1. if the pKa of HCHO2 is 3.74 and the pH of an HCHO2/NaCHO2 solution is 3.00, which of the following is TRUE?
a. [HCHO2] > [NaCHO2]
b. [HCHO2] < [NaCHO2]
c. [HCHO2] << [NaCHO2]
d. it is not possible to make a buffer of this ph from HCHO2 and NaCHO2.
e. [HCHO2] = [NaCHO2]

(a) Ionic compounds have strong electrostatic forces of attraction and Covalent compounds have weaker forces of attraction between their molecules. (b) Ionic compounds are generally soluble in water, while covalent compounds are often insoluble in water. (c) Ionic compounds conduct electricity when dissolved in water or melted and Covalent compounds, do not conduct electricity.

Ionic and covalent compounds are two types of chemical compounds that differ in their properties. Here are the differences between ionic and covalent compounds based on three properties:

(a) Strength of forces between constituent elements: Ionic compounds are formed by the transfer of electrons from one atom to another, resulting in the formation of ions with opposite charges that are held together by strong electrostatic forces of attraction. Covalent compounds are formed by the sharing of electrons between atoms, resulting in the formation of molecules that are held together by weaker intermolecular forces.

(b) Solubility of compounds in water: Ionic compounds are generally soluble in water because they can dissociate into ions that can be hydrated by water molecules. Covalent compounds are generally insoluble in water because they do not dissociate into ions and are not attracted to water molecules.

(c) Electrical conduction in substances: Ionic compounds can conduct electricity when dissolved in water or when melted because the ions are free to move and carry an electric charge. Covalent compounds do not conduct electricity because they do not have free ions to carry an electric charge.

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Calcium on reaction with water form calcium hydroxide which is sparingly soluble whereas the hydroxides of sodium and potassium are soluble in water.

Water reacts with calcium, magnesium, potassium, and sodium to form its hydroxide compounds. The amount of calcium hydroxide in water has changed.

Due to the compound's extremely poor solubility, calcium hydroxide appears opaque. In comparison to other oxides, calcium hydroxide has an extremely low Ksp (solubility product).

Magnesium, potassium, and sodium hydroxides are soluble in water. Therefore, these substances don't cause water to get hazy.

Because phenolphthalein is a basic substance, the solution turns pink when it is added. In a base, phenolphthalein has a pink colour; in an acid, it has no colour.

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1,2-dibromo-1,2-diphenylethane is prepared through the bromination of trans-stilbene, a reaction involving an electrophilic addition mechanism.

The reaction starts with the generation of a bromine radical (Br•) by a free-radical initiator. This radical reacts with trans-stilbene, producing a brominated stilbene radical (Ph-CH=CH-Ph•Br). The brominated radical further reacts with another bromine radical to form the final product, 1,2-dibromo-1,2-diphenylethane (Ph-CHBr-CHBr-Ph).

Arrow pushing in the mechanism:
1. The π bond of trans-stilbene donates an electron pair to Br•, forming a bond between the carbon and bromine.
2. The brominated stilbene radical donates an electron pair to another Br•, forming a bond between the second carbon and bromine.

A potential undesired side reaction is the formation of 1,1-dibromo-1,2-diphenylethane, a regioisomer. This occurs when the brominated stilbene radical reacts with another bromine molecule (Br₂) instead of a bromine radical. The carbon-bromine bond in the intermediate species can break, forming a carbocation (Ph-CHBr-CH⁺-Ph) and a bromide ion (Br⁻). The carbocation then captures the bromide ion, resulting in the undesired product (Ph-CHBr₂-CHBr-Ph).

Arrow pushing in the side reaction:
1. The brominated stilbene radical donates an electron pair to Br₂, forming a bond between the second carbon and one bromine.
2. The carbon-bromine bond in the intermediate species breaks, producing a carbocation and a bromide ion.
3. The carbocation captures the bromide ion, forming the undesired product.

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The federal agency is responsible for overseeing the disposal of hazardous waste is the Environmental Protection Agency b. (EPA). The EPA is an essential government organization in the United States that protects human health and the environment by developing and enforcing regulations related to pollution and waste management.

It operates under the authority of federal laws, such as the Resource Conservation and Recovery Act (RCRA), which provides guidelines for the proper management and disposal of hazardous waste.
While the EPA is the primary agency in charge of hazardous waste management, other federal agencies like the Occupational Safety and Health Administration (OSHA), the Chemical Safety Board (CSB), and the Department of Energy (DOE) also play significant roles in ensuring the safe handling of hazardous materials.
OSHA is responsible for establishing and enforcing workplace safety standards, including those for handling hazardous waste. It aims to protect the health and safety of workers who may come into contact with dangerous substances.
The CSB is an independent federal agency that investigates industrial chemical accidents, with the goal of improving chemical safety and preventing similar incidents. While it does not have regulatory authority, its findings and recommendations help inform regulations and best practices.
The DOE is primarily focused on energy policy and research, but it also manages nuclear waste disposal and works to ensure the safe storage and handling of nuclear materials.
In summary, the EPA is the primary federal agency overseeing the disposal of hazardous waste, while OSHA, CSB, and DOE play essential roles in regulating and ensuring the safe handling of hazardous materials.

The complete question is:-

What federal agency oversees the disposal of hazardous waste?

a. OSHA

b. EPA

c. CSB

d. DOE

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The incomplete combustion of ethane, there should have been 2 ethane molecules (C₂H₆) and 5 oxygen molecules (O₂) present in the reactants.

The incomplete combustion of ethane producing carbon monoxide, we first need to determine the balanced chemical equation for this process. An incomplete combustion typically involves a limited supply of oxygen. The general equation for the incomplete combustion of ethane (C₂H₆) can be represented as:

C₂H₆ + O₂ → CO + H₂O

To balance the equation, we would have:

2C₂H₆ + 5O₂ → 4CO + 6H₂O

In this balanced equation, the reactants include 2 particles of ethane (C₂H₆) and 5 particles of oxygen (O₂).

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Answer: Wrote the answers below

Explanation:

The balanced equation for Number 1 is:

Fe2O3(s) + 3H2(g) --> Fe(s) + 3H2O(l)

Step 1:

moles ratio of iron (III) oxide and hydrogen is 1:3

step 2:

work out mr (molar mass) of fe2o3: 111.68+ 48 = 159.68

moles of  iron (III) oxide: 33.5g  divided by 159.68 = 0.21 mol

Step 3:

1:3 ratio so 0.21 times 3 = 0.63 mol of hydrogen

Step 4:

mass of hydrogen = mol times mr

0.63 times 2 = 1.26g

mass of hydrogen = 1.26g or 1.27g depending on whether you used 1.00 or 1.01 for the mr of hydrogen

The pressure (in mmHg) of the 9.62 L container having 4.95 moles of gas at 592.84 is 19022.77 mmHg

First, we shall list out the given parameters from the question. This is shown below:

Ideal gas equation states as follow:

PV = nRT

Inputting the give parameters, we can obtain the pressure as follow:

P × 9.62 = 4.95 × 62.36 × 592.84

P × 9.62 = 182999.03688

Divide both sides by 9.62

P = 182999.03688 / 9.62

P = 19022.77 mmHg

Thus, we can conclude from the above calculation that the pressure of the container is 19022.77 mmHg

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Alcohol and carboxylic acid combinations to make each of the given esters based on their structural formulae.

Ester with benzene ring attached to carbonyl carbon and a four carbon chain attached to the singly-bonded oxygen:

The necessary alcohol and carboxylic acid combinations to make this ester are:

Alcohol: 2-phenylethanol ([tex]C_8H_{10}O[/tex])

Carboxylic acid: 3-phenylpropionic acid [tex](C_8H_{10}O_2)[/tex]

Ester with a two carbon chain containing the carbonyl carbon and a phenyl ring attached to the singly-bonded oxygen:

The necessary alcohol and carboxylic acid combinations to make this ester are:

Alcohol: 2-phenylethanol ([tex]C_8H_{10}O[/tex])

Carboxylic acid: 3-phenylpropionic acid [tex](C_8H_{10}O_2)[/tex]

Ester with a three carbon chain containing the carbonyl and a two carbon chain attached to the singly-bonded oxygen:

The necessary alcohol and carboxylic acid combinations to make this ester are:

Alcohol: 2-phenylethanol ([tex]C_8H_{10}O[/tex])

Carboxylic acid: 3-phenylpropionic acid [tex](C_8H_{10}O_2)[/tex]

Ester with a four carbon chain containing the carbonyl carbon and a one carbon chain attached to the singly-bonded oxygen:

The necessary alcohol and carboxylic acid combinations to make this ester are:

Alcohol: 2-methyl-2-phenylethanol ([tex]C_8H_{10}O[/tex])

Carboxylic acid: 3-methyl-3-phenylpropionic acid [tex](C_8H_{10}O_2)[/tex]

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Answer:

The balanced chemical equation is: 4Al + 3O2 → 2Al2O3

From the equation, we can see that 4 moles of aluminum react with 3 moles of oxygen to form 2 moles of aluminum oxide.

Therefore, to find out how many moles of aluminum will react with 1.35 moles of oxygen, we can set up a proportion:

4 moles Al / 3 moles O2 = x moles Al / 1.35 moles O2

Cross-multiplying, we get:

4 moles Al × 1.35 moles O2 = 3 moles O2 × x moles Al

5.4 = 3x

x = 5.4 / 3

x = 1.8 moles Al

Therefore, 1.8 moles of aluminum will be used when reacted with 1.35 moles of oxygen

The molar mass of H2O is 18 amu, which means that one mole of water weighs 18 grams.

The molar mass of H2O, also known as water.
To determine the molar mass of H2O, we need to consider the molecular weight of its constituent elements, hydrogen (H) and oxygen (O). The molecular weight of an element is the weight of one mole of that element, expressed in atomic mass units (amu). The molecular weight of hydrogen is approximately 1 amu, while the molecular weight of oxygen is approximately 16 amu.
Now, let's calculate the molar mass of H2O. In a water molecule, there are two hydrogen atoms and one oxygen atom, so we'll need to add the molecular weights of these elements together.
Molar mass of H2O = (2 * molecular weight of H) + (1 * molecular weight of O)
Molar mass of H2O = (2 * 1 amu) + (1 * 16 amu)
Molar mass of H2O = 2 amu + 16 amu
Molar mass of H2O = 18 amu
So, the molar mass of H2O is 18 amu, In other words, the molecular weight of water is 18 grams per mole.

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You with writing ionic compound formulas. An ionic compound consists of a positive ion (cation) and a negative ion (anion) bonded together through electrostatic forces. To write the formula of an ionic compound, you need to balance the charges of the cation and anion to ensure the compound is neutral.

In your example, you have a sodium ion (Na+) and a fluoride ion (F-). The sodium ion has a positive charge of +1, while the fluoride ion has a negative charge of -1. To write the formula for the ionic compound formed by these two ions, you simply combine them in a way that balances their charges. Since the charges are already equal and opposite, you just need to put them together:
Na+ & F- → NaF
The resulting ionic compound is sodium fluoride (NaF). To write formulas for other ionic compounds, follow the same process:
1. Identify the cation and anion involved.
2. Determine the charges of each ion.
3. Balance the charges by adjusting the number of ions as needed.
4. Write the formula, placing the cation first followed by the anion.
Remember to always ensure that the charges are balanced, and the resulting compound is neutral. This method will allow you to write ionic compound formulas effectively.

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The chemical reaction known as bromination of p-chlorophenyl isopropyl ether acts replacing a hydrogen atom on the aromatic ring of the ether molecule with a bromine atom. A Lewis acid catalyst, like aluminum bromide (AlBr₃), is typically used to initiate the reaction.

The rate law for the given reaction is:

Rate =[tex]k_1[A][B]^2[/tex]

According to the given stoichiometry, 2 moles of A respond with 1 mole of B to give 2 moles of C. In this manner, 0.02 mol of A will respond with 0.01 mol of B to give 0.02 mol of C.

We need to figure out how long it takes to convert 65 percent of A, which means that 35 percent of A won't react. As a result, at a conversion rate of 65%, the concentration of A will be:

0.35 ˣ 0.02 = 0.007mol/L

We can involve the incorporated rate regulation for the second-request response to make the opportunity taken for the given change:

t = [tex]1/((k_1/k_2)[/tex] ˣ [tex](1/[A] - 1/[A]0))[/tex]

Where,

[tex]k_1 = 2 L/mol-mink_2 = 9200 (L/mol)^2/min[/tex]

[A]0 = initial concentration of A = 0.02mol/L

[A] = concentration of A at 65% conversion = 0.007mol/L

Plugging in the values, we get:

t = [tex]1/((2/9200)[/tex] ˣ [tex](1/0.007 - 1/0.02))[/tex] = [tex]145.4[/tex]min

Therefore, the time taken for the given reaction to reach 65% conversion of p-chlorophenyl isopropyl ether is approximately 145.4 minutes.

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The rate of formation of [tex]H_{2}[/tex] is 0.45 min⁻¹

The balanced chemical equation is:

2 [tex]NH_{3}[/tex]→ [tex]N_{2}[/tex] + 3[tex]H_{2}[/tex]

From the equation, we can see that for every 2 moles of [tex]NH_{3}[/tex] consumed, 3 moles of H2 are formed. Therefore, the ratio of the rate of formation of [tex]H_{2}[/tex]to the rate of consumption of [tex]NH_{3}[/tex] is 3/2.

Given that [tex]NH_{3}[/tex] is being consumed at a rate of 0.30 min⁻¹, the rate of formation of [tex]H_{2}[/tex] can be calculated as follows:

Rate of formation of [tex]H_{2}[/tex] = (3/2) × Rate of consumption of [tex]NH_{3}[/tex]

Rate of formation of [tex]H_{2}[/tex] = (3/2) × 0.30 min⁻¹

Rate of formation of [tex]H_{2}[/tex] = 0.45 min⁻¹

Therefore, the rate of formation of [tex]H_{2}[/tex] is 0.45 min⁻¹

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The selectivity of a promoted reaction versus an unpromoted reaction can depend on the location, orientation, and shape of the active sites within the zeolite cavities, as well as the size and shape of the reactant molecules.

The shape of the Na-Y zeolite cavities can affect product selectivity.

Na-Y zeolites are characterized by their uniform, microporous structure, which consists of interconnected channels and cages of fixed sizes. These cavities can act as molecular sieves, selectively allowing smaller molecules to pass through while restricting the movement of larger ones. The size and shape of the cavities can determine which molecules can enter and interact with the active sites on the catalyst surface.

In the case of 2-bromotoluene and 4-bromotoluene, since they have similar sizes, they can enter and leave the zeolite with approximately equal ability. However, the location and orientation of the active sites within the cavities can affect the reaction pathway and product selectivity. For example, if the active sites are located in a cavity that is more accessible to 2-bromotoluene than to 4-bromotoluene, then the promoted reaction may favor the production of products that arise from the 2-bromotoluene pathway.

Similarly, if the cavities have different shapes, such as one cavity being more elongated or twisted than the other, the selectivity may be affected by how the reactants and intermediates fit within the cavity. For instance, a more elongated cavity may favor reactions that occur along a linear pathway, whereas a twisted cavity may promote reactions that involve more complex rearrangements.

The selectivity of a promoted reaction versus an unpromoted reaction can depend on the location, orientation, and shape of the active sites within the zeolite cavities, as well as the size and shape of the reactant molecules.

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Between the same two atoms, the strongest covalent bond is the triple bond and the weakest is the single bond.

A single bond is formed when two atoms share one pair of electrons. This type of bond is the most stable and secure of all covalent bonds as the shared electrons are firmly held in place by the two atoms. The weakest covalent bond between two atoms is a triple bond. A triple bond is formed when two atoms share three pairs of electrons. This type of bond is less stable than a single bond as the three shared electrons are more loosely held and may be more easily lost or broken.

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The reaction takes place in acidic solution, which provides the necessary H+ ions for the reduction of Ni₂+ to Ni. The reaction takes place in acidic solution, which provides the necessary H+ ions for the oxidation of MnO₂ to MnO₄-, and in the presence of hydroxide ions, which are required for the reduction of MnO₄- to MnO₂. The reaction takes place in acidic solution, which provides the necessary H+ ions for the reduction of H₂SO₃ to HSO₄-. The reaction takes place in basic solution, which provides the necessary OH- ions for the oxidation of Cl₂ to ClO₃-.

A. Ni+(aq) ? Ni₂+(aq) + Ni(s) (acidic solution)

This disproportionation reaction involves nickel ions in both +1 and +2 oxidation states. The balanced equation for the reaction is:

2Ni+(aq) + 2H₂O(l) ? Ni₂+(aq) + Ni(s) + 4H+(aq) + O₂(g)

In this reaction, Ni₂+ is reduced to Ni, while Ni+ is oxidized to Ni2+ and O₂ is also produced. The reaction takes place in acidic solution, which provides the necessary H+ ions for the reduction of Ni₂+ to Ni.

B. MnO₂(s) ? MnO₄-(aq) + MnO₂(s) (acidic solution)

This disproportionation reaction involves manganese ions in both +4 and +7 oxidation states. The balanced equation for the reaction is:

3MnO₂(s) + 4H₂O(l) + 2H+(aq) ? 2MnO₄-(aq) + MnO₂(s) + 8OH-(aq)

In this reaction, MnO₂ is both oxidized to MnO₄- and reduced to MnO₂. The reaction takes place in acidic solution, which provides the necessary H+ ions for the oxidation of MnO₂ to MnO₄-, and in the presence of hydroxide ions, which are required for the reduction of MnO₄- to MnO₂.

C. H₂SO₃(aq) ? S(s) + HSO₄-(aq) (acidic solution)

This disproportionation reaction involves sulfur in both +4 and +6 oxidation states. The balanced equation for the reaction is:

H₂SO₃(aq) + 2H₂O(l) ? S(s) + 2HSO₄-(aq) + 4H+(aq) + 2e-

In this reaction, H₂SO₃ is oxidized to S and reduced to HSO₄-. The reaction takes place in acidic solution, which provides the necessary H+ ions for the reduction of H₂SO₃ to HSO₄-.

D. Cl₂(aq) ? Cl-(aq) + ClO-(aq) (basic solution)

This disproportionation reaction involves chlorine in both 0 and -1 oxidation states. The balanced equation for the reaction is:

3Cl₂(aq) + 6OH-(aq) ? 5Cl-(aq) + ClO₃-(aq) + 3H2O(l)

In this reaction, Cl2 is reduced to Cl-, while Cl₂ is oxidized to ClO₃-. The reaction takes place in basic solution, which provides the necessary OH- ions for the oxidation of Cl₂ to ClO₃-.

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The temperature of the bromine reaction mixture during reflux, it typically depends on the boiling point of the solvent being used.

For example, if the solvent is chloroform, the reflux temperature would be around 61-62°C. If the solvent is carbon tetrachloride, the reflux temperature would be around 76-77°C.

As for the condenser, a water condenser is typically used during reflux to prevent the loss of solvent and/or reagents due to evaporation. Air condensation is not likely to be sufficient, especially for reactions that require longer reflux times.

Regarding the bromination of alkenes, the substituents do remain on opposite sides of the molecule after the reaction, resulting in a trans product. The absolute configurations of the carbons depend on the starting configuration of the alkene. For example, if the starting alkene is (Z)-2-butene, the product of bromination would be (2R,3S)-2,3-dibromobutane, as shown in the following diagram:

   H     Br

   |     |

H -- C=C -- C -- H

   |     |

   Br    H

Note that the stereochemistry of the product is determined by the anti-addition mechanism of bromination, which results in the formation of a meso compound with two chiral centers.

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For a given fluorophore, absorption < fluorescence < phosphorescence is correct.

Fluorescence, absorption, and phosphorescence are all related to the way light interacts with matter.

Absorption occurs when a molecule absorbs a photon of light, which causes an electron to jump to a higher energy level. This process occurs at a specific wavelength, which is unique to each molecule. The absorbed energy is usually converted into heat or used to drive a chemical reaction.

Fluorescence occurs when a molecule that has been excited by absorbing light emits a photon of light as it returns to its ground state. This process occurs at a longer wavelength than the absorbed light, and the emitted light is usually of a different color than the absorbed light. Fluorescence occurs quickly, typically within nanoseconds of the molecule being excited.

Phosphorescence is a type of delayed fluorescence that occurs when a molecule remains in an excited state for a longer period of time, typically microseconds to seconds. During this time, the molecule can emit light as it returns to its ground state. Phosphorescence occurs at an even longer wavelength than fluorescence.

The order of these processes in terms of wavelength is absorption < fluorescence < phosphorescence. When a molecule absorbs light, it does so at a specific wavelength. When it fluoresces, it emits light at a longer wavelength. When it phosphoresces, it emits light at an even longer wavelength. Therefore, the wavelength increases as we move from absorption to fluorescence to phosphorescence.

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Answer:

a. The given molecular formula of the akene is C22H40. When it undergoes ozonolysis, it gives two products with the same molecular formula C22H40, which are functional isomers of each other. These two isomers are 1-octene and 9-octene.

The two-step chemical equation for the conversion of the akene into 1-octene and 9-octene is as follows:

Step 1: Ozonolysis of the akene to form ozonides

C22H40 + 3O3 → C22H40O3 + 3O2

Step 2: Reduction of the ozonides to form 1-octene and 9-octene

C22H40O3 + 6H2O → 3C8H16 + 3C8H18O

b. The structural formula of the two isomers 1-octene and 9-octene are as follows:

1-octene: CH3(CH2)6CH=CH2

9-octene: CH3(CH2)4CH=CH(CH2)2CH3

They are called functional isomers because they have the same molecular formula but different functional groups. In this case, both isomers have an alkene functional group, but they differ in the position of the double bond.

c. When hydrogen gas in the presence of a nickel catalyst is passed over X, it undergoes hydrogenation to form a saturated compound Y. The chemical equation for the reaction is as follows:

X + H2 → Y

d. One way to prove that the compound X is unsaturated is by performing the bromine water test. Bromine water is a reddish-brown solution of bromine in water. When added to an unsaturated compound, it undergoes decolorization due to the addition of bromine across the double bond. If X is unsaturated, then it will decolorize bromine water, indicating the presence of a double bond.

an experiment where a Gummy Bear is sacrificed for the sake of science The 2nd part of the experiment involves tossing a Gummy Bear into molten potassium chlorate. As a result, the sucrose reacts with oxygen and generates purple sparks and a lot of heat and The balanced reaction looks like :

C₁₂H₂₂O₁₁ (s) + 8KClO₃ (s) =  12CO₂ (g) + 11H₂O (g) + 8KCl (s)

When the potassium chlorate is heated, it decomposes into potassium chloride and oxide, as seen below:

2KClO₃(s) =  2KCl(s) + 3O₂(g)

When the gummy bear is dropped, the oxide from the decomposition of potassium chlorate reacts with the glucose molecule in sucrose. This reaction is a spontaneous combustion reaction:

C₆H₁₂O₆ (s) + 6O₂(g) = 6CO₂(g) + 6H₂O (g)

The overall reaction is seen below:

C₁₂H₂₂O₁₁ (s) + 8KClO₃ (s) =  12CO₂ (g) + 11H₂O (g) + 8KCl (s)

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Calculating the mass of 6.80 x 1023 calcium chlorite, Ca(ClO2)2 molecules requires multiplying the number of molecules by the compound's molar mass.

Calcium chlorite has a molar mass of 117.98 g/mol. As a result, the provided calcium chlorite molecules have a mass of 8.09 x 1024 g. This is obtained by dividing the compound's molar mass (117.98 g/mol) by the number of molecules (6.80 x 1023).

One mole of calcium chlorite weighs 117.98 grammes. This is the case because the atomic masses of all the atoms that make up a compound are added to determine its molar mass. Chlorine and calcium have atomic weights of 35.45 g/mol and 40.08 g/mol, respectively.

Oxygen has a molar mass of 16.00 g/mol.  The atomic masses of all the atoms in the compound are added to determine the molar mass of calcium chlorite, which equals 40.08 + 35.45 + (4 x 16.00) = 117.98 g/mol.

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Delta S greater than 0 refers to a positive change in entropy, where the entropy of a system increases.

This means that the system becomes more disordered or random, and there is a greater number of possible arrangements or configurations of its particles. This can occur due to various factors, such as an increase in temperature, a phase transition, mixing of different substances, or chemical reactions that produce more products than reactants. A positive Delta S value is important in thermodynamics, as it indicates the direction of spontaneous processes that are favorable in terms of energy and entropy.

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The final molarity of sodium cation in the aqueous solution of potassium carbonate after dissolving 0.5 moles of sodium chloride is 0.5 M.

To calculate the final molarity of sodium cation in the solution, we first need to determine how much sodium chloride dissociates into sodium cations and chloride anions in the aqueous solution of potassium carbonate. Sodium chloride is a strong electrolyte and completely dissociates in water. Therefore, one mole of sodium chloride will yield one mole of sodium cations and one mole of chloride anions.
The next step is to calculate the total number of moles of sodium cations in the solution. We know that 0.5 moles of sodium chloride is dissolved in the solution. Since one mole of sodium chloride yields one mole of sodium cations, we have 0.5 moles of sodium cations in the solution.
Finally, we need to determine the final volume of the solution to calculate the final molarity of sodium cation. However, the question states that the volume of the solution doesn't change when the sodium chloride is dissolved in it. Therefore, the final volume of the solution is the same as the initial volume.
Using the formula for molarity, we can calculate the final molarity of sodium cation in the solution. Molarity is defined as the number of moles of solute per liter of solution. Therefore, the final molarity of sodium cation is 0.5 moles / 1 liter, which simplifies to 0.5 M.
In conclusion, the final molarity of sodium cation in the aqueous solution of potassium carbonate after dissolving 0.5 moles of sodium chloride is 0.5 M.

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The diene that reacts with one equivalent of HBr to form the two bromoalkene products described in the question can be drawn as follows:

    H          H

    |          |

H3C-C=C-CH2-CH=CH2

    |          |

    H          H

In this diene, there are two double bonds, one between carbons 2 and 3 and another between carbons 4 and 5. When one equivalent of HBr is added to this diene, an electrophilic addition reaction occurs in which the H and Br add to the two double bonds to form two different products, as described in the question.

The effect of increasing temperature on the relative amount of each product can be explained by considering the mechanism of the reaction. The reaction proceeds through a carbocation intermediate, which is formed by protonation of the diene with HBr. The carbocation intermediate can then react with Br- to form the bromoalkene products.

Product 1 is formed by the addition of HBr to the double bond between carbons 2 and 3, which results in the formation of a more stable tertiary carbocation intermediate. As the temperature increases, the reaction rate increases, which can lead to a higher proportion of product 1 being formed. However, at very high temperatures, the reaction rate can become too fast, leading to increased side reactions such as rearrangements, which can decrease the relative amount of product 1.

Product 2 is formed by the addition of HBr to the double bond between carbons 4 and 5, which results in the formation of a less stable secondary carbocation intermediate. As the temperature increases, the reaction rate also increases, which can lead to a higher proportion of product 2 being formed. However, at very high temperatures, the reaction rate can become too fast, leading to increased side reactions such as elimination, which can decrease the relative amount of product 2.  Therefore, the answer to the question is that as the temperature increases, the relative amount of product 1 is expected to increase, while the relative amount of product 2 is expected to decrease due to side reactions. However, at very high temperatures, both products can be affected by side reactions, and the relative amounts may not change significantly.

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The molarity of the sodium hydroxide solution is 0.253 M. This means that there are 0.253 moles of NaOH in 1 liter of the solution.

To determine the molarity of the sodium hydroxide solution, we can use the equation:

Molarity of NaOH = (Molarity of HCl) x (Volume of HCl) / (Volume of NaOH)

First, we need to calculate the number of moles of HCl used in the titration. We can do this using the formula:

Number of moles of HCl = Molarity x Volume

Substituting the given values, we get:

Number of moles of HCl = 0.1985 M x 0.03896 L = 0.00774356 moles

Now, let's calculate the volume of NaOH used in the titration by subtracting the initial buret reading from the final buret reading:

Volume of NaOH = 31.93 ml - 1.24 ml = 30.69 ml = 0.03069 L

Substituting these values in the equation, we get:

Molarity of NaOH = (0.1985 M) x (0.03896 L) / (0.03069 L) = 0.253 M

Therefore, the molarity of the sodium hydroxide solution is 0.253 M. This means that there are 0.253 moles of NaOH in 1 liter of the solution.

It is important to note that standardizing a solution is a crucial step in ensuring accurate and precise results in chemical analysis. By standardizing the NaOH solution, we can determine its exact concentration and use it for future titrations with confidence.

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You need to add 42 grams of glucose to make 1500 grams of a 2.8% glucose solution.

To calculate the amount of glucose you need to add to make 1500 grams of a 2.8% glucose solution, you first need to determine what 2.8% by mass means.

2.8% by mass means that there are 2.8 grams of glucose per 100 grams of solution.

So, if you want to make 1500 grams of a 2.8% glucose solution, you can use the following equation:

mass of glucose = (percent by mass / 100) x total mass of solution

mass of glucose = (2.8 / 100) x 1500

mass of glucose = 42 grams

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Ethylene is the smallest member in the family of hydrocarbons containing carbon-carbon double bond.

Ethylene is a simple organic compound with the chemical formula as C₂H₄ and it is a colorless, flammable gas with a sweet odor. Ethylene is an important industrial chemical that can be used in the production of plastics, synthetic rubber, and other chemicals.

Molecule of ethylene consists of two carbon atoms connected by double bond and each carbon atom is bonded to two hydrogen atoms. Double bond in ethylene is a type of covalent bond formed by the overlapping of two sp² hybrid orbitals on each of the carbon atom.

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Here's a list with the first 20 elements, elements 24-30, 35, 53, 54, 37, 38, 55, and 56:1. Hydrogen (H);2. Helium (He);3. Lithium (Li);4. Beryllium (Be);5. Boron (B);6. Carbon (C);7. Nitrogen (N);8. Oxygen (O);9. Fluorine (F);10. Neon (Ne);11. Sodium (Na);12. Magnesium (Mg);13. Aluminum (Al);14. Silicon (Si);15. Phosphorus (P);16. Sulfur (S);17. Chlorine (Cl);18. Argon (Ar);19. Potassium (K);20. Calcium (Ca).

24. Chromium (Cr);25. Manganese (Mn);26. Iron (Fe);27. Cobalt (Co);28. Nickel (Ni);29. Copper (Cu);30. Zinc (Zn);35. Bromine (Br);53. Iodine (I);54. Xenon (Xe);37. Rubidium (Rb);38. Strontium (Sr);55. Cesium (Cs);56. Barium (Ba)
To memorize these names and symbols, try making flashcards with the element name on one side and the symbol on the other side. Review them regularly, and quiz yourself by trying to recall the symbols when given the names, and vice versa.

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