Give the IUPAC name for the following alkyl group, and classify it as primary, secondary, or tertiary.

CH3(CH2)9CH2

Answer:

If 7.9 moles of C₅H₁₂ reacts with excess O₂, 39.5 moles of CO₂ will be produced.

Explanation:

The balanced reaction is:

C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Then you can apply the following rule of three: if by stoichiometry 1 mole of C₅H₁₂ produces 5 moles of CO₂, then 7.9 moles of C₅H₁₂ will produce how many moles of CO₂?

[tex]amount of moles of CO_{2} =\frac{7.9 moles of C_{5}H_{12}*5 moles of CO_{2} }{1 mole of C_{5}H_{12} }[/tex]

amount of moles of CO₂= 39.5 moles

If 7.9 moles of C₅H₁₂ reacts with excess O₂, 39.5 moles of CO₂ will be produced.

Answer:

That means Cu2O is limiting reagent and C is excess reagent

Explanation:

Based on the reaction, 1 mole of Cu2O reacts per mole of C. The ratio of reaction is 1:1.

To solve this question we need to convert the mass of each reactant to moles. The reactant with the lower amount of moles is limiting reactant and the excess reactant is the reactant with the higher number of moles.

Moles Cu2O -Molar mass: 143.09 g/mol-

114.2g Cu2O * (1mol / 143.09g) = 0.798 moles Cu2O

Moles C -Molar mass: 12.01g/mol-

11.1g C * (1mol / 12.01g) = 0.924 moles C

Answer:

please mark as brainliest

Explanation:

The sun is shining on the shallow ocean water where the grasses and dugongs live. What is happening to the carbon in the water around the grasses and the dugongs? Is carbon moving into the water, moving out of the water, or both? Carbon is not moving into the water; it is only moving out of the water.

Answer:

a. NaNO3, ionic - water

b. I2, nonpolar - CCl4

c. sucrose (table sugar), polar - water

d. gasoline, nonpolar - CCl4

e. vegetable oil, nonpolar - CCl4

f. benzene, nonpolar -CCl4

g. LiCl, ionic - water

h. Na2SO4, ionic - water

Explanation:

Water is a polar substance. This means that it has the ability to dissolve other polar substances. Furthermore, water, even being made by covalent bonds, manages to dissolve ionic substances, because it is a molecule with a partial positive charge on one side, due to hydrogen, and a partial negative charge on the other side, due to the two molecules of oxygen. In this case, any polar or ionic substance has the ability to be dissolved in water, while any non-polar substance needs a non-polar liquid to be able to be dissolved, such as CCI4.

Answer:

2-ethyl-3-methylbut-2-ene

Explanation:

The whole idea of IUPAC nomenclature is to devise a universally accepted system of writing the name of a compound from its structure.

According to IUPAC nomenclature, the root of the compound is the longest carbon chain. The substituents are named in alphabetical order and in such a way as to give each one the lowest number. The position of the functional group is indicated accordingly.

For the compound in question, its correct IUPAC name is 2-ethyl-3-methylbut-2-ene.

Answer:

6.88 mg

Explanation:

Step 1: Calculate the mass of ³²P in 175 mg of Na₃³²PO₄

The mass ratio of Na₃³²PO₄ to ³²P is 148.91:31.97.

175 mg g Na₃³²PO₄ × 31.97 g ³²P/148.91 g Na₃³²PO₄ = 37.6 mg ³²P

Step 2: Calculate the rate constant for the decay of ³²P

The half-life (t1/2) is 14.3 days. We can calculate k using the following expression.

k = ln2/ t1/2 = ln2 / 14.3 d = 0.0485 d⁻¹

Step 3: Calculate the amount of P, given the initial amount (P₀) is 37.6 mg and the time elapsed (t) is 35.0 days

For first-order kinetics, we will use the following expression.

ln P = ln P₀ - k × t

ln P = ln 37.6 mg - 0.0485 d⁻¹ × 35.0 d

P = 6.88 mg

Explanation:

strong electrolyte- Nacl HCL NAOH

weak electrolyte- c12H22O, NH3

NaCl,HCl and NaOH  are strong electrolytes while ammonia is a weak electrolyte and sucrose is a non-electrolyte.

It is a solution which consists of ions which are electrically conducting as a result of movement of ions.Class of electrolytes include most soluble salts,acids and bases which are dissolved in a polar solvent.On dissolution, they separate into the constituent ions.

There are 3 classes according to the nature of substance which results upon dissolution:

1) Strong electrolytes- Substances which on dissolution in a medium dissociate completely are strong electrolytes. eg: NaCl,HCl

2) Weak electrolytes-  Substances which on dissolution in a medium dissociate partially are weak electrolytes. eg: NH₃

3)Non-electrolytes- Substances which do not dissociate on dissolution are non-electrolytes. eg: sucrose

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Answer:

what is the hybridization of each of the carbon atoms in the compounds

а) CH₃ CH = CH2

b) CH₂=CH2

c) Acetylene. (CH triple bond CH).​

Explanation:

The hybridization of carbon atom in the organic compounds can be determined by counting the number of surrounding atoms that are in bond with it.

If carbon is directly bonded with two atoms, then it has sp hybridization.

If carbon is directly bonded with three atoms, then it has [tex]sp^2[/tex] hybridization.

If carbon is directly bonded with four atoms, then it has [tex]sp^3[/tex] hybridization.

For the given molecules, the hybridization of each carbon atom is shown below:

Answer:

As the temperature increases, the average kinetic energy increases as does the velocity of the gas particles hitting the walls of the container. The force exerted by the particles per unit of area on the container is the pressure, so as the temperature increases the pressure must also increase.

I hope this will help you if not soo sorry :)

Answer:

D.) ±0.01 cm?​

Explanation:

Since 32.23 cm has two decimal places, the uncertainty is taken as one-half the last decimal pace.

The last decimal place is 0.03. Half of this is 0.03 cm/2 = 0.015 cm.

Since we cannot go below two decimal places, we ignore the 5 in 0.015 cm.

So, we have our uncertainty as 0.01 cm.

So, the best expression of the uncertainty in the measurement 32.23 cm is ± 0.01 cm.

So, the answer is D. which is ± 0.01 cm.

An OH group attached to a hydrocarbon is called an alkyl group whereas hydroxide is a polyatomic ion with a charge of -1.

OH group is also called hydroxyl group. Alcohol is a type of organic compound that is characterized by one or more hydroxyl (―OH) groups attached to a carbon atom of an hydrocarbon chain so we can conclude that an OH group attached to a hydrocarbon is called an alkyl group whereas hydroxide is a polyatomic ion with a charge of -1.

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Answer:

Newton's first and third law of Motion

Explanation:

The laws applying in the example Newton's first and third laws of Motion.

Answer:

It Newtons first, second, and third laws

Explanation:

Answer:

attached below

Explanation:

Structure of two acyclic compounds with 3 or more carbons that exhibits one singlet in 1H-NMR spectrum

a) Acetone CH₃COCH₃

Attached below is the structure

b) But-2-yne (CH₃C)₂

Attached below is the structure

AAnswer:A

Explanation:

Answer:

Gains

Explanation:

It gets more electrons

Answer and Explanation:

We have to identify the anion (negatively charged ion) and the positive ion to form each compound. The sum of the positive and negative charges will be equal to 0 for a neutral compound.

Chloride: the anion is Cl⁻ (1 negative charge).

Magnesium (Mg²⁺) + Chloride (Cl⁻) : MgCl₂

Sodium (Na⁺) + Chloride (Cl-): NaCl

Zinc (Zn²⁺) + Chloride (Cl-): ZnCl₂

Lithium (Li⁺) + Chloride (Cl-) : LiCl

Lead(II) (Pb²⁺) + Chloride (Cl⁻): PbCl₂

Calcium (Ca²⁺) + Chloride (Cl⁻): CaCl₂

Iron(II) (Fe²⁺) + Chloride (Cl⁻): FeCl₂

Iron(III) (Fe³⁺) + Chloride (Cl⁻): FeCl₃

Potassium (K⁺) + Chloride (Cl): KCl

Nitrate: the anion is NO₃⁻ (1 negative charge).

Magnesium (Mg²⁺) + Nitrate (NO₃⁻) : Mg(NO₃)₂

Sodium (Na⁺) + Nitrate (NO₃⁻): NaNO₃

Zinc (Zn²⁺) + Nitrate (NO₃⁻): Zn(NO₃)₂

Lithium (Li⁺) + Nitrate (NO₃⁻) : LiNO₃

Lead(II) (Pb²⁺) + Nitrate (NO₃⁻): Pb(NO₃)₂

Calcium (Ca²⁺) + Nitrate (NO₃⁻): Ca(NO₃)₂

Iron(II) (Fe²⁺) + Nitrate (NO₃⁻): Fe(NO₃)₂

Iron(III) (Fe³⁺) + Nitrate (NO₃⁻): Fe(NO₃)₃

Potassium (K⁺) + Nitrate (NO₃⁻): KNO₃

Sulphate: SO₄²⁻ (2 negative charges)

Magnesium (Mg²⁺) + Sulphate (SO₄²⁻) : MgSO₄

Sodium (Na⁺) + Sulphate (SO₄²⁻): Na₂SO₄

Zinc (Zn²⁺) + Sulphate (SO₄²⁻): ZnSO₄

Lithium (Li⁺) + Sulphate (SO₄²⁻) : Li₂SO₄

Lead(II) (Pb²⁺) + Sulphate (SO₄²⁻): PbSO₄

Calcium (Ca²⁺) + Sulphate (SO₄²⁻): CaSO₄

Iron(II) (Fe²⁺) + Sulphate (SO₄²⁻): FeSO₄

Iron(III) (Fe³⁺) + Sulphate (SO₄²⁻): Fe₂(SO₄)₃

Potassium (K⁺) + Sulphate (SO₄²⁻): K₂SO₄

Carbonate: CO₃²⁻ (2 negative charges)

Magnesium (Mg²⁺) + Carbonate (CO₃²⁻) : MgCO₃

Sodium (Na⁺) + Carbonate (CO₃²⁻): Na₂CO₃

Zinc (Zn²⁺) + Carbonate (CO₃²⁻): ZnCO₃

Lithium (Li⁺) + Carbonate (CO₃²⁻): Li₂CO₃

Lead(II) (Pb²⁺) + Carbonate (CO₃²⁻): PbCO₃

Calcium (Ca²⁺) + Carbonate (CO₃²⁻): CaCO₃

Iron(II) (Fe²⁺) + Carbonate (CO₃²⁻): FeCO₃

Iron(III) (Fe³⁺) + Carbonate (CO₃²⁻): Fe₂(CO₃)₃

Potassium (K⁺) + Carbonate (CO₃²⁻): K₂CO₃

Hydroxide: OH⁻ (1 negative charge)

Magnesium (Mg²⁺) + Hydroxide (OH⁻): Mg(OH)₂

Sodium (Na⁺) + Hydroxide (OH⁻): NaOH

Zinc (Zn²⁺) + Hydroxide (OH⁻): Zn(OH)₂

Lithium (Li⁺) + Hydroxide (OH⁻): LiOH

Lead(II) (Pb²⁺) + Hydroxide (OH⁻): Pb(OH)₂

Calcium (Ca²⁺) + Hydroxide (OH⁻): Ca(OH)₂

Iron(II) (Fe²⁺) + Hydroxide (OH⁻): Fe(OH)₂

Iron(III) (Fe³⁺) + Hydroxide (OH⁻): Fe(OH)₃

Potassium (K⁺) + Hydroxide (OH⁻): KOH

Phosphate: PO₄³⁻ (3 negative charges)

Magnesium (Mg²⁺) + Phosphate (PO₄³⁻): Mg₃(PO₄)₂

Sodium (Na⁺) + Phosphate (PO₄³⁻): Na₃PO₄

Zinc (Zn²⁺) + Phosphate (PO₄³⁻): Zn₃(PO₄)₂

Lithium (Li⁺) + Phosphate (PO₄³⁻): Li₃PO₄

Lead(II) (Pb²⁺) + Phosphate (PO₄³⁻): Pb₃(PO₄)₂

Calcium (Ca²⁺) + Phosphate (PO₄³⁻): Ca₃(PO₄)₂

Iron(II) (Fe²⁺) + Phosphate (PO₄³⁻): Fe₃(PO₄)₂

Iron(III) (Fe³⁺) + Phosphate (PO₄³⁻): FePO₄

Potassium (K⁺) + Phosphate (PO₄³⁻): K₃PO₄

Answer:

h2o

Explanation:

Explanation:

Compressions and Rarefactions

A vibrating tuning fork is capable of creating such a longitudinal wave. As the tines of the fork vibrate back and forth, they push on neighboring air particles. The forward motion of a tine pushes air molecules horizontally to the right and the backward retraction of the tine creates a low-pressure area allowing the air particles to move back to the left.



Because of the longitudinal motion of the air particles, there are regions in the air where the air particles are compressed together and other regions where the air particles are spread apart

Regions of compressed air caused by the sound of an explosion correspond to the compressions and rarefactions of the sound waves.

A sound wave is any distortion in the movement of energy transported by a suitable environment (e.g., air).

In conclusion, regions of compressed air caused by the sound of an explosion correspond to the compressions and rarefactions of the sound waves.

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Answer:

Explanation:

Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.

Metal X can form 2 oxides (A and B).

A + B = 3g

The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.

The mass of metal X in the two oxides will be the same because it's the same metal.

Thus, we represent the mass of the metal in the two oxides as 2X.

2X + 0.72 + 1.16 = 3

2X + 1.88 = 3

2X = 3 - 1.88

2X = 1.12

X = 0.56

Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.

Thus, mass of metal (X) in 1g of oxygen in A is

0.56g ⇒ 0.72g

X ⇒ 1

X = 1 × 0.56/0.72

X = 0.78 g

Hence, 0.78g of the metal will combine with 1g of oxygen for A

Also, mass of metal (X) in 1g of oxygen in B is

0.56g ⇒ 1.16g

X ⇒ 1g

X = 1×0.56/1.16

X = 0.48 g

Thus, 0.48g of the metal will combine with 1g of oxygen for B

Answer:

For A: The wavelength of the light is [tex]1.052\times 10^4nm[/tex]

For B: The number of photons per joule is [tex]2.063\times 10^{18}[/tex]

For C: The binding energy of a metal is 1.197 eV.

Explanation:

The equation used to calculate the energy of a photon follows:

[tex]E=\frac{hc}{\lambda}[/tex]            ......(1)

where,

E = energy of a photon

h = Planck's constant = [tex]6.626\times 10^{-34}J.s[/tex]

c = speed of light = [tex]3\times 10^{8}m/s[/tex]

[tex]\lambda[/tex] = wavelength

Given values:

E = [tex]1.89\times 10^{-20}J[/tex]

Putting values in equation 1, we get:

[tex]\lambda=\frac{(6.626\times 10^{-34}J.s)\times (3\times 10^8m/s)}{1.89\times 10^{-20}J}\\\\\lambda=1.052\times 10^{-5}m[/tex]

Converting the wavelength into nanometers, the conversion factor used is:

[tex]1m=10^9nm[/tex]

So, [tex]\lambda=1.052\times 10^{-5}m\times \frac{10^9nm}{1m}=1.052\times 10^4nm[/tex]

Hence, the wavelength of the light is [tex]1.052\times 10^4nm[/tex]

Given values:

[tex]\lambda=410nm=410\times 10^{-9}m[/tex]

Putting values in equation 1, we get:

[tex]E=\frac{(6.626\times 10^{-34}J.s)\times (3\times 10^8m/s)}{410\times 10^{-9}m}\\\\E=4.848\times 10^{-19}J[/tex]

To calculate the number of photons, we use the equation:

[tex]\text{Number of photons}=\frac{\text{Total energy}}{\text{Energy of a photon}}[/tex]

Total energy = 1 J

Energy of a photon = [tex]4.848\times 10^{-19}J[/tex]

Putting values in the above equation:

[tex]\text{Number of photons}=\frac{1J}{4.848\times 10^{-19}J}\\\\\text{Number of photons}=2.063\times 10^{18}[/tex]

Hence, the number of photons per joule is [tex]2.063\times 10^{18}[/tex]

To calculate the binding energy of a metal, we use the equation:

[tex]E=K+B[/tex]             .....(2)

E = Total energy

K = Kinetic energy of a photon

B = Binding energy of metal

Converting the energy from joules to eV, the conversion factor used is:

[tex]1eV=1.602\times 10^{-19}J[/tex]

Using the above conversion factor:

[tex]K=2.93\times 10^{-19}J=1.829eV\\\\E=4.848\times 10^{-19}J=3.026eV[/tex]

Putting values in equation 2:

[tex]B=(3.026-1.829)eV=1.197eV[/tex]

Hence, the binding energy of a metal is 1.197 eV.

Answer:

a

Explanation:

Easily broken in chemical reactions

Answer:

The product of an organic reaction is analyzed by column chromatography using silica as the stationary phase and toluene as the mobile phase.

Explanation:

The given statement is true.

In chromatography silica gel is used as the predominant stationary phase.

Since silica gel is a good adsorbent.

It is a polar adsorbent.

In order to remove polar components, silica gel is used as the stationary phase.

Answer is a.true.

Answer:

A physical change is a change in form.

A Chemical change is a change in materials.

Explanation:

Example of a physical change would be an ice cube meting.

Example of a chemical change would be mixing food coloring into a cup of water.

Answer:

Kindly check explanation

Explanation:

The force applied is directly proportional to the distance moved by an object, the larger the applied force, the greater the distance moved.

a = f/m

a = acceleration ; f = applied force ; m = mass

From the relation, we can see that acceleration is directly proportional to force applied.

The ball will travel farthest with the greatest applied force while, nearest distance will be attained with the smallest applied force.

The distance covered is affected by both the mass of the object and the applied force

Answer:

C₄H₄N₂

Explanation:

Given that:

M+ = 80.

It implies that the number of nitrogen present in the molecule must also be even according to the Nitrogen rule.

So from the Formula CHN, the nitrogen will have to be 2 because if we make use of 4, it will exceed the given M+ which is 80.

∴

C₄ = 4 × 12 = 48

H₄  = 4 × 1   = 4

N₂ =  2 × 14 = 28      

                      80      

As such, the molecular formula of the compound is C₄H₄N₂

it is moss

Explanation:

Answer:

the answer is letter D, Dilute solution

I think it's 9 ................'

Explanation:

Iron atom is been oxidised as it losses 2 electron to form 2 + ion.

Answer:

yes the experimental result is the expected result .

Explanation:

When Titrating acetate buffer with acid the PH will decrease gradually from a more neutral PH to a more acidic level and this is because buffer solutions are prepared with weak acids and its conjugate base.

The results gotten from the continuous addition of base NaOH to the acetate buffer is the expected result because the base is been absorbed by the buffer solution and it is converted to a conjugate base of the buffer solution which will gradually increase the PH level of the solution as more conjugate base is formed due to the addition of more NaOH.