What mass of hcl gas must be added to 1.00 l of a buffer solution that contains [aceticacid]=2.0m and [acetate]=1.0m in order to produce a solution with ph = 4.11?

Answers

Answer 1
pH of acidic buffer = pKa + log [CH₃COONa - HCl] / [CH₃COOH + HCl]
pKa of CH₃COOH = 4.74
Concentration of acetic acid in buffer = 2.0 M
Concentration of sodium acetate = 1.0 M
Concentration of HCl must add = x
pH = 4.74 + log (1-x) / (2+x) = 4.11
x = concentration of HCl must be added = 0.43 M
number of moles of HCl = M * V = 0.43 * 1 = 0.43 mol
mass of HCl must be added = 0.43 * 36.5 = 15.7 g
 
Answer 2

To achieve a pH of 4.11 in the given buffer solution, approximately 15.71 grams of HCl gas need to be added.

To determine the mass of HCl gas needed to achieve the desired pH in the buffer solution, we can use the Henderson-Hasselbalch equation:

pH = pKₐ + log([A⁻] / [HA])

For acetic acid (CH₃COOH), the pKₐ is approximately 4.76. Given: [HA] = 2.0 M (acetic acid), [A-] = 1.0 M (acetate), and the target pH is 4.11, we can set up the equation:

4.11 = 4.76 + log([A⁻] / [HA])

Rearranging to solve for the ratio [A⁻]/[HA]:

4.11 - 4.76 = log([A⁻] / [HA])
-0.65 = log([A⁻] / [HA])
[A⁻] / [HA] =  [tex]10^-^0^.^6^3[/tex] ≈ 0.234

Next, let x be the number of moles of HCl gas added. HCl will convert acetate (A⁻) to acetic acid (HA):

(1.0 - X) / (2.0 + X) = 0.234

Solving for X: multiply both sides by (2.0 + X):

1.0 - x = 0.234 x (2.0 + X)
1.0 - x = 0.468 + 0.234X
1.0 - 0.448 = 0.224X + X
0.532 = 1.234X
X ≈ 0.431

Thus, moles of HCl needed are approximately 0.431 moles. Using the molar mass of HCl (36.46 g/mol), we can calculate the mass:

Mass of HCl = 0.431 mol x 36.46 g/mol ≈ 15.71g


Related Questions

What is the standard gibbs free energy of formation, ∆gºf, of nh3(g) at 298 k?

Answers

The standard Gibbs free energy of formation of NH3(g) at 298 K is approximately -16.5 kJ/mol.

The standard Gibbs free energy of formation (ΔG°f) for ammonia (NH3(g)) at 298 K is approximately -16.5 kJ/mol. The standard Gibbs free energy of formation for a substance is the change in Gibbs free energy when one mole of the substance is formed from its elements in their standard states. The standard state for nitrogen gas (N2) is N2(g) at 1 bar pressure, and for hydrogen gas (H2) it is H2(g) at 1 bar pressure.

please note that the given value of -16.5 kJ/mol is an approximation, and it is recommended to refer to the latest thermodynamic data sources for the most accurate and up-to-date information.

What is the ground-state electron configuration for the element cobalt (z = 27)?

Answers

Final answer:

The ground-state electron configuration for cobalt (Z = 27) is [Ar] 3d7 4s2.

Explanation:

The ground-state electron configuration for the element cobalt (Co), which has an atomic number (Z) of 27, is as follows:

Co: [Ar] 3d74s2

This configuration indicates that cobalt has 2 electrons in the outermost 4s subshell and 7 electrons in the 3d subshell after the [Ar] noble gas core.

The ground-state electron configuration for the element cobalt (Z = 27) is 1s²2s²2p⁶3s²3p⁶4s²3d⁷. Co has 27 protons, 27 electrons, and 33 neutrons. The electron configuration represents the distribution of electrons in the energy levels and orbitals of an atom.

Question 7 what is the volume of 28.0 g of nitrogen gas at stp? 44.8 l 33.6 l 11.2 l 22.4 l none of the above

Answers

At STP, 1 mol of gas corresponds to 22.4 L. The molar mass of nitrogen is 14.0g so the molar mass of nitrogen gas (N2) is 28.0g. The moles of nitrogen in this sample of nitrogen gas is
                           28.0 g / (1 mol / 28.0 g) = 1.00 mol
If we convert to liters, we get
                           (1.00 mol)(22.4 L / 1mol) = 22.4 L
Therefore, your answer is 22.4 L.

The equilibrium constant for the reaction sr(s) + mg2+(aq) ⇌ sr2+(aq) + mg(s) is 3.69 × 1017 at 25°c. calculate e o for a cell made up of sr / sr2+ and mg / mg2+ half-cells.

Answers

we are going to use this formula:

E° = 0.0592/ n * ㏒K

when E° is the standard state cell potential

and n is number of moles of electrons transferred in the balanced equation

for the reaction of the cell.

Sr(s) + Mg2+ (aq) ↔ Sr2+(aq) + Mg(s)

we can get it by using the given balanced equation, we here have 2 electrons

transferred so, n = 2 

and K the equilibrium constant = 3.69 x 10^17

so, by substitution:

∴ E° = 0.0592 / 2 * ㏒ (3.69 x 10^17)

        = 0.52 V

To calculate the standard electrode potential (E°) for a cell composed of Sr/Sr²⁺ and Mg/Mg²⁺ half-cells, use the Nernst equation with the given equilibrium constant value. The resulting E° is approximately 1.038 V.

To find the standard electrode potential (E°) for the cell, we can use the Nernst equation and the relationship between the equilibrium constant (Keq) and the standard electrode potential:

Nernst equation:

E° = (RT / nF) * ln(Keq)

Where:

R = 8.314 J/(mol·K)T = 298 K (since the temperature is 25°C)n = 2 (number of electrons transferred, as each Sr replaces an Mg)F = 96485 C/mol

Plugging in the values, we get:

E° = (8.314 * 298 / (2 * 96485)) * ln(3.69 × 1017)

Simplify the constants:

E° = (0.0257 V) * ln(3.69 × 1017)

Evaluating the natural logarithm:

E° = 0.0257 * 40.38

E° ≈ 1.038 V

Therefore, the standard electrode potential (E°) for the cell is approximately 1.038 V.

A solution with a volume of 1.00 l is 0.450 m in ch 3 cooh(aq) and 0.550 m in ch 3 coona(aq). what will the ph be after 0.0800 mol of hcl is added to the solution?

Answers

Before addition of HCl,

conc. of CH3COOH = 0.450 M
conc. of CH3COONa = 0.550 M

After addition of 0.08 M HCl, following reaction occurs in system:
HCl + CH3COONa ↔  CH3COOH + NaCl

Thus, in reaction system conc. of CH3COOH will increase to 0.53 M (0.08M + 0.450M)
And, conc to CH3COONa will reduce to 0.47 M (0.550M - 0.08M)

Now, conc. of H+ ions = ka [tex] \frac{[acid]}{[conjugated base]}[/tex]
where ka = dissociation constant for acid = 10^-5 for Ch3COOH

∴ conc. of H+ ions = [tex] \frac{0.53}{0.47} [/tex]
                              = 1.1277 x 10^-5

Now, pH = -log [H+] = -log (1.1227 x 10^-5) = 4.94

what is the molarity of 10.0 g of kcl in 0.500 l of solution

Answers

Molarity is defined as number of moles of the solute in 1 liter of solution.

Molarity = Moles of solute (mol) / Volume of the solution (L)

moles of KCl = mass (g) / molar mass (g/mol)
                      = 10.0 g / 74.5 g/mol
                      = 0.134 mol

Hence molarity = 0.134 mol / 0.500 L
                         = 0.268 mol/L

A mixture of 100 g of k2cr207 and 200 g of water is stirred at 60 °c until no more of the salt dissolves. the resulting solution is poured off, leaving the undissolved solid behind. the solution is now cooled to 20°c. what mass of k2cr207 crystallizes from the solution during the cooling?

Answers

One example of a solution is salt water which is a mixture of water and salt. You cannot see the salt and the salt and water will stay a solution if left alone. Parts of aSolution. Solute - The solute is the substance that is being dissolved by another substance.

No K₂Cr₂O₇ will crystallize out of the solution during cooling.

What is crystallization?

Crystallization is a separation technique used to purify a solid substance by selectively dissolving it in a suitable solvent at a high temperature and then cooling the solution to obtain pure crystals of the solute.

Given:

The solubility of K₂Cr₂O₇ in water at 60°C is 127 g/100 mL, and at 20°C is 13.9 g/100 mL.

Dissolved 100 g of  K₂Cr₂O₇ in 200 g of water, which is 200 mL of water.

At 60°C, the solution can dissolve 127 g/100 mL × 2 L

= 254 g of  K₂Cr₂O₇.

100 g of  K₂Cr₂O₇, the solution is saturated and no more of the salt can dissolve.

When the solution is cooled to 20°C, the solubility of K₂Cr₂O₇ is only 13.9 g/100 mL.

The amount of water in the solution at 20°C is 200 mL. The maximum amount of K₂Cr₂O₇ that can remain in solution at this temperature is:

13.9 g/100 mL × 2 L = 278 g

Dissolved 100 g of K₂Cr₂O₇ in the solution, the amount that will crystallize out is:

100 g - 278 g = -178 g

This result is negative, indicating that all the K₂Cr₂O₇ will remain in solution at 20°C.

Therefore, no K₂Cr₂O₇ will crystallize out of the solution during cooling.

To learn more about the crystallization, follow the link:

https://brainly.com/question/13008800

#SPJ2

Find the new concentration of a solution if 25.0 mL of water is added to 125.0 mL of 0.150 M NaCl solution. What is the final volume? mL

Answers

Answer:

150.0 is the final volume

next question is a

and the next is 0.125 M

Explanation:

got it right in edge in 2020

Final answer:

To calculate the new concentration of NaCl solution after dilution, the final volume of the solution is determined to be 150.0 mL by adding the initial volume and the volume of water added. Using the molarity equation, the new concentration is found to be 0.125 M.

Explanation:

To find the new concentration of the NaCl solution after dilution, we use the concept of molarity and the principle of conservation of mass. The amount of solute (NaCl in this case) stays the same, but the volume increases when more solvent (water) is added.

The initial molarity (Mi) of the NaCl solution is 0.150 M and the initial volume (Vi) is 125.0 mL. The volume of water added is 25.0 mL.

The final volume (Vf) is the sum of the initial volume and the volume of water added:

Vf = Vi + volume of water added

Vf = 125.0 mL + 25.0 mL

Vf = 150.0 mL

The new concentration Cf can be calculated using the formula:

Ci × Vi = Cf × Vf

(0.150 M) × (125.0 mL) = Cf × (150.0 mL)

Cf = (0.150 M × 125.0 mL) / 150.0 mL

Cf = 0.125 M

The final volume of the solution is 150.0 mL.

which of the following is a balanced chemical equation for the reaction of magnesium with nitrogen gas to form magnesium nitride

Answers

So the unbalanced equation would be Mg + N^2 --> Mg^3N^2

Which means the balanced equation would be 3Mg + N^2 --> Mg^3N^2

This is balance the equation out since you now has 3 magnesium and 2 nitrogen on the left side, and 3 magnesium on 2 nitrogen on the right. Double check my work though, it's been awhile.

What volume of 0.182 m na3po4 solution is necessary to completely react with 90.7 ml of 0.103 m cucl2?

Answers

Answer is: 34.2 mL sodium phosphate solution is necessary to completely react.
Chemical reaction: 2Na₃PO₄ + 3CuCl₂ → 6NaCl + Cu₃(PO₄)₂.
n(CuCl₂) = c(CuCl₂) · V(CuCl₂).
n(CuCl₂) = 0.103 M · 0.0907 L.
n(CuCl₂) = 0.00934 mol.
From chemical reaction: n(CuCl₂) : n(Na₃PO₄) = 3 : 2.
n(Na₃PO₄) = 2 · 0.00934 mol ÷ 3.
n(Na₃PO₄) = 0.00622 mol.
V(Na₃PO₄) = n(Na₃PO₄) ÷ c(Na₃PO₄).
V(Na₃PO₄) = 0.00622 mol ÷ 0.182 M.
V(Na₃PO₄) = 0.0342 L · 1000 mL/L = 34.2 mL.

Consider 2NH3(g)+3CuO(s)→N2(g)+3Cu(s)+3H2O(g) 2NH 3 ​ (g)+3CuO(s)→N 2 ​ (g)+3Cu(s)+3H 2 ​ O(g). What volume (in mL) of NH3 NH 3 ​ is required to completely react 45.2 g of CuO at STP? Enter your solution as a numerical value with no units.

Answers

the volume of  in Ml  of NH3  required  to completely  react  with  45.2 g  of CUO   at  STP  is 8489.6 Ml

 calculation
calculate the moles  of CUO  used
moles= mass/molar mass
45.2 g/79.5 g/mol =0.569 moles

Reacting  equation 
2NH3  + 3CUO = N2  +3CU  +3H2O
by  use of reacting  ratio  between  NH3 to CUO  which jis  2:3 the moles of  NH3  = 0.569 x2/3 = 0.379 moles

At STP  1  mole  = 22.4 L
what about  0.379 moles

=0.379  x22.4  = 8.4896  L

in  Ml  = 8.4896  x1000 =8489.6 ml

Among the alkali earth metals, the tendency to react with other substances

A) increases from bottom to top within the group.

B) is shown by the ways they react with water.

C) varies in an unpredictable way within the group.

D) does not vary among the members of the group.

Answers

The answer would be letter A.Among the alkali earth metals, the tendency to react with other substances increases from bottom to top within the group. We know that when there is an increase of metallic property, there will also be an increase of reactivity. 

When 5-hydroxypentanal is treated with methanol in the presence of an acid catalyst, 2-methoxytetrahydropyran is formed. draw the structure of the second intermediate in this reaction?

Answers

The conversion of 5-Hydroxypentanal to 2-methoxytetrahydropyran takes place in two steps. In first step the carbonyl group of 5-hydroxypentanal gets protonated and is being attacked by the hydroxyl group of its own molecule and results in the formation of six membered hetrocyclic ring called tetrahydro-2H-pyran-2-ol
The tetrahydro-2H-pyran-2-ol on protonation looses the water molecule results in the formation of second intermediate called as 2,3,4,5-tetrahydropyrylium (Shown in red). This intermediate when attacked by Methanol molecule results in the formation of 2-methoxytetrahydropyran as shown below.

n comparison to molecules that interact by London dispersion forces only, the melting point of similar-sized molecules forming hydrogen bonds would most likely be about the same. unpredictable. lower. higher.

Answers

Hydrogen bonds are the strongest intermolecular bonds, so the melting points would be much higher. 

Answer:

d

Explanation:

Calculate the freezing point of a 0.08500 m aqueous solution of nano3. the molal freezing-point-depression constant of water is 1.86°c/m. remember to include the value of i.

Answers

Depression in freezing point (Δ[tex] T_{f} [/tex]) = [tex] K_{f} [/tex]×m×i,
where, [tex] K_{f} [/tex] = cryoscopic constant = [tex] 1.86^{0} C/m[/tex],
m= molality of solution = 0.0085 m
i = van't Hoff factor = 2 (For [tex] NaNO_{3} [/tex])

Thus, (Δ[tex] T_{f} [/tex]) = 1.86 X 0.0085 X 2 = [tex] 0.03162^{0}C [/tex]

Now, (Δ[tex] T_{f} [/tex]) = [tex] T^{0} [/tex] - T
Here, T = freezing point of solution
[tex] T^{0} [/tex] = freezing point of solvent = [tex] 0^{0}C [/tex]
Thus, T = [tex] T^{0} [/tex] - (Δ[tex] T_{f} [/tex]) = -[tex] 0.03162^{0}C [/tex]

Answer : The freezing point of a solution is, [tex]0.32^oC[/tex]

Explanation :

First we have to calculate the Van't Hoff factor (i) for [tex]NaNO_3[/tex].

The dissociation of [tex]NaNO_3[/tex] will be,

[tex]NaNO_3\rightarrow Na^++NO_3^-[/tex]

So, Van't Hoff factor = Number of solute particles = [tex]Na^++NO_3^-[/tex] = 1 + 1 = 2

Now we have to calculate the freezing point of a solution.

Formula used for lowering in freezing point :

[tex]\Delta T_f=i\times k_f\times m[/tex]

or,

[tex]T_f^o-T_f=i\times k_f\times m[/tex]

where,

[tex]\Delta T_f[/tex] = change in freezing point

[tex]T_f[/tex] = temperature of solution = ?

[tex]T^o_f[/tex] = temperature of pure water = [tex]0^oC[/tex]

[tex]k_f[/tex] = freezing point constant  = [tex]1.86^oC/m[/tex]

m = molality  = 0.08500 m

i = Van't Hoff factor = 2

Now put all the given values in this formula, we get the freezing point of a solution.

[tex]0^oC-T_f=2\times (1.86^oC/m)\times 0.08500m[/tex]

[tex]T_f=0.32^oC[/tex]

Therefore, the freezing point of a solution is, [tex]0.32^oC[/tex]

How many carbon-carbon sigma bonds are present in each of the following molecules: benzene, cyclobutane?

Answers

The first bond which forms between two atoms is known as Sigma Bond. It is considered as the strongest covalent bond which forms by the head-on overlapping between two orbitals.

In Benzene there are six sp² hybridized carbon atoms. Each carbon atom is bonded to two carbons atoms and one hydrogen atom. Each carbon atom is bonded to one carbon atom via a double bond and through single bond to another carbon atom. Hence, there are Six sigma bonds between carbon atoms in Benzene.

In Cyclobutane there are four sp³ hybridized carbon atoms. Each carbon atom is singly bonded to two other carbon atoms and two hydrogen atoms. Therefore, there are FOUR sigma bonds between the carbon atoms of cyclobutane.

The sigma bonds are highlighted BLUE in given structures,

Final answer:

Benzene has six carbon-carbon sigma bonds because each carbon atom forms two sigma bonds with its adjacent carbon atoms. Cyclobutane has four carbon-carbon sigma bonds since it is a four-membered ring with each carbon bonded to two other carbons.

Explanation:

The question asks about the number of carbon-carbon sigma bonds in benzene and cyclobutane. Let's examine each molecule separately.

Benzene (C₆H₆)

In benzene, the carbons form a hexagonal ring and each carbon atom is bonded to two other carbon atoms. The structure of benzene is often depicted with alternating single and double bonds. However, due to resonance, the electrons are delocalized and each carbon-carbon bond is equivalent, resembling a structure with one-and-a-half bonds. Despite the presence of alternating double bonds, each carbon is involved in two sigma bonds with adjacent carbons and one sigma bond with a hydrogen atom, making a total of six carbon-carbon sigma bonds in benzene.

Cyclobutane (C₄H₈)

Cyclobutane is a four-membered ring where each carbon atom is bonded to two other carbons forming single bonds. Thus, for cyclobutane, there are four carbon-carbon sigma bonds, one between each pair of adjacent carbon atoms.

In the lewis structure of ch3oh, how many bonding pairs of electrons are there?

Answers

Answer:
            There are five bonding pairs of electrons in Methanol.

Explanation:
                   Those electron pairs which are being shared between two atoms in molecule are called as bonding pair electrons. While, those electron pairs which are not involved in bond formation and are not shared between two atoms are called as Non-Bonding electron pairs.
                   In molecule of Methanol as shown below, it can be seen that carbon atom is forming four bonds with three hydrogen atoms and one oxygen atom by sharing four electron pairs and oxygen is forming two bonds, one with carbon atom and one with hydrogen atom. There are two lone pair of electrons present on oxygen atom which are not taking part in and formation.

the number of atoms bonded to the central atom is 4

Explanation:

The solubility product for chromium(iii) fluoride is ksp = 6.6 × 10–11. what is the molar solubility of chromium(iii) fluoride?

Answers

Final answer:

To determine the molar solubility of chromium(III) fluoride, we need to use the solubility product constant (Ksp) of 6.6 × 10⁻¹¹. The molar solubility of chromium(III) fluoride is 8.184 × 10⁻⁴M.

Explanation:

To determine the molar solubility of chromium(III) fluoride, we need to use the solubility product constant (Ksp) of 6.6 × 10⁻¹¹ for chromium(III) fluoride. The Ksp expression is Ksp = [Cr³⁺][F⁻]³.

Let's assume the molar solubility of chromium(III) fluoride is x M. Since the stoichiometry of the balanced equation for its dissolution is 1:3 (one chromium ion per three fluoride ions), the equilibrium expression becomes Ksp = x(3x)³.

Now, we can set up the expression and calculate the molar solubility by solving the equation: Ksp = 108x⁴ = 6.6 × 10⁻¹¹. Solving for x gives x = 8.184 × 10⁻⁴ M. Therefore, the molar solubility of chromium(III) fluoride is 8.184 × 10⁻⁴ M.

A sample of a compound contains 32.0g C and 8.0g H. It’s molar mass is 30.0/mol. What is the compounds molecular formula?

Answers

Answer:
 
Given mass of C is 32.0 grams
Given mass of H is 8.0 grams
Molar mass of C is 1212 g/mol 
Molar mass of H is 1.01 g/mol

Thus
32.0 grams of C 8.0 grams of H →3212=2.66 moles of C→81.01=7.92 moles of H32.0 grams of C →3212=2.66 moles of C8.0 grams of H →81.01=7.92 moles of H

When we divide 7.92 by 2.66 we obtain 2.977 which is approximately 3. This means that the ratio of atoms of CC to the atoms of HH is 1:3.

Thus, empirical formula for the compound is CH3CH3.

Molar mass of CH3CH3 is 1⋅12+3⋅1.01=15.031⋅12+3⋅1.01=15.03. Since molar mass of the compound we have to find is 3030 g/mol we have tu multiply subscripts by 2.

Thus, final compound is C2H6C2H6.

If 0.158 g of a white, unknown solid carbonate of a group 2A metal (M) is heated and the resulting CO2 is transferred to a 285 ml sealed flask and allowed to cool to 25 degrees Celsius, the pressure in the flask is 69.8 mmHg. What is the identity of the carbonate?

Answers

using the ideal gas law equation we can find the number of moles of CO₂ formed 
PV = nRT
where 
P - pressure - 69.8 mmHg x 133 Pa/mmHg = 9 283 Pa
V - volume - 285 x 10⁻⁶ m³
n - number of moles 
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 25 °C + 273 = 298 K
substituting these values in the equation 
9283 Pa x 285 x 10⁻⁶ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 298 K
n = 1.067 x 10⁻³ mol

decomposition of metal carbonate is as follows 
MCO₃ ---> MO + CO₂
stoichiometry of MCO₃ to CO₂ is 1:1
therefore number of moles of MCO₃ heated = number of CO₂ moles formed 
number of MCO₃ moles = 1.067 x 10⁻³ mol
molar mass = mass / number of moles 
molar mass = 0.158 g / 1.067 x 10⁻³ mol = 148 g/mol 
since carbonate molar mass is known -
 (molar mass of C x 1 C atom) + (molar mass of O x 3  O atoms)  
12 + 16x 3 = 12 + 48 = 60 
then mass of metal M - 148 - 60 = 88
group II metal with molar mass of 88 is Ra - Radium 

Select all that apply.

Which statements concerning this diagram are correct?



X = -ΔH
X = ΔH
Z = A.E.

Answers

Correct answer is X = ΔH

Reason:
1) The graph of enthalpy Vs reaction coordinate  suggest the reaction is endothermic in nature. For endothermic reaction, energy if product is more than that of reactant. Hence,  option 1 i.e. X = -ΔH cannot be correct.
2) Since the reaction is endothermic in nature, energy if product is more than that of reactant. Hence,  option 2 i.e. X = ΔH is correct.
3) Activation energy is energy difference between Reactant (A) and transition state (B). However, as per option C, activation energy (A.E.) is energy difference between product (C) and transition state (B), which is incorrect. 

Answer:

Correct answer is X = ΔH

Explanation:

Which statement describes a gas condensing into a liquid?

Answers

condensation / hope this helps !

Answer:

The molecules get closer together and move more slowly.

Explanation:

from a gas to a liquid, the atoms or molecules of the substance do not change. Instead, the motion of the atoms or molecules in the substance determines its state of matter.

gas condenses to form a liquid, the molecules get closer together and move more slowly. If the molecules continue to slow down until they move so slowly that they can only vibrate against one another, then the substance becomes a solid.

Define atom. list the subatomic particles, describe their relative masses,charges, and positions in the atom.

Answers

atom is a small indivisible particle that can only be seen under a microscope

help!!
The electron sea is responsible for which of the following properties?
A. All of these
B. Malleable
C. Ductile
D. Conducting electricity and heat

Answers

A. All of these, because these are properties of metals, and the electron sea presets in metals.

Answer:

D.Conducting electricity and heat

Explanation:

Metal : It is defined as that substance which is good conductor of electricity and heat. Metal have  ductile and malleable property.In metal, atoms are held together by metallic bonds .The valence electrons from s and p orbital are delocalised and they form sea of electrons that surround the positively charged nuclei of the interacting metal ion.Then , the electrons are freely move throughout the space between atomic nuclei.

Due to availability of free  electrons, metal conduct electricity and heat.

Therefore, the electron sea is responsible for conducting electricity and heat .

2,2-dimethyl-4-propyloctane has how many secondary carbons? view available hint(s) 2,2-dimethyl-4-propyloctane has how many secondary carbons? five nine six seven

Answers

Structure of 2,2-dimethyl-4-propyloctane is in the Word document below.
Answer is: six secondary carbons.
Secondary carbon (2°) is attached to two other carbons. Secundary carbons are third, fifth, sixth and seventh in octane and first and second in propyl.
Primary carbon (1°) is  attached to only one carbon.
Tertiary carbon (3°) is attached to three other carbons.
Quaternary carbon (4°) is attached to four other carbons.

"at what gas temperature t e would the average translational kinetic energy of a helium atom be equal to that of an oxygen molecule in a gas of temperature 300 k"

Answers

Based on equipartition theorem, the average kinetic energy of molecules is mathematically expressed as,
<K.E.> = [tex] \frac{3kT}{2} [/tex]
where, k = Boltzmann's Constant.
T = temperature in Kelvins.

Thus from above expression, it can be seen that average kinetic energy of molecules is independent of mass and, it is dependent only on temperature.

Hence, at temperature of 300 K the average kinetic energy associated with both helium and oxygen will be same.  

The average translational kinetic energy of helium s equal to that of oxygen at [tex]\boxed{300\;{\text{K}}}[/tex] .

Further Explanation:

One of the states of matter is gas. In gases, the atoms and molecules have space between them and can easily move over each other hence gases are compressible. Gases neither have fixed shape nor volume. It occupies the shape and volume of the container. The examples of matter that are gases are nitrogen and carbon dioxide.

The kinetic theory is based on the following postulates:

1. Gas molecules have a large collection of individual particles with empty space between them and the volume of each particle is very small as compared to the volume of the whole gas.

2. The gas particles are in straight-line motion or random motion until they are not collided with the wall of the container or with each other.

3. The collision between the gas particles and the wall of the containers are an elastic collision that means molecules exchange energy but they don’t lose any energy during the collision. So the total kinetic energy is constant.

The formula to calculate the average translational kinetic energy of helium is as follows:

[tex]{{\text{E}}_{{\text{He}}}} = \frac{3}{2}{\text{k}}{{\text{T}}_{{\text{He}}}}[/tex]           …… (1)

Here,

[tex]{{\text{E}}_{{\text{He}}}}[/tex] is the average translational kinetic energy of helium gas.

k is the Boltzmann constant.

[tex]{{\text{T}}_{{\text{He}}}}[/tex]  is the absolute temperature of helium gas.

The formula to calculate the average translational kinetic energy of oxygen is as follows:

[tex]{{\text{E}}_{{{\text{O}}_{\text{2}}}}} = \frac{3}{2}{\text{k}}{{\text{T}}_{{{\text{O}}_{\text{2}}}}}[/tex]                                                                                                …… (2)

Here,

[tex]{{\text{E}}_{{{\text{O}}_{\text{2}}}}}[/tex] is average translational kinetic energy of oxygen gas.

k is the Boltzmann constant.

[tex]{{\text{T}}_{{{\text{O}}_{\text{2}}}}}[/tex]  is absolute temperature of oxygen gas.

Since both gases have same average translational energy. So left-hand side of equation (1) and (2) becomes equal, and therefore right-hand side of both equations can be compared as follows:

[tex]\frac{3}{2}{\text{k}}{{\text{T}}_{{{\text{O}}_2}}} = \frac{3}{2}{\text{k}}{{\text{T}}_{{\text{He}}}}[/tex]                                                                                                       …… (3)

Rearrange equation (3) to calculate  [tex]{{\text{T}}_{{{\text{O}}_{\text{2}}}}}[/tex]  

[tex]{{\text{T}}_{{{\text{O}}_2}}} = {{\text{T}}_{{\text{He}}}}[/tex]                                                        …… (4)

The value of [tex]{{\text{T}}_{{\text{He}}}}[/tex] is 300 K. So according to equation (4),  [tex]{{\text{T}}_{{{\text{O}}_{\text{2}}}}}[/tex] also becomes 300 K.

Learn more:

1. What is the kinetic energy of electrons? https://brainly.com/question/5031462

2. Calculate the frequency of yellow light: https://brainly.com/question/5882803

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Ideal gas equation

Keywords: 300 K, helium, oxygen, average translational kinetic energy, k, Boltzmann constant, absolute temperature, gas, kinetic theory, 3/2 kT, same, equal.

What crossed-aldol product results when butanal is heated in the presence of excess benzaldehyde and sodium hydroxide? draw the molecule on the canvas by choosing buttons from the tools (for bonds), atoms, and advanced template toolbars. the single bond is active by defa?

Answers

Final answer:

When butanal is heated with excess benzaldehyde and sodium hydroxide, the crossed-aldol product formed is 4-phenyl-2-butanol, which is obtained through the aldol condensation reaction.

Explanation:

Crossed-aldol product formation occurs when two different carbonyl compounds, such as aldehydes or ketones, react in an aldol condensation. The products are a mixture of different aldol adducts, resulting from the nucleophilic addition of one carbonyl compound to the other. This reaction allows for the creation of complex molecules with varied functionalities.

When butanal is heated in the presence of excess benzaldehyde and sodium hydroxide, the crossed-aldol product formed is 4-phenyl-2-butanol. The product is obtained by the aldol condensation reaction between butanal and benzaldehyde. In this reaction, a molecule of water is eliminated, and the resulting product contains both the aldehyde and the alcohol functional groups.

Learn more about Crossed-aldol product formation here:

https://brainly.com/question/32134011

#SPJ12

How many unpaired electrons are present in a neon atom?

Answers

None. Neon has two layers of electrons, both filled completely. This means that there cannot be unpaired electrons in this atom.

Electrons are a  subatomic particle with a negative charge. They are also found present in atoms.

The number of unpaired electrons in a neon atom is zero.

A neon atom is part of the first twenty element and it is the tenth element in the series. Its chemical symbol is represented as Ne, its atomic number is 10 while the atomic mass is approximately 20.

It belongs to period 2 and found in the noble gas group. As the number of protons is the same as the electrons, therefore the number of electrons is 10.

In a neon atom, its electronic configuration is [He] 2S2 2p6. All electrons here are paired (in twos) with none unpaired. Therefore, the number of unpaired electrons in a neon atom is zero.

learn more: https://brainly.com/question/13653359

A 3.5 gram sample of a radioactive element was formed in a 1960 explosion of an atomic bomb at Johnson Island in the Pacific test site. The half-life of the radioactive element is 28 years. How much of the original sample will remain in the year 2030? Choose the closest.

DO NOT GUESS, ONLY COMMENT IF YOU KNOW

Answers

0.50 g

That should be the correct answer :)

Answer : The correct answer for amount of radioisotope remain in 2030 is 0.619 g .

Radioactive Decay is emission of radiations ( in form of alpha , beta particle etc ) by unstable atom .

Radioactive decay is FIRST ORDER reaction . So , the equation of first order can be used to find decay constant , amount of radioisotopes or half life .

The equation for radioactive decay is given as :

[tex] ln (\frac{N}{N_0}) = -k * t [/tex]

Where : N = amount of radioisotope at time t

N₀ = amount of radioisotope initially present

k = decay constant t = time

Half life :

It is time when amount of radioisotope decrease to 50 % of its original amount . Half life [tex] (T_\frac{1}{2} ) [/tex] and decay constant can be related :

[tex] T_\frac{1}{2} = \frac{ln 2 }{k} = \frac{0.693}{k} [/tex]

Following are the steps can be used to determine amount of radioisotope (N) :

1) To find decay constant :

Given : [tex] (T_\frac{1}{2} ) [/tex] = 28 yrs

Decay constant can be calculated using half life by plugging value in half life formula :

[tex] 28 yrs = \frac{0.693}{k} [/tex]

On multiplying both side by k

[tex] 28 yrs * k= \frac{0.693}{k} *k [/tex]

On dividing both side by 28 yrs

[tex] \frac{28 yrs * k}{28 yrs} = \frac{0.693}{28 yrs} [/tex]

k = 0.02475 yrs⁻¹

2) To find amount of radioisotope (N):

Given : Amount of radioisotope originally present = 3.5 g

Time = 2030 - 1960 = 70 yrs

decay constant = 0.02475 yrs⁻¹

Amount of radioisotope (N) = ?

Plugging these values in the formula as:

[tex] ln (\frac{N}{3.5 } ) = - 0.02475 yrs^-^1 * 70 yrs [/tex]

[tex] ln (\frac{N}{3.5 } ) = - 1.7325 [/tex]

[tex] ln\frac{N}{No} [/tex] can be converted using the formula ( [tex] ln (\frac{a}{b} ) = ln a - ln b [/tex] )

ln N - ln (3.5 ) = - 1.7325

(ln 3.5 = 1.253 )

ln N -1.253 = -1.7325

Adding both side 1.253

ln N -1.253 + 1.253 = -1.7325 + 1.253

ln N = -0.4795

Taking anti ln of -0.4795

N = 0.619 g

Hence amount of radioisotope remained in 2030 is 0.619 g



A solution of ammonia and water contains 2.60×1025 water molecules and 6.90×1024 ammonia molecules. how many total hydrogen atoms are in this solution

Answers

1 molecule of NH3 has 3 atoms of H.
6.90*10²⁴ molecules NH3 have 3*6.90*10²⁴ =20.7*10²⁴ atoms H=2.07*10²⁵ atoms H

1 molecule H2O of has 2 atoms of H.
2.60*10²⁵ molecules of H2O have 2*2.60*10²⁵ =5.20 *10²⁵ atoms of H

2.07*10²⁵ + 5.20 *10²⁵ =7.27*10²⁵ atoms of H altogether.
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