Calculate how many moles of FeSO4 • 7H2O were added to the Erlenmeyer flask in trial 2

The equations (C + O2 → C2O), C (4Cu + O2 → 2Cu2O), and D (2H2 + O2 → 2H2O) demonstrate the law of conservation of mass. Option B.

The law of conservation of mass states that in a chemical reaction, the total mass of the reactants is equal to the total mass of the products. Let's analyze each equation to determine if it demonstrates the conservation of mass:

A Mg + S → MgS2:

This equation does not demonstrate the conservation of mass. The reactants contain one magnesium atom and one sulfur atom, while the product contains one magnesium atom and two sulfur atoms.

The number of atoms on the left side is not equal to the number of atoms on the right side, violating the law of conservation of mass.

B C + O2 → C2O:

This equation demonstrates the conservation of mass. The reactants contain one carbon atom and two oxygen atoms, while the product contains two carbon atoms and two oxygen atoms. The number of atoms on the left side is equal to the number of atoms on the right side, satisfying the law of conservation of mass.

C 4Cu + O2 → 2Cu2O:

This equation demonstrates the conservation of mass. The reactants contain four copper atoms and two oxygen atoms, while the product contains four copper atoms and two oxygen atoms.

The number of atoms on the left side is equal to the number of atoms on the right side, satisfying the law of conservation of mass.

D 2H2 + O2 → 2H2O:

This equation demonstrates the conservation of mass. The reactants contain four hydrogen atoms and two oxygen atoms, while the product contains four hydrogen atoms and two oxygen atoms. The number of atoms on the left side is equal to the number of atoms on the right side, satisfying the law of conservation of mass.

E H2SO4 + Zn → 4ZnSO + H2:

This equation does not demonstrate the conservation of mass. The reactants contain one sulfur atom, while the products contain four sulfur atoms.

The number of atoms on the left side is not equal to the number of atoms on the right side, violating the law of conservation of mass. So Option B is correct.

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Answer:

2.282 atm

P1V1/T1 = P2V2/T2

2.70atm / (50+273) = X/ 273

make x subject of formula

:. X = 2.28 atm

or 2.28 * 1.01 *10⁵ N/m²

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The number of moles of [tex]MnO_2[/tex] required is  3.345 moles.

In the given reaction, the balanced equation shows that for every 4 moles of [tex]MnO_2[/tex], 4 moles of [tex]KMnO_4[/tex] are produced. Therefore, we can use the stoichiometry of the reaction to calculate the moles of [tex]MnO_2[/tex] and the moles of [tex]KMnO_4[/tex]

Given:

Mass of [tex]MnO_2[/tex] = 291 g

Molar mass of[tex]MnO_2[/tex] = 87 g/mol

To find the moles of [tex]MnO_2[/tex], we use the formula:

Moles = Mass / Molar mass

Moles of [tex]MnO_2[/tex] = 291 g / 87 g/mol = 3.345 mol

Now, since the stoichiometry of the reaction tells us that the ratio of [tex]MnO_2[/tex]to [tex]KMnO_4[/tex] is 4:4, we can conclude that 3.345 moles of [tex]MnO_2[/tex]will produce an equal number of moles of [tex]KMnO_4[/tex]

Therefore, the moles of [tex]KMnO_4[/tex] produced will also be 3.345 mol.

However, the question asks for the moles of NaOH, which is not directly related to the given reaction. We cannot determine the moles of NaOH based on the information provided.

To find the moles of NaOH, we would need additional information or another relevant equation that includes NaOH.

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To calculate the pH of a solution of NaOH (sodium hydroxide), we need to consider that NaOH is a strong base that dissociates completely in water, producing hydroxide ions (OH⁻).

Given:

Concentration of NaOH = 0.005 M

Since NaOH dissociates into one hydroxide ion (OH⁻) per molecule, we can determine the concentration of hydroxide ions in the solution, which will allow us to calculate the pOH. Then, we can convert the pOH to pH using the relationship: pH + pOH = 14.

1. Calculate the concentration of hydroxide ions (OH⁻):

The concentration of OH⁻ ions will be the same as the concentration of NaOH since NaOH dissociates completely.

Concentration of OH⁻ = 0.005 M

2. Calculate the pOH:

pOH = -log[OH⁻]

pOH = -log(0.005)

Using logarithm properties, we can determine the pOH value:

pOH = -log(0.005)

pOH = -(-2.301)

pOH = 2.301

3. Calculate the pH:

pH = 14 - pOH

pH = 14 - 2.301

pH ≈ 11.699

Therefore, the pH of a 0.005 M NaOH solution is approximately 11.699.

The pH of a 0.005 M concentration of NaOH ( sodium hydroxide ) solution is approximately 11.70.

The pH of a solution is defined as the logarithm of the reciprocal of the hydrogen ion concentration  [H+] of the given solution.

From the formula;

pH = -log[ H⁺ ]

pOH = -log[ OH⁻ ]

pH + pOH = 14

Given that; the concentration of solution (molarity) ( OH⁻ ) is 0.005 M.

First, we determine the pOH.

pOH = -log[ OH⁻ ]

Plug in ( OH⁻ ) = 0.005

pOH = -log[ 0.005 ]

pOH = 2.30

Now, plug pOH = 2.30 into the above formula and solve for the pH:

pH + pOH = 14

pH + 2.30 = 14

Subtract 2.30 from both sides:

pH + 2.30 - 2.30 = 14 - 2.30

pH = 14 - 2.30

pH = 11.7

Therefore, the pH of the solution is 11.7.

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We can use the following formula to determine how many grams of AgNO3 are needed to make a 0.30 M solution with a volume of 750 ml:

moles = volume (L) x concentration (M)

The volume provided must first be converted from milliliters to liters:

Volume = 750 ml ÷ 1000 ml/L = 0.75 L

Now we can find the molarity of AgNO3:

moles = 0.30 M × 0.75 L = 0.225 moles

To find the grams of AgNO3, we need to use the molar mass of AgNO3, which is calculated as follows:

Ag: 1 atom × 107.87 g/mol = 107.87 g/mol

N: 1 atom × 14.01 g/mol = 14.01 g/mol

O: 3 atoms × 16.00 g/mol = 48.00 g/mol

Total molar mass of AgNO3:

107.87 g/mol + 14.01 g/mol + 48.00 g/mol = 169.88 g/mol

Now, we can calculate the grams of AgNO3 needed:

grams = moles × molar mass

grams = 0.225 moles × 169.88 g/mol = 38.22 grams

Therefore, approximately 38.22 grams of AgNO3 are needed to prepare 750 ml of a 0.30 M solution.

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The volume of the gas at standard temperature and pressure conditions is approximately 12.77 liters.

To find the volume of the gas at STP, we can use the combined gas law:

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]

Note that: at STP (Standard Temperature and Pressure) is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atm.

Given that:

P₁ = initial pressure = 4.71 atm

V₁ = initial volume = 2.99 L

T₁ = initial temperature = 28.10 °C = ( 28.10 + 273.15 ) = 301.25 K

P₂ = final pressure (STP pressure ) = 1 atm

T₂ = final temperature (STP temperature)  = 0°C = 273.15 K

V₂ = final volume = ?

Substituting the given values into the formula:

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\\\\\frac{4.71\ *\ 2.99 }{301.25} = \frac{1\ *\ V_2}{273.15 }\\\\V_2 = 12.77\ L[/tex]

Therefore, the final volume is 12.77 litres.

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The boiling point of a solution is influenced by the presence of solute particles, which can cause a change in the boiling point compared to the pure solvent. This phenomenon is known as boiling point elevation.

The magnitude of boiling point elevation depends on the concentration of the solute and the nature of the solute particles. In general, the greater the concentration of solute particles, the greater the boiling point elevation.

Comparing a 0.5m sodium chloride (NaCl) solution to a 0.3m aluminum sulfate ([tex]Al_2(SO_4)_3[/tex]) solution, we can determine the relative boiling point elevation.

Sodium chloride (NaCl) dissociates into two ions in solution (Na+ and Cl-), while aluminum sulfate ([tex]Al_2(SO_4)_3[/tex])dissociates into three ions (2[tex]Al_3[/tex]+ and 3[tex]SO_4[/tex]2-). This means that the aluminum sulfate solution will have a greater concentration of solute particles per mole than the sodium chloride solution.

Therefore, the boiling point of the 0.5m sodium chloride solution will be lower than the boiling point of the 0.3m aluminum sulfate solution.

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If a compound is found to contain 3.622 % carbon and 96.38 % bromine by weight. The molecular formula for the compound is CBr4.

First, get the empirical formula in order to calculate the molecular formula of the chemical. The empirical formula shows the atoms of a compound in their most straightforward whole number ratio.

Suppose 100 grams of the substance. To determine the mass of carbon and bromine in the compound using the provided percentages.

Mass of C = 3.622% of 100g

= 3.622g

Mass of Br = 96.38% of 100g

= 96.38g

The next step is to determine the atomic masses of carbon and bromine in order to determine the number of moles for each.

Atomic mass of carbon = 12.01 g/mol

Atomic mass of bromine = 79.90 g/mol

Moles of C = (mass of carbon) / (atomic mass of carbon)

= 3.622g / 12.01 g/mol

= 0.3016 mol

Moles of Br = (mass of bromine) / (atomic mass of bromine)

= 96.38g / 79.90 g/mol

= 1.205 mol

Divide the moles of each element by the fewest number of moles obtained, in this case the moles of carbon, to arrive at the empirical formula.

Empirical formula ratio:

C: (0.3016 mol) / (0.3016 mol)

= 1

Br: (1.205 mol) / (0.3016 mol)

= 4

The empirical formula for the compound is C₁Br4.

To determine the molecular formula, it is required to know the molecular weight of the compound. The molecular weight is  331.61 g/mol.

To find the number of empirical formula units in the molecular formula, divide the molecular weight by the empirical formula weight.

Empirical formula weight:

C = 12.01 g/mol × 1

= 12.01 g/mol

Br= 79.90 g/mol × 4

= 319.60 g/mol

Empirical formula weight = 12.01 + 319.60

= 331.61 g/mol

Now find the number of empirical formula units in the molecular formula:

Number of empirical formula units

= (molecular weight) ÷ (empirical formula weight)

Number of empirical formula units

= 331.61 g/mol / 331.61 g/mol

= 1

The number of empirical formula units is 1, the empirical formula C₁Br4 is would be  the molecular formula for this compound.

Thus, the molecular formula for the compound is CBr₄.

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When the temperature is lowered to -76 °C and the pressure is 0.561 atm, the volume of the balloon will be approximately 179 L.

To solve this problem, we can use the combined gas law equation, which relates the initial and final conditions of a gas sample:

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

where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.

Substituting the given values:

(P1 * 266 L) / (38 + 273.15 K) = (0.561 atm * V2) / (-76 + 273.15 K)

Simplifying the equation:

(0.995 atm * 266 L) / (311.15 K) = (0.561 atm * V2) / (197.15 K)

Solving for V2:

V2 = [(0.995 atm * 266 L) / (311.15 K)] * (197.15 K / 0.561 atm)

V2 ≈ 179 L

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Balance the following chemical equations:

1. N2 + 3 H2 → 2 NH3

Ex: Synthesis reaction

2. 2 KClO3 → 2 KCl + 3 O2

Single Replacement reaction

3. 2 NaF + ZnCl2 → ZnF2 + 2 NaCl

Decomposition reaction

4. 2 AlBr3 + 3 Ca(OH)2 → Al2(OH)6 + 6 CaBr2

Double Replacement reaction

5. 2 H2 + O2 → 2 H2O

Combustion reaction

6. 2 AgNO3 + MgCl2 → 2 AgCl + Mg(NO3)2

Synthesis reaction

7. 2 Al + 6 HCl → 2 AlCl3 + 3 H2

Decomposition reaction

8. C3H8 + 5 O2 → 3 CO2 + 4 H2O

Combustion reaction

9. 2 FeCl3 + 6 NaOH → Fe2O3 + 6 NaCl + 3 H2O

Double Replacement reaction

10. 4 P + 5 O2 → 2 P2O5

Synthesis reaction

11. 2 Na + 2 H2O → 2 NaOH + H2

Single Replacement reaction

12. 2 Ag2O → 4 Ag + O2

Decomposition reaction

13. C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

Combustion reaction

14. 2 KBr + MgCl2 → 2 KCl + MgBr2

Double Replacement reaction

15. 2 HNO3 + Ba(OH)2 → Ba(NO3)2 + 2 H2O

Double Replacement reaction

16. C5H12 + 8 O2 → 5 CO2 + 6 H2O

Combustion reaction

17. 4 Al + 3 O2 → 2 Al2O3

Synthesis reaction

18. Fe2O3 + 2 Al → 2 Fe + Al2O3

Single Replacement reaction

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At the usual working conditions of 101.5 kPa and 21°C, the chemical oxygen generator system would generate roughly 220.84 litres of oxygen using 500 g of LiClO4.

We may use the ideal gas law and stoichiometry to calculate how many litres of oxygen are created by the chemical oxygen generator system employing 500 g of LiClO4.

We must first determine the moles of LiClO4. LiClO4 has a molar mass of approximately 106.39 g/mol. As a result, 4.704 mol of LiClO4 are produced from 500 g of LiClO4 using the formula: 500 g / 106.39 g/mol

We can see from the chemical equation that 1 mole of LiClO4 results in 2 moles of O2. 4.704 mol of LiClO4 will therefore result in:

2 mol O2 / 1 mol LiClO4 4.704 mol LiClO4 = 9.408 mol O2

The moles of O2 under the specified conditions must then be converted to volume. The ideal gas law, which goes as follows:

PV = nRT

Where:

P = pressure = 101.5 kPa

V = volume (in liters)

n = moles of gas = 9.408 mol

R = ideal gas constant = 8.314 J/(mol·K)

T = temperature = 21°C = 294 K (converted to Kelvin)

Rearranging the equation to solve for V:

V = (nRT) / P

V = (9.408 mol × 8.314 J/(mol·K) × 294 K) / (101.5 kPa × 1000 Pa/kPa)

Simplifying the units:

V = (9.408 × 8.314 × 294) / 101.5

V ≈ 220.84 liters

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Answer:

[tex]\huge\boxed{\sf 1.869\ L}[/tex]

Explanation:

Given that,

Mass = m = 2.67 g

Molar mass (O₂) = 16 × 2 = 32 g/mol

We know that,

No. of moles = 2.67 / 32

No. of moles = 0.08 moles

Also, we know that:

1 × 0.08 moles of O₂ at STP = 22.4 × 0.08 L

So, the volume of 0.08 moles of oxygen gas at STP will be 1.869 L.

[tex]\rule[225]{225}{2}[/tex]

The molarity of the solution, if 44g of [tex]CaCl_{2}[/tex] is dissolved in 95 ml of water is 4.1733 M

To calculate the molarity (M) of a solution, we use the formula:

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

As per the question:

Mass of [tex]CaCl_{2}[/tex] = 44 g

Volume of water = 95 mL = 0.095 L

To find molarity, we need to determine the number of moles of [tex]CaCl_{2}[/tex] by dividing the given mass by its molar mass.

Molar mass of [tex]CaCl_{2}[/tex] = 40.08 g/mol (for [tex]Ca[/tex]) + (2 × 35.45 g/mol) (for [tex]Cl[/tex])

Molar mass of [tex]CaCl_{2}[/tex] = 110.98 g/mol

Number of moles of [tex]CaCl_{2}[/tex] = Mass of [tex]CaCl_{2}[/tex] / Molar mass of [tex]CaCl_{2}[/tex]

Number of moles of [tex]CaCl_{2}[/tex] = 44 g / 110.98 g/mol

Number of moles of [tex]CaCl_{2}[/tex] ≈ 0.3965 mol

Now, to calculate the molarity of the solution, we can use this formula:

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

Molarity (M) = 0.3965 mol / 0.095 L

Molarity (M) ≈ 4.1733 M

Therefore, the molarity of the solution is approximately 4.1733 M when 44 g of [tex]CaCl_{2}[/tex] is dissolved in 95 mL of water.

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Option B of 1 is correct, option B of 2 is correct, option B of 3 is correct, and option A of 4 is correct. 1. Since the current energy level (200 J) is less than the minimum energy required (240 J), the reaction will not proceed.

2. Increasing the temperature by 20 degrees Celsius adds 50 J of energy. However, even with this additional energy, the total energy level (200 J + 50 J = 250 J) is still less than the minimum energy required (240 J). Therefore, the reaction will not proceed.

3. Increasing the concentration by 1.5 M adds 30 J of energy. However, the energy level contribution from concentration is usually negligible compared to temperature. Therefore, even with this additional energy, the total energy level will still be less than the minimum energy required. The reaction will not proceed.

4. Increasing the temperature by 20 degrees Celsius adds 50 J of energy, and increasing the concentration by 1.5 M adds 30 J of energy. The total energy added is 50 J + 30 J = 80 J. The current energy level (200 J + 80 J = 280 J) is now higher than the minimum energy required (240 J). Therefore, with these additional changes, the reaction will proceed.

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There are approximately 2.8939 x[tex]10^2^4[/tex] carbon atoms in 0.37 mol of procaine.  The molecular formula of procaine (C₁₃H₂₀N₂₀₂), one can see that there are 13 carbon atoms (C13) in one molecule of procaine.

Avogadro's number (6.022 x [tex]10^2^3[/tex]) represents the number of particles (atoms, molecules, or formula units) in one mole of a substance

The number of molecules of procaine in 0.37 mol:

Number of molecules = 0.37 mol x (6.022 x[tex]10^2^3[/tex] molecules/mol)

Number of carbon atoms = Number of molecules x 13 carbon atoms/molecule

Number of molecules = 0.37 mol x (6.022 x [tex]10^2^3[/tex]molecules/mol)

= 2.22614 x [tex]10^2^3[/tex]molecules

Number of carbon atoms = 2.22614 x [tex]10^2^3[/tex] molecules x 13 carbon atoms/molecule

= 2.8939 x [tex]10^2^4[/tex]carbon atoms

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The dissociation of sodium chloride in water allows it to act as an electrolyte, conducting electricity through the movement of ions.

The bonding between sodium and chlorine is classified as ionic bonding. In this type of bonding, electrons are transferred from one atom to another, resulting in the formation of ions. Sodium (Na) readily donates one electron from its outermost shell to achieve a stable electron configuration, while chlorine (Cl) accepts this electron to fill its outermost shell. As a result, sodium forms a positively charged ion (Na+), known as a cation, while chlorine forms a negatively charged ion (Cl-), known as an anion. The electrostatic attraction between these oppositely charged ions creates a strong bond between sodium and chlorine, forming sodium chloride (NaCl) as the resulting compound.

When sodium chloride is dissolved in water, the compound dissociates into separate sodium cations and chloride anions. Water molecules, which have a polar nature, surround the individual ions due to their attraction to opposite charges. This process is called hydration or solvation. The water molecules effectively separate the sodium and chloride ions, leading to the compound's dissolution. This is because water molecules have a higher affinity for the charged ions compared to the ionic bond holding the compound together.

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The underlying foundation of all these emissions can be traced back to the burning of fossil fuels, making it the dominant and pervasive cause of human-induced greenhouse gas emissions.

The human activity that is intricately connected to every piece of the pie chart representing greenhouse gas emissions is the burning of fossil fuels. Fossil fuel combustion, including coal, oil, and natural gas, is the primary contributor to the rise in greenhouse gas concentrations over the past 150 years. When these fuels are burned for energy generation, transportation, industrial processes, and residential use, carbon dioxide (CO2) is released into the atmosphere. CO2 is the most significant greenhouse gas, accounting for approximately 75% of total emissions. The other greenhouse gases, such as methane (CH4) and nitrous oxide (N2O), are also released as byproducts of certain human activities, such as agriculture, deforestation, and waste management.

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The following chemical equations must be balanced:

1. N2 + 3 H2 → 2 NH3

Type: Synthesis reaction

2. 2 KClO3 → 2 KCl + 3 O2

Type: Single Replacement reaction

3. 2 NaF + ZnCl2 → ZnF2 + 2 NaCl

Type- Decomposition reaction

4. 2 AlBr3 + 3 Ca(OH)2 → Al2(OH)6 + 6 CaBr2

Type- Double Replacement reaction

5. 2 H2 + O2 → 2 H2O

Type: Combustion reaction

6. 2 AgNO3 + MgCl2 → 2 AgCl + Mg(NO3)2

Type: Synthesis reaction

7. 2 Al + 6 HCl → 2 AlCl3 + 3 H2

Type: Decomposition reaction

8. C3H8 + 5 O2 → 3 CO2 + 4 H2O

Type: Combustion reaction

9. 2 FeCl3 + 6 NaOH → Fe2O3 + 6 NaCl + 3 H2O

Type: Double Replacement reaction

10. 4 P + 5 O2 → 2 P2O5

Type: Synthesis reaction

11. 2 Na + 2 H2O → 2 NaOH + H2

Type: Single Replacement reaction

12. 2 Ag2O → 4 Ag + O2

Type: Decomposition reaction

13. C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

Type: Combustion reaction

14. 2 KBr + MgCl2 → 2 KCl + MgBr2

Type: Double Replacement reaction

15. 2 HNO3 + Ba(OH)2 → Ba(NO3)2 + 2 H2O

Type: Double Replacement reaction

16. C5H12 + 8 O2 → 5 CO2 + 6 H2O

Type: Combustion reaction

17. 4 Al + 3 O2 → 2 Al2O3

Type: Synthesis reaction

18. Fe2O3 + 2 Al → 2 Fe + Al2O3

Type: Single Replacement reaction

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The amount of time it will take for the reaction to consume 6.02 x 10²³ molecules of reactant is 3.67 × 10² seconds.

The amount of time it will take for a molecule to react can be calculated by dividing the number of molecules in the substance by the rate of time as follows;

Time taken = no of molecules ÷ no of molecules/seconds

According to this question, if a chemical reaction consumes reactants at a steady rate of 1.64 x 10²¹ molecules per second, the amount of time it will take for the reaction to consume 6.02 x 10²³ molecules of reactant is as follows!

Time = 6.02 x 10²³ molecules ÷ 1.64 x 10²¹ molecules per second

Time = 3.67 × 10² seconds

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An impact of the specific heat of the water on the planet is that islands and coastal places have moderately pleasant climates. Therefore, option A is correct.

The specific heat of water is relatively high compared to other substances. This means that water requires a significant amount of heat energy to increase its temperature. As a result, water has a stabilizing effect on the climate of coastal and island regions.

The high specific heat of the water helps to moderate temperature changes, resulting in milder and more pleasant climates in these areas.

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Answer:

Overall, energy is released in exothermic and absorbed in endothermic reactions. Therefore, "a" is the correct statement. The other statements are all incorrect.

Explanation:

u suck

Answer:

85.99 grams of AgCl will be formed.

Explanation:

To determine the grams of AgCl formed in the reaction, we need to consider the stoichiometry of the reaction between silver nitrate (AgNO3) and barium chloride (BaCl2):

AgNO3 + BaCl2 -> AgCl + Ba(NO3)2

The balanced equation shows that the molar ratio between AgNO3 and AgCl is 1:1. This means that 1 mole of AgNO3 produces 1 mole of AgCl.

Given:

Volume of BaCl2 solution = 1500 ml = 1.5 L

Molarity of BaCl2 solution = 0.400 M

First, we need to calculate the number of moles of BaCl2 present in the solution:

moles of BaCl2 = volume of BaCl2 solution * molarity of BaCl2 solution

= 1.5 L * 0.400 M

= 0.600 moles

Since the molar ratio between BaCl2 and AgNO3 is 1:1, the number of moles of AgNO3 needed for complete reaction is also 0.600 moles.

Now, using the molar mass of AgCl, which is 143.32 g/mol, we can calculate the grams of AgCl formed:

grams of AgCl = moles of AgNO3 * molar mass of AgCl

= 0.600 moles * 143.32 g/mol

= 85.99 grams

Therefore, by adding enough AgNO3 to react fully with the 1500 ml of 0.400 M BaCl2 solution, approximately 85.99 grams of AgCl will be formed.

Theoretically, if 20 grams of hydrogen reacts then 112.5 grams of ammonia is produced.

The balanced chemical equation can be given as:

N₂+3H₂→ 2NH₃

From stoichiometry, 2 mol of NH₃is produced from 3 mol of H₂

5 mol of NH₃ will be produced from = 3/2×5 = 7.5 mol of H₂

∴mass of H₂=7.5×2= 15gm of H₂.

Excess reagents are those reactants in a chemical reaction that are not exhausted at the end of the reaction. A completely exhausted or reacted reagent is called a limiting reagent because its amount limits the number of products formed. In this reaction, the excess reagent is Nitrogen as 35 grams of nitrogen and 15 grams of hydrogen react to produce 34 grams of ammonia.

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Answer:

Thermal Conduction

Explanation:

Answer:

Particulate matter pollution decreases the amount of sunlight plants can absorb for photosynthesis.

Explanation:

These are 10 unique questions you can ask your fellow classmates about the things they have learned in your special ed classroom:

A special education classroom is a classroom designed to meet the needs of students with disabilities. These classrooms are staffed by specially trained teachers who are able to provide individualized instruction and support to students with a variety of disabilities.

These questions are designed to get your classmates thinking about the things they have learned in your special ed classroom and how those things have impacted them. The answers to these questions can be used to create a powerful and informative ad for your classroom.

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The extraction of iron from haematite ore involves a process called reduction. Reduction is the chemical reaction in which oxygen is removed from a compound, resulting in the formation of a new substance.

In the case of haematite, the reduction process involves removing the oxygen from the iron oxide (Fe2O3) to obtain elemental iron (Fe). This is typically achieved through a process called smelting, which is carried out in a blast furnace. Before the haematite ore is reduced in a blast furnace, it needs to undergo a series of steps to separate impurities and prepare it for the reduction process. The first step is crushing and grinding the ore into smaller particles. This is done to increase the surface area of the ore, allowing for better contact with the reducing agent. After crushing and grinding, the ore is then subjected to a process called beneficiation, where it is separated from gangue materials and other impurities.

Beneficiation techniques vary, but commonly involve processes such as gravity separation, magnetic separation, and flotation. These methods exploit the differences in physical and chemical properties between the haematite ore and the impurities, allowing for their separation. Once the ore is purified and separated, it is ready to be reduced in a blast furnace, where the smelting process takes place.

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