2) A student was working on an investigation to measure the relative activity
of an enzyme at various pH values. He collected the following data:
pH 2, enzyme activity 10; pH 8, enzyme activity 50;
pH 12, enzyme activity 10; pH 4, enzyme activity 20;
pH 6, enzyme activity 40; pH 10, enzyme activity 40

Answer:

See below

Explanation:

From periodic table

mole weight of Fe   is   55.845  gm / mole

76.15 gm  /  55.845 gm/mol   *  6.022 x 10^23  atoms/mole =

                8.212 x 10^23   atoms    ( four signif. digits)

Answer:

Protons, electrons

Explanation:

Elements in a groups have the same number of protons and the same electrons.

Answer:

Alkynes

Explanation:

Alkynes are hydrocarbons which contain carbon-carbon triple bonds. Their general formula is CnH2n-2 for molecules with one triple bond (and no rings). Alkynes undergo many of the same reactions as alkenes, but can react twice because of the presence of the two p-bonds in the triple bond.

Answer:

The alcohol cleans the penny due to the acid in the vinegar. A chemical reaction between the alcohol removes copper oxide. Copper oxide is what causes pennies to become dull. Salt is also a key ingredient in the other cleaning agents, thus making them effective in cleaning the pennies.

The excess reactant is Al, and the mass of the remaining excess Al after the reaction is 16.5 grams.

To determine the excess reactant and the mass of the remaining excess reactant, we need to first write and balance the chemical equation for the reaction between \(Fe_{2} O _{3}\) and Al:

2 \(Fe_{2} O _{3}\) + 2 Al → \(Al_{2} O_{3}\) + 4 Fe

From the balanced equation, we can see that the stoichiometric ratio of \(Fe_{2} O _{3}\)to Al is 2:2, or 1:1. This means that both reactants are consumed in equal amounts, and neither is in excess.

To confirm this, we can calculate the amount of product formed by each reactant and compare it to the actual amount of product formed:

For \(Fe_{2} O _{3}\) : Using its molar mass of 159.69 g/mol, the number of moles of Fe2O3 is 35.0 g / 159.69 g/mol = 0.2192 mol. According to the balanced equation, 2 moles of \(Fe_{2} O _{3}\) produces 4 moles of Fe. Therefore, the number of moles of Fe produced from 0.2192 mol of \(Fe2O _{3}\) is (4/2) x 0.2192 = 0.4384 mol.

For Al: Using its molar mass of 26.98 g/mol, the number of moles of Al is 30.0 g / 26.98 g/mol = 1.111 mol. According to the balanced equation, 2 moles of Al produces 4 moles of Fe. Therefore, the number of moles of Fe produced from 1.111 mol of Al is (4/2) x 1.111 = 2.222 mol.

We can see that Al produces twice as much Fe as \(Fe_{2} O _{3}\), indicating that \(Fe_{2} O _{3}\) is the limiting reactant and Al is in excess. However, we also need to calculate the mass of the remaining excess Al:

The number of moles of Al that reacted is 1.111 - 0.5 (since 2 moles of Al react with 2 moles of \(Fe_{2} O _{3}\)) = 0.611 mol.

The mass of the remaining excess Al is therefore 0.611 mol x 26.98 g/mol = 16.5 g (rounded to one decimal place).

Therefore, the excess reactant is Al, and the mass of the remaining excess Al after the reaction is 16.5 grams.

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The ideal gas law is equivalent to Charles's law when number of moles and pressure are constant (option D)

PV = nRT

Where

Charles' law states as follow:

V₁ / T₁ = V₂ / T₂

Where

PV = nRT

Divide both side by T

PV / T = nR

Divide both side by P

V / T = nR / P

Let nR / P be costant

V / T = constant

Thus,

V₁ / T₁ = V₂ / T₂

From the above illustrations, we can conclude that the ideal gas equation will become Charles' law equation if the number of mole and pressure are constant (Option D)

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Answer:

1. summer 2 spring 3 winter 4 fall/autumn

Answer:

Ionic

Na has one valence electron

Chlorine has seven

Sodium loses one electron

Chlorine gains one electron

• The factors that influence vapour pressure is amongst others surface area.

• So the most suitale nswer will be the mole raction of the liquid will influence vapour pressure.

• Option C is the most closest answer,.

The equilibrium concentrations of the reactant (CoCl2(g)) and products (Co(g) and Cl2(g)) when 0.363 moles of CoCl2(g) are introduced into a 1.00 L vessel at 600 K can be expressed as [CoCl2(g)] = (0.363 - x) moles/L, [Co(g)] = x moles/L, and [Cl2(g)] = x moles/L

To calculate the equilibrium concentrations of reactant and products, we need to use the equilibrium constant (K) expression and the stoichiometry of the balanced chemical equation.

First, let's write the balanced chemical equation for the reaction:

CoCl2(g) ⇌ Co(g) + Cl2(g)

Next, we need the value of the equilibrium constant (K) at 600 K. Unfortunately, the equilibrium constant value is not provided in the question. Without the equilibrium constant, we cannot determine the exact equilibrium concentrations of the reactant and products.

However, we can still calculate the equilibrium concentrations using the ICE (Initial, Change, Equilibrium) table method. We start by writing down the initial concentrations of the reactant and products, which is 0.363 moles of CoCl2(g) in a 1.00 L vessel.

Next, we assume x moles of Co(g) and Cl2(g) are formed or consumed at equilibrium. Using the stoichiometry of the balanced equation, we know that the change in concentration of Co(g) and Cl2(g) is x moles.

Therefore, the equilibrium concentrations are as follows:

[CoCl2(g)] = (0.363 - x) moles/L
[Co(g)] = x moles/L
[Cl2(g)] = x moles/L

Without the value of the equilibrium constant, we cannot calculate the exact equilibrium concentrations. However, we can express the concentrations in terms of x, which represents the change in moles at equilibrium.

In summary, the equilibrium concentrations of the reactant (CoCl2(g)) and products (Co(g) and Cl2(g)) when 0.363 moles of CoCl2(g) are introduced into a 1.00 L vessel at 600 K can be expressed as [CoCl2(g)] = (0.363 - x) moles/L, [Co(g)] = x moles/L, and [Cl2(g)] = x moles/L.

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In an ecosystem, strike-slip type of basin  are formed at transform boundaries.

Ecosystem is defined as a system which consists of all living organisms and the physical components with which the living beings interact. The abiotic and biotic components are linked to each other through nutrient cycles and flow of energy.

Energy enters the system through the process of photosynthesis .Animals play an important role in transfer of energy as they feed on each other.As a result of this transfer of matter and energy takes place through the system .Living organisms also influence the quantity of biomass present.By decomposition of dead plants and animals by microbes nutrients are released back in to the soil.

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The limiting reactant is Ca₃(PO₄)₂.

To determine the limiting reactant in a chemical reaction, we need to compare the amount of moles of each reactant and see which one is completely consumed first.

The balanced chemical equation for the reaction between calcium phosphate (Ca3(PO4)2) and sulfuric acid (H₂SO₄) is:

3Ca₃(PO₄)₂ + 2H₂SO₄ → 6CaSO₄ + H₄P₂O₇

First, let's convert the given masses of each reactant to moles:

moles of Ca₃(PO₄)₂ = 129 g / (3 x 310.18 g/mol) = 0.139 mol

moles of H₂SO₄ = 97.4 g / (2 x 98.08 g/mol) = 0.993 mol

According to the balanced chemical equation, it takes 3 moles of Ca₃(PO₄)₂ and 2 moles of H₂SO₄ to produce the products. So, we need to multiply the amount of moles of each reactant by the appropriate stoichiometric coefficient in the balanced equation to see which reactant is completely consumed first:

For Ca₃(PO₄)₂: 0.139 mol x (2/3) = 0.093 mol of H₂SO₄ required

For H₂SO₄: 0.993 mol x (3/2) = 1.49 mol of Ca₃(PO₄)₂ required

From the above calculations, we can see that 0.139 mol of Ca₃(PO₄)₂ require 0.093 mol of H₂SO₄ for complete reaction. But we have 0.993 mol of H₂SO₄ available which is much greater than the required amount of H₂SO₄. Therefore, the limiting reactant is Ca₃(PO₄)₂.

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Fires are classified according to their properties, fires fueled by metals fall under Class D fire category.

Fires are classified into different classes based on the nature of the fuel. One of the classes is Class D fire, which involves fires fueled by metals. Metals such as magnesium, sodium, potassium, and titanium can ignite and burn under certain conditions. Class D fires require specific extinguishing agents, such as dry powder extinguishers, to effectively control and suppress them.

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0.08 gram mass of magnesium oxide will be formed.

Mass is an intrinsic property in the mass of a body. It was the  traditionally believed to the  be related to the quantity of matters in the  physical body, until it is  discovery of the atom and  particle in physics.

Mass is quantitative measure of inertia a fundamental quantity of all matters.

2Mg + O 2→2MgO

Molar mass of Mg =24 g

Molar mass of MgO =40 g

No. of moles of O2=6.022×10^20/6.022×10^23

=0.001 mole of O2

No. of moles of Mg=0.486/24

=0.02 moles  of Mg

As moles of O 2

​ are less than 0.01, O 2

 is limiting reagent and it is  0.002 moles of MgO is formed.

⇒0.002×40=0.08 g of MgO

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Answer:

D. Ruler

Explanation:

Use the formula to find the volume of a cuboid by measuring the length, width, and height and multiplying all of them by each other.

The presence of impurities in a sample can be detected by observing changes in the melting point of the substance.

When an impure sample is heated, its melting point range becomes broader and its melting point decreases. This is because the impurities disrupt the crystal lattice structure of the substance, making it easier to break apart and melt. In contrast, a pure substance will melt at a sharp and well-defined melting point.

Therefore, if the melting point range of a sample is broad and has a lower melting point than the expected value of the pure substance, it may contain impurities. Additionally, the appearance of a plateau or depression on the melting point curve is also indicative of impurities.

To confirm the presence of impurities, one can perform a mixed melting point test. This involves mixing a small amount of the sample with a known pure compound and taking the melting point of the mixture. If the melting point of the mixture is depressed and has a lower range than that of the pure compound alone, then the sample contains impurities. If the melting point range remains the same and is sharp, then the sample is likely pure.

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Approximately 777.6 grams of water is needed to make 7.2 moles of glucose based on the balanced equation.

The balanced equation provided is:

6CO2 + 6H2O -> C6H12O6 + 6O2

From the equation, we can see that for every 6 moles of water (H2O), 1 mole of glucose (C6H12O6) is produced. Therefore, we need to determine the amount of water required to produce 7.2 moles of glucose.

The mole ratio between water and glucose is 6:1. This means that for every 6 moles of water, we obtain 1 mole of glucose. To find the amount of water needed for 7.2 moles of glucose, we set up a proportion using the mole ratio:

(6 moles H2O / 1 mole glucose) = (x moles H2O / 7.2 moles glucose)

Solving for x, we can cross-multiply:

6 moles H2O * 7.2 moles glucose = x moles H2O * 1 mole glucose

43.2 moles H2O = x moles H2O

Therefore, we need 43.2 moles of water to produce 7.2 moles of glucose.

To convert moles of water to grams, we need to know the molar mass of water, which is approximately 18 g/mol. Using the molar mass, we can calculate the mass of water needed:

Mass of water = moles of water * molar mass of water

Mass of water = 43.2 moles * 18 g/mol

Mass of water = 777.6 g

Therefore, approximately 777.6 grams of water is needed to make 7.2 moles of glucose based on the balanced equation.

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INFORMATION:

We know that:

- sodium and oxygen react to produce Sodium Oxide

-

Option E) In the reaction A+ 2B-C that is found to be first order with respect to both a (0.200 m concentration and b (0.400 m concentration) the numeric value of k is 0.0875

Concentration in chemistry terms is calculated by dividing a constituent's abundance by the mixture's present total volume.

Mass concentration, molar concentration, number concentration, and volume concentration are four different categories that can be described in a mathematical description.

A + 2B → C

Rate = K [A] [B]

Here,

[A] = 0.2 M [B]

⇒ [A] = 0.4 M

rate = 0.007 M/s

Therefore, rate = K [A] [B]

⇒ 0.007 = K (0.2) (0.4)

⇒ K = 0.0875 M^(-1) S^(-1)

Thus, option E is correct.

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Complete question

The reaction A+ 2B-C is found to be first order with respect to both A and B. When the concentration of A is 0.200 M and the concentration of B is 0.400 M, the reaction rate is 0.00700 M/s. What is the numerical value of k for this reaction (with appropriate units where time is in seconds and concentration is in M, as needed).

A) 0.0438

B) 0.0140

C) 11.4

D) 5.60 x 10^(-4)

E) 0.0875

Answer:

Why does D-block start on the fourth row of the periodic table?

Explanation:

The require more energy to reach 3d than 4s, and they fill up 4p before 4d, and so forth. ... The energy is lower for 6s than 4f, and 4f and 5f are lower in energy than the final D level.

To calculate the Ka for the monoprotic acid, we can use the relationship between the concentration of the acid, the concentration of the conjugate base, and the pH of the solution. The Ka value for the monoprotic acid is approximately 5.7 × 10⁻⁵.

In a monoprotic acid, the acid (HA) donates one proton (H+) to water, forming the conjugate base (A-).

The dissociation of the acid can be represented as follows:

HA(aq) ⇌ H+(aq) + A-(aq)

Given that the pH of the solution is 2.71, we know that the concentration of H+ is 10^(-pH), which is 10⁻²°⁷¹ in this case.

Let's assume that the initial concentration of the acid HA is represented by [HA]_0.

At equilibrium, the concentration of H+ is equal to the concentration of A-. Therefore, [H+] = [A-].

Using the given concentration of the acid solution (1.78 M), we can assume that the concentration of the acid HA at equilibrium is (1.78 - [H+]) M.

The equilibrium expression for the acid dissociation is given by the equation:

Ka = ([H+][A-])/[HA]

Substituting the values we have:

Ka = ([H+][H+])/[(1.78 - [H+])]

Now, we can substitute [H+] = 10⁻²°⁷¹ into the equation and solve for Ka:

Ka = (10⁻²°⁷¹ * 10⁻²°⁷¹)/[(1.78 - 10⁻²°⁷¹)]

Ka ≈ 5.7 × 10⁻⁵

Therefore, the Ka value for the monoprotic acid is approximately 5.7 × 10⁻⁵.

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Answer:

The mass of chlorine gas that reacted ≅ 24.03 g

Explanation:

From the given information:

The number of moles of NaCl formed = mass of NaCl/molar mass

mass of NaCl = 39.6 g

molar mass = (23 + 35.5) g/mol = 58.5 g/mol

∴

The number of moles of NaCl formed = 39.6 g / 58.5 g/mol

The number of moles of NaCl formed = 0.6769 g/mol

Thus, the mass of chlorine gad that reacted = 0.6769 × 35.5

The mass of chlorine gas that reacted = 24.02995  g

The mass of chlorine gas that reacted ≅ 24.03 g

Answer: 1.00 m

Explanation:

Ignoring electron repulsion, the ground state energy of helium can be related to that of hydrogen by a factor of 4.

This is due to the fact that the ground state energy of an atom is determined by the total energy of its electrons, which is primarily determined by their positions and motions around the nucleus. In the case of hydrogen, the ground state energy is determined by the electron's interaction with the positively charged nucleus. This interaction results in a specific energy level that the electron can occupy.

However, in the case of helium, there are two electrons that occupy the same space and interact with the same nucleus. This results in the phenomenon known as electron repulsion, which makes the energy level of helium higher than that of hydrogen. To ignore electron repulsion means to assume that the two electrons in helium do not interact with each other, which allows us to treat helium as if it had only one electron.

By doing this, the energy level of helium becomes comparable to that of hydrogen, and we can relate them by a factor of 4, which is the ratio of the charge of the helium nucleus to that of the hydrogen nucleus. In summary, ignoring electron repulsion allows us to relate the ground state energy of helium to that of hydrogen by a factor of 4, which is a useful approximation in certain situations.

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Answer: C.  Waves don't move matter, just energy.

Explanation: