va compound is found to be made of 82.659% carbon and 17.341% hydrogen. which choice is the empirical formula for this compound?

Answer:

B- Candium

Explanation:

Because it has larger atomic radius than sulfur.

Answer:

Cadmium has larger atomic radius than sulfur.

Explanation:

Down a period, atomic radii decrease from left to right due to the increase in the number of protons and electrons across a period: when a proton is added the pull of the electrons towards the nucleus is larger, so the size of the atom decreases.

Hence, you can compare the elements that belong to a same period and predict that the atom with lower atomic number (number of protons) will haver larger atomic radius. With that:

Oxygen and fluorine are in the period 3, being oxygen to the left of fluorine, so oxygen is larger than fluorine.

Sulfur and chlorine are in the period 4, being sulfur to the left of chlorine, so sulfur is larger than chlorine.

Now see whan happens down a group. Atomic radius increases from top to bottom within a group due to electron shielding. That permits you to compare the size of the elements in a group:

Fluorine and chlorine are in the same group (17), with chlorine directly below fluorine, so the atomic radius of chlorine is larger than the atomic radius of fluorine.

Sulfur and oxygen are in the same group (16), with sulfur directlly below oxygen, so sulfur the atomic radius of sulfur is larger than the atocmi radius of oxygen.

So far, you can rank the atomic radius of sulfur, chlorine, fluorine, and oxygen, in increasing order as:

O < F < Cl < S, concluding that O, F, and Cl have smaller atomic radius than S.

Cadmiun, Cd, is to the left and below sulfur, so both electron shielding (down a group) and increase of the number of protons (down a period) lead to predict the cadmium has a larger atomic radius than sulfur.

The distance between the two points ( 8 , − 5 ) and ( 1 , 1 ) is determined as 9.2 units.

The distance between two points is calculated using the following equation.

for a given points (x1, y1), (x2, y2), the distance, d is;

d = √[(x2-x1)² + (y2 - y1)²]

From the given question;

x1 = 8, y1 = -5

x2 = 1, y2 = 1

distance = √[(1-8)² + (1 - -5)²]

distance = √[(-7)² + (16)²]

distance = 9.22 units

Thus, the distance between the two points ( 8 , − 5 ) and ( 1 , 1 ) is determined as 9.2 units.

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The ICE table depicts the change in the equilibrium position, which implies that whenever the change displays an increase in reactant concentration, there will be a corresponding drop in product concentration.

Initial, Change, Equilibrium is referred to as ICE. The fluctuating concentrations of components and reactants in (dynamic) equilibrium processes can be calculated using an ICE table. Before any modifications take place, this approach first reports the reactant and product concentrations for each sample.

The ICE table depicts the change in the equilibrium position, which implies that whenever the change displays an increase in reactant concentration, there will be a corresponding drop in product concentration.

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Answer:

d is the answer.........

If a solid sample of copper is an excellent conductor of electric current, it is likely due to the presence of metallic bonds in the sample.

Good conductors are the materials which offer very low resistance to the flow of electric current.

Solid copper is a metal, and metals have metallic bonds. Metallic bonds are a type of chemical bond that occur between atoms of metallic elements. In a metallic bond, valence electrons of the metal atoms are delocalized and shared among all the atoms in metallic solid, creating "sea" of electrons that are free to move throughout the structure.

This gives metals their unique properties such as high electrical conductivity and ductility. So, if solid sample of copper is an excellent conductor of electric current, it is likely due to the presence of metallic bonds in sample.

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Note: The question given on the portal is incomplete. Here is the complete question.

Question: A solid sample of copper is an excellent conductor of electric current . Which type of chemical bonds are in the sample?

Answer:

20°C and 2 atm is correct

Explanation:

Answer:

Ancient CO₂ levels can be measured from

1) Air bubbles trapped in deep ancient ice cores (up to 800,000 years ago) or

2) Tree rings (up to 10,000 years ago)

Explanation:

The percentages of CO₂ concentration in the atmosphere from previous or ancient times are found by analyzing the CO₂ content in trapped air located in the deep ancient ice cores present in Antarctica and Greenland such as the EPICA Dome C cores

As the ice is drilled though, the air released from trapped bubbles are captured for analysis and the time frame for the captured bubbles can date back up to 800,000 years

Ancient CO₂ levels can also be measured from tree rings which provide ancient CO₂ levels up to 10,000 years

Answer:

Pathway #1 will give a higher yield because there is much less steric hindrance for primary substrates than for secondary substrates in the SN2 reaction, which is the first mechanism in each of the two pathways.

The SN2 reaction proceeds via a concerted, bimolecular, one-step mechanism, where the nucleophile attacks the electrophilic carbon center, and the leaving group is simultaneously displaced. In pathway #1, the SN2 reaction involves a primary substrate with a less hindered leaving group, which will react faster and give a higher yield.

On the other hand, pathway #2 involves a secondary substrate with a more hindered leaving group, which will react slower and give a lower yield. Hence, pathway #1 is more efficient because of the higher reactivity of the primary substrate and the less steric hindrance in the reaction mechanism.

Therefore, the pathway #1 will give a higher yield than pathway #2 in a multistep reaction pathway because there is much less steric hindrance for primary substrates than for secondary substrates in the SN2 reaction, which is the first mechanism in each of the two pathways.

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Answer:

74

Explanation:

The law of conservation of mass states that in a chemical reaction, the total mass of reactants is equal to the total mass of products

hope this helps!

The total mass of the products will also be 74 g

From the above law, we understood that in any chemical reaction, the total mass of reactants must be equal to the total mass of products i.e

From the question given above,

Total mass of reactants = 74 g

Total mass of reactants = Total mass of products

74 g = Total mass of products

Thus,

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First, we need to determine the number of moles of hydroiodic acid in 100mL of 8.5M solution. Therefore, 1.98 g of calcium iodide forms in the reaction.

8.5 moles/L x 0.100 L = 0.85 moles of hydroiodic acid

Next, we need to determine the number of moles of calcium hydroxide used in the reaction.

0.500 g Ca(OH)2 x (1 mol Ca(OH)2/74.09 g) = 0.00674 moles Ca(OH)2

Since the stoichiometry of the reaction is 1:2 between Ca(OH)2 and HI, we need twice as many moles of hydroiodic acid to completely react with all the Ca(OH)2.

0.00674 moles Ca(OH)2 x 2 = 0.0135 moles HI needed

Since we have 0.85 moles of HI, which is more than enough to react with 0.0135 moles of Ca(OH)2, the reaction is complete.

To determine how much salt (calcium iodide) forms, we need to use the stoichiometry of the reaction, which is:

Ca(OH)2 + 2HI ⇒ CaI2 + 2H2O

Since two moles of hydroiodic acid react with one mole of calcium hydroxide to form one mole of calcium iodide, the amount of salt formed is:

0.00674 moles Ca(OH)2 x (1 mol CaI2/1 mol Ca(OH)2) = 0.00674 moles CaI2

Finally, we can calculate the mass of calcium iodide formed using its molar mass:

0.00674 moles CaI2 x 293.88 g/mol = 1.98 g CaI2

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After 10 seconds, 88 grams of unobtanium-31 are present, after 20 seconds, 44 grams are present, after 30 seconds, 22 grams are present, after 40 seconds, 11 grams are present, and after 50 seconds, 5.5 grams are present.

The amount of unobtanium-31 remaining after certain intervals of time, given that the half-life of the radioactive element is 10 seconds is calculated below.

Initial quantity of the element is given as 176 grams.  

After 10 seconds:

Let's first figure out the fraction of the sample remaining after each time period because the half-life of a radioactive element tells us the fraction that decays over a certain period of time.

Using the half-life equation:

amount remaining = original amount x (1/2)^(time elapsed / half-life)

After 10 seconds, the time elapsed is equal to the half-life of the element,

So:

amount remaining = 176 x (1/2)^(10/10)

                               = 88 grams

After 20 seconds:

amount remaining = 176 x (1/2)^(20/10)

                               = 44 grams

After 30 seconds:

amount remaining = 176 x (1/2)^(30/10)

                               = 22 grams

After 40 seconds:

amount remaining = 176 x (1/2)^(40/10)

                               = 11 grams

After 50 seconds:

amount remaining = 176 x (1/2)^(50/10)

                               = 5.5 grams

Therefore, after 10 seconds, 88 grams of unobtanium-31 are present, after 20 seconds, 44 grams are present, after 30 seconds, 22 grams are present, after 40 seconds, 11 grams are present, and after 50 seconds, 5.5 grams are present.

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Answer:

N2O is an empirical formula

Explanation:

A.) P4O10   NO:  Can be reduced to P2O5

B.) C2H4    NO:  Can be reduced to CH2

C.) H2O2     NO:  Can be reduced to H O

D.) N2O     YES, the formula cannot be reduced any further.

Decreasing vapour pressure for the following compounds is as follow:

CH4

CH3CH-CH3CH2CH3

CH3CH2CH2CH2CH3  

CH3CH2CH2CH2OH  

Vapor Pressure of the compound:

The pressure characteristic of a pure compound's vapour at any given temperature when it is in equilibrium with its liquid or solid state is known as the vapour pressure. Compound molecules that bond well with one another will have a low vapor pressure (less inclination to escape to the vapor phase), whereas compounds that connect poorly with one another would have a high vapor pressure. Vapor pressure is a measure of a compound's capacity to bond with itself.

The number of solvent molecules escaping into vapour phase gets reduced and as a result the pressure exerted by the vapour phase is also reduced. This is known as relative lowering of vapour pressure.

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The ph of the basic solution will always be having the ph value above 7 in the ph scale.

what is the ph scale  ?

The pH scale is a numerical scale ranging from 0 to 14. It indicates whether an aqueous solution is acidic or alkaline. The pH scale is used to determine whether a solution is acidic, alkaline, or neutral.

Neutral solutions have a pH of 7.

The pH of acidic liquids is less than 7. The closer a solution is near pH 0, the more acidic it is.

The pH of alkaline solutions is greater than 7. The closer a solution is near pH 14, the more alkaline it is.

Acidic, alkaline, or neutral solutions are all possible. Pure water, for example, has a pH of 7 and is neutral. Because a solution of common salt (sodium chloride) is neutral, it has a pH of 7.

The word aqueous solution is frequently used in science. It is used to define a water-based solution. Only water-based solutions may be used to assess pH.

As a result, we may determine the pH of a solution whic is basic will be having the ph value above seven.

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You are probably asked to convert the given number of methane (CH4) molecules into moles.

1.5 x 10^20 molecules of CH4 is  to 0.0249 moles of CH4

The atomic mass of carbon =  12.01 g/mol,

the atomic mass of hydrogen=  1.008 g/mol.

The molecular weight of CH4 is shown below:

Molecular weight CH4 = (1 x 12.01 g/mol) + (4 x 1.008 g/mol)

Molecular weight CH4   = 16.04 g/mol

Number of moles = number of molecules / Avogadro's number

Number of moles = (1.5 x 10^20) / (6.022 x 10^23 molecules/mol)

Number of moles = 0.0249 moles

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Answer:

C

Explanation:

Answer: decrease I believe

Explanation:

Just because if you replaced copper with iron it gets stronger

The value of the solubility product constant (Ksp) is 1.6352 × 10^-39 mol^2/L^2.

The solubility of T3U2 is given as 4.04 × 10^-20 mol/L. This means that for every liter of the solution, 4.04 × 10^-20 moles of T3U2 dissolves.

The dissociation of T3U2 can be represented by the equation:

T3U2 ⇌ T2+ + U3-

Let's assume that x moles of T3U2 dissolve, resulting in the formation of x moles of T2+ and x moles of U3-. Since the stoichiometry of the reaction is 1:1, the concentration of T2+ and U3- ions in the solution will also be x mol/L.

The solubility product constant (Ksp) is defined as the product of the concentrations of the dissociated ions raised to the power of their respective stoichiometric coefficients:

Ksp = [T2+]^1 × [U3-]^1

Substituting the values, we have:

Ksp = (x)^1 × (x)^1 = x^2

We know that the solubility of T3U2 is 4.04 × 10^-20 mol/L, so x = 4.04 × 10^-20 mol/L.

Therefore, the value of the solubility product constant (Ksp) is:

Ksp = (4.04 × 10^-20 mol/L)^2 = 1.6352 × 10^-39 mol^2/L^2.

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a. Toluene chlorination in sunlight forms chloro methyl benzene via free radicals. b. Phenol reacts with concentrated sulfuric acid to produce phenol-4-sulfonic acid, with odor-based differentiation. EILCILCHCICOOH is stronger due to a stronger electron-withdrawing group, while CH CHFCOOH is weaker due to an electron-donating group.

a. Chlorination of toluene in the presence of sunlight:

The structural formula for the chlorination of toluene is as follows:

\(CH_3-C_6H_5\) + \(Cl_2\) -> \(CH_3-C_6H_4-Cl\) + HCl

Reaction conditions: Sunlight

Mechanism:

The reaction proceeds through a free radical mechanism. In the presence of sunlight, the chlorine molecule (\(Cl_2\)) undergoes homolytic cleavage to form chlorine radicals (Cl•).

One of these radicals abstracts a hydrogen atom from the methyl group of toluene, resulting in the formation of a methyl radical and HCl.

The methyl radical then reacts with another chlorine molecule to form the product, chloro methyl benzene, and regenerate a chlorine radical.

b. Sulphonation of phenol:

The structural formula for the sulphonation of phenol is as follows:

\(C_6H_5OH + H_2SO_4 - > C_6H_4(OH)SO_3H + H_2O\)

Reaction conditions: Concentrated sulfuric acid (\(H_2SO_4\))

Observations for differentiation between the compounds:

a. buntanal, benzaldehyde, ethanol:

- Buntanal: It is an aldehyde and exhibits a fruity odor.

- Benzaldehyde: It is an aromatic aldehyde and has a distinct almond-like smell.

- Ethanol: It is a colorless liquid with a characteristic alcoholic odor.

b. 2-pentanone, 3-hexanone, hexanoyl chloride:

- 2-pentanone: It is a ketone with a fruity odor.

- 3-hexanone: It is a ketone and has a slightly sweet, fruity smell.

- Hexanoyl chloride: It is an acyl chloride and has a pungent odor similar to that of chlorine.

Explanation of stronger acid pairs:

- EILCILCHCICOOH: This compound has an electron-withdrawing group (chlorine) attached to the carboxylic acid functional group, making it more acidic.

- CI CHCICH COOH: This compound has a stronger electron-withdrawing group (trichloromethyl) attached to the carboxylic acid functional group, making it even more acidic than the first compound.

- CH CHCICOOH: This compound has an electron-donating group (methyl) attached to the carboxylic acid functional group, which reduces its acidity compared to the next compound.

- CH CHFCOOH: This compound has a weaker electron-withdrawing group (trifluoromethyl) attached to the carboxylic acid functional group, making it less acidic than the previous compound.

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Answer:

0.136mol

Explanation:

Using the general gas law equation as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (Litres)

n = number of moles (mol)

R = gas constant (0.0821 Latm/mol K)

T = temperature (K)

K = °C + 273

K = 50°C + 273

K = 323K

According to this question, P = 3.6atm, T = 323K, V = 1.00L, R = 0.0821 Latm/mol K, n = ?

3.6 × 1 = n × 0.0821 × 323

3.6 = 26.52n

n = 3.6/25.52

n = 0.1357

n = 0.136mol

Answer:

high

Explanation:

The amount of the original isotope that remains after 30 days is 11.14.

The half life of a radioactive nuclide has to do with the time that it takes for only half of the original number of nuclides to remain.

Now;

No = 150 grams

N = ?

t1/2 = 8 days

t = 30 days

Then;

N/No = (1/2)^t/t1/2

N/150 = (1/2)^30/8

N/150 = (1/2)^3.75

N = (1/2)^3.75 * 150

N = 11.14

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Given the half-life, the amount remaining is 11.14 g

The half-life of idine-131 is 8 days.

Using the half-life formula:

Nt = No * (0.5)^(t/t1/2)

Amount remaining = 150 * (05)^(30/8)

Amount remaining = 11.14 g

In conclusion, the amount remaining is determined from the half-life of the substance.

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The pH of a solution with an [OH⁻] concentration of 6.9 x 10⁻¹¹ is (a) 3.84 M.

To determine the pH of a solution with an [OH⁻] concentration of 6.9 x 10⁻¹¹, we can first find the pOH using the formula:

pOH = -log₁₀([OH⁻])

pOH = -log₁₀(6.9 x 10⁻¹¹) ≈ 10.16

Next, we can use the relationship between pH and pOH:

pH + pOH = 14

Now, we can calculate the pH:

pH = 14 - pOH = 14 - 10.16 ≈ 3.84

Thus, the pH of the solution is approximately 3.84 (option A). This indicates that the solution is weakly acidic, as pH values less than 7 represent acidic conditions. Keep in mind that the pH scale ranges from 0 to 14, with 7 being neutral (neither acidic nor basic). The given [OH⁻] concentration reflects a higher concentration of H⁺ ions, resulting in a lower pH value.

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Answer:

I dont quit understand this question.

Explanation:

Try typing it again

The correct statement(s) regarding the van der Waals equation for gases are a low value for a reflects weak intermolecular forces among the gas molecules and Among the gases H2, N2, CH4, and CO2, H2 has the lowest value for a.

The van der Waals equation is used to describe the behavior of real gases by taking into account their intermolecular forces and non-zero molecular volumes, which are ignored in the ideal gas law. The equation is given by (P + a(n/V)^2)(V - nb) = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, T is the temperature, a is a constant that reflects the strength of the intermolecular forces, and b is a constant that reflects the size of the molecules.

A low value for a indicates weak intermolecular forces among the gas molecules, while a high value for a indicates strong intermolecular forces. Therefore, statement 1 is correct.

Among the gases H2, N2, CH4, and CO2, H2 has the lowest value for a because it has the weakest intermolecular forces among the gases listed. Therefore, statement 3 is also correct.

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At a temperature of 554.3 K, a 0.00330 m solution of glucose and water has an osmotic pressure of 0.150 atm.

Osmotic pressure is a colligative property of a solution that is it is dependent on the concentration of the solution.

Osmotic pressure = icRT

where c is the concentration

R is the gas constant

T is the temperature

i is the van't hoff factor

Given,

c = 0.0033 m

R = 0.082 L atm mol⁻¹ K⁻¹

osmotic pressure = 0.150 atm

i = 1 for glucose as it neither associates nor dissociates

0.150 = 0.0033 * 0.082 * T

T = 554.3 K

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Answer:

2 x 10^5 liters

Explanation:

P1V1 = P2 V2

P1V1 / P2  = V2  =  .05 *4 x 10^6  / 1  = .2 x 10^6  liters