What is the difference between collecting and analysing data​

The mass of the HI that is produced is  1843 g

Stoichiometry is a branch of chemistry that focuses on the quantitative relationships in chemical reactions between reactants and products.

What is the number of moles?

Number of moles of [tex]N_{2} H_{4}[/tex] = 115.7 g/32 g/mol

= 3.6 moles

Now we have that the balanced reaction equation is;

[tex]N_{2} H_{4} + 2I_{2} --- > N_{2} + 4 HI[/tex]

If 1 mole of [tex]N_{2} H_{4}[/tex] produces 4 moles of HI

3.6 moles of [tex]N_{2} H_{4}[/tex] will produce 3.6 * 4/1

= 14.4 moles of HI

Mass of HI produced = 14.4 moles * 128 g/mol

= 1843 g

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Chemical reactions nearly always include a change in energy between the products and reactants.

Thus, Energy is released when chemical bonds are created and is released when chemical bonds are destroyed.

The overall energy of a system, however, must remain constant according to the Law of Conservation of Energy, and chemical reactions frequently absorb or release energy in the form of heat, light, or both.

The difference in the amounts of chemical energy that are stored in the products and reactants accounts for the energy change in a chemical reaction. Enthalpy refers to the system's heat content or stored chemical energy.

Thus, Chemical reactions nearly always include a change in energy between the products and reactants.

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Answer:

Vestigial

Explanation:

The retention of genetically determined traits or structures that have partially or completely lost their ancestral purpose in a specific species is known as vestigiality. In most cases, evaluating the vestigality requires comparison with comparable traits in closely related species.

Answer:

Explanation:

Both 2CH4 and C2H8 have the same number and kind of elements. But practically, 2CH4 will be existing but C2H8 cannot exist.

We can use the combined gas law to determine the pressure of the gas at the final state. 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 = 1.01 atm), volume (V1 = 2.31 L), and temperature (T1 = 279 K) of the gas, and the final volume (V2 = 1.09 L), and temperature (T2 = 308 K) of the gas. We can solve for P2, the final pressure of the gas:

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

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

P2 = (1.01 atm x 2.31 L x 308 K) / (1.09 L x 279 K)

P2 = 2.41 atm (rounded to three significant figures)

Therefore, the pressure of the gas when the volume is 1.09 L and the temperature is 308 K is approximately 2.41 atm.

Using the Ideal Gas Law, the pressure expected for 30 mol of  [tex]C_2H_4[/tex] gas in a 6.00-L container at 20 °C is 1210.07 atm, and using the van der Waals equation, the pressure is 1179.71 atm.

To calculate the pressure expected for 30 mol of [tex]C_2H_4[/tex] gas in a 6.00-L container at 20 °C, we will use both the Ideal Gas Law and van der Waals equation.

Ideal Gas Law: PV = nRT
P = pressure
V = volume (6.00 L)
n = moles (30 mol)
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature (20 °C + 273.15 = 293.15 K)

Solve for P (pressure):

P = nRT / V

P = (30 mol)(0.0821 L•atm/mol•K)(293.15 K) / 6.00 L

P = 1210.07 atm

Van der Waals equation:

(P + a(n/V)²)(V - nb) = nRT
a = 4.612 atm•L²/mol²
b = 0.0582 L/mol

Solve for P (pressure):
(P + (4.612)(30/6)²) (6 - 0.0582 * 30) = (30)(0.0821)(293.15)

P = 1179.71 atm

Using the Ideal Gas Law, the pressure is 1210.07 atm, and using the van der Waals equation, the pressure is 1179.71 atm.

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The best statement that describes the three enzymes is: The enzymes have different structures because they have a different sequence of amino acids. Opton A is correct.

Enzymes are biological catalysts that facilitate chemical reactions in living organisms. Enzymes are proteins, and their function is determined by their three-dimensional structure, which is determined by the sequence of amino acids that make up the protein.

In this scenario, the three enzymes have different structures, which suggests that they have a different sequence of amino acids. This difference in amino acid sequence results in different folding patterns and ultimately different shapes of the enzymes. The specific shape of an enzyme determines its ability to catalyze a particular chemical reaction. Hence, the different structures of these enzymes indicate that they may perform different functions or catalyze different chemical reactions.

Option A, "The enzymes have different structures because they have a different sequence of amino acids," is the correct answer as it aligns with the fundamental principle of protein structure and function.

Option B, "The enzymes have the same sequence of amino acids because they are all digestive enzymes," is incorrect because enzymes can have different sequences of amino acids even if they perform the same function.

Option C, "The enzymes perform different functions because they have the same sequence of amino acids," is incorrect because the sequence of amino acids determines the enzyme's structure and thus its function.

Option D, "The enzymes break down the same molecules because they have a different sequence of amino acids," is also incorrect because different amino acid sequences can result in different substrate specificity, which means that the enzymes can break down different molecules. Therefore option A is correct.

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Answer:

Explanation:

The chemical equation for the production of manganese from manganese (II) carbonate, oxygen, and aluminium can be represented as follows:

3MnCO3(s) + 3O2(g) + 4Al(s) → 3Mn(s) + 3CO2(g) + 2Al2O3(s)

In this equation, manganese (II) carbonate (MnCO3) reacts with oxygen (O2) and aluminium (Al) to produce manganese (Mn), carbon dioxide (CO2), and aluminium oxide (Al2O3). The equation is balanced with three molecules of manganese carbonate, three molecules of oxygen, and four molecules of aluminium reacting to produce three molecules of manganese, three molecules of carbon dioxide, and two molecules of aluminium oxide.

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To solve this problem, we can use the formula:

q = m × c × ΔT

where q is the heat absorbed or released, m is the mass of the substance, c is its specific heat, and ΔT is the change in temperature.

First, let's calculate the mass of water:

m = 225.0 g

Next, let's calculate the heat absorbed by the water:

q_water = m × c × ΔT

q_water = 225.0 g × 4.184 J/(g·°C) × (24.60°C - 20.53°C)

q_water = 3749.8 J

Since the metal released 4274 J of heat, the heat absorbed by the calorimeter can be calculated by subtracting the heat absorbed by the water from the total heat released by the metal:

q_calorimeter = - (q_water + q_metal)

q_calorimeter = - (3749.8 J + 4274 J)

q_calorimeter = - 8023.8 J

Therefore, the heat absorbed by the calorimeter is -8023.8 J, which is approximately equal to -8000 J or -8.0 kJ. The answer is (c) -339 J, since it is the closest to the calculated value when rounded to the nearest integer. Note that the negative sign indicates that the calorimeter absorbed the heat, which is expected since the reaction involved a release of heat.

Answer:

90g

Explanation:

Ans. 90 gram

we know that,

n = wt/m.wt

where, n=  moles

wt.= weight

m.wt = molecular weight

putting values we get

5 = wt./18 ( molecular weight of water is 18

wt.= 90

hence  ans.= 90 gram

There are 15.4 equivalents of chloride ion present in the solution

We need to multiply the total number of moles of chloride ion in the solution by the number of equivalents.

The molar mass of NaCl is 58.44 g/mol, so 3.5 mol of NaCl contains :

3.5 mol NaCl x 2 mol Cl⁻/1 mol NaCl = 7 mol Cl⁻

Similarly, the molar mass of MgCl₂ is 95.21 g/mol, so 4.2 mol of MgCl₂ contains:

4.2 mol MgCl₂ x 2 mol Cl⁻/1 mol MgCl₂ = 8.4 mol Cl⁻

Therefore, the total number of moles of chloride ion in the solution is:

7 mol Cl⁻ + 8.4 mol Cl⁻ = 15.4 mol Cl⁻

By dividing the total number of moles by the number of equivalents per mole, we can finally determine how many equivalents of the chloride ion there are. There is one equivalent of the chloride ion per mole since it has a valency of -1.

15.4 mol Cl⁻ x 1 eq/mol = 15.4 eq

So there are 15.4 equivalents of chloride ion present in the solution.

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The equation that shows that the total mass during a chemical reaction stays the same is 2) H₂ + O₂ → H₂O.

This is an example of a balanced chemical equation where the number of atoms of each element on both the reactant and product side is equal. In other words, the total number of atoms of each element is conserved, and therefore the total mass is conserved. In the other reactions, either the number of atoms on the product side is different from the reactant side or there is no reaction at all.

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Answer:

Explanation:

a) The balanced equation for the double replacement reaction between lead (II) nitrate and potassium bromide is:

Pb(NO₃)₂(aq) + 2KBr(aq) → PbBr₂(s) + 2KNO₃(aq)

In this reaction, lead (II) bromide (PbBr₂) will precipitate, while potassium nitrate (KNO₃) will remain in solution.

b) To determine the amount of precipitate produced, we need to first determine the limiting reactant. We can do this by calculating the number of moles of each reactant and comparing it to the stoichiometry of the balanced equation.

The molar mass of lead (II) nitrate is 331.21 g/mol and the molar mass of potassium bromide is 119.00 g/mol.

The number of moles of lead (II) nitrate is 32.5 g / 331.21 g/mol = 0.0981 mol The number of moles of potassium bromide is 38.75 g / 119.00 g/mol = 0.3256 mol

According to the balanced equation, one mole of lead (II) nitrate reacts with two moles of potassium bromide to produce one mole of lead (II) bromide. This means that if all the lead (II) nitrate were to react, it would require 0.0981 mol * 2 = 0.1962 mol of potassium bromide.

Since we have more than enough potassium bromide (0.3256 mol > 0.1962 mol), lead (II) nitrate is the limiting reactant.

The number of moles of lead (II) bromide produced will be equal to the number of moles of lead (II) nitrate consumed, which is 0.0981 mol.

The molar mass of lead (II) bromide is 367.01 g/mol, so the mass of lead (II) bromide produced will be 0.0981 mol * 367.01 g/mol = 36.0 g.

c) To determine the amount of excess reactant remaining, we need to subtract the amount consumed from the initial amount.

The number of moles of potassium bromide consumed is half the number of moles of lead (II) nitrate consumed, which is 0.0981 mol / 2 = 0.04905 mol.

The mass of potassium bromide consumed is 0.04905 mol * 119.00 g/mol = 5.84 g.

The mass of potassium bromide remaining is 38.75 g - 5.84 g = 32.91 g.

Answer:

Br, Se, Sr, Rb

Explanation:

Atomic radius increases as you move to the left and down the periodic table. The increase in radius as you move left is due to decreasing effective nuclear charge (the pull an electron feels from the nucleus) since the number of protons decrease. The increase in radius as you move down is due to a higher number of principle energy levels (orbital in which the electron is located relative to the atom's nucleus), causing the electrons to be farther from the nucleus.

The volume of C₂H₂ that is collected over water at 23 ∘C by the reaction of 1.52 g of CaC₂ is 0.61 L

First, we shall determine the mole of CaC₂ that reacted. Details below:

Mole = mass / molar mass

Mole of CaC₂ = 1.52 / 64

Mole of CaC₂ = 0.024 mole

Next, we shall determine the mole of C₂H₂. obtained. Details below:

CaC₂ + 2H₂O -> C₂H₂ + Ca(OH)₂

From the balanced equation above,

1 mole of CaC₂ reacted to produced 1 mole of C₂H₂

Therefore,

0.024 mole of CaC₂ will also react to produce 0.024 mole of C₂H₂

Finally, we shall determine the volume of C₂H₂ collected. This is shown below:

PV = nRT

729.93 × V = 0.024 × 62.36 × 296

Divide both sides by 729.93

V = (0.024 × 62.36 × 296) / 729.93

V = 0.61 L

Thus, the volume of the C₂H₂ gas collected is 0.61 L

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The complete equation for the neutralization reactions for the LiOH + HNO₂ is as :

LiOH  +  HNO₂ ---->  LiNO₂  +  H₂O

The Neutralization reaction is the reaction as in the chemical reaction in which the acid will reacts with the base and to produce the salt and the water molecule. The general equation of the chemical reaction is as :

HX  +  BOH  -->  BX  + H₂O

The reaction with the LiOH and the HNO₂ is :

LiOH  +  HNO₂ ---->  LiNO₂  +  H₂O

There is the combination of the H⁺ ions and OH⁻ ions that will form the water.

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The 210.00 g sample of the water with the initial temperature of the 29.0°C absorbs the 7,000.0 J of heat. The final temperature of the water is the 36.9  °C .

The mass of the water = 210 g

The initial temperature = 29.0 °C

The final temperature = ?

The heat energy = 7000 J

The specific heat capacity = 4.184 J/g  °C

The heat energy is expressed as :

Q = m c ΔT

Where,

The m is mass of water = 210 g

The c is specific heat of water = 4.184 J/g  °C

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

The  ΔT is change in temperature = T - 29.0 °C

7000 = 210 × 4.184 ( T - 29.0  )

7000 = 878.64 ( T - 29.0  )

( T - 29.0  ) = 7.966

T = 36.9  °C

The final temperature is 36.9  °C .

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the answer is 2.5 according to me

For strong electrolytes, molar conductivity decreases as the solution is diluted because the concentration of ions decreases.

For weak electrolytes, molar conductivity increases as the solution is diluted because as the solution is diluted, the concentration of ions increases.

The expression for the dilution law is A = εcb

The conductivity of a solution containing one mole of an electrolyte when placed between two electrodes spaced one centimeter apart is known as the molar conductivity of the electrolyte. The strength of the electrolyte affects how molar conductivity changes with dilution.

At infinite dilution, the molar conductivity of an electrolyte reaches its maximum value because the electrolyte's ions are so far apart that they no longer interact with one another.

The dilution law or Beer-Lambert law states that the absorbance of a solution is directly proportional to the concentration of the solution and the path length of the light through the solution.

A ∝ cb

Adding a proportionality constant gives:

A = εcb

where;

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The molarity of the solution that has 2.0 moles of solute in 3.0 L of solution is 0.67 mol/L

Molarity is  described as a measure of the concentration of a chemical species, in particular of a solute in a solution, in terms of amount of substance per unit volume of solution.

Molarity = moles of solute / liters of solution

we then substitute the given values, and have

Molarity = 2.0 moles / 3.0 L

Molarity = 0.67 mol/L

Molarity is  very important because the ration used to express the concentration of  any  solution.

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The correct molar mass for nickel chloride is 94.14 g/mol (option C).

Molar mass is the mass of a given substance divided by its amount, measured in moles. It is commonly expressed in grams (sometimes kilograms) per mole.

The molar mass of a substance can be calculated by summing up the atomic masses of the element components.

According to this question, the atomic mass of nickel is 58.693 amu while that of chlorine gas is 35.45 amu. The molar mass of nickel chloride can be calculated as follows;

molar mass = 35.45 amu + 58.693 amu = 94.14 g/mol

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To determine the change in enthalpy associated with the generation of 155.3 g of [tex]C_5H_1_0O[/tex] we must first calculate the moles of [tex]C_5H_1_0O[/tex] produced using its molar mass.

The molar mass of C₅H₁₀O is:

5(12.01 g/mol) + 10(1.01 g/mol) + 16.00 g/mol = 88.15 g/mol

Moles of C₅H₁₀O produced:

155.3 g / 88.15 g/mol = 1.763 mol C₅H₁₀O

The balanced chemical equation states that the formation of 1 mol of C₅H₁₀O results in an enthalpy change of -191.3 kJ.

As a result, the enthalpy change during the formation of 1.763 mol of C₅H₁₀O is: -191.3 kJ/mol x 1.763 mol = -337.8 kJ

The enthalpy change for the production of 155.3 g of C5H10O is -337.8 kJ.

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The chemical formula for the ionic compound barium iodide is [tex]BaI_2[/tex] .

Barium iodide is composed of one barium ion (Ba2+) and two iodide ions (I-), which together form a neutral compound. Barium iodide is a white crystalline solid with a high melting point and is soluble in water. It is commonly used in the manufacture of photographic paper and in the production of cathode ray tubes for televisions and computer monitors.

Barium iodide has a variety of other uses, including in medicine as a contrast agent for X-ray imaging and in the synthesis of organic compounds.

The compound has several different crystal structures, including hexagonal and cubic, and can be prepared by reacting barium carbonate with hydroiodic acid. Overall, barium iodide is an important and versatile compound with a range of practical applications.

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The electronic configuration of the d-block metal ions in the 3d series is represented by electronic configuration of Argon (Ar), 3d and 4s sub orbitals.

The electronic configuration of the d-block metal ions in the 3d series is as follows:

Scandium (Sc): [Ar] 3d¹ 4s²

Titanium (Ti): [Ar] 3d² 4s²

Vanadium (V): [Ar] 3d³ 4s²

Chromium (Cr): [Ar] 3d⁵ 4s¹

Manganese (Mn): [Ar] 3d⁵ 4s²

Iron (Fe): [Ar] 3d⁶ 4s²

Cobalt (Co): [Ar] 3d⁷ 4s²

Nickel (Ni): [Ar] 3d⁸ 4s²

Copper (Cu): [Ar] 3d¹⁰ 4s¹

Zinc (Zn): [Ar] 3d¹⁰ 4s²

Thus, the above illustration shows the electronic configuration of all the metal ions in the d-blocks (3d series)​.

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In the titration between HCl and NaOH, the medium is neutral at the end point because of complete neutralization of a strong acid by a strong base.

Neutralization is a chemical reaction in which acid and base react to form salt and water. Hydrogen (H⁺) ions and hydroxide (OH⁻ ions) react with each other to form water.

The strong acid and strong base neutralization have a pH value of 7.

The beaker gets warm which indicates that the reaction between acid and base is an exothermic reaction releasing heat energy into the surroundings.

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The standard cell potential is found as +1.95 V and is a  spontaneous  reaction.

The standard cell potential (E°cell) of an electrochemical cell is given by the difference between the standard reduction potentials of the two half-cells involved.

E°cell = E°reduction (cathode) - E°reduction (anode)

The half-reactions given are:

Pb4+ + 2e− → Pb2+ (reduction)

Co3+ + e− → Co2+ (reduction)

The standard reduction potentials for these half-reactions are:

E°reduction(Pb4+/Pb2+) = -0.13 V

E°reduction(Co3+/Co2+) = +1.82 V

We then calculate as:

E°cell = E°reduction (Co3+/Co2+) - E°reduction (Pb4+/Pb2+)

E°cell = (+1.82 V) - (-0.13 V)

E°cell = +1.95 V

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Answer: Moss

Explanation: It's a flowerless plant