please help show i need help​

To calculate the amount of aluminum produced from 9.00 tons of Al2O3, we need to use stoichiometry. First, we'll convert the mass of Al2O3 to moles, and then use the balanced chemical equation to find the moles of aluminum. Finally, we'll convert the moles of aluminum back to mass.

1. Convert mass of Al2O3 to moles:
9.00 tons = 9,000 kg
Molar mass of Al2O3 = (2 * 26.98) + (3 * 16.00) = 101.96 g/mol
9,000 kg * (1000 g/kg) = 9,000,000 g
moles of Al2O3 = 9,000,000 g / 101.96 g/mol = 88,258 moles

2. Use balanced chemical equation to find moles of aluminum:
The balanced chemical equation is:
2 Al2O3 → 4 Al + 3 O2
Using stoichiometry, we find the ratio of Al2O3 to Al is 2:4 or 1:2.
moles of Al = 88,258 moles Al2O3 * (2 moles Al / 1 mole Al2O3) = 176,516 moles

3. Convert moles of aluminum back to mass:
Molar mass of Al = 26.98 g/mol
Mass of Al = 176,516 moles * 26.98 g/mol = 4,762,984 g
Mass of Al in tons = 4,762,984 g / (1000 g/kg) / (1000 kg/ton) = 4.76 tons

So, 4.76 tons of aluminum can be produced from 9.00 tons of Al2O3.

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When an equilibrium system is put under stress, Le Chatelier's principle can be used to forecast changes in equilibrium concentrations.

Thus, The adjustments required to reach equilibrium might not be as obvious if we have a mixture of reactants and products that have not yet reached equilibrium.

In this situation, we can compare the Q and K values for the system to forecast changes.

By adding or withdrawing one or more of the reactants or products, an equilibrium chemical system can be momentarily moved out of equilibrium. After that, additional adjustments are made to the reactant and product concentrations in order to bring the system back to equilibrium.

Thus, When an equilibrium system is put under stress, Le Chatelier's principle can be used to forecast changes in equilibrium concentrations.

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In comparison to towns located inland, cities close to water features like lakes or oceans typically experience cooler summer temperatures.

Since water has a higher specific heat capacity than land, this is the case. The quantity of energy needed to raise a substance's temperature by a specific amount is known as its specific heat capacity. Compared to land, raising the temperature of water requires more energy because water has a higher specific heat capacity.

The summer sun warms both land and water, but due to land's lower specific heat capacity, land warms up more quickly than water. As a result, communities farther from water bodies tend to be hotter than cities closer to water.

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If a gaseous sample has a density of 0.978 g/L at 30 °C and 615 torr,  the molar mass of the compound is 24.8 g/mol.

To calculate the molar mass of the compound, we first need to calculate the number of moles present in the gaseous sample using the ideal gas law:

PV = nRT

Where P is the pressure in atm, V is the volume in L, n is the number of moles, R is the gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin.

Converting the given pressure of 615 torr to atm:
615 torr = 0.811 atm
Converting the given temperature of 30°C to Kelvin:
30°C + 273.15 = 303.15 K

Rounding off to two significant figures, we get:
P = 0.81 atm
T = 303 K

Now, rearranging the ideal gas law equation to solve for n:
n = PV/RT

Substituting the given values:

n = (0.978 g/L) x (1 L) / (0.081 atm x 0.0821 L atm/mol K x 303 K)

n = 0.0394 mol

Next, we can calculate the molar mass of the compound using the formula:

molar mass = mass / mole
molar mass = (0.978 g/L) x (1 L) / 0.0394 mol
molar mass = 24.8 g/mol
Therefore, 24.8 g/mol is the molar mass of the compound.

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The Gibbs free energy change for the given reaction at 25°C and the given concentrations is -25.5 kJ/mol

The Gibbs free energy change (∆G) of a reaction can be calculated using the equation:

∆G = -RT ln(K)

Where R is the gas constant (8.314 J/molK), T is the temperature in Kelvin, and K is the equilibrium constant.

The equilibrium constant (K) can be calculated from the acid dissociation constant (Ka) as:

K = [C] ÷ ([A] × [B])

Substituting the given values, we get:

K = (5.00 x 10⁻⁵) ÷ (1.50 x 1.00) = 3.33 x 10⁻⁵

Therefore,

∆G = - (8.314 J/molK) × (298 K) × ln(3.33 x 10⁻⁵)

= 25.5 kJ/mol

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_______________________________

2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g)

Moles of NA = Given Mass (g) ÷ Molecular Mass (g/mol)

= 27.5 ÷ 22.9897

= 1.196 mol

Moles of H2 Produced = Mol of NA × 1 mol H2 ÷ 2 Mol NA

= 1.196 × 1 ÷ 2

= 0.60 mol

Number of Molecules = Moles × Avogadro's Number

= 0.60 × 6.023 × 10²³ mol - 1

= 3.61 × 10²³

The Number of Molecules of Hydrogen Gas Produced When Added To Water Is 3.61 × 10²³

_________________________________

Answer:

The answer is 966.67 mm of Hg.

Explanation:

To solve this problem, we can use Boyle's Law which states that the pressure of a gas is inversely proportional to its volume when the temperature and the amount of gas are kept constant. The formula for Boyle's Law is:

P1V1 = P2V2

where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, and V2 is the final volume.

Using the given values:

P1 = 580 mmHg

V1 = 500 cm³

V2 = 300 cm³

We can solve for P2:

P1V1 = P2V2

580 mmHg x 500 cm³ = P2 x 300 cm³

290,000 mmHg·cm³ = P2 x 300 cm³

P2 = 290,000 mmHg·cm³ / 300 cm³

P2 = 966.67 mmHg (rounded to the nearest hundredth)

Therefore, the pressure when the volume is reduced to 300 cm³ is approximately 966.67 mmHg.

1. The star that has a temperature of approximately 9000 K and luminosity about 20 times greater than the Sun’s luminosity is Vega.

2. The type of star that is considered part of the main sequence is red dwarfs.

3. The star that is cooler than the Sun is Barnard’s Star.

4. The Sun is classified as a main sequence star.

5. The force most responsible for the formation of stars is gravity.

6. The star that has a higher luminosity and a lower temperature than the Sun is Aldebaran.

7. Compared to the temperature and luminosity of the star Polaris, the star Sirius is hotter and more luminous.

Answer:

The main difference between practical work inside and outside a laboratory is that the practical work inside the lab includes good equipment and chemicals which are very advanced and the practical outside a laboratory is more about the safety of life.

Explanation:

Practicals are set up at stations with lab equipment and chemicals, where students can learn, and researchers can experiment and find different new things.

Thus, the practical work inside the lab includes lab equipment and chemicals, and the practical outside a laboratory is more about conserving nature.  

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The number is converted to 60. 2 × 10²²

Index forms are simply described as mathematical forms that are used in the representation of numbers that are too small or too large in more convenient forms.

These index forms are also referred to as scientific notation or standard forms.

Some rules of index forms are;

From the information given, we have that;

0. 0602 × 10 ²⁵

Subtract three from the exponent value and move three spaces right, we have;

60. 2 × 10²⁵⁻³

60. 2 × 10²²

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Plants rely on their leaves to produce food through a process called photosynthesis. This involves converting light energy into organic compounds, like sugars, using chloroplasts that contain the pigment chlorophyll. By combining carbon dioxide and water with light energy, plants create glucose, which serves as an energy source and building material. Along with stomata, which help regulate gas exchange with the environment, the leaf acts as the plant's primary kitchen for food production.

The partial pressure of each gas are:

First, we shall determine the mole of 8.00 g of methane, CH₄ and 18.0 g of ethane, C₂H₆. Details below:

For methane, CH₄

Mole = mass / molar mass

Mole of CH₄ = 8 / 16

Mole of CH₄ = 0.5 mole

For ethane, C₂H₆

Mole = mass / molar mass

Mole of C₂H₆ = 18 / 30

Mole of C₂H₆ = 0.6 mole

Next, we shall determine the total mole. Details below:

PV = nRT

3.70 × 10 = n × 0.0821 × 293

Divide both sides by (0.0821 × 293)

n = (3.70 × 10) / (0.0821 × 293)

n = 1.52 mole

Finally, we shall determine the partial pressure of each gas. Details below:

For methane, CH₄

Partial pressure = (Mole / total mole) × total pressure

Partial pressure of CH₄ = (0.5 / 1.52) × 3.70

Partial pressure of CH₄ = 1.22 atm

For ethane, C₂H₆

Partial pressure = (Mole / total mole) × total pressure

Partial pressure of C₂H₆ = (0.6 / 1.52) × 3.70

Partial pressure of C₂H₆ = 1.46 atm

For propane, C₃H₈

Partial pressure of C₃H₈ = Total pressure - (Partial pressure of CH₄ + Partial pressure of C₂H₆)

Partial pressure of C₃H₈ = 3.7 - (1.22 + 1.46)

Partial pressure of C₃H₈ = 1.02 atm

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The overall charge on the structure is negative one (-1).

The central boron atom in the structure is bonded to two hydrogen atoms. Boron has three valence electrons, and it has formed only two bonds, so it has a formal charge of +1.

Each of the hydrogen atoms has one valence electron, and each is bonded to the boron atom, so each hydrogen atom has a formal charge of -1. The sum of the formal charges in the structure is equal to the charge of the ion, which is -2. Adding up the formal charges of the atoms, we get:

B: +1

H: -1 (two times)

Overall charge = sum of formal charges = +1 - 1 - 1 = -1

Therefore, the overall charge on the structure is negative one (-1).

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The total mass of products obtained when 130 g of zinc react completely with HCl is 274 g (3rd option)

First, we shall determine the mass of each product obtained. Details below:

For ZnCl₂

2HCl + Zn -> ZnCl₂ + H₂

From the balanced equation above,

65 g of Zn reacted to produce 135 g of ZnCl₂

Therefore,

130 g of Zn will react to produce = (130 × 135) / 65 = 270 g of ZnCl₂

Thus, the mass of ZnCl₂ obtained is 270 g

For H₂

2HCl + Zn -> ZnCl₂ + H₂

From the balanced equation above,

65 g of Zn reacted to produce 2 g of H₂

Therefore,

130 g of Zn will react to produce = (130 × 2) / 65 = 4 g of H₂

Thus, the mass of H₂ obtained is 4 g

Finally, we shall determine the total mass of the product produced. Details below:

Total mass of product = mass of ZnCl₂ + mass of H₂

Total mass of product = 270 + 4

Total mass of product = 274 g (3rd option)

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The information given can be used to construct the empirical formula for a tin oxide. We must first determine the mass of tin in the sample. This may be achieved by deducting the mass of the crucible, cover, and sample (21.76 g) from the mass of the empty crucible and cover (19.66 g).

This gives us a mass of 2.10 g of tin in the sample. The mass of oxygen in the sample must then be determined. To achieve this, we must deduct the mass of the crucible, cover, and sample (21.76 g) from the mass of the same components (22.29 g) prior to protracted heating. This provides us with an oxygen mass of 0.53 g.

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The value of K at equilibrium, for the reaction is 0.0049

First, we shall list out the given parameters from the question. This is shown below:

Equilibrium constant is defined as:

Equilibrium constant = [Product]ᵐ / [Reactant]ⁿ

Where

With the above formula, we can obtain the equilibrium constant, K as follow:

Equilibrium constant, K = [B₂][AC] / [AB₂C]

K = (0.007 × 0.0118) / 0.0168

K = 0.0049

Thus, the equilibrium constant, K for the reaction is 0.0049

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We can use the combined gas law to determine the volume of the balloon at a higher altitude. The combined gas law relates the pressure, volume, and temperature of a gas:

(P1 x V1) / T1 = (P2 x V2) / T2

where P1, V1, and T1 are the pressure, volume, and temperature of the gas at the initial state, and P2, V2, and T2 are the pressure, volume, and temperature of the gas at the final state.

We are given the initial pressure (P1 = 761 mmHg), volume (V1 = 56.0 L), and temperature (T1 = 23.1 °C = 296.25 K) of the gas, and the final pressure (P2 = 0.0772 atm), and temperature (T2 = -6.97 °C = 266.18 K) of the gas. We can solve for V2, the final volume of the gas:

(P1 x V1) / T1 = (P2 x V2) / T2

V2 = (P1 x V1 x T2) / (P2 x T1)

V2 = (761 mmHg x 56.0 L x 266.18 K) / (0.0772 atm x 296.25 K)

V2 = 2,040 L (rounded to three significant figures)

Therefore, the volume of the weather balloon at the higher altitude is approximately 2,040 L.

15.66% of the first cation is still in solution as the second cation is just beginning to precipitate.

One of the three main nutrients that are most frequently used in fertilisers is phosphorous, which is obtained from processing phosphate rock (the other two are nitrogen and potassium).

You can also make phosphoric acid into phosphoric acids, which are utilised in everything from food and skincare to animal feed and electronics. Over the course of millions of years, organic matter accumulates to form the sedimentary rock known as phosphate.

When [tex]Na_{3}Po_{4}[/tex] added to the solution of Ca and Mg, [tex]Ca_{3}(Po_{4})_{2}[/tex] and [tex]Ag_{3}Po_{4}[/tex] are formed.

Ksp of [tex]Ca_{3}(Po_{4})_{2} = 2.07*10^{-33}[/tex]

Ksp of [tex]Ag_{3}Po_{4} = 0.09*10^{-17}[/tex]

Concentration of [tex][Ca^{2+}][/tex] = 0.040 M

Concentration of [tex][Ag^{+}][/tex] = 0.0990 M

[tex]Ag_{3} Po_{4} - > 3Ag^{+} + Po_{4}^{3-}[/tex]

Ksp = [tex][Ag^{+}]^{3} [Po4^{3-}][/tex]

[tex]0.09*10^{-17} = (0.099)^{3} [Po_{3-}][/tex]

[tex][Po_{3-}] = 9.16*10^{-14}M[/tex]

[tex]Ksp = [Ca^{2+}]^{3} [Po_{3-}][/tex]

[tex]2.07*10^{-33} = (0.040)^{3} [Po_{4}^{3-}]^{2}[/tex]

[tex][Po_{4}^{3-}] = 5.68*10^{-15} M[/tex]

[tex][Po_{4}^{3-}][/tex] is smaller in [tex]Ca_{3}(Po_{4})_{2}[/tex]

[tex]Ca_{3}(Po_{4})_{2}[/tex] will start precipitating first

[tex]Ksp = [Ca^{2+}]^{3} [Po_{4}^{3-}]^{2}[/tex]

[tex]2.07*10^{-33} = [Ca^{2+}]^{3} (9.16*10^{-14})^{2}[/tex]

[tex][Ca^{2+}] = 6.27*10^{-3} M[/tex]

[tex]\%\ of\ Ca^{2+}[/tex] remaining [tex]= 6.27*10^{-3}/0.040 * 100[/tex]

= 15.66 %

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Initially, there was a 266.8 mm Hg pressure.

solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas sample. The formula is:
(P1 × V1) ÷ (T1) = (P2 × V2) ÷ (T2)
where P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, and T1 and T2 are the initial and final temperatures.

We are given that:
- V1 = 436 mL
- V2 = 612 mL
- T1 = 24 C + 273.15 = 297.15 K (convert from Celsius to Kelvin)
- T2 = 97 C + 273.15 = 370.15 K
- P2 = 526 mm Hg

We want to find P1, the original pressure.
Plugging in the values, we get:
(P1 × 436 mL) ÷ (297.15 K) = (526 mm Hg × 612 mL) ÷ (370.15 K)

Solving for P1, we get:
P1 = (526 mm Hg × 612 mL × 297.15 K) ÷ (436 mL × 370.15 K) = 266.8 mm Hg

Therefore, the original pressure was 266.8 mm Hg.

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Answer:

Explanation:

Here are the labels for each situation:

1.Usually conducts electricity & heat well - Metal (1)

2.At room temperature these are gases or liquids - Nonmetal (2)

3.Will lose valance electrons to form compounds - Metal (1)

4.Can be used as semiconductors - Metalloid (3)

5.Will gain valance electrons to form compounds - Nonmetal (2)

The maximum mass of water that can be produced by the reaction is 43.3 g, rounded to three significant figures.

The balanced chemical equation for the reaction between butane and oxygen is:

C4H10 + 13/2 O2 → 4 CO2 + 5 H2O

From the equation, we can see that 1 mole of butane reacts with 13/2 moles of oxygen to produce 5 moles of water.

moles of butane = 34. g / 58.12 g/mol = 0.585 mol

moles of oxygen = 200. g / 32.00 g/mol = 6.25 mol

Determining the limiting reactant.

butane : oxygen = 0.585 mol : 6.25 mol

= 0.0936 : 1.00

stoichiometric ratio = 1 : 13/2

= 0.7692 : 1.00

Since the actual ratio is lower than the stoichiometric ratio for oxygen, it is the limiting reactant.

The maximum amount of water that can be produced is determined by the amount of limiting reactant (oxygen).

moles of water = 5/13 * 6.25 mol

= 2.403 mol

Finally, we can convert the moles of water to grams:

mass of water = 2.403 mol * 18.015 g/mol

= 43.3 g

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Diffrences in water temperature in the ocean create movement because bodies of water at different temperatures have different densities.

Water that is colder is generally denser than water that is warmer, so when a body of water with colder, denser water is next to a body of water with warmer, less dense water, a density gradient is established. This gradient creates a difference in pressure between the two bodies of water, with the colder, denser water being at a higher pressure than the warmer, less dense water.

This difference in pressure creates a force that drives the movement of water from the denser, colder region to the less dense, warmer region. This movement of water is known as convection, and it can occur both vertically and horizontally in the ocean. Vertical convection occurs when differences in temperature cause water to rise or sink, while horizontal convection occurs when water moves laterally due to differences in temperature.

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missing options:

1. as water heats up, the atoms of water more faster.

2. warm water is pulled more by gravity than cold water.

3. warm and cold water mix and reach the same temperature.

4. bodies of water at different temperatures have different densities.

The pH of a 0.100M NaClO solution is 1.

pH, meaning power of hydrogen, is a measure of how acidic/basic a solution is. The range goes from 0 - 14, with 7 being neutral. pHs of less than 7 indicate acidity, whereas a pH of greater than 7 indicates a base.

pH is really a measure of the relative amount of free hydrogen and hydroxyl ions in the water. It can be estimated using the following formula;

pH = - log {H+}

Where;

pH = - log {0.100}

pH = 1

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The equilibrium constant is written as;

Keq = [tex][D]^2 [C]^5/[A] [B]^3[/tex]

The equilibrium constant's value is influenced by the reaction's chemical make-up and temperature.

The product of the product concentrations, each raised to the power of their stoichiometric coefficient, divided by the product of the reactant concentrations, each raised to the power of their stoichiometric coefficient, is known as the equilibrium constant.

The equilibrium constant is Keq = [tex][D]^2 [C]^5/[A] [B]^3.[/tex]

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The product that is formed first is product B and the product that will be the major product if given enough time would also be product B.

In an energy diagram, the vertical axis represents the overall energy of the reactants, while the horizontal axis is the ‘reaction coordinate’, tracing from left to right the progress of the reaction from starting compounds to final products.

The activation energy of the reaction can be shown on a diagram as the energy between the reactants that the transition state.

The product with lesser activation energy is the product that is formed first and the major product is decided by the stability of the product which depends on the energy. Lesser is the energy of the product, greater is its stability.

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A sunscreen made with Benzophenone-5 ([tex]C_1_4H_1_1NaO_6S[/tex]) would be expected to be more waterproof than benzophenone-3 ([tex]C_1_4H_1_2O_3[/tex]).

This is due to the presence of a sodium salt group (Na) and a sulfonic acid group ([tex]SO_3H[/tex] ) in benzophenone-5, which makes it more polar than benzophenone-3. Polar molecules interact more strongly with water molecules and are less likely to dissolve in nonpolar solvents such as oils.

Because sunscreen is designed to be water-resistant, the more polar benzophenone-5 should have stronger interactions with water and give more water resistance than benzophenone-3.

Moreover, the sulfonic acid group in benzophenone-5 may allow it to make stronger hydrogen bonds with water, increasing its water resistance even further.

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Answer:

solution resists changes in pH

Explanation:

the inherent property of buffers is to resist change to ph even when acids and bases are added. when the base is added, it is quickly neutralized by the conjugate acid, so the ph won't change.

Answer:

B: The solutions resists changes in pH.

Explanation:

Buffer reactions maintain stable pH of solutions.

The average mass of chromium is 52.1. Isotopic mass is defined as the average mass of all the isotopes of a specific element.

The average atomic mass of an element is referred to as the sum of the masses of its isotopes, each multiplied by its natural abundance which can be also explained as the decimal associated with the percent of atoms of that element that are of a given isotope. Average atomic mass is equal to f1M1 + f2M2 and so on. Hydrogen, chromium, lithium, cobalt, oxygen, boron, plutonium, and carbon are some examples.

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