Explain why water is polar. Check all that apply. View Available Hint(s) Water is known as a polar molecule because Check all that apply. the hydrogen atom attracts electrons much more strongly than the oxygen atom. the oxygen atom has a greater attraction for electrons than the hydrogen atom does. the oxygen and hydrogen atoms have the same electronegativity. the electrons of the covalent bond are not shared equally between the hydrogen and oxygen atoms. water dipole moment is equal to zero.

Answer:

a. 90.288 kJ.

b. - 54.06 kJ/mol.

Explanation:

a. Calculate the energy released per 10 g of CH₃COOH.

Q = m.c.ΔT,

where, Q is the amount of heat released to water (Q = ??? J).

m is the mass of water (m = 2000.0 g).

c is the specific heat capacity of solution (c = 4.18 J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = 34.3°C - 23.5°C = 10.8°C).

∴ Q = m.c.ΔT = (2000.0 g)(4.18 J/g.°C)(10.8°C) = 90288 J = 90.288 kJ.

b. Calculate the energy released per mole of CH₃COOH.

∵ ΔH = Q/n

no. of moles of CH₃COOH (n) = mass/atomic mass = (10.0 g)/((60.052 g/mol) = 0.167 mol.

∴ ΔH = - Q/n = - (90.288 kJ)/(0.167 mol) = - 54.06 kJ/mol.

The negative sign is not from calculation, but it is an indication that the reaction is exothermic.

Answer: The nuclear binding energy of the given element is [tex]2.938\times 10^{12}J/mol[/tex]

Explanation:

For the given element [tex]_3^6\textrm{Li}[/tex]

Number of protons = 3

Number of neutrons = (6 - 3) = 3

We are given:

[tex]m_p=1.00728amu\\m_n=1.00866amu\\A=6.015126amu[/tex]

M = mass of nucleus = [tex](n_p\times m_p)+(n_n\times m_n)[/tex]

[tex]M=[(3\times 1.00728)+(3\times 1.00866)]=6.04782amu[/tex]

Calculating mass defect of the nucleus:

[tex]\Delta m=M-A\\\Delta m=[6.04782-6.015126)]=0.032694amu=0.032694g/mol[/tex]

Converting this quantity into kg/mol, we use the conversion factor:

1 kg = 1000 g

So,  [tex]0.032694g/mol=0.032694\times 10^{-3}kg/mol[/tex]

To calculate the nuclear binding energy, we use Einstein equation, which is:

[tex]E=\Delta mc^2[/tex]

where,

E = Nuclear binding energy = ? J/mol

[tex]\Delta m[/tex] = Mass defect = [tex]0.032694\times 10^{-3}kg/mol[/tex]

c = Speed of light = [tex]2.9979\times 10^8m/s[/tex]

Putting values in above equation, we get:

[tex]E=0.032694\times 10^{-3}kg/mol\times (2.9979\times 10^8m/s)^2\\\\E=2.938\times 10^{12}J/mol[/tex]

Hence, the nuclear binding energy of the given element is [tex]2.938\times 10^{12}J/mol[/tex]

The binding energy of the lithium nucleus is 2.94 * 10^14 J/mol.

The term binding energy refers to the energy that hold the nucleons in the atom together.

We know that the atomic mass of the Li is 6.015126 amu. Note that there are three protons and three neutrons. Hence;

Mass of protons= 3(1.00728 amu) = 3.02184

Mass of neutrons = 3(1.00866 amu)  = 3.02598

Mass defect = (3.02184 + 3.02598) - 6.015126 amu = 0.032694 amu =  0.032694 g/mol

Now;

E = mc^2 = (0.032694 g/mol * (3 * 10^8)^2) = 2.94 * 10^14 J/mol

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Answer:

- 431.15 kJ/mol.

Explanation:

Q = m.c.ΔT,

where, Q is the amount of heat released from the solution (Q = ??? J).

m is the mass of solution (m = 1.5 g + 100 g = 101.5 g).

c is the specific heat capacity of solution (c = 4.18 J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = 33.1°C - 22°C = 11.1°C).

∴ Q = m.c.ΔT = (101.5 g)(4.18 J/g.°C)(11.1°C) = 4709.4 J.

∵ ΔH = Q/n

no. of moles of Ba (n) = mass/atomic mass = (1.50 g)/(137.3270 g/mol) = 0.011 mol.

∴ ΔH = - Q/n = (4709.4 J)/(0.011 mol) = - 431.15 kJ/mol.

The negative sign is not from calculation, but it is an indication that the reaction is exothermic.

Answer:

7.186

Explanation:

The mean is the average of some given data from a sample point.

To calculate the mean, we use the formula below:

             Mean = ∑fx/∑f

Where f = frequency

           x = sample data

From the given pH of the solutions, we can form a table:

      x                                     f                                          fx

    7.15                                  1                                         7.15

    7.16                                  1                                         7.16

    7.18                                  1                                         7.18

    7.19                                  1                                         7.19

    7.20                                 3                                        21.6

    7.21                                  1                                         7.21

Now ∑fx = 7.15 +7.16 +7.18 + 7.19 + 7.20 + 7.21 = 57.49

         ∑f = 1 + 1 + 1 + 1 + 3 + 1 = 8

The mean = [tex]\frac{57.49}{8}[/tex] = 7.18625 = 7.186

The first reaction is a redox reaction, the second and third reactions are none of the above, and the fourth reaction is a precipitation reaction.

The reactions asked in the question can be classified as follows:

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The first reaction is an oxidation-reduction (combustion) reaction, the second reaction is a precipitation reaction, the third reaction is an acid-base neutralization reaction, and the fourth reaction is also an acid-base neutralization reaction.

The first reaction, Ba(ClO3)2(s)⟶BaCl2(s)+3O2(g), is a decomposition reaction also known as oxidation-reduction (combustion). The solid compound breaks down into a solid product and a gas.

The second reaction, 2NaCl(aq)+K2S(aq)⟶Na2S(aq)+2KCl(aq), is a precipitation reaction. The combination of two aqueous solutions forms an insoluble product.

The third reaction, CaO(s)+CO2(g)⟶CaCO3(s), is an acid-base neutralization reaction. The solid oxide reacts with a gas to form a solid carbonate.

The fourth reaction, KOH(aq)+AgNO3(aq)⟶KNO3(aq)+AgOH(s), is also an acid-base neutralization reaction. The aqueous solutions react to form a solid hydroxide and an aqueous salt.

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Answer:

[Ar] 4s² 3d¹⁰ 4p³

Explanation:

Arsenic has an atomic number of 33, thus 33 protons and 33 electrons. The electron configuration is:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p³

The shorthand notation includes the previous noble gas to represent the inner electrons. The gas previous to arsenic is argon, with 18 electrons. The shorthand notation for arsenic is:

[Ar] 4s² 3d¹⁰ 4p³

Arsenic (Ar) is located in the 5th period (row) and 15th group (column) of the periodic table. Its atomic number is 33, which means it has 33 electrons. The electrons' configuration is: [tex]1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^3[/tex].

This is the electron configuration of arsenic (Ar). It indicates that arsenic has two electrons in the 1s orbital, two electrons in the 2s orbital, six electrons in the 2p orbital, two electrons in the 3s orbital, six electrons in the 3p orbital, two electrons in the 4s orbital, ten electrons in the 3d orbital, and three electrons in the 4p orbital.

The electronic configuration of Arsenic is [tex]1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^3[/tex].

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Answer:

D => pH will be equal to 7 at equivalence point

Explanation:

For Strong Acid + Strong Base titrations, pH = 7 as neither ion of the salt produced will undergo hydrolysis as would weak electrolyte titrations.

BOTH CASES ARE PRESENTED FOR CONTRAST ...

Stong monoprotic acid being titrated with NaOH ...

=> HX + NaOH => NaCl + H₂O

=> NaCl => Na⁺ + Cl⁻

=> Na⁺ + H₂O => No Rxn ( formation of NaOH will not occur as a strong electrolyte prefers to remain 100% ionized)

=> X⁻ + H₂O => No Rxn (formation of HX will not occur as a strong electrolyte prefers to remain 100% ionized)

This leaves only the Auto Ionization of Water as the reaction affecting the pH of the solution at the equivalence point of a strong acid + strong base titration. That is ...

HOH ⇄ H⁺ + OH⁻ & [H⁺] = [OH⁻] = 1 x 10⁻⁷M

pH = -log[H⁺] = -log(1 x 10⁻⁷) = -(-7) = 7  

----------------------------------

Weak Acid + Strong Base titration => pH > 7 at equivalence point

Assume => HA = weak acid

=> HA + NaOH => NaA + H₂O

=> NaA => Na⁺ + A⁻  &  A⁻ is the conjugate base of a weak acid HA

=> Na⁺ + H₂O => No Rxn ( formation of NaOH will not occur as a strong electrolyte prefers to remain 100% ionized)

=> A⁻ +  H₂O => HA + OH⁻  => Excess OH⁻ at equivalence pt => pH > 7

-------------------------------------

Weak Base + Strong Acid titration => pH < 7 at equivalence point

Weak Bases => ammonia (NH₃) or ammonia derivatives (RNH₂)* in water.

(ammonia in water) => :NH₃ + H₂O => NH₄OH ⇄ NH₄⁺ + OH⁻

(ammonia derivative in water) => RN:-H₂ + H₂O => RNH₃OH ⇄ RNH₃⁺ + OH⁻

Titration of weak base with strong acid ...

=> NH₄OH + HX => NH₄X + H₂O

=> NH₄X => NH₄⁺ + X⁻

=> X⁻ + H₂O => No Rxn (formation of HX will not occur as a strong electrolyte prefers to remain 100% ionized)

=> NH₄⁺ + HOH ⇄ NH₄OH + H⁺ => Weak base is in molecular form with excess hydronium ions (H₃O⁺ = H⁺) at equivalence point => pH < 7.

----------------------

*RNH₂ is a primary amine used in the illustration, but the above process will also occur for secondary (R₂N:-H) and tertiary amines (R₃N:) in water also.

Answer: The number of moles of [tex]Na_2SO_4[/tex] is 0.05 moles.

Explanation:

To calculate the molarity of solution, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

We are given:

Molarity of solution = 0.10 mol/L

Volume of solution = 1 L

Putting values in above equation, we get:

[tex]0.10mol/L=\frac{\text{Moles of sodium}}{1.0L}\\\\\text{Moles of sodium}=0.10mol[/tex]

The chemical reaction for the ionization of sodium sulfate follows the equation:

[tex]Na_2SO_4(s)\rightarrow 2Na^+(aq.)+SO_4^{2-}(aq.)[/tex]

By Stoichiometry of the reaction:

2 moles of sodium ions are produced by 1 mole of sodium sulfate

So, 0.10 moles of sodium ions will be produced by = [tex]\frac{1}{2}\times 0.1=0.05moles[/tex] of sodium sulfate.

Hence, the number of moles of [tex]Na_2SO_4[/tex] is 0.05 moles.

Answer:

4B + 3O₂ => 2B₂O₃; ΔH° = -3673Kj

Explanation:

Work these type problems in pairs of rxns… That is, add Rxn-1 & Rxn-2 => Rxn-1,2; then add Rxn-3 to Rxn-1,2 => Rxn- 1,2,3. Rxn-4 is not needed to obtain target rxn.  

Target Rxn => 4B + 3O₂ => 2B₂O₃

Given …

(1) B₂O₃ + 2H₂O => 3O₂ + B₂H₆

       => reverse and double

       => 2B₂H₆ + 6O₂ => 2B₂O₃ + 6H₂O

(2) 2B + 3H₂ => B₂H₆ => double and add to Rxn-1 => 4B + 6H₂ => 2B₂H₆

         2B₂H₆ + 6O₂ => 2B₂O₃ + 6H₂O

              4B + 6H₂ => 2B₂H₆

              ________________________

∑(1,2)    4B + 6O₂ + 6H₂ => 2B₂O₃ + 6H₂O; ΔH°₁₂ = -2035Kj + (+72Kj) = -1963Kj

(3) => H₂ + ½O₂ => H₂O

       => reverse and multiply by 6, then add to (1,2) => 6H₂O => 6H₂ + 3O₂

                           6H₂O => 6H₂ + 3O₂

           4B + 6O₂ + 6H₂ => 2B₂O₃ + 6H₂O  

   ________________________

∑[(1,2,3) 4B + 3O₂ => 2B₂O₃; ΔH°₁₂₃ = -1963Kj + 6(-285Kj) = -3673Kj

A possible structure for compound A is γ-Bromoglutamic acid (2-amino-4-bromobutanedioic acid).

What is this structure?

Ninhydrin test: A positive ninhydrin test indicates the presence of a primary or secondary amine. Glutamine has a primary amine, which would explain the positive test.

Specific optical rotation: The specific rotation (37.5 mL/(g·dm)) suggests chirality, which glutamine also possesses.

Not one of the 20 essential amino acids: This eliminates most common amino acids.

Isoelectric point (pI) of 9.4: This is consistent with γ-bromoglutamic acid, as the additional bromine group introduces a new acidic side chain (pKa ≈ 4.5) that lowers the pI compared to glutamine (pI ≈ 5.6).

Synthesis from L-glutamine and Br2/NaOH: This reaction is known to selectively brominate at the γ-carbon of glutamine, making it a plausible route for obtaining γ-bromoglutamic acid.

Answer:

M (third main energy level)

Explanation:

The third main energy level bears the first appearance of the 'd' sublevel. The principal quantum number(n) depicts the main energy levels in which an orbital is located. It takes values of n=1,2,3,4,5..... and it can be represented by the shells k,l,m,n.......

The subshells in these main orbitals are represented by s,p,d and f. For the K shell, the principal quantum number is m and its sublevel notations are s,p and d. This is where the d-sublevel first appears.

The 'd' sublevel first appears in the third main energy level (M). This is because the electron configuration in atoms is organized into main energy levels and sublevels, which define an electron's distance from the nucleus and energy.

The 'd' sublevel first appears in the third main energy level, also labeled as M. The electron configuration in atoms is organized into main energy levels and sublevels, which help define an electron's relative distance from the nucleus and its energy. The main energy levels are generally labeled K, L, M, N, etc., starting from the nucleus. Each energy level has one or more sublevels: s, p, d, f. The 's' sublevel appears in all main energy levels, the 'p' sublevel starts from the second (L), and the 'd' sublevel commences from the third (M) main energy level. Therefore, the 'd' sublevel does not exist in the first (K) or second (L) main energy levels.

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Answer:

15L of 0.40M NaCl(aq) solution

Explanation:

2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g)

2Na⁺(aq) + 2Clˉ(aq) + 2H₂O(l) → 2Na⁺(aq) + 2OHˉ(aq) + Cl₂(g)

Na⁺(aq) is a spectator ion in the given reaction and does not enter into the reaction process…

Net Ionic Equation is then 2Clˉ(aq) + 2H₂O(l) → 2OHˉ(aq) + Cl₂(g)

From Rxn, 2 moles Clˉ(aq) is needed to produce 1 mole of  Cl₂(g)

Therefore, 6 moles Clˉ(aq) is needed to produce 3 moles of Cl₂(g)

That is, 6 moles NaCl(aq) → 6 moles Clˉ(aq) = 0.40M x V(NaCl)liters  

V(NaCl) liters = 6 moles Clˉ(aq)/(0.40mole/liter) = 15 liters of 0.40M NaCl(aq)

7.5 liters of 0.4M sodium chloride solution are needed to produce 3.0 moles of chlorine gas, and 0.134 moles of water are needed to produce 3.0 liters of chlorine gas at STP.

The question is related to molarity and stoichiometry. Let's solve part (a) and (b).

Molarity is represented by moles of solute / liters of solution. The balanced equation gives a 1:1 ratio between NaCl and Cl2. The question gives us 3.0 moles of Cl2 gas, implying that we need the same moles of NaCl.

So, Moles = Molarity x Volume, Volume = Moles / Molarity, Volume = 3.0 moles / 0.4 M = 7.5 liters of NaCl solution are needed.

For part (b), we see from the stoichiometry of the balanced reaction that the number of moles of water is equal to the number of moles of chlorine. According to the ideal gas law (PV=nRT), at standard temperature and pressure (STP), 1 mole of any gas occupies 22.4L. So, 3.0L of Cl2 gas corresponds to 3.0 L / 22.4 L/mole = 0.134 moles. Therefore, you'll need 0.134 moles of water.

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Answer : The mass of sodium iodide used to produced must be, 105.22 grams.

Explanation : Given,

Mass of [tex]I_2[/tex] = 89.1 g

Molar mass of [tex]I_2[/tex] = 253.8 g/mole

Molar mass of [tex]NaI[/tex] = 149.89 g/mole

First we have to calculate the moles of [tex]I_2[/tex].

[tex]\text{Moles of }I_2=\frac{\text{Mass of }I_2}{\text{Molar mass of }I_2}=\frac{89.1g}{253.8g/mole}=0.351moles[/tex]

Now we have to calculate the moles of [tex]NaI[/tex].

The balanced chemical reaction is,

[tex]2NaI(aq)+Cl_2(g)\rightarrow I_2(s)+2NaCl(aq)[/tex]

From the balanced chemical reaction, we conclude that

As, 1 mole of [tex]I_2[/tex] obtained from 2 moles of [tex]NaI[/tex]

So, 0.351 moles of [tex]I_2[/tex] obtained from [tex]2\times 0.351=0.702[/tex] moles of [tex]NaI[/tex]

Now we have to calculate the mass of [tex]NaI[/tex].

[tex]\text{Mass of }NaI=\text{Moles of }NaI\times \text{Molar mass of }NaI[/tex]

[tex]\text{Mass of }NaI=(0.702mole)\times (149.89g/mole)=105.22g[/tex]

Therefore, the mass of sodium iodide used to produced must be, 105.22 grams.

To produce 89.1 grams of iodine (I2), 178.2 grams of sodium iodide (NaI) must be used.

To find the number of grams of sodium iodide (NaI) needed to produce 89.1 g of iodine (I2), we need to use the stoichiometric coefficients from the balanced chemical equation:

2NaI(aq) + Cl2(g) ⟶ I2(s) + 2NaCl(aq)

From the equation, we can see that 2 moles of NaI react to produce 1 mole of I2. The molar mass of NaI is 149.89 g/mol. We can set up a proportion:

(2 mol NaI / 1 mol I2) = (x g NaI / 89.1 g I2)

Solving for x:

x = (2 mol NaI / 1 mol I2) * (89.1 g I2 / 1 mol I2) = 178.2 g NaI

Therefore, 178.2 grams of sodium iodide (NaI) must be used to produce 89.1 grams of iodine (I2).

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Answer : The equilibrium concentrations of [tex]COCl_2,CO\text{ and }Cl_2[/tex] are, 0.2646, 0.0584 and 0.0584.

Explanation : Given,

Moles of  [tex]COCl_2[/tex] = 0.323 mole

Volume of solution = 1 L

Initial concentration of [tex]COCl_2[/tex] = 0.323 M

Let the moles of [tex]CO\text{ and }Cl_2[/tex] be, 'x'. So,

Concentration of [tex]CO[/tex] = x M

Concentration of [tex]Cl_2[/tex] = x M

The given balanced equilibrium reaction is,

                             [tex]COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)[/tex]

Initial conc.          0.323 M          0          0

At eqm. conc.    (0.323-x) M    (x M)       (x M)

The expression for equilibrium constant for this reaction will be,

[tex]K_c=\frac{[CO][Cl_2]}{[COCl_2]}[/tex]

Now put all the given values in this expression, we get :

[tex]1.29\times 10^{-2}=\frac{(x)\times (x)}{(0.323-x)}[/tex]

By solving the term 'x', we get :

x = 0.0584

Thus, the concentrations of [tex]COCl_2,CO\text{ and }Cl_2[/tex] at equilibrium are :

Concentration of [tex]COCl_2[/tex] = (0.323-x) M  = (0.323-0.0584) M = 0.2646 M

Concentration of [tex]CO[/tex] = x M = 0.0584 M

Concentration of [tex]Cl_2[/tex] = x M = 0.0584 M

Therefore, the equilibrium concentrations of [tex]COCl_2,CO\text{ and }Cl_2[/tex] are, 0.2646, 0.0584 and 0.0584.

In this chemistry problem, we first calculate the heat produced by the combustion of C6H6, and then we use the heat equation to find the final temperature of the water which comes out to be 53.14 °C.

The subject of this question pertains to the field of chemistry, specifically physical chemistry dealing with energy changes in chemical reactions, and this seems to be a high school-level problem based on the complexities involved. The principle used here is q=mc∆T, which allows us to calculate how much heat is transferred when the water temperature changes.

First, the amount of heat produced by the combustion of 8.6 g of C6H6 should be calculated based on its enthalpy change (heat produced per mole). Given the heat of combustion, -6542 kJ for 2 moles of C6H6, the heat of combustion for 8.6g of C6H6 can be calculated by (8.6 g / (78.11 g/mol) mol ) * (-6542 kJ / 2 mol) = -227.11 kJ.

The heat absorbed by the water can then be calculated using the heat equation: q=mc∆T. Here, m = mass of water = 5691 g, c = specific heat capacity of water = 4.18 J/g°C. The heat is converted to kilojoules: q = -227.11 kJ * 1000 = -227110 J. Hence, the equation becomes -227110 = 5691*4.18*∆T. After rearranging and solving, the final temperature (T) will be 53.14 °C.

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Answer:

40.g CaH2

Explanation:

1. ideal gas law(PV = nRT) → use ideal gas law first when volume is given

P = 0.995atm

V = 48.0L H2

n = ?

R = 0.0821L atm/molK

T = 32 + 273 = 305K

n = (0.995atm x 48.0L H2)/(0.0821L atm/molK x 305K) → do not simplify as small decimals might change the answer

2. Conversions

2 H2 and 1 CaH2 → 1/2

(mole of H2) x 1/2 x (molar mass of CaH2)

(0.995atm x 48.0L H2)/(0.0821L atm/molK x 305K) x 1/2 x (40.08 + 2.02) = 40.g CaH2

the longer answer will be 40.14887882 but as the minimum sigfig given in the question is 2, it is 40.g CaH2.

Hope it helped!

Answer:

1,081.1 units of heparin should be given to 298 lb person.

Explanation:

We are given:

Weight of the person = 298 lb = 135.143 kg

Conversion factor used:

1 lb = 0.4535 kg

Number of unit of heparin to be given in an hour = 8.0 units/kg

Number of units given to the patient weighing 135.143 kg :

[tex]8.0 units/kg\times 135.143 kg=1,081.144 units\approx 1,081.1 units[/tex]

1,081.1 units of heparin should be given to 298 lb person.

Answer:

Explanation:

The biochemistry that takes place inside cells depends on various elements, such as sodium, potassium, and

calcium, that are dissolved in water as ions. These ions enter cells through narrow pores in the cell membrane

known as ion channels. Each ion channel, which is formed from a specialized protein molecule, is selective

for one type of ion. Measurements with microelectrodes have shown that a 0.30-nm-diameter potassium ion (K+) channel carries a current of 1.8 pA.

Part A. How many potassium ions pass through if the ion channel opens for 1.0 ms?

Part B. What is the current density in the ion channel?

Solution: In 1.0 ms, the charge that passes through is

Q = I ∆t =

( 1.8 × 10−12 A ) (1.0 × 10−3 s )

= 1.8 × 10−15 C

Since each ion has a +1 charge (measured in electron charges), this represents

NK+ = Q e = (1.8 × 10−15 C ) (1.60 × 10−19 C) = 11250

The current density is calculated from the current and the size of the channel.

J = I A = ( 1.8 × 10−12 A ) ( π (0.30 × 10−9 m/2)2 ) = 2.55 × 107 A/m2

A) The number of potassium ions that will pass through the ion channel; 11250

B) The current density in the ion channel = 2.55 × 10⁷ A/m²

Given data;

Measurement with microelectrodes = 0.30-nm- diameter

K⁺ = 1.8 pA

a) calculate the number of potassium ions that passes through the ion channel

given that the channel opens for 1.0 ms

first step ; determine the value of charge ( Q )

Q = I*Δt = ( 1.8 × 10⁻¹² A )*(1.0 × 10⁻³ s ) = 1.8 × 10⁻¹⁵ C

where ; I = 1.8 × 10⁻¹² A

           Δt =  1.0 × 10⁻³ s

next step : determine number of K⁺ ions passing through

NK⁺ = Q*e = ( 1.8 × 10⁻¹⁵ C )*( 1.60 × 10⁻¹⁹ C)  = 11250

∴ number of K⁺ ions passing through = 11250 ions.

B) determine the current density in the ion channel

current density = current  * size of channel

                         = ( 1.8 * 10⁻¹² ) * ( π * ( 0.30 * 10⁻⁹ )²

                         = 2.55 × 10⁻⁷ A/m²

Hence we can conclude that the  number of potassium ions that will pass through the ion channel; 11250 and  The current density in the ion channel = 2.55 × 10⁷ A/m²

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Explanation:

Since, it is given that density is 1.16 grams per milliliter and molar mass of copper sulfate pentahydrate is 249.68 g/mol.

Now, as we known that density is the amount of mass present in a unit volume.

Mathematically,         Density = [tex]\frac{\text{molar mass}}{volume}[/tex]

Hence, calculate the volume as follws.

                    Density = [tex]\frac{\text{molar mass}}{volume}[/tex]

                  1.16 g/mL = [tex]\frac{249.68 g/mol}{volume}[/tex]          

                 volume = 215.24 mL

As, 215.24 mL can dissolve in 249.68 g/mol of complex. So, in 100 mL volume amount dissolved will be calculated as follows.

                       [tex]\frac{249.68 g/mol}{215.24 mL} \times 100[/tex]

                    = 116 grams        

Thus, we can conclude that 116 grams of copper sulfate pentahydrate that will dissolve in 100 g of water at 0[tex]^{o}C[/tex].                              

To calculate the grams of copper sulfate pentahydrate that dissolve in 100 g of water at 0°C, solubility data at that temperature is needed, which is not provided in the question.

The student is asking to calculate the grams of copper sulfate pentahydrate that will dissolve in 100 grams of water at 0°C, given that the density of the saturated solution is 1.16 g/mL and the molar mass of copper sulfate pentahydrate is 249.68 g/mol. Since the volume of solution can be derived from the mass of the water and the density of the solution, we can calculate the amount of copper sulfate pentahydrate dissolved in this volume by using dimensional analysis. However, to perform this calculation, we need the solubility data for copper sulfate pentahydrate at 0°C, which is not provided in the question. Without this information, we cannot proceed with the calculation.

Answer: Option (4) is the correct answer.

Explanation:

According to the reactivity series, a more reactive metal has the ability to replace a less reactive metal in a chemical reaction.

It is known that calcium is more reactive than magnesium. So, in a chemical reaction magnesium can never replace calcium.

For example, [tex]Mg(s) + Ca(OH)_2(aq) \rightarrow Ca(s) + Mg(OH)_2(aq)[/tex]

Therefore, this given reaction is not possible. But rest all given reactions are possible.

Thus, we can conclude that out of the given options [tex]Mg(s) + Ca(OH)_2(aq) \rightarrow Ca(s) + Mg(OH)_2(aq)[/tex] reaction will not occur as written.

OPTION D.

The reaction Mg (s) + Ca(OH)₂ (aq) → Ca (s) + Mg(OH)₂ (aq) will not occur as written because in the activity series of metals, magnesium (Mg) is more reactive than calcium (Ca), and hence, Mg cannot displace Ca from its compound.

In the given set of reactions, Mg (s) + Ca(OH)₂ (aq) → Ca (s) + Mg(OH)₂ (aq) will not occur as written. The explanation is based on activity series of metals. In this series, magnesium (Mg) is more reactive than calcium (Ca). So, Mg cannot displace Ca from its compound. In other words, a more reactive metal can displace a less reactive metal from its compound in a solution, but not the other way round. The other reactions would occur as represented since in each of them the free metal is more reactive than the one in the compound.

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The empirical formula and the molecular formula for the given compound will be [tex]C_2H_4O[/tex] and [tex]C_6H_{12} O_3[/tex] respectively.

The simplest whole-number ratio of atoms in a molecule is the empirical formula of a chemical compound in chemistry.

Let us consider that, we have 100g of the sample, then carbon would be 54.53 g, hydrogen would be 9.15g and oxygen would be 36.32g and the number of moles for each element will be = [tex]\frac{GivenMass}{Molar Mass}[/tex]

Number of moles of carbon = [tex]\frac{54.53}{12} = 4.544[/tex] mol

Similarly, the number of moles of Hydrogen and Oxygen would be:

Number of moles of Oxygen = [tex]\frac{36.32}{16} = 2.27\\[/tex] mol

Number of moles of Hydrogen = [tex]\frac{9.15}{1} = 9.15 mol[/tex]

Molar ratio of the elements = C:H:O = 4.54 : 9.15 : 2.27

The molar ratio = 2:4:1,

So, the empirical formula = [tex]C_2H_4O[/tex] and empirical mass = [tex]2*12 + 4*1 + 16 = 44[/tex]

The empirical formula and the molecular formula are often related in the following way: (Molecular Formula = n [tex]*[/tex] Empirical Formula). This may be used to determine the compound's empirical formula as well as its molecular formula. The term “n” in the relationship denotes the proportion between the compound's molecular mass and empirical mass.

[tex]n = \frac{Molecular Mass}{Empirical Mass} = \frac{132}{44} = 3[/tex]

So, molecular formula = [tex]3* C_2H_4O = C_6 H_{12} O_3[/tex]

Thus, the molecular formula and empirical formula for the given compound are  [tex]C_2H_4O[/tex] and [tex]C_6H_{12} O_3[/tex].

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The empirical and molecular formula of a compound with 54.53% carbon, 9.15% hydrogen, and 36.32% oxygen and molar mass 132 amu, is CH₄O₂.

The process to find the empirical and molecular formula of a compound uses percentage composition of its constituent elements and molar mass. First, we convert the percentage to grams (assuming 100g of the compound) which gives: C 54.53g, H 9.15g, and O 36.32g. Using atomic masses of C(12.01), H(1.01), and O(16.00), we find moles of C, H, and O.

Dividing the obtained values by the smallest number of moles, gives us the ratio of the elements. In this case, the empirical formula comes out to be CH₄O₂. We then calculate the molecular mass of the empirical formula and divide the molar mass of the compound by the empirical formula mass. The quotient gives the number of empirical formula units present in the compound's molecular formula. If this quotient is anything other than '1', we must multiply the subscript of each element in the empirical formula by this number to get the molecular formula. However, in this case, the empirical and molecular formulas are the same, i.e., CH₄O₂.

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These are two questions and two answers

Answer:

    Question 1:

    Question 2:

Explanation:

Question 1:

The neutralization reaction that occurs between H₂SO₄ and KOH is an acid-base reaction.

The products of an acid-base reaction are salt and water.

This is the sketch of such neutralization reaction:

1) Word equation:

                 ↑                               ↑                              ↑                       ↑

               acid                          base                        salt                   water

2) Skeleton equation (unbalanced)

#) Balanced chemical equation (including phases)

Question 2:

1) Mol ratio:

Using the stoichiometric coefficients of the balanced chemical equation you get the mol ratio:

2) Moles of H₂SO₄:

3) Moles of KOH:

4) Determine the limiting reagent:

a) Stoichiometric ratio:

   1 mol H₂SO₄ / 2 mol NaOH = 0.500 mol H₂SO4 / mol NaOH

b) Actual ratio:

   0.360 mol H₂SO4 / 0.203 mol NaOH = 1.77 mol H₂SO₄ / mol NaOH

Since hte actual ratio of H₂SO₄  is greater than the stoichiometric ratio, you conclude that H₂SO₄ is in excess.

5) Amount of H₂SO₄ that reacts:

         x / 0.203 mol KOH = 1 mol H₂SO₄ / 2 mol KOH ⇒

         x = 0.203 / 2 = 0.0677 mol of H₂SO₄

6) Concentration of H₂SO₄ remaining:

        M = n / V = 0.292 mol / 1.450 liter = 0.201 M ← answer

The balanced neutralization reaction between H2SO4 and KOH is H2SO4 + 2KOH → K2SO4 + 2H2O. To calculate the concentration of sulfuric acid remaining after neutralization, use the moles of KOH reacted and the mole ratio in the balanced equation.

(a) The neutralization reaction between H2SO4 (sulfuric acid) and KOH (potassium hydroxide) in aqueous solution can be represented as:

H2SO4 + 2KOH → K2SO4 + 2H2O

(b) To determine the concentration of sulfuric acid remaining after neutralization, we need to calculate the moles of KOH reacted using the given volumes and concentrations. Since H2SO4 has a 1:2 mole ratio with KOH in the equation, half of the moles of KOH reacted will be the moles of sulfuric acid neutralized. Therefore, the concentration of sulfuric acid remaining can be calculated by subtracting the moles of sulfuric acid neutralized from the initial concentration.

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Answer:

The initial concentration of ethanal was 0.1590 mol/L.

Explanation:

Integrated rate law for second order kinetic:

[tex]k=\frac{1}{t}(\frac{1}{[A]}-\frac{1}{[A]_o})[/tex]

k = Rate constant =[tex]6.73\times 10^{-3} L mol s[/tex]

t = Time elapsed  = 50.0 s

[tex][A]_o[/tex] =initial concentration of ethanal

[A] = Concentration of ethanal left after time t = 0.151 mol/L

On substituting the value:

[tex]6.73\times 10^{-3} L mol s=\frac{1}{50.0 s}(\frac{1}{0.151 mol/L}-\frac{1}{[A_o]})[/tex]

[tex][A]_o=0.1590 mol/L[/tex]

The initial concentration of ethanal was 0.1590 mol/L.

Answer : The net ionic equation will be,

[tex]Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)[/tex]

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The same number of ions present on reactant and product side which do not participate in a reactions.

The given balanced ionic equation will be,

[tex]MgSO_4(aq)+Pb(NO_3)_2(aq)\rightarrow Mg(NO_3)_2(aq)+PbSO_4(s)[/tex]

The ionic equation in separated aqueous solution will be,

[tex]Mg^{2+}(aq)+SO_4^{2-}(aq)+Pb^{2+}(aq)+2NO^{3-}(aq)\rightarrow PbSO_4(s)+Mg^{2+}(aq)+2NO^{3-}(aq)[/tex]

In this equation, [tex]Mg^{2+}\text{ and }NO_3^-[/tex] are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

[tex]Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)[/tex]

Answer: The chemical reactions are given below.

Explanation:

Combustion reaction is defined as the chemical reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.

[tex]\text{hydrocarbon}+O_2\rightarrow CO_2+H_2O[/tex]

If supply of oxygen gas is limited, it is known as incomplete combustion and carbon monoxide gas is also produced as a product.

Here, complete combustion reaction takes place. The chemical equation follows:

[tex]2C_{10}H_{22}+31O_2\rightarrow 20CO_2+22H_2O[/tex]

Here, incomplete combustion takes place and carbon monoxide is also formed.

[tex]C_{10}H_{22}+13O_2\rightarrow 5CO+5CO_2+11H_2O[/tex]

Here, incomplete combustion takes place and only carbon monoxide with water are formed as the products.

[tex]2C_{10}H_{22}+21O_2\rightarrow 20CO+22H_2O[/tex]

When a compound is burned in air, it means that unlimited supply of oxygen is there. So, complete combustion reaction takes place and carbon dioxide gas is formed as a product.

[tex]2C_{10}H_{22}+31O_2\rightarrow 20CO+22H_2O[/tex]

Hence, the chemical reactions are given below.

Answer:

WAIT SORRY I CANT DECLICK ill try to answer doe a.

Explanation:

Answer:

CH4 + 2 O2 --> CO2 + 2 H2O

Explanation:

CH4 + 2 O2 --> CO2 + 2 H2O is the only reaction where an element (oxygen) undergoes a change in oxidation state. In this reaction oxygen changes disproportionately to O⁻². That is ...

O₂ → CO₂ + 4e⁻ ==> oxidation

O₂ + 4e⁻  →  H₂O ==> reduction

2O₂ + 4e⁻  →  CO₂ + H₂O + 4e⁻  ==> Net oxidation-reduction

=>  4e⁻ gained by one mole O₂ in formation of CO₂ = 4e⁻ lost by the other mole O₂ in forming H₂O.

Then...

Including CH₄ (whose elements do not undergo changes in oxidation states) requires doubling reaction to balance by mass thus giving ...

2CH₄ + 2O₂ + 8e⁻  →  2CO₂ + 2H₂O + 8e⁻

Cancelling 8 reduction electrons on left with 8 oxidation electrons on right gives...

2CH₄ + 2O₂  →  2CO₂ + 2H₂O

Answer:

CH₄ + 2O₂ ⟶ CO₂ + 2H₂O  

Explanation:

To identify an oxidation-reduction reaction, you must determine the oxidation number of every atom involved in the reaction and see if it changes.

The only reaction where two elements change oxidation number is the oxidation of methane.

Here's the oxidation number of every atom involved:.

[tex]\stackrel{\hbox{-4}}{\hbox{C}}\stackrel{\hbox{+1}}{\hbox{H}}_{4} +\stackrel{\hbox{0}}{\hbox{O}}_{2} \, \longrightarrow \, \stackrel{\hbox{+4}}{\hbox{C}}\stackrel{\hbox{-2}}{\hbox{O}}_{2} + \stackrel{\hbox{+1}}{\hbox{H}}_{2}\stackrel{\hbox{-2}}{\hbox{O}}[/tex]

We see that some elements change oxidation numbers.

C:  -4 ⟶ +4; increase in oxidation number = oxidation

O:  0 ⟶  -2; decrease in oxidation number = reduction

H: +1 ⟶   +1; no change.

The reaction is an oxidation-reduction reaction, because carbon is oxidized, and oxygen is reduced.

Answer: Option (A) is the correct answer.

Explanation:

Lattice energy is defined as the energy needed by an ionic solid to break its constituent components into gaseous ions.

For example, [tex]NaCl(s) \rightarrow Na^{+}(g) + Cl^{-}(g)[/tex]

Two factors that help in determining lattice energy are changes on the ions and size of ions.

So, larger is the charge on ions, smaller will be the size of ions. As a result, more energy is required to break the bond between ions. Hence, lattice energy will also increase.

Whereas when smaller is the charge on ions, larger will be the size of ions. As a result, less energy is required to break the bond between ions. Hence, lattice energy will also decrease.

Lattice energy is the energy required to convert a mole of ionic solid into its constituent ions in the gas phase. It is an important factor in the ionic bond strength and varies depending on the charges of the ions and the distances between them. The lattice energy can be calculated via Coulomb's law.

Lattice energy is the energy required to convert a mole of ionic solid into its constituent ions in the gas phase, as option A suggests. Just to give an example, for the ionic solid MX, the lattice energy is the enthalpy change of the process: AH lattice MX (s) Mn+ (g) + X(g). Lattice energy gives us an insight into ionic bond strength, which varies depending on the charges of the ions and the distances between them.

The lattice energy of an ionic crystal can be expressed by an equation derived from Coulomb's law, which governs the forces between electric charges. The lattice energy increases as the charges of the ions increase and the sizes of the ions decrease. For instance, the lattice energy of LiF is 1023 kJ/mol, whereas that of MgO is significantly higher at 3900 kJ/mol due to increased charges.

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Final answer:

The net ionic equation for the reaction between barium hydroxide and sulfuric acid is Ba2+(aq) + SO42-(aq) → BaSO4(s), which shows the formation of an insoluble precipitate of barium sulfate.

Explanation:

The correct net ionic equation for the reaction between barium hydroxide and sulfuric acid is as follows:

Ba2+(aq) + SO42−(aq) → BaSO4(s)

In this equation, Ba2+(aq) represents the barium ion in aqueous solution, SO42−(aq) denotes the sulfate ion in aqueous solution, and BaSO4(s) is the precipitate of barium sulfate. It is important to remember that net ionic equations must be balanced by both mass and charge, and in this case, the equation follows these rules with a product that is insoluble in water.

Answer : The metal used was iron (the specific heat capacity is [tex]0.44J/g^oC[/tex]).

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

[tex]q_1=-q_2[/tex]

[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]

where,

[tex]C_1[/tex] = specific heat of metal = ?

[tex]C_1[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]

[tex]m_1[/tex] = mass of metal = 47.1 g

[tex]m_2[/tex] = mass of water = 120 g

[tex]T_f[/tex] = final temperature of water = [tex]24.5^oC[/tex]

[tex]T_1[/tex] = initial temperature of metal = [tex]99^oC[/tex]

[tex]T_2[/tex] = initial temperature of water = [tex]21.4^oC[/tex]

Now put all the given values in the above formula, we get

[tex]47.1g\times c_1\times (24.5-99)^oC=-120g\times 4.18J/g^oC\times (24.5-21.4)^oC[/tex]

[tex]c_1=0.44J/g^oC[/tex]

Form the value of specific heat of metal, we conclude that the metal used in this was iron.

Therefore, the metal used was iron (the specific heat capacity is [tex]0.44J/g^oC[/tex]).