What do not have true leaves or roots

A silver bar with the mass of the 300 grams is heated from the 30 °C to 55 °C. The amount heat does the silver bar absorb in the joules is 1762.5 J.

The mass of the silver bar = 300 g

The initial temperature = 30 °C

The final temperature = 55 °C

The heat energy is expressed as :

Q = mc ΔT

Where,

The m is mass of the silver bar = 300 g

The c is the specific heat capacity = 0.235 J/g °C

The ΔT is the change in the temperature = final temperature - initial temperature

The ΔT is the change in the temperature = 55 °C - 30 °C

The ΔT is the change in the temperature = 25 °C

The heat energy, Q = 300 × 0.235 × 25

The heat energy, Q = 1762.5 J

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The given reaction equation tells us that for every 1 mol of C₅H₁₂, 5 moles of CO₂ will be produced. Since 8.25 mol of C₅H₁₂ is given, 8.25 mol C₅H₁₂ x 5 moles CO₂/1 mol C₅H₁₂ = 41.25 moles CO₂ will be produced.

Reaction is the process of responding to an event or stimulus in a particular way. It can occur at the physical, cognitive, or emotional level. Physically, a reaction could be as simple as a reflex or as complex as a multi-step process. Cognitively, it could involve forming a judgment or understanding. Emotionally, it could involve feelings of fear, shock, anger, or joy. In the context of science, reactions are often chemical or physical processes that involve the conversion of one set of substances into another.

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Answer:

56.11 g/mol

Explanation:

To determine the molar mass of potassium hydroxide, we need to find the atomic mass of each element in the compound and add them up.

The atomic mass of potassium (K) is 39.10 g/mol, the atomic mass of oxygen (O) is 16.00 g/mol, and the atomic mass of hydrogen (H) is 1.01 g/mol.

So, the molar mass of potassium hydroxide (KOH) is:

Molar mass of K = 39.10 g/mol

Molar mass of O = 16.00 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of KOH = Molar mass of K + Molar mass of O + Molar mass of H

= 39.10 g/mol + 16.00 g/mol + 1.01 g/mol

= 56.11 g/mol

Therefore, the molar mass of potassium hydroxide (KOH) is 56.11 g/mol.

Under the specified conditions, the cell potential of the given galvanic cell is 3.11 V.

The standard reduction potentials for the half-reactions involved in the cell reaction are:

Ti²⁺(aq) + 2e- ⇆ Ti(s) E° = -1.63 V

Au³⁺(aq) + 3e- ⇆ Au(s) E° = +1.50 V

The cell potential (Ecell) is given by:

Ecell = E°cathode - E°anode

where E°cathode = standard reduction potential of the cathode (reduction half-reaction) and E°anode = standard reduction potential of the anode (oxidation half-reaction).

In this case, Ti²⁺ is oxidized (anode) and Au³⁺ is reduced (cathode). Therefore:

E°anode = -1.63 V

E°cathode = +1.50 V

So, Ecell = +1.50 V - (-1.63 V) = +3.13 V

The Nernst equation can be used to calculate the cell potential (Ecell) under non-standard conditions:

Ecell = E°cell - (RT/nF) ln(Q)

where R = gas constant (8.314 J/(molK)), T = temperature in Kelvin (25 °C = 298 K), n = number of electrons transferred in the balanced equation (3 in this case), F = Faraday constant (96,485 C/mol), and Q = reaction quotient.

For the given concentrations:

[Ti²⁺] = 0.00140 M

[Au³⁺] = 0.887 M

The reaction quotient Q can be written as:

Q = ([Ti²⁺]³/[Au³⁺]²)

Substituting the values into the Nernst equation:

Ecell = E°cell - (RT/nF) ln([Ti²⁺]³/[Au³⁺]²)

Ecell = 3.13 V - (8.314 J/(molK) × 298 K / (3 × 96,485 C/mol)) ln(0.00140³/0.887²)

Ecell = 3.13 V - 0.0217 V

Ecell = 3.11 V

Therefore, the cell potential for the given galvanic cell under the given conditions is 3.11 V.

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Answer: Mixtures can be physically separated by using methods that use differences in physical properties to separate the components of the mixture, such as

Explanation:

The volume of oxygen gas, O₂ collected at 1.05 atm pressure and 44.0 °C when 42.5 g of KClO₃ decomposed is 13.01 L

We shall begin by obtaining the mole in 42.5 g of KClO₃. Details below:

Mole = mass / molar mass

Mole of CaC₂ = 42.5 / 122.5

Mole of CaC₂ = 0.35 mole

Next, we shall determine the mole of oxygen gas, O₂. produced. Details below:

2KClO₃ -> 2KCl + 3O₂

From the balanced equation above,

2 moles of KClO₃ decomposed to produced 3 mole of O₂

Therefore,

0.35 mole of KClO₃ will decompose to produce = (0.35 × 3) / 2 = 0.525 mole O₂

Finally, we shall determine the volume of oxygen gas, O₂ collected. Details below:

PV = nRT

1.05 × V = 0.525 × 0.0821 × 317

Divide both sides by 1.05

V = (0.525 × 0.0821 × 317) / 1.05

Volume of oxygen gas = 13.01 L

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There are 0.25 moles of HCl present in 250 mL of 1.0 M HCl.

We have to calculate the number of moles of HCl present in some mL of 1.0M HCl.  A mole is defined as the amount of substance in a system that contains as many elementary entities as there are atoms in 0.012 kg of carbon 12. We represent mole by the symbol 'mol'. Now, we will see how to calculate the number of moles.

We can calculate the number of moles of a substance using the following expression;

Molarity = no of moles of an element/volume

According to this question, we were given 250. mL of 1.0 M HCl. The number of moles will be calculated by the formula as follows;

no of moles of HCl = 0.250L × 1.0M

no of moles of HCl = 0.250 moles.

Therefore, 0.25 moles are present in 250 mL of 1.0 M HCl.

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For P₂O₅ the intermolecular forces such as van der Waals forces, dipole-dipole interactions, and hydrogen bonding must be overcome.

For HI the intermolecular forces that must be overcome are as van der Waals forces, dipole-dipole interactions, and hydrogen bonding.

P₂O₅ is a covalent compound and it is solid. To convert P₂O₅ from a solid to a gas, intermolecular forces such as van der Waals forces, dipole-dipole interactions, and hydrogen bonding must be overcome.

HI is a covalent compound that is a gas at room temperature and pressure. To convert HI from a liquid to a gas, intermolecular forces such as van der Waals forces, dipole-dipole interactions, and hydrogen bonding must be overcome.

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Answer:

The enthalpy of combustion of ethanol is -1430 kJ/mol, which means that for every mole of ethanol that is burned, 1430 kJ of heat is released.

To determine the amount of heat given off from the combustion of 1 dm³ of ethanol, we need to first calculate the number of moles of ethanol in 1 dm³.

1 dm³ is equivalent to 1000 cm³. Since the density of ethanol is 0.79 g/cm³, the mass of 1 dm³ of ethanol can be calculated as:

mass = density x volume

mass = 0.79 g/cm³ x 1000 cm³

mass = 790 g

To convert this mass to moles, we need to divide by the molar mass of ethanol:

moles = mass / molar mass

moles = 790 g / 46 g/mol

moles = 17.17 mol

Therefore, 1 dm³ of ethanol contains 17.17 moles of ethanol.

To calculate the heat given off from the combustion of 1 dm³ of ethanol, we can use the following equation:

heat = enthalpy of combustion x moles of ethanol

heat = -1430 kJ/mol x 17.17 mol

heat = -24,551 kJ

Therefore, the heat given off from the combustion of 1 dm³ of ethanol is -24,551 kJ, or approximately 24,551 kJ of heat is released.

a) The amount of N₂O₅ is lowered to 2.5 g during the course of around 4.41 × 10⁴  seconds or 12.25 hours.

b) 9.71 L of O₂ are generated at 745 mmHg and 45 °C.

a) To solve for the time required for the quantity of N₂O₅ to be reduced to 2.5 g, use the first-order integrated rate law:

ln[N₂O₅]t/[N₂O₅]0 = -kt

where [N₂O₅]t = concentration of N₂O₅ at time t, [N₂O₅]0 = initial concentration of N₂O₅, k = rate constant, and t = time.

Find the initial concentration of N₂O₅:

n(N₂O₅) = m/M = 80.0 g / 108.01 g/mol = 0.7413 mol

[N₂O₅]0 = n/V = 0.7413 mol / 0.153 L = 4.846 M

where M = molar mass of N₂O₅ and V = volume of the solution.

Substituting the given values into the equation:

ln([N₂O₅]t / 4.846 M) = -6.2×10⁻⁴ s⁻¹ × t

When the quantity of N₂O₅ is reduced to 2.5 g, the concentration is:

n(N₂O₅) = m/M = 2.5 g / 108.01 g/mol = 0.02314 mol

[N₂O₅]t = n/V = 0.02314 mol / 0.153 L = 0.151 M

Substituting this concentration into the equation and solving for t:

ln(0.151 M / 4.846 M) = -6.2×10⁻⁴ s⁻¹ × t

t = 4.41 × 10⁴ s

Therefore, it takes approximately 4.41 × 10⁴ seconds or 12.25 hours for the quantity of N₂O₅ to be reduced to 2.5 g.

b) The balanced equation for the reaction shows that 1 mole of N₂O₅ produces 1/2 mole of O₂:

N₂O₅ → N₂O₄ + 1/2 O2

Therefore, the number of moles of O₂ produced can be calculated using the stoichiometry:

n(O₂) = 1/2 × n(N₂O₅) = 1/2 × 0.7413 mol = 0.3707 mol

The ideal gas law can be used to calculate the volume of O₂ produced at 745 mmHg and 45°C:

PV = nRT

where P = pressure, V = volume, n = number of moles, R = gas constant, and T = temperature in Kelvin.

Convert the pressure to atm and the temperature to Kelvin:

P = 745 mmHg / 760 mmHg/atm = 0.980 atm

T = 45°C + 273.15 = 318.15 K

Substituting the values and solving for V:

V = nRT/P = (0.3707 mol) × (0.08206 L·atm/mol·K) × (318.15 K) / (0.980 atm) = 9.71 L

Therefore, the volume of O₂ produced at 745 mmHg and 45°C is 9.71 L.

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The S-P difference (sec) is used to calculate the distance (km) between an earthquake epicenter and a seismic station, while the magnitude (M) is a measure of the energy released during the earthquake.

These parameters are important for understanding the severity and impact of an earthquake, as well as for predicting future seismic activity.

The S-P difference (sec) refers to the time difference between the arrival of the primary (P) waves and the secondary (S) waves at a seismic station. This time difference is used to calculate the distance (km) between the earthquake epicenter and the seismic station, using the equation: distance (km) = S-P difference (sec) x 8 km/sec. This calculation assumes that the waves travel at a constant speed through the Earth's interior.
The magnitude (M) of an earthquake is a measure of the energy released during the earthquake, and is usually determined using a seismometer. The magnitude scale is logarithmic, meaning that each increase of one unit represents a tenfold increase in seismic energy. For example, an earthquake with a magnitude of 5.0 is ten times more powerful than one with a magnitude of 4.0, and 100 times more powerful than one with a magnitude of 3.0.

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As the hour of her new job approached, Emma could feel her excitement peak.

"Peak" is a synonym to "intensify" in this context because it means to reach the highest point or level of something. In the given passage, Emma's excitement is growing stronger and stronger as the time for her volunteer job approaches.

When her excitement "peaks," it means that it has reached the highest point of intensity, just like when something is intensified, it becomes stronger or more intense.

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Answer:

D is correct.(1s22s22p63s2)

The limiting reagent for the reaction between 0.150 g of salicylic acid and 0.350 mL of acetic anhydride is salicylic acid, C₇H₆O₃

First, we shall determine the mass of the acetic anhydride. Details below:

Mass = density × volume

Mass of acetic anhydride, C₄H₆O₃ = 1.082 × 0.350

Mass of acetic anhydride, C₄H₆O₃ = 0.3787 g

Finally, we shall determine the limiting reagent. Details below:

C₇H₆O₃ + C₄H₆O₃ -> C₉H₈O₄ + CH₃COOH

From the balanced equation above,

138.121 g of C₇H₆O₃ reacted with 102.09 g of C₄H₆O₃

Therefore,

0.150 g of C₇H₆O₃ will react with = (0.150 × 102.09) / 138.121 = 0.11089 g of C₄H₆O₃

We can see from the above that only 0.11089 g of acetic anhydride, C₄H₆O₃ out of 0.3787 g is needed to react with 0.150 g of salicylic acid, C₇H₆O₃

Thus, the limiting reagent is salicylic acid, C₇H₆O₃

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Answer:

Leaves are green due to the presence of an organelle chloroplast (in abundance) which contains the pigment chlorophyll

Explanation:

Now saying chlorophyll pigment  is a green pigment might be slightly incorrect. The two famous types (Chlorophyll a, Chlorophyll b) only absorb red and blue light from the atmosphere and reflect green light hence giving the pigment a green appearance and lastly giving the leaves a green color too

Answer:

Chlorophyll

Explanation:

Plants are often seen as green to the human eye due to the presence of chlorophyll, which is the primary pigment used in photosynthesis. Chlorophyll absorbs light in the red and blue-violet parts of the spectrum, but reflects or transmits green light, resulting in the characteristic green color of leaves.

The identity of element X in the compound X2Fe(CN)6 · 12H2O is sodium (Na).

To find the identity of element X in the compound X2Fe(CN)6 · 12H2O, we can start by determining the molar mass of the compound.

The molar mass of X2Fe(CN)6 is:

2 × molar mass of X + molar mass of Fe + 6 × molar mass of C + 6 × molar mass of N

= 2 × atomic mass of X + atomic mass of Fe + 6 × 12.01 g/mol + 6 × 14.01 g/mol

= 2 × atomic mass of X + 55.85 g/mol + 432.72 g/mol + 84.06 g/mol

= 2 × atomic mass of X + 572.63 g/mol

The molar mass of 12H2O is:

12 × (atomic mass of H + atomic mass of O) = 12 × (1.01 g/mol + 16.00 g/mol) = 216.24 g/mol

The total molar mass of the compound is:

2 × atomic mass of X + 572.63 g/mol + 216.24 g/mol = 2 × atomic mass of X + 788.87 g/mol

Now we can use the given information that the compound is 45.34% water by mass. This means that the mass of water in the compound is 45.34% of the total mass of the compound, and the mass of the rest of the compound (X2Fe(CN)6) is 100% - 45.34% = 54.66% of the total mass of the compound.

Let's assume we have 100 g of the compound. Then the mass of water in the compound is:

45.34 g water = 0.4534 × 100 g compound

The mass of the rest of the compound (X2Fe(CN)6) is:

54.66 g rest of the compound = 0.5466 × 100 g compound

We can now use the mass of the rest of the compound (X2Fe(CN)6) to find the number of moles of the compound:

moles of X2Fe(CN)6 = (54.66 g) / (2 × atomic mass of X + 572.63 g/mol)

We can also use the mass of water to find the number of moles of water:

moles of H2O = (45.34 g) / 18.02 g/mol

Since the compound has 12 moles of water per mole of X2Fe(CN)6, we have:

moles of X2Fe(CN)6 = 1/12 × moles of H2O

We can now set these two expressions for moles of the compound equal to each other and solve for the atomic mass of X:

(54.66 g) / (2 × atomic mass of X + 572.63 g/mol) = 1/12 × (45.34 g) / 18.02 g/mol

Simplifying this equation and solving for the atomic mass of X gives:

atomic mass of X = 22.99 g/mol

The atomic mass of X is very close to the atomic mass of sodium (22.99 g/mol), so it is likely that X is sodium. Therefore, the identity of element X in the compound X2Fe(CN)6 · 12H2O is sodium (Na).

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The amount of heat evolved when 25 moles of Sulfur is burned in excess oxygen is -74000 kJ.

The balanced reaction is given that is:

[tex]S_8(s) + 8 O_2(g) \rightarrow 8 SO_2(g)[/tex]

We can see that 1 mole of [tex]S_8[/tex] reacts with 8 moles of [tex]O_2[/tex] to produce 8 moles of [tex]So_2[/tex].

If 25.0 moles of [tex]S_8[/tex] reacts with excess Oxygen, then the amount of [tex]O_2[/tex] which is required in the reaction will be:

8 moles [tex]O_2[/tex] / 1 mole S8 × 25.0 moles S8 = 200 moles [tex]O_2[/tex]

We can use the enthalpy change and calculate the amount of heat evolved:

[tex]\Delta H[/tex] = -2368 kJ/ 8 moles [tex]SO_2[/tex]

The heat evolved = [tex]\Delta H[/tex] × moles of [tex]SO_2[/tex] produced

Moles of [tex]SO_2[/tex] produced =  8 moles [tex]SO_2[/tex] / 1 mole [tex]S_8[/tex] × 25.0 moles [tex]S_8[/tex]

= 200 moles [tex]SO_2[/tex].

Therefore, Heat evolved= -2368 kJ/ 8 moles [tex]SO_2[/tex] × 200 moles [tex]SO_2[/tex]

= -74000 kJ

The amount of heat evolved when 25 moles of Sulfur is burned in excess oxygen is -74000 kJ, the negative sign here indicates that the reaction is exothermic.

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The molar enthalpy for the reaction 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) is 266.4 kJ/mol.

The molar enthalpy is determined from Hess's law as follows:

Equation 1 x2:

2 N₂ (g) + 2 O₂ (g) → 4 NO (g) ∆H° = 361.2 kJ/mol

Equation 3 x3, :

6 H₂ (g) + 3 O₂ (g) → 6 H₂O (g) ∆H° = -1451.1 kJ/mol

Equation 2 x -4:

-8 N₂ (g) - 12 H₂ (g) → -8 NH₃ (g) ∆H° = 367.2 kJ/mol

Adding the equations together:

-6 N₂ (g) - 6 H₂ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) - 8 NH₃ (g) ∆H° = 266.3 kJ/mol

Multiplying the equation above by -1/2:

3 N₂ (g) + 3 H₂ (g) - 5/2 O₂ (g) → -2 NO (g) - 3 H₂O (g) + 4 NH₃ (g) ∆H° = -133.2 kJ/mol

Multiplying the above equation by -2:

4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g) ∆H° = 266.4 kJ/mol

This is the molar enthalpy of the given reaction

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The height of fall of the ice is determined as 313.25 m.

The total heat energy of the ice during the fall is calculated as follows;

Q = ml + mcΔT

where;

Q = 334000 x 0.4  +  0.4 x 4200 x (10)

Q = 150,400 J

The energy converted into potential energy is calculated as;

3%K.E = 150,400 J

0.03K.E = 150,400 J

K.E = 5,013,333.33 J

¹/₂mv² = 5,013,333.33 J

v = √(2 x 5,013,333.33)/(0.4)

v = 5,006.67 m/s

The height of fall is calculated as;

h = √2gh

h = √(2 x 5,006.67 x 9.8)

h = 313.25 m

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Answer:

the pH of the solution is approximately 5.857.

Explanation:

The pH of a solution can be calculated using the formula:

pH = -log[H+]

where [H+] is the concentration of hydrogen ions in moles per liter (M).

In this case, [H+] = 1.39x10^-6 M, so:

pH = -log(1.39x10^-6)

= 5.857

Therefore, the pH of the solution is approximately 5.857.

The concentration of the solution is  0.000435 mol/dm³.

The concentration of a solution is calculated as follows;

Concentration = (Absorbance) / (Molar absorptivity x path length)

the path length =  3cm

the molar absorptivity (E) = 400 dm/mol/cm.

if the solution transmits 30% of the light, it absorbs 70% of the incident light.

Absorbance = log (1/Transmittance)

Absorbance  = log (1/0.3)

Absorbance  = 0.523

Concentration = (0.523) / (400 dm/mol/cm x 3 cm)

= 0.000435 mol/dm³

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The volume of a 1.0 M solution that has 4.0 moles of solute is 4.0 liters.

A mole is a unit of measurement used in chemistry to represent the quantity of a chemical.

We can use the following formula to get the volume of a 1.0 M solution containing 4.0 moles of solute:

moles of solute = molarity x volume of solution

To determine the volume of the solution, we can rearrange this formula as follows:

Volume of solution = molarity / moles of solute.

By entering the specified values, we obtain:

4.0 moles / 1.0 M is the solution's volume.

Solution volume = 4.0 L

Therefore, the volume of a 1.0 M solution that has 4.0 moles of solute is 4.0 liters.

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The mass of Fe₂O₃ produced is 101.9 g.

The balanced chemical equation for the combustion of iron pyrite is:

4FeS₂(s) + 11O₂(g) → 2Fe₂O3(s) + 8SO₂(g)

From the equation, 11 moles of oxygen are required to produce 2 moles of Fe₂O₃. Convert the given volume of oxygen to moles:

n(O2) = PV/RT = (2.97 atm)(74 L)/(0.0821 L·atm/mol·K)(161 + 273 K) = 3.51 mol

Since the reaction requires 11 moles of O₂ for every 2 moles of Fe₂O₃, calculate the moles of Fe₂O₃ produced:

n(Fe₂O₃) = (2/11) × n(O₂) = (2/11) × 3.51 mol = 0.638 mol

Finally, use the molar mass of Fe₂O₃ to convert moles to grams:

m(Fe₂O₃) = n(Fe₂O₃) × M(Fe₂O₃) = 0.638 mol × 159.69 g/mol = 101.9 g

Therefore, the mass of Fe₂O₃ produced is 101.9 g.

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when pressure of the reactant mixture at the equilibrium and with the fewer moles in reactant side will be increased and the equilibrium will be shift to the side in the reaction where the fewer moles of the gas.

According to the Le Chartelier, when the reaction is in the equilibrium phase and the one of the constraints which will affect the rate of the reactions, and the equilibrium will be shift to the cancel out  this effect that the constraint had.

Therefore, If the pressure of the system or the reaction is in the equilibrium is change, the equilibrium of the reaction will be change that is depending on the side of the reaction with the highest number of the molecules.

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Hydrazine, reacts with the oxygen to form the nitrogen gas and the water. The percent yield of the reaction is 3.18 %.

The balanced reaction is :

N₂H₄  + O₂  --->  N₂ + 2H₂O

The mass of the N₂H₄  = 3.55 g

The moles of N₂H₄ = mass / molar mass

The moles of N₂H₄ = 3.55 / 32

The moles of N₂H₄ = 0.110 mol

The theoretical yield = 0.110 mol × 28 g/mol

The theoretical yield = 3.08 g

The gas equation is :

P V = n R T

n = P V / R T

n = (1 × 0.850 ) / ( 0.0823 ×295 )

n = 0.0035 mol

The actual yield = 0.0035 × 28

The actual yield = 0.098 g

The percent yield = ( 0.098 / 3.08 ) × 100 %

The percent yield = 3.18 %.

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5.221 grams of Al are required to react with 352 mL of 1.65 M HCl solution.

Molarity (M) is defined as the moles of solute per liter of the solution.

Balanced chemical equation is : 2Al + 6HCl → 2AlCl₃ + 3H₂

From the equation, we can see that 2 moles of Al react with 6 moles of HCl to produce 2 moles of AlCl₃ and 3 moles of H₂.

As moles of HCl = Molarity × Volume

moles of HCl = 1.65 mol/L × 0.352 L

moles of HCl = 0.58128 mol

and moles of Al = (2/6) × moles of HCl

moles of Al = (1/3) × 0.58128 mol

moles of Al = 0.19376 mol

mass of Al = moles of Al × molar mass of Al

mass of Al = 0.19376 mol × 26.98 g/mol

mass of Al = 5.221 g

So, 5.221 grams of Al are required to react with 352 mL of 1.65 M HCl solution.

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Boiling point of solution will be 100 C. The incorrect statement is C)

An aqueous solution is one in which water serves as the solvent. One or more substances are dissolved in water to create such a solution, and the water molecules surround and separate the individual solute particles to create a homogeneous mixture.

Therefore, A solvent's boiling point and freezing point change when a solute is dissolved in it, respectively. Boiling point elevation and freezing point depression are two terms used to describe this occurrence. The concentration of the solute determines how much of an impact it has.

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