The acid dissociation constant, Ka, of HNO3 is 4.0 x104. What does the ka
value indicate about this compound?
A. HNO3 is neither an acid nor a base,
B. HNO3 is a strong acid.
C. HNO3 is a weak acid
D. HNO3 is a strong base.

Answer:

(a) The moles of CuSO₄ is 3.125 × 10⁻³ moles.

(b) The moles of Cu is 3.125 × 10⁻³ moles.

(c) The mass of Cu is 0.2 g.

Explanation:

Given:

Mass of CuSO₄ = 0.5 g

Molar mass of CuSO₄ = 160 g/mol

The given balanced chemical equation is:

[tex]2Al+3CuSO_4\rightarrow 3Cu+Al_2(SO_4)_3[/tex]

Part (a):

Calculating the moles of CuSO₄.

[tex]\text{Moles of } CuSO_4=\frac{\text{Mass of }CuSO_4}{\text{Molar mass of }CuSO_4}\\\\\text{Moles of } CuSO_4=\frac{0.5g}{160g/mol}\\\\\text{Moles of } CuSO_4=3.125\times 10^{-3}mol[/tex]

Thus, the moles of CuSO₄ is 3.125 × 10⁻³ moles.

Part (b):

Calculating the moles of Cu.

From the balanced chemical equation, we conclude that:

As, 3 moles of CuSO₄ reacts to give 3 moles of Cu

So, 3.125 × 10⁻³ moles of CuSO₄ reacts to give 3.125 × 10⁻³ moles of Cu

Thus, the moles of Cu is 3.125 × 10⁻³ moles.

Part (c):

Calculating the mass of Cu.

Mass of Cu = Moles of Cu × Molar mass of Cu

Molar mass of Cu = 64 g/mol

Mass of Cu = (3.125 × 10⁻³ mole) × (64 g/mol)

Mass of Cu = 0.2 g

Thus, the mass of Cu is 0.2 g.

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: The mass of [tex]CO_2[/tex] produced is 0.44 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of calcium carbonate = 10 g

Molar mass of calcium carbonate = 100 g/mol

Plugging values in equation 1:

[tex]\text{Moles of calcium carbonate}=\frac{10g}{100g/mol}=0.1 mol[/tex]

The formula used to calculate molarity:

[tex]\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}[/tex] .....(2)

Molarity of HCl = [tex]0.2mol/dm^3=0.2mol/L[/tex]     (Conversion factor: [tex]1L=1dm^3[/tex]

Volume of solution = [tex]100cm^3=100mL[/tex]          (Conversion factor: [tex]1mL=1cm^3[/tex]

Putting values in equation 2, we get:

[tex]0.2=\frac{\text{Moles of HCl}\times 1000}{100}\\\\\text{Moles of HCl}=\frac{0.2\times 100}{1000}=0.02mol[/tex]

For the given chemical reaction:

[tex]CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)+CO_2(g)[/tex]

By stoichiometry of the reaction:

If 2 moles of HCl reacts with 1 mole of calcium carbonate

So, 0.02 moles of HCl will react with = [tex]\frac{1}{2}\times 0.02=0.01mol[/tex] of calcium carbonate

As the given amount of calcium carbonate is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, HCl is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 2 moles of [tex]HCl[/tex] produces 1 mole of [tex]CO_2[/tex]

So, 0.02 moles of [tex]HCl[/tex] will produce = [tex]\frac{1}{2}\times 0.02=0.01mol[/tex] of [tex]CO_2[/tex]

We know, molar mass of [tex]CO_2[/tex] = 44 g/mol

Putting values in equation 1, we get:

[tex]\text{Mass of }CO_2=(0.01mol\times 44g/mol)=0.44g[/tex]

Hence, the mass of [tex]CO_2[/tex] produced is 0.44 g

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:

aasjajiakjka

Explanation:

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:

Chlorine

Chlorine has two naturally occurring isotopes. In a sample of chlorine, 75.77% of the atoms are Cl-35, with a mass of 34.97amu.

Explanation:

hope it helps thanks❤

Chlorine has two naturally occurring isotopes. In a sample of chlorine, 75.77% of the atoms are Cl-35, with a mass of 34.97amu.

List the known and unknown values ​​and edit the problem. Convert each percentage value into a decimal by dividing it by 100. Multiply this number by the atomic mass of that spot. Mix together each isotope to find the median atomic mass.

To calculate the maximum percentage of each isotope in a sample of something, chemists usually divide the atomic number of a particular isotope by the total number of atoms of all that isotope and multiply the result by 100.

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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

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:

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 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

To make the steel less likely to rust, element like chromium is added to it. Therefore, the given statement is correct.

Steel can be described as an alloy composed of iron with a few tenths of a percent of carbon to enhance its strength and fracture resistance. In steel, alloying elements inclusions within the iron act as hardening agents as they stop the movement of dislocations.

Stainless steels are corrosion- and oxidation-resistant but they need an additional 11% chromium. This type of steel is used in buildings, infrastructure, ships, trains, machines, electrical appliances, weapons, and rockets because of its high tensile strength and low cost.

It can take two crystalline forms body-centered cubic and face-centered cubic depending on the temperature. The interaction of the allotropes of iron with elements such as carbon gives the steel its range of unique properties.

Therefore, the given statement is true that chromium is used in steel formation.

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Explanation:

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

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.

Coal and Energy obtained from coal are formed as a result of decomposed life matter such as plants

A. Compressed ancient plants

It should be noted that coal is also a type of fossil fuel formed by prolong compaction of fossil material e.g dead plants over millions of years, forest lives etc. and as a result of high pressure and heat acting on these fossil matter they are transformed to fuel.

In summary, fossil fuels are formed as a result of age long compressed matter.

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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

Answer:

the answer is letter D, Dilute solution

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:

it is moss

Explanation:

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:

an estimate of the strength of a bond.

Explanation:

The bond order is given as;

1/2(number of bonding electrons - number of anti bonding electrons)

The bond order tells us about the strength of bond. As the bond order increases, so does the strength of the bond because atoms involved in bonding come closer to each other as the bond length decreases.

Hence, bond order is an index of bond strength. Triple bonds are stronger than double bonds which are stronger than single bonds.

Answer:

Gains

Explanation:

It gets more electrons

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:

answer is b. : It has been balanced by using the half-reaction method.

Explanation:

edge 2021

Answer:b

Explanation:cuz it is

Answer:

a

Explanation:

Easily broken in chemical reactions

Answer:

66s^-1 will be 1/66

then to convert to minute you multiply by 69

1/66 x 60 = 3960 mins

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.

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