Which substance will not react with hydrochloric acid to form a salt?AgZnCuCO3Ca(OH)2

Answer:

Answer is D

Explanation:

an increase in the number of protons

The mass of oxygen dissolved in water has been 49.4 mg.

The partial pressure of oxygen in the air:

Partial pressure = Mole fraction  Pressure of gas

Partial pressure = 0.21  1.13 atm

Partial pressure of oxygen = 0.237 atm.

The concentration of oxygen can be given by Henry's law.

Concentration of Oxygen = Henry's constant (\(K_{H}\))  Partial pressure

Concentration = \(1.3 * 10^{-3} M/atm * 0.237 atm\)

Concentration of oxygen = \(3.08 * 10 ^{-4} M\)

Moles can be given by:

Moles = Molarity * Volume

Moles of oxygen = \(3.08 * 10^{-4} M * 5\)

Moles of oxygen = 1.542 \(* 10^{-3} mol\)

Moles = \(\frac{weight}{molecular weight}\)

Weight of oxygen = Moles  Molecular weight

Weight of oxygen = 1.542\(* 10^{-3} mol * 32 g/mol\)

Weight of oxygen = 0.0494 grams

Weight of oxygen = 49.4 mg.

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An example of an aromatic nucleophilic substitution would be this. The mechanism is as described previously.

Nucleophilic describes a type of chemical reaction in which a nucleophile (a molecule or ion with an electron-rich nucleus) attacks an electron-deficient species, such as an electrophile or a molecule with a partially positive charge. In a nucleophilic reaction, the nucleophile donates electrons to form a covalent bond with the electrophilic species, resulting in the formation of a new molecule.

The second portion of the query seems a bit hazy. But I'm assuming it's a comparison of the reactivity of 2, 3, and 4-substituted chloro pyridine. Essentially, the reaction is a nucleophilic substitution. Therefore, it only occurs in electron-poor centres. When we look at the magnetic structures of pyridine, the 2nd and 4th positions are electron deficient centres.

In order to create the corresponding pyridine methoxide, sodium methoxide and 2 and 4 chloropyridine will react. In contrast, sodium methoxide won't react with 3 chloropyridine.

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Answer:

50°F,  70°F

Explanation:

Answer:

50 and 70 degrees F

Explanation:

got it right on the quiz on edge

Answer:

the higher the temperature the more molecules are spread out and the lower density,high pressure means high density please make it a brainlist answer

The entropy change of the universe is zero, which is consistent with the second law of thermodynamics. Therefore, the entropy of the universe remains constant or the change in entropy is zero.

Answer:AScoffee = -1.71 J/K, ASroom = 1.71 J/K, ASuniverse = 0.

According to the second law of thermodynamics, entropy of an isolated system always increases. In the given question, we need to calculate the entropy change of coffee, room, and universe. Let's solve this problem.

Given: Mass of coffee = 170 g (density of water = 1 g/cm³)

Specific heat of coffee = specific heat of water = 4.18 J/(g·°C)

Temperature of coffee = 88°C

Temperature of room = 20°C

The initial temperature of coffee = 88°C

The final temperature of coffee = 20°C

Change in temperature of coffee (ΔT) = final temperature - initial temperature = 20°C - 88°C = -68°C

Let's calculate the entropy change of coffee. Entropy change of coffee:

AScoffee = -[170 g/(1000 g/1 kg)](4.18 J/(g·°C))ln[(20°C + 273 K)/(88°C + 273 K)]

AScoffee = -[0.17 kg](4.18 J/(g·°C))ln(293 K/361 K)

AScoffee = -1.71 J/K

Now, let's calculate the entropy change of the room. The change in entropy of the room would be equal and opposite to the change in entropy of coffee (based on the principle of energy conservation).

ASroom = -AScoffeeASroom = 1.71 J/K

The entropy change of the universe would be the sum of entropy change of coffee and entropy change of the room.

ASuniverse = AScoffee + ASroomASuniverse = -1.71 J/K + 1.71 J/KA

Suniverse = 0

The entropy change of the universe is zero, which is consistent with the second law of thermodynamics. Therefore, the entropy of the universe remains constant or the change in entropy is zero.

Answer:AScoffee = -1.71 J/K, ASroom = 1.71 J/K, ASuniverse = 0.

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In the esterification experiment, the apparatus was set up at an angle to ensure proper mixing and efficient heat distribution. This angled setup promotes contact between the reactants and allows the esterification process to occur effectively, leading to the formation of the desired ester product.

In the esterification experiment, the apparatus was set up on an angle to ensure that the reaction mixture could be stirred efficiently. Esterification is a slow reaction, and it requires constant stirring to improve the contact between the reactants and to prevent the formation of a boundary layer around the reaction flask. Setting up the apparatus on an angle allows the reactants to move freely and come into contact with each other more effectively, thus improving the rate of the reaction. Additionally, the angle also helps to prevent any potential backflow of the reaction mixture into the condenser, which could cause contamination or damage to the apparatus.

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One of the crucial tests for sand is the moisture content test. Formula for calculating sand's moisture content: [(W-Wd) / W] x 100% = 20%

The term "moisture content" (MC) refers to how much moisture is present in a given material. This value is frequently expressed as a percentage of the mass of the substance (e.g., X% MC). A variety of techniques, including oven-dry tests and moisture meters, can be used to determine how much moisture is present in a given product.

Calculate the difference between the initial dried sample weight and the sample weight after secondary drying. Next, double this number by the sample's weight after secondary drying. Afterward, multiply this number by 100.

The process relies on oven-drying a soil sample until the weight stays constant to remove soil moisture. The sample weight before and after drying is used to compute the moisture content (%). Special equipment. For soils with fine grained (maximum particle.

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A substitution reaction has two steps, and doubling the concentration of the alkyl halide is found to double the reaction rate. This is a bimolecular SN2 reaction is a false statement.

Substitution reaction is a single step reaction in which one functional group is substitute by another functional group. An example is halogenation reactions, in which a functional group is replaced by a halogen.

Thus, the statement is false.

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The strawberry Koolaid solution has a pH of 5.4, making it mildly acidic.

The pH scale is 0 to 14, with 7 indicating neutral. Any pH number less than 7 is considered acidic, whereas any pH value greater than 7 is considered basic or alkaline. Pure water has a pH of 7, which is considered neutral. As a result, a pH value less than 7 suggests that the solution contains more hydrogen ions (H+) than hydroxide ions (OH-). To put it a different way, the solution is acidic.

Citric acid, which is added to strawberry Koolaid to give it a tangy flavor, is a flavoring for drink mixes. Citric acid is a weak organic acid used as a food preservation and flavoring ingredient. Because citric acid is present, when the Koolaid mix is mixed in water, it produces a somewhat acidic solution.

As a result of the pH of 5.4, we can deduce that the strawberry Koolaid solution is mildly acidic.

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Answer: The enthalpy of the reaction is -1791.31 kJ.

Explanation:

Enthalpy change is the difference between the enthalpies of products and the enthalpies of reactants each multiplied by its stoichiometric coefficients. It is represented by the symbol \(Delta H^o_{rxn}\)

\(\Delta H^o_{rxn}=\sum (n \times \Delta H^o_{products})-\sum (n \times \Delta H^o_{reactants})\)        .....(1)

For the given chemical reaction:

\(3MnO_2(s)+4Al(s)\rightarrow 2Al_2O_3(s)+3Mn(s)\)

The expression for the enthalpy change of the reaction will be:

\(\Delta H^o_{rxn}=[(2 \times \Delta H^o_f_{(Al_2O_3(s))}) + (3 \times \Delta H^o_f_{(Mn(s))})] - [(3 \times  \Delta H^o_f_{(MnO_2(s))}) + (4 \times \Delta H^o_f_{(Al(s))})]\)

Taking the standard heat of formation values:

\(\Delta H^o_f_{(Al_2O_3(s))}=-1675.7kJ/mol\\\Delta H^o_f_{(Al(s))}=0kJ/mol\\\Delta H^o_f_{(MnO_2(s))}=-520.03kJ/mol\\\Delta H^o_f_{(Mn(s))}=0kJ/mol\)

Plugging values in the above expression:

\(\Delta H^o_{rxn}=[(2 \times (-1675.7))+(3 \times 0)] - [(3 \times (-520.03))+(4 \times 0)]\\\\\Delta H^o_{rxn}=-1791.31 kJ\)

Hence, the enthalpy of the reaction is -1791.31 kJ.

Answer:

D is the answer

Explanation:

Answer:

This should help you with your problem

The symbol that is used to indicate  “produces” or “yields” is "→"

Forward arrow is used to indicate the reaction moving towards products.

H₂SO₄ (aq) + 2NaOH(aq) →  Na₂SO₄(s) + 2H₂O (l)

This reaction yields or produces sodium sulphate and water when sulphuric acid and sodium hydroxide reacts.

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The number of moles present in 175ml of Ba(OH)₂ is 0.105 moles.

To find the moles of Ba(OH)₂ present in 175 mL of 0.600 M Ba(OH)2 solution, you can use the formula:

Moles = Molarity × Volume (in liters)

Molarity = moles of solute/liters of solution

where "moles of solute" refers to the amount of substance being dissolved, measured in moles, and "liters of the solution" refers to the volume of the solution in liters.

First, convert the volume from mL to L:
175 mL = 175 / 1000 = 0.175 L

Next, put the values into the formula:
Moles = 0.600 M × 0.175 L

Moles = 0.105 moles of Ba(OH)₂

Therefore, there are 0.105 moles of Ba(OH)₂ present in 175 ml of 0.600 M Ba(OH)₂.

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Since the temperature is kept constant, the new volume of the oxygen gas is 33.17 mL.

In order to calculate the new volume of this gas oxygen, we would apply Boyle's law. Mathematically, Boyle's law is given by this formula:

PV = k

P₁V₁ = P₂V₂

Where:

By making V₂ the subject of formula, we have the following:

V₂ = (P₁V₁)/P₂

By substituting the given parameters into the formula for Boyle's law, we have the following:

V₂ = (4.50 × 14.3)/1.94

V₂ = 64.35/1.94

V₂ = 33.17 mL.

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Answer:

Human beings are anatomically similar and related to the great apes but are distinguished by a more highly developed brain and a resultant capacity for articulate speech and abstract reasoning.

Explanation:

Yes, tert-butoxide anion (t-BuO-) is a strong enough base to react with water. A solution of potassium tert-butoxide can be prepared in water.

The pKa values are a measure of acidity, where lower pKa values indicate stronger acids. Conversely, higher pKa values indicate weaker acids. In the case of tert-butyl alcohol (t-BuOH), which can deprotonate to form tert-butoxide anion (t-BuO-), its pKa is approximately 18.

Comparing the pKa of t-BuOH with the pKa of water (15.74), we can see that water is a weaker acid than t-BuOH. Therefore, t-BuO- can act as a stronger base than water.

When a strong base like t-BuO- is added to water, it will react with water to form hydroxide ions (OH-) through the following equilibrium reaction:

t-BuO- + H2O ⇌ t-BuOH + OH-

This reaction results in an increase in the concentration of hydroxide ions (OH-) in the solution, making it basic.

Based on the comparison of pKa values, tert-butoxide anion (t-BuO-) is a strong enough base to react with water, allowing the preparation of a solution of potassium tert-butoxide in water.

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Answer:

BARIUM HYDROXIDE IS THE COMPOUND

Pure elements has Gibbs free energy equal to 0.

The Gibbs free energy is a concept in chemistry specially of thermodynamics. The maximum amount of work that can be accomplished at a constant temperature and pressure by a closed system can be calculated using the Gibbs free energy (also known as Gibbs energy; symbol: denoted as delta G). Additionally, it offers a prerequisite for any processes like chemical reactions that might take place in such circumstances.

When a system achieves equilibrium without being pushed by an input electrolytic voltage, the Gibbs energy is the thermodynamic potential that is reduced. At the equilibrium point, its derivative w.r.t. the system's reaction coordinate vanishes.

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The half-life of a radioisotope if 25.0 grams of an original 200. -gram sample of the isotope remains unchanged after 11.46 days is 3.82 days.

The radioactive isotopes of an element are known as radioisotopes. They can also be described as atoms with an excessive amount of energy in their nucleus or unstable neutron-proton combinations.

A radioisotope's unstable nucleus can form either naturally or as a result of atomic manipulation. Radioisotope production can take place using a cyclotron or a nuclear reactor, depending on the situation. Molybdenum-99, which has a high neutron content, can be produced most efficiently in nuclear reactors, whereas fluorine-18, which has a high proton content, can be produced most efficiently in cyclotrons.

Uranium is the most well-known illustration of a naturally-occurring radioisotope. Uranium-238 makes up 99.9% of naturally occurring uranium; the remaining 0.7% is the less stable and more radioactive uranium-235, which has three fewer neutrons in its nucleus.

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The provided statement states that a radioisotope's half-life is 3.82 days if 25.0 milligrams of a 200.0-gram initial sample of something like the isotope remain intact after 11.46 days.

An unstable chemical element that releases radiation as it breaks down and becomes more stable. Both natural habitat and then a laboratory can manufacture radioisotopes. They are used in medical treatments and imaging examinations. also called a radionuclide.

Since we now know that first phase kinetics governs radioactive decay,

Therefore, the following is the equation for the rate rule for first order kinetics:

\(k=\frac{2.303}{t} \log \frac{a}{a-x}\)

where,

k = rate constant

t = time taken for decay process  = 11.46 days

a = initial amount of the reactant  = 200 g

After the decay process, the quantity remaining is equal to 25 g.

By plugging data into the equation above, we can get the rate constant.

\(k=\frac{2.303}{11.46} \log \frac{200}{25}=0.1814\)

We now need to determine a radioisotope's half life.

Formula used : \(t_{1 / 2}=\frac{0.693}{k}\)

We obtain the half life by entering the value of "k" into this calculation.

\(t_{1 / 2}=\frac{0.693}{0.1814}=3.820\)

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The complete question is-

What is the half-life of a radioisotope if 25.0 grams of an original 200.-gram sample of the isotope remains unchanged after 11.46 days?

(1) 2.87 d

(2) 11.46 d

(3) 3.82 d

(4) 34.38 d

The volume of the dry nitrogen gas at STP is 57.4 mL

From the question given above, the following data were obtained:

Initial volume (V₁) = 77 mL

Initial temperature (T₁) = 50 °C = 50 + 273 = 323 K

Initial pressure (P₁) = 763 – 92.5 = 670.5 mmHg

Final temperature (T₂) = STP = 273 K

Final pressure (P₂) = 760 mmHg

The volume of the dry nitrogen can be obtained by using the combine gas equation as follow:

\( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \\ \\ \frac{670.5 \times 77}{323} = \frac{760 \times V_2}{273} \\ \\ cross \: multiply \\ \\ 323 \times 760 \times V_2 = 670.5 \times 77 \times 273 \\ \\ divide \: both \: side \: by \: 323 \times 760 \\ \\ V_2 = \: \frac{670.5 \times 77 \times 273 }{323 \times 760} \\ \\ V_2 = 57.4 \: ml\)

Therefore, the volume of the dry nitrogen gas is 57.4 mL

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Answer: There is a couple different ways to determine if a bond is ionic or covalent. By definition, an ionic bond is between a metal and a nonmetal, and a covalent bond is between 2 nonmetals. So you usually just look at the periodic table and determine whether your compound is made of a metal/nonmetal or is just 2 nonmetals.

Answer:

Protons:

- positive

- aka cation

- in the nucleus along with the neutrons

Electrons:

- negative

- aka anion

- situated in the orbital shells/configuration levels (there are many names)

Answer: Not possible to find average amu with only the information provided.

Explanation:

Is there more information about the other isotopes?  I don't see how one could calculate an overall atomic mass with only one isotope is known.  An isotope of 186.956 with abundance of 62.6% isn't enough information.  We're not told anything about the other isotopes:  are they larger, smaller, and at what percentage.  If one has an amu of  204.956 at 37.4%, the overall atomic mass would be quite different than if it had an atomic mass of 182.956 at 37.4%.  More infornmation is required.

When an ideal gas is isobarically compressed to one-third of its initial volume, the resulting pressure is four times its initial pressure. Here option D is the correct answer.

When an ideal gas is compressed isobarically, the pressure remains constant while the volume changes. As per Boyle's law, the product of pressure and volume is constant at a constant temperature.

Therefore, if the volume of an ideal gas is compressed to one-third of its initial volume, the pressure must increase by a factor of three to maintain the product of pressure and volume constant. This is because the final volume is one-third of the initial volume, so the pressure must be three times larger to keep the product of pressure and volume the same.

Therefore, if the initial pressure is P, the final pressure after compression will be 3P. However, the question asks for the resulting pressure, which is the final pressure after compression. Therefore, the resulting pressure will be three times the initial pressure, i.e., 3P.

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Complete question:

Which of the following is the resulting pressure when an ideal gas is compressed isobarically to one-third of its initial volume?

A) One-third of its initial pressure

B) Two-thirds of its initial pressure

C) Three times its initial pressure

D) Four times its initial pressure

The appearance of the protist differ from that of the onion sample due to the presence of motile structures.

This is an eukaryote which could be unicellular or multicellular and examples include protozoa etc.

The major difference between a protist and onion sample is that protists have motile structures such as flagella, cilia etc while plant cells such as onions don't have.

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C/ If Fe is greater than Fg, the negatively charged oil droplet will move freely toward the negatively charged plate.

In the Millikan oil droplet experiment, the negatively charged oil droplets are subjected to an electric field created by the charged plates. The electric force (Fe) acts on the oil droplet in a direction opposite to the gravitational force (Fg). When Fe is greater than Fg, the electric force overcomes the gravitational force, causing the negatively charged oil droplet to experience an upward force. As a result, the oil droplet moves freely upward toward the negatively charged plate.

Option B is incorrect because if Fe is equal to Fg, the forces balance each other, resulting in a stationary droplet. However, the question states that Fe is increased until it is greater than Fg, implying that the droplet is no longer stationary but moves in response to the electric force.

Therefore, option C is the correct answer, as it describes the effect of an electric field on the motion of a negatively charged oil droplet in the Millikan oil droplet experiment.

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The mass of a cube of magnesium : 322.2 g

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

The unit of density can be expressed in g / cm³ or kg / m³

Density formula:

\(\large{\boxed {\bold {\rho~=~ \frac {m} {V}}}}\)

ρ = density , g / ml

m = mass , g

v = volume , ml

mass of aluminum = 500 g

density of aluminium = 2.7 g/ml

there is data that needs to be added⇒ density of magnesium : 1.74 g/ml

the same dimensions = the same volume

\(\tt V=\dfrac{m}{\rho}\\\\V=\dfrac{500}{2.7}=185.19~ml\)

V magnesium = V aluminum

\(\tt m=\rho\times V\\\\m=1.74\times 185.19\\\\m=\boxed{\bold{322.2~g}}\)

The cube of same dimensions of magnesium and aluminum have same volume. The mass of cube of magnesium has been 321.85 grams.

The density has been given as the mass of substance per unit volume. The density of aluminum cube has mass 500 grams, and density 2.7 g/mL. The volume of the aluminum cube has been:

\(\rm Density=\dfrac{Mass}{Volume}\\\\2.7\;g/mL=\dfrac{500\;g}{Volume} \\\\Volume=185.18\;mL\)

The volume of the magnesium and aluminum cube has been the same. The density of magnesium has been 1.735 g/mL. Thus, the mass of Mg cube has been given as:

\(\rm 1.738\;g/mL=\dfrac{Mass}{185.18\;mL}\\\\ Mass=321.85\;g\)

Thus, the mass of magnesium cube has been 321.85 grams.

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