We can use bond-line formulas to represent alkenes in much the same way that we use them to represent alkanes. Consider the following alkene: h5ch5e4 How many carbon atoms are sp2−hybridized in this alkene?

Answer:

Sr is the more metallic element

Bi is the more metallic element

O is the more metallic element

As is the more metallic element

Explanation:

One thing should be clear; metallic character increases down the group but decreases across the period.

Hence, as we move across the period, elements become less metallic. As we move down the group elements become more metallic.

This is the basis upon which decisions were made about the metallic character of each of the elements listed above.

Answer:

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Answer:

a. Na(s); b. Al(s); c. Zn(s)

Explanation:

Let's consider the following redox reactions.

a. ZnCl₂ (aq) + 2 Na(s) → Zn(s) + 2 NaCl(aq)

Na is oxidized because its oxidation number increases from 0 to +1 (in NaCl) whereas Zn is reduced because its oxidation number decreases from 2+ (in ZnCl₂) to 0.

b. Al(s) + FeBr₃ (aq) → AlBr₃ (aq) + Fe(s)

Al is oxidized because its oxidation number increases from 0 to +3 (in AlBr₃) whereas Fe is reduced because its oxidation number decreases from 3+ (in FeBr₃) to 0.

c. FeSO₄ (aq) + Zn(s) → Fe(s) + ZnSO₄(aq)

Zn is oxidized because its oxidation number increases from 0 to +2 (in ZnSO₄) whereas Fe is reduced because its oxidation number decreases from 2+ (in FeSO₄) to 0.

Answer:

5.20 grams of Br₂

Explanation:

From our previous knowledge;

We understand that:

The number of moles of a given element = mass of the element divided by its molar mass.

Mathematically:

[tex]\mathbf{no \ of \ moles =\dfrac{ mass}{ molar \ mass}}[/tex]

From the given information, let's assume that the 0.065 moles of liquid -bromine partake in the reaction.

From the periodic table, the molar mass of Bromine is = 79.9 g/mol

As such, the mass of liquid that partakes is calculated as:

0.065 mol  = mass/ 79.9 g/mol

mass = 0.065 mol × 79.9 g/mol

mass of liquid that partakes in the reaction = 5.20 grams of Br₂

Answer: The thermal energy that would be released by 276.0g of sodium acetate trihydrate is 71.8kJ.

Explanation:

Supercooling is the process of lowering the temperature a liquid below its freezing point, without it becoming solid. A liquid below its freezing point will crystallize in the presence of a seed crystal because it serves as a structure for formation of crystals. From the question,

The given mass of sodium acetate trihydrate

(CH3COONa.3H2O)= 276.0g

Molar mass of sodium acetate

trihydrate= 136.08g/mol

Thermal heat of fusion of sodium acetate

trihydrate = 35.9 kJ/mol

From the given mass the number of moles present= 276.0/ 136.08

= 2.0moles

Therefore the heat (thermal) energy of the given mass of sodium acetate

trihydrate = 2.0 × 35.9

= 71.8kJ

Therefore, upon addition of a small seed crystal, the solution temperature increases as sodium acetate trihydrate crystallizes.

Answer:

c. 29 J

Explanation:

Step 1: Given data

Step 2: Calculate the temperature change

ΔT = 37 °C - 22 °C = 15 °C

Step 3: Calculate the heat (Q) required to raise the temperature of the lead piece

We will use the following expression.

Q = c × m × ΔT

Q = 0.13 J/g.°C × 15 g × 15 °C = 29 J

Answer:

11.9 pH

Explanation:

First, we need to find pOH

To find that, we use the formula -log[OH]

-log[7.7x10^-3] = 2.11351

To find the pH, we'll use this formula: 14 = pH + pOH

14 = pH + 2.11351

Subtract boths sides by 2.11351

14 = pH + 2.11351

-2.11351  -2.11351

pH = 11.88649

In a REDOX titration, one specie is oxidized while the other is reduced. The concentration of methanol is 0.012  mol L-1. Methanol is the oxidizing agent while permanganate is the reducing agent.

The average titre value is; [tex]\frac{7.95 + 7.10 + 7.00 + 7.10}{4}[/tex] = 7.29 mL

Equation of the reaction is:

[tex]4MnO4- + 12H+ + 5CH3OH ----> 4Mn2+ + 11H2O + 5HCOOH[/tex]

Concentration of oxidizing agent = CA = ?

Concentration of reducing agent = CB = 0.0021 mol L-1

Volume of oxidizing agent =  VA= 7.29 mL

Volume of reducing agent = VB = 50.00 mL

Number of moles of oxidizing agent NA = 4

Number of moles of reducing agent NB = 5

Note that NA and NB are obtained from the balanced reaction equation

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CA =  CBVBNA/VANB

CA = 0.0021 mol L-1 * 50.00 mL * 4/7.29 mL * 5

CA= 0.012  mol L-1

For a comprehensive definition of redox titration see

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Explanation:

Sulfuric Acid (H2SO4) is a strong mineral acid that has is colorless when pure. This chemical is used as a chemical intermediate to manufacture other chemicals and cleaning metal surfaces. The formula for sulfuric acid is H2SO4. The molar mass of sulfuric acid is 98.07848 g mol.

The statement that "substance A is rice, and substance B is sugar" is correct.

WHAT IS A MIXTURE?

According to this question, a mixture contains two substances A and B. Water is added to this mixture for easy separation.

Therefore, in the mixture of two substances A and B, substance A is rice, and substance B is sugar.

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The kinetic-molecular theory of gases assumes that ideal gas molecules

(1) are constantly moving;

(2) have negligible volume;

(3) have negligible intermolecular forces;

(4) undergo perfectly elastic collisions; and

(5) have an average kinetic energy proportional to the ideal gas's absolute temperature.

The five main postulates of the KMT are as follows:

(1) the particles in a gas are in constant, random motion,

(2) the combined volume of the particles is negligible

(3) the particles exert no forces on one another,

(4) any collisions between the particles are completely elastic.

(5) the average kinetic energy of the particles is proportional to the temperature in kelvins.

Answer:

A redox reaction in which a change of oxidation number of the reacting species occurs either by oxidation or by reduction which occur simultaneously and to the same extent during the reaction.

Silver atoms are oxidized while oxygen atoms are reduced by a loss of electrons and a gain of electrons respectively.

Explanation:

A redox reaction in which a change of oxidation number of the reacting species occurs either by oxidation or by reduction which occur simultaneously and to the same extent during the reaction.

In a redox reaction,the two reactions occurring simultaneously can be divided into two half reactions; an oxidation half-reaction and a reduction half-reaction.

The oxidation half-reaction involves losing electrons and thus an increase in oxidation number of the species being oxidized. Whereas, the reduction half reaction involves gaining electrons and thus, a reduction innthe oxidation number of the species being reduced.

The species which oxidizes another species is known as an oxidizing agent and isnitself reduced due to its accepting electrons from the species being oxidized. Th reducing agent reduces another species and is itself oxidized as it loses electrons to the oxidized agent.

In the given reaction as shown below:

4 Ag (s) + 2 H₂S (g) + O₂ (g) ---> 2 Ag₂S (s) + 2 H₂0 (g)

The reaction is a redox reaction as a change innthe oxidation number of the reacting species; both oxidation and reduction occurs simultaneously and to the same extent.

The metallic silver atoms, have an oxidation number of zero initially. However, each of the four moles of atoms give up one mole of a electrons each to become oxidized to silver (i) ions, Ag+.

On the other hand, molecular oxygen gas also having oxidation number of zero becomes reduced to oxygen ion, O²-. Each of the two moles of atom in the oxygen gas molecule accept two electrons each donated by the metallic silver atoms to become reduced to oxygen ion, O²-.

The oxidation numbers of hydrogen ion and sulfide ion do not change.

Answer:

A BEC ( Bose - Einstein condensate ) is a state of matter of a dilute gas of bosons cooled to temperatures very close to absolute zero is called BEC.

Examples - Superconductors and superfluids are the two examples of BEC.

Explanation:

Water, mercury chloride and nitrogen oxide.

Water, mercury chloride and nitrogen oxide will present in the precipitate when we slowly add a solution of mercury(II) nitrate to a solution of aqueous hydrochloric acid having halide ions both in equal concentrations. The equation of this reaction is Hg2(NO3)2 + 4 HCl ----> 2 HgCl2 + 2 H2O + 2 NO so it is concluded that from this reaction we get precipitate of water, mercury chloride and nitrogen oxide.

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Answer:

56%

Explanation:

If I use 25 grams of iron (III) phosphate and obtain 18.5 g of iron (III) sulfate, what is my percent yield?

Step 1: Write the balanced equation

2 FePO₄ + 3 Na₂SO₄ ⇒ Fe₂(SO₄)₃ + 2 Na₃PO₄

Step 2: Calculate the theoretical yield of Fe₂(SO₄)₃

The mass ratio of FePO₄ to Fe₂(SO₄)₃ is 301.64:399.88.

25 g FePO₄ × 399.88 g Fe₂(SO₄)₃/301.64 g FePO₄ = 33 g Fe₂(SO₄)₃

Step 3: Calculate the percent yield of Fe₂(SO₄)₃

We will use the following expression.

%yield = (experimental yield/theoretical yield) × 100%

%yield = (18.5 g/33 g) × 100% = 56%

Answer:

1:2:1 is the correct ratio of carbon hydrogen to oxygen in glucose.

Answer:

White phosphorusRed PhosphorusIt is insoluble in water but soluble in carbon disulphide.It is insoluble in both water and carbon disulphide.It undergoes spontaneous combustion in air.It is relatively

Explanation:

I hope it will help you

Answer:

Molar mass = 180 g/mol

Explanation:

Relative Atomic Mass of C = 12

of H = 1

of O =16

Let Molar mass be mm

mm of C6H12O6 = 6(12) + 12(1) + 6(16)

= 72 + 12 + 96 = 180 g/mol

Answer:

the rate of a reaction is measured by how fast a REACTANT is used up or how fast a PRODUCT is formed

Answer:

number of moles of gas increases the volume also increases.

Answer:

a measure of concentration equal to the gram equivalent weight per liter of solution.

Explanation:

Gram equivalent weight is the measure of the reactive capacity of a molecule. The solute's role in the reaction determines the solution's normality. Normality is also known as the equivalent concentration of a solution.

hope it helped

Answer:

Tapeworm, also called cestode, any member of the invertebrate class Cestoda (phylum Platyhelminthes), a group of parasitic flatworms containing about 5,000 species. ... Tapeworms also lack a circulatory system and an organ specialized for gas exchange.

Answer:

A

Explanation:

1-ethyl-4-methylcyclohexane

i believe it is A

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Plants play a vital role in the environment because photosynthesis uses atmospheric carbon dioxide and water to produce sugars and oxygen. Option C is correct.

Photosynthesis is the process by which plants, algae, and some bacteria convert light energy into chemical energy in the form of glucose (sugar). This process occurs in the chloroplasts of plant cells and requires the presence of sunlight, carbon dioxide (CO₂), and water (H₂O).

During photosynthesis, plants absorb carbon dioxide from the atmosphere through tiny openings called stomata in their leaves. At the same time, they take in water from the soil through their roots. Through a series of complex chemical reactions, the energy from sunlight is used to combine carbon dioxide and water, resulting in the production of glucose (sugar) and the release of oxygen (O₂) as a byproduct.

This process is represented by the equation:

6CO₂ + 6H₂O + sunlight → C₆H₁₂O₆ (glucose) + 6O₂

Plants use the glucose produced through photosynthesis as a source of energy for various metabolic processes and growth. The released oxygen is essential for the survival of other organisms, including humans, as it is used in cellular respiration to release energy from glucose.

Hence, C. is the correct option.

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Answer:

Its C  you can check

Answer:

1. 17.5 g of CO₂

2. The limiting reactant is carbon (graphite), and its formula is C(graphite)

3. 3.7 g of O₂

Explanation:

First, we have to write the chemical equation for the reaction. For this, we have to know the chemical formula of each reactant and product:

Thus, we write the chemical equation:

C(graphite) + O₂(g) → CO₂(g)

The equation is already balanced because it has the same number of C and O atoms on both sides. Thus, we can see that 1 mol of C(graphite) reacts with 1 mol of O₂ and produce 1 mol of CO₂ (mole-to-mole reaction).

Now we convert the grams of reactants to moles by using the molecular weight (Mw) of each compound:

Mw(C) = 12 g/mol

moles of C(graphite) = 4.77g/(12 g/mol) = 0.3975 mol

Mw(O₂) = 16 g/mol x 2 = 32 g/mol

moles of O₂ = 16.4 g/(32 g/mol) = 0.5125 mol

Now, we can compare the stoichiometric ratio (given by the moles of reactants in the equation) with the actual ratio (given by the mass of reactants we have):

stoichiometric ratio ⇒ 1 mol C(graphite)/mol O₂

actual ratio ⇒ 0.3975 mol C(graphite)/0.5125 mol O₂

We can see that we need 0.3975 moles of O₂ to react with C(graphite) and we have more moles (0.5125 mol) so the excess reactant is O₂. Thus, the limiting reactant is C(graphite).

The amount of product (CO₂) that is formed is calculated from the amount of limiting reactant. We can see in the chemical equation that 1 mol of CO₂ is produced from 1 mol of C(graphite) ⇒ stoichiometric ratio = 1 mol CO₂/mol C(graphite).

Thus, we multiply the moles of C(graphite) we have by the stoichiometric ratio to calculate the moles of CO₂ produced:

moles of CO₂ = 0.3975 mol C(graphite) x 1 mol CO₂/mol C(graphite) = 0.3975 mol CO₂

Now, we convert the moles of CO₂ to mass by using the Mw:

Mw(CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol

mass of CO₂ = 0.3975 mol CO₂ x 44 g/mol CO₂ = 17.5 g

Therefore, the maximum amount of carbon dioxide (CO₂) formed is 17.5 g.

Since this is a mole-to-mole reaction, the moles of excess reactant that remains after the reaction is complete is calculated as the difference between the moles of excess reactant and limiting reactant:

remaining moles of O₂ = 0.5125 mol - 0.3975 mol = 0.115 mol O₂

Finally, we convert the moles of O₂ to mass with the Mw (32 g/mol) :

mass of O₂ = 0.115 mol O₂ x 32 g/mol = 3.68 g

Therefore, the mass of the excess reagent that remains after the reaction is complete is 3.7 g.

Answer:

True

Explanation:

Aromatic rings undergo nucleophillic substitution reactions in the presence of a electron withdrawing group which stabilizes the Meisenheimer complex.

When the nucleophile attacks the ring carbon atom carrying the eventual leaving group. A resonance-stabilized carbanion intermediate called a Meisenheimer complex is formed.

Subsequent loss of the leaving group from the intermediate complex yields the product of the reaction.