polysilicon and SiO2 layers cannot be less than:

1000 Å + 50 Å + 1.67 Å = 1051.67 Å

So, the maximum polysilicon film thickness is:

1051.67 Å - 1000 Å = 51.67 Å

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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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Answer: Reduction

Explanation: When an atom or ion experiences a decrease in its oxidation state, it gains electrons.

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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To make a 1.75 M solution of Ba(NO₃)₂ with a volume of 2.50 L, you will need 1141.72 grams of the solute.

Firstly, we need to understand that Molarity (M) is defined as the number of moles of solute per liter of solution. Thus, we can use the formula:

Molarity (M) = (Number of moles of solute) / (Volume of solution in liters)

We have been given the volume of the solution (V) as 2.50 L and the Molarity (M) as 1.75 M. We need to find out the number of moles of solute (n) required to prepare this solution.

Rearranging the above formula, we get:

Number of moles of solute = Molarity × Volume of solution in liters

Substituting the given values, we get:

Number of moles of solute = 1.75 mol/L × 2.50 L = 4.375 mol

The molecular weight of Ba(NO₃)₂ can be calculated by adding the atomic weights of its constituents, which are Ba=137.33 g/mol, N=14.01 g/mol, O=16.00 g/mol. Thus, the molecular weight of Ba(NO₃)₂ comes out to be:

Molecular weight of Ba(NO₃)₂ = (137.33 g/mol) + 2 × (14.01 g/mol + 3 × 16.00 g/mol) = 261.34 g/mol

Now we can use the formula:

Mass of solute (in grams) = Number of moles of solute × Molecular weight of solute

Substituting the values, we get:

Mass of solute (in grams) = 4.375 mol × 261.34 g/mol = 1141.72 g

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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

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) The neutralization reaction between sulfuric acid and potassium hydroxide can be written as follows:

[tex]H_{2}SO_{4} + 2KOH - > K_{2}SO_{4} + 2H_{2}O[/tex]

b) Mass of [tex]K_{2}SO_{4}[/tex]= 104.6 g; mass of [tex]H_{2}O[/tex]= 5.4 g

c) Volume of 1.6 M [tex]H_{2}SO_{4}[/tex] needed to produce 250 g of [tex]K_{2}SO_{4}[/tex]= 0.896 L or 896 mL.

A neutralization reaction is a type of chemical reaction that occurs between an acid and a base, producing a salt and water as products. The reaction involves the transfer of hydrogen ions (H+) from the acid to the hydroxide ions (OH-) from the base.

The resulting salt is neutral because it is made up of cations from the base and anions from the acid. The reaction can be represented by the general equation: acid + base → salt + water.

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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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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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