Addition of _____________ to pure water causes the least increase in conductivity. A) weak bases B) acetic acid C) ionic compounds D) organic molecules

1)Delta H=(Delta H of reactants)-(Delta H of products)

2)And we know CO have 3 bond CO and CO2 have 2 bond that each of them are 2 bond, please see the picture!

so lets answer it:

[tex](3 \times 358) + (2 \times 463) - (4 \times 358) - 436 = 132[/tex]

The total energy change for the given chemical reaction, CO + H2O -> CO2 + H2, is +132 kJ/mol. This is calculated by subtracting the total energy released when the product bonds are formed from the total energy required to break the reactant bonds.

To determine the total energy change for the given reaction: CO + H2O -> CO2 + H2, you first need to calculate the total energy required to break the reactant bonds (C-O and H-O), and then subtract from this the total energy released when the product bonds are formed (C=O and H-H). The given energies are:

C-O bond: 358 kJ/mol; H-O bond: 463 kJ/mol; H-H bond: 436 kJ/mol.

Energy required to break reactant bonds: (1 x C-O bond) + (2 x H-O bonds) = (1 x 358 kJ/mol) + (2 x 463 kJ/mol) = 1284 kJ/mol Energy released when the product bonds form: (2 x C=O bonds) + (1 x H-H bond) = (2 x 358 kJ/mol) + (1 x 436 kJ/mol) = 1152 kJ/mol

Total energy change = Energy required - Energy released = 1284 kJ/mol - 1152 kJ/mol = 132 kJ/mol.

So, the total energy change for the reaction is +132 kJ/mol, which corresponds to option A.

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Answer:

Explanation:

1) Data:

a) % w/v = 2.5%

b) compound: H₂O₂ (from a table molar mass = 34.0147 g/mol)

c) M = ?

2) Formulae:

a) % w/v = (mass of soulte / volume of solution) × 100

b) numer of moles, n = mass in grams / molar mass

c) M = number of moles of solute / liters of solution

3) Solution:

a) Take a base of 100 ml of solution (0.100 liter):

b) Calculate the number of moles of solute, n:

c) Calculate the molarity, M:

Round to two significant figures: 0.74 M ← answer

Each of the four quantum numbers for an electron in an atom represents a specific characteristic and has specific rules regarding which values are allowed. The principal quantum number n can be any positive integer, the azimuthal quantum number l ranges from 0 to n - 1, the magnetic quantum number ml ranges from -l to +l, and the spin quantum number ms can either be +1/2 or -1/2. The quantum numbers 2, 1, 0, 1; 3, 1, 0, -1/2; 4, 2, -1, -1/2 from the provided list are valid.

To evaluate the possible set of quantum numbers, it is important to understand the rules that govern the values for each quantum number. The quantum numbers are expressed in the form (n, ℓ, mℓ, ms). Each one of these represents a certain feature of a given electron in an atom.

The first quantum number n, known as the principal quantum number, denotes the electron's energy level and can be any positive integer starting from 1.

The second quantum number ℓ, known as the azimuthal or angular momentum quantum number, is responsible for the shape of the electron's orbital and can have values ranging from 0 to n - 1.

The third quantum number, mℓ, known as the magnetic quantum number, describes the orientation of the electron's orbital. It can have values between -ℓ and +ℓ including 0.

Lastly, the fourth quantum number ms, the electron spin quantum number, can have one of two values: +1/2 or -1/2, denoting the two possible spin states of an electron.

Using these rules, we can verify the following quantum numbers: 4, 2, 3, -1/2, 2, 1, 0, 1, 3, 1, 0, -1/2, 4, 3, -2, 1/2, -3, 2, 2, -1/2, 4, 2, -1, -1/2, 2, 2, 2, 1/2, 3, 2, -3, 1/2. It is evident the following quantum numbers are valid: 2, 1, 0, 1; 3, 1, 0, -1/2; 4, 2, -1, -1/2.

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Final answer:

The natural abundance of Cl-35 is found by solving for the fraction x in the average atomic mass equation, which corresponds to the percentage. The solution shows that the natural abundance of Cl-35 is 75.77%.

Explanation:

The question is asking us to determine the natural abundance of Cl-35 given the average atomic mass of chlorine and the exact masses of its two stable isotopes, Cl-35 and Cl-37. To find this, we can use the formula for calculating the average atomic mass:

Average atomic mass = (fraction of isotope 1 × mass of isotope 1) + (fraction of isotope 2 × mass of isotope 2)

Let x be the fraction of Cl-35 and (1-x) the fraction of Cl-37. Since the percentages must add up to 100, we have:

35.45 amu = (x × 34.9689 amu) + [(1 - x) × 36.9695 amu]

Rearrange to solve for x:

x = (35.45 - 36.9695) / (34.9689 - 36.9695)

Calculate x and then convert x to a percentage:

x × 100 = percent abundance of Cl-35

Using x = 0.7577 (from given information), we find:

Percent abundance of Cl-35 = 0.7577 × 100 = 75.77%

Therefore, the natural abundance of Cl-35 is 75.77%.

To determine the number of molecules of sodium oxide created, we can use the given balanced chemical equation, molar mass, and Avogadro's number. The reaction equation tells us that for every 4 moles of sodium, 2 moles of sodium oxide are formed. From the given mass of oxygen, we can calculate the number of moles of sodium oxide and then convert that to molecules using Avogadro's number.

To determine the number of molecules of sodium oxide created, we need to use the given balanced chemical equation and the molar mass of sodium oxide (Na2O). From the balanced equation, we can see that 4 moles of sodium (Na) react with 1 mole of oxygen (O2) to form 2 moles of sodium oxide (Na2O). Thus, the stoichiometry of the reaction tells us that for every 4 moles of sodium, 2 moles of sodium oxide are formed.

Using the molar mass of sodium oxide, which is 61.98 grams/mol, we can calculate the number of moles of sodium oxide formed from the given mass of oxygen. First, convert the mass of oxygen to moles by dividing it by the molar mass of oxygen, which is 32.0 grams/mol. So, 187 grams of oxygen is equal to 5.84 moles. Based on the stoichiometry of the reaction, for every 1 mole of oxygen, 2 moles of sodium oxide are formed. Therefore, 5.84 moles of oxygen will form 11.68 moles of sodium oxide.

Finally, to determine the number of molecules of sodium oxide, we can use Avogadro's number, which states that 1 mole of any substance contains 6.02 x 10^23 particles (atoms or molecules). So, multiplying the number of moles of sodium oxide by Avogadro's number, we get:

11.68 moles x 6.02 x 10^23 molecules/mol = 7.03 x 10^24 molecules of Na2O

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Answer:

[tex]\boxed{\text{0.20 g}}[/tex]

Explanation:

We know we will need a balanced chemical equation with molar masses, volumes, and concentrations, so, let's gather all the information in one place.

M_r:                       84.01

                 HCl + NaHCO₃ ⟶ NaCl + H₂O + CO₂

V/mL:        200.

c/mol·L⁻¹:  0.012

(a) Moles of HCl

[tex]\text{Moles of HCl} =\text{0.200 L HCl} \times \dfrac{\text{0.012 mol HCl}}{\text{1 L HCl}}\\\\=\text{0.0024 mol HCl}[/tex]

(b) Moles of NaHCO₃

The molar ratio is 1 mol NaHCO₃ = 1 mol HCl

[tex]\text{Moles of NaHCO$_{3}$}= \text{0.0024 mol HCl} \times \dfrac{\text{1 mol {NaHCO$_{3}$}}}{ \text{1 mol HCl}}\\\\= \text{0.0024 mol NaHCO$_{3}$}[/tex]

(c) Mass of NaHCO₃

[tex]\text{Mass of NaHCO$_{3}$}= \text{0.0024 mol NaHCO$_{3}$} \times \dfrac{\text{84.01 g {NaHCO$_{3}$}}}{ \text{1 mol NaHCO$_{3}$}}\\\\= \textbf{0.20 g NaHCO$_{3}$}\\\\\text{The man would need to ingest }\boxed{\textbf{0.20 g}} \text{ of NaHCO$_{3}$}.[/tex]

Answer:

P = 3.8 Atm = 2900 mmHg  (2 sig.figs.)

Explanation:

PV = nRT => P = nRT/V = (1.9 moles)(0.08206LAtm/molK)(228K)/9.45L  = 3.76 ~ 3.8 Atm x 760mmHg/Atm =2888 mmHg ~ 2900 mmHg (2 sig.figs.)

2900 mmHg is the pressure of 1.9 mols of nitrogen gas in a 9.45 L tank and at a temperature of 228 K.

"Pressure" is defined as the thrust (force) applied to a surface per area. The force to area ratio is another way to describe it (over which the force is acting). The height of a mercury column that precisely balanced the mass of the columns of atmosphere above the barometer is used to measure atmospheric pressure, which is also referred to as barometric pressure.

It can be stated using a number of various measurement methods, including millimeters (or inches) of mercury, pounds every square inch (psi), dynes per square centimeter (dyn/sq cm), millibars (mb), standard atmospheres, as well as kilopascals.

PV = nRT

P = nRT/V

(1.9 moles)(0.08206LAtm/molK)(228K)/9.45L  = 3.76

3.8 Atm x 760mmHg/Atm =2888 mmHg

=2900 mmHg

Therefore, the pressure is 2900 mmHg.

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Answer:

Explanation:

To answer this kind of questions your best tool is a periodic table.

The periodic table shows the elements ordered by increasing atomic number (number of protons), in an arragement of rows and columns. In such arrangement, the elements appear classified as metals, non-metals, and metaloids.

Roughly metals are on the left side of the table, nonmetals are on the right side, and metalloids are a reduced group that are in a kind of step ladder dividing the metals and nonmetals.

With that, you can follow this procedure for each of the answer choices:

a) Barium is an alkali earth metal:

TRUE. Barium, Ba, has atomic number 56, is in the column (group) number 2, which is the group of the alkali earth metals.

b) Manganese is a transition metal:

TRUE. Manganese, Mn, has atomic number 25 and is in the column 7. The columns 3 through 12 enclose the transition metals. So, manganese is a transition metal.

c) Sulfur is considered a metalloid.

FALSE. Sulfur, S, has atomic number 16, is in the column 16, (right below oxygen) and is classified as a nonmetal.

d) Krypton is one of the noble gasses

TRUE. Krypton, Kr, has atomic number 36, and is in the column 18. This column includes all the noble gases, which are elements whose valence shells are complete (2 electrons in the case of He and 8 electrons in all the other cases).

This group is named noble gases because the elements have very low reactivity, so they are almos inert.

e) Iodine is a halogen

TRUE. Iodine, I, is the element with atomic number 53, and is in the column 17. This column includes all the halogens (F, Cl, Br, I, At, and the most recently discovered Ts).

The false statement is that sulfur is considered a metalloid; sulfur is actually a nonmetal. Barium is an alkali earth metal, manganese is a transition metal, krypton is a noble gas, and iodine is a halogen.So,option c is correct.

The statement that sulfur is considered a metalloid is false; sulfur is actually a nonmetal. Below is a clarification of each element's classification:

Barium is an alkali earth metal and is in Group 2 of the periodic table.

Manganese is indeed a transition metal, found in Group 7 of the transition metals.

Sulfur is a nonmetal and is part of the chalcogens family, also known as the oxygen family.

Krypton is one of the noble gases, which are known for having their outer energy levels full, making them very unreactive.

Iodine is a halogen, part of Group 17 on the periodic table, which is known for being highly reactive.

Answer:

ΔH(mix'g)= 42.3Kj/mole

Explanation:

ΔH = (mcΔT)water/moles X

m = mass(g) = 20ml x 1.00g/ml = 20 g

c = 4.184 j/g⁰C

ΔT = 25°C- 15°C = 10°C

moles X = (1.5g)/(76g/mole) = 0.0197 mole X

ΔH = (20g)(4.184j/g°C)(10°C)/(0.0197 mole X) = 42,300J/mole = 42.3Kj/mole (3 sig.figs.)  

The enthalpy change for the reaction per mole of substance X, which dissolves in water and increases the temperature from 15.0 °C to 25.0 °C, is calculated to be 41.8 kJ/mol. The positive sign indicates the reaction is exothermic.

The first step is to calculate the amount of heat (q) released during the reaction. We use the formula q = mcΔT, where m is the mass of the water (density x volume = 1 g/mL x 20.0 mL = 20.0 g), c is the specific heat capacity of water (4.18 J/g⋅∘C ), and ΔT is the change in temperature (25.0 ∘C - 15.0 ∘C = 10.0 ∘C). Therefore, q = 20.0 g x 4.18 J/g⋅∘C x 10.0 ∘C = 836 J.

Next, we calculate the number of moles of X using the formula n = mass / molar mass = 1.50 g / 76.0 g/mol = 0.02 mol.

Finally, we calculate the enthalpy change (ΔH) per mole of X using the formula ΔH = q/n, therefore ΔH = 836 J / 0.02 mol = 41800 J/mol. Since the standard unit for enthalpy change is kJ/mol, we convert it from J/mol to kJ/mol by dividing by 1000 to get ΔH = 41.8 kJ/mol.

Notice that the answer is positive indicating that heat energy is released to the surroundings, which means the reaction is exothermic.

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Answer:

B

Explanation:

When work is done, energy is transferred between systems, or transformed from one type of energy into another type. Energy shares the same units of measure as work.

Work and energy are directly related in physics - the amount of work done is the same as the energy transferred. Energy can be defined as the capacity to do work.

Work and energy are directly related concepts in physics. The amount of work done is equivalent to the amount of energy transferred. This is because, by definition, work is done when a force acts on an object to move it some distance, and this process requires energy. Therefore, energy can be viewed as the capacity to do work. If an entity does not contain energy, it cannot perform work. This concept is fundamental to understanding the physics of energy transfer and conservation.

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Answer:

ΔE(work)= 692 Kj/mole e⁻ = 4.32 x 10²⁴ eV/mole e⁻

Explanation:

ΔE = hc/λ = (6.63 x 10⁻³⁴j·s)(2.99792 x 10⁸m/s)/(1.73 x 10⁻⁷m) = 1.15 x 10⁻¹⁸j/e⁻ x 6.023 x 10²³ e⁻/mole = 691,999 j/mole e⁻ = 692 Kj/mole e⁻ x 6.24 x 10²¹ eV/KJ = 4.32 x 10²⁴ eV/mole e⁻

Answer:

D. Dependent and Independent Variable

Explanation:

Since the independent variable is already the variable being changed, that is one of the things that can be changed and fromt the independent variable being changed, that changes the dependent variable.

The only things that can change in a valid experiment are dependent and independent variables.

when someone is conducting an experiment, the independent variable is what they change, and the dependent variable is what changes comes because of independent variables, like the independent variable as the cause and the dependent variable as the effect.

so when one is changing the independent variable and dependent varible automatically being changed.

Therefore the correct answer is Dependent and Independent Variable (option D)

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Answer:

ΔH = 57.04 Kj/mole H₂O

Explanation:

60ml(0.300M Ba(OH)₂(aq) + 60ml(0.600M HCl(aq)

=> 0.06(0.3)mole Ba(OH)₂(aq) + 0.60(0.6)mole HCl(aq)

=> 0.018mole Ba(OH)₂(aq) + 0.036mole HCl(aq)

=> 100% conversion of reactants => 0.018mole BaCl₂(aq) + 0.036mole H₂O(l) + Heat

ΔH = mcΔT/moles H₂O <==> Heat Transfer / mole H₂O

=(120g)(4.0184j/g°C)(27.74°C - 23.65°C)/(0.036mole H₂O)

ΔH = 57,042 j/mole H₂O = 57.04 Kj/mole H₂O

ΔH = 57.04 Kj/mole H₂O
The symbol "Δ" stands for the change in enthalpy; (Hproducts -Hreactants). A positive value suggests an endothermic reaction or that the products have a higher enthalpy (heat is required) If the value is negative, the reaction is exothermic or the reactants have a higher enthalpy (heat is produced).

        ΔH = 57,042 j/mole

         H₂O = 57.04 Kj/mole

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Answer:

C. That atoms made up the smallest form of matter

Explanation:

The crux of the Dalton's atomic theory is that atoms are the smallest form of matter. He propositioned that atoms is an indivisible particle and beyond an atom, no form of matter exists.

Series of discoveries through time have greatly shaped the Dalton's atomic theory. The discovery of cathode rays by J.J Thomson in 1897 opened up the atom. Atoms were now seen to be made up of some negatively charged particles. Ernest Rutherford through his gold foil experiment proposed the nuclear model of the atom.

Answer:

b. 137 g.

Explanation:

Fe₂O₃(s) + 6HCl(aq) → 2FeCl₃(s) + 3H₂O(l),

It is clear that 1.0 mole of Fe₂O₃ react with 6.0 mol of HCl to produce 2.0 moles of FeCl₃ and 3.0 moles of H₂O.

n = mass/molar mass = (100.0 g)/(159.69 g/mol) = 0.6262 mol.

Using cross multiplication:

1.0 mol of Fe₂O₃ react completely with → 6.0 mol of HCl, from stichiometry.

0.6262 mol of Fe₂O₃ produced with → ??? mol of HCl.

∴ The no. of moles of HCl = (6.0 mol)(0.6262 mol)/(1.0 mol) = 3.757 mol.

∴ The mass of HCl needed = no. of moles x molar mass = (3.757 mol)(36.46 g/mol) = 137.0 g.

So, the right choice is: b. 137 g.

To find the required mass of HCl to react with 100 g of rust, calculate the moles of rust, use the stoichiometry of the reaction to find moles of HCl needed, and then convert those moles to grams. The answer is approximately 137 g of HCl.

To determine the mass of hydrochloric acid (HCl) required to react with 100 g of rust (iron (III) oxide, Fe₂O₃), we use stoichiometry based on the balanced chemical equation: Fe₂O₃ (s) + 6HCl (aq) → 2FeCl₃ (aq) + 3H₂O (l).

First, calculate the molar mass of Fe₂O₃ (55.85 g/mol for Fe and 16.00 g/mol for O):

2(55.85) + 3(16.00) = 159.70 g/mol.

Next, calculate the moles of Fe₂O₃ in 100 g:

100 g ÷ 159.70 g/mol = 0.626 moles of Fe₂O₃.

According to the equation, 1 mole of Fe₂O₃ reacts with 6 moles of HCl. Therefore, 0.626 moles of Fe₂O₃3 will react with 0.626 * 6 = 3.756 moles of HCl.

Molar mass of HCl = 36.46 g/mol, so the mass of HCl needed is:

3.756 moles * 36.46 g/mol = 136.93 g, or approximately 137 g.

Thus, option (b) 137 g HCl is the correct answer for the mass of HCl required to react with 100 g of rust.

Answer:

48 c

Explanation:

its b on usa test prep

I think so 4.2, we get the answer when we add together the pressure of each of the gases.

Answer: The total pressure of a mixture is 1.247 atm.

Explanation:

To calculate the total pressure of the mixture of the gases, we use the equation given by Raoult's law, which is:

[tex]p_T=\sum_{i=1}^n(\chi_{i}\times p_i)[/tex]

where,

[tex]p_T[/tex] = total pressure of the mixture

[tex]\chi_{i}[/tex] = mole fraction of i-th species

[tex]p_i[/tex] = partial pressure of i-th species

We are given:

Mole fraction of nitrogen = 0.5

Partial pressure of nitrogen = 1.7 atm

Mole fraction of oxygen = 0.23

Partial pressure of oxygen = 1.1 atm

Mole fraction of argon = 0.12

Partial pressure of argon = 0.7 atm

Mole fraction of methane = 0.10

Partial pressure of methane = 0.5 atm

Mole fraction of water vapor = 0.05

Partial pressure of water vapor = 0.2 atm

Putting values in above equation, we get:

[tex]p_T=[(0.5\times 1.7)+(0.23\times 1.1)+(0.12\times 0.7)+(0.10\times 0.5)+(0.05\times 0.2)][/tex]

[tex]p_T=1.247atm[/tex]

Hence, the total pressure of a mixture is 1.247 atm.

Answer:

D.Cardboard pieces

Explanation:

Cardboard pieces would a good representation of the crust.

The crust of just a thin layer in terms of thickness. We can picture the crust as an apple skin or an orange peel. It's thickness is averages about 5-10km for thinner oceanic crusts and about 30-50km for thicker continental crust.

The crust is the thinnest layer of the earth.

Below the crust is the region of the mantle. The mantle is made of the upper mantle, asthenosphere and mesosphere. The asthenosphere is in a weak plastic form and it is very thick. It averages a thickness of about 190km,about 100 times the oceanic crust and 4 times the continental crust. The mantle is the thickest region of the earth.

The core is the innermosr layer of the earth. It is about 2300km thick.

It would be very ideal to relatively represent the crust as pieces of cardboard floating on an oily asthenosphere.

Answer:

At least four such rinses.

Explanation:

Let [tex]c_n[/tex] denotes the concentration of the residue after the [tex]n[/tex]th rinse. [tex]n[/tex] can be any non-negative integer (which includes zero.)

The initial concentration of the solution in the flask is [tex]\rm 0.900\;M[/tex]. That is:

[tex]c_0 = \rm 0.900\;M[/tex].

[tex]\rm 1.00\;mL[/tex] of this [tex]\rm 0.900\;M[/tex] solution is left in the flask after it is emptied for the first time.

The [tex]\rm 9.00\;mL[/tex] solvent will increase the volume of the solution from [tex]\rm 1.00\;mL[/tex] to [tex]\rm 10.00\;mL[/tex]. At this point the number of moles of solute in the flask stays unchanged. The concentration of the residue will drop to [tex]1/10[/tex] of the initial value. That is:

[tex]\displaystyle c_1 = \rm \frac{1}{10}\;c_0 = \frac{1}{10}\times 0.900\;M[/tex].

Repeat this process, and the concentration of the residue will drop by a factor of [tex]1/10[/tex] again. That is:

[tex]\displaystyle c_2 = \rm \frac{1}{10}\;c_1 = \frac{1}{10}\times (\frac{1}{10}\times 0.900\;M) = {\left(\frac{1}{10}\right)}^{2}\times 0.900\;M[/tex].

Summarize these values in a table:

[tex]\begin{array}{l|cccc}\text{Number of Rinses} & 0 & 1 & 2 & \dots \\[0.5em]\displaystyle\text{Concentration}\atop\displaystyle\text{of the residue}& \rm 0.900\;M & \rm\dfrac{1}{10}\times 0.900\;M &\rm {\left(\dfrac{1}{10}\right)}^{2}\times 0.900\;M \dots \end{array}[/tex].

The trend in [tex]c_{n}[/tex] is similar to that of a geometric series with

Again, refer to the trend in [tex]c_{n}[/tex] as the value of [tex]n[/tex] increases. The general formula for [tex]c_{n}[/tex], the concentration after the [tex]n[/tex]th rinse, will be:

[tex]\displaystyle c_{n} = \rm {\left(\frac{1}{10}\right)}^{n}\; 0.900\;M[/tex].

As a side note, [tex]0 < 1/10 < 1[/tex]. As a result, the value of [tex]c_{n}[/tex] will decrease but stay positive as the value of [tex]n[/tex] increases. Increasing the number of rinses will indeed reduce the concentration of the residue.

In other words, what's the minimum value of [tex]n[/tex] for which [tex]\c_{n} \le \rm 0.000100\;M[/tex]?

Recall that

[tex]\displaystyle c_{n} = \rm {\left(\frac{1}{10}\right)}^{n}\; 0.900\;M[/tex].

As a result, [tex]n[/tex] should satisfy the condition:

[tex]\displaystyle \rm {\left(\frac{1}{10}\right)}^{n}\; 0.900\;M \le 0.000100\;M[/tex].

Multiply both sides by

[tex]\displaystyle \frac{1}{\rm 0.900\;M}[/tex],

which is positive and will not change the direction of the inequality:

[tex]\displaystyle \rm {\left(\frac{1}{10}\right)}^{n} \le \frac{0.000100}{0.900}[/tex].

Take the natural logarithm [tex]\ln[/tex] of both sides of the inequality. The function [tex]\ln(x)}[/tex] is increasing as [tex]x[/tex] increases on the range [tex]x > 0[/tex]. This function will not change the direction of the inequality, either.

[tex]\displaystyle \rm \ln\left[{\left(\frac{1}{10}\right)}^{n}\right] \le \ln\left(\frac{0.000100}{0.900}\right)[/tex].

Apply the power rule of logarithms: [tex]n[/tex] is an exponent inside the logarithm. That will be equivalent to an expression where [tex]n[/tex] is a coefficient in front of the logarithm operator:

[tex]\displaystyle \rm \ln\left[{\left(\frac{1}{10}\right)}^{n}\right]  = n\cdot \left[\ln{\left(\frac{1}{10}\right)}\right][/tex].

[tex]\displaystyle n\cdot \left[\ln{\left(\frac{1}{10}\right)}\right]\le \ln\left(\frac{0.000100}{0.900}\right)[/tex].

Multiply both sides by

[tex]\displaystyle \frac{1}{\ln \left(\dfrac{1}{10}\right)}}[/tex].

Keep in mind that logarithms of numbers between 0 and 1 (excluding 0 and 1) are negative. The reciprocal of a negative number is still negative. The direction of the inequality will change. That is:

[tex]\displaystyle n\cdot \left[\ln{\left(\frac{1}{10}\right)}\right]\times \frac{1}{\ln \left(\dfrac{1}{10}\right)}}\ge \ln\left(\frac{0.000100}{0.900}\right)\times \frac{1}{\ln \left(\dfrac{1}{10}\right)}}[/tex].

The right-hand side is approximately 3.95. However, [tex]n[/tex] has to be an integer. The smallest integer that is greater than 3.95 is 4. In other words, it takes at least four such rinses to reduce the residual concentration to under 0.000100 M (0.000090 M to be precise.) Three such rinses will give only 0.00900 M.

Final answer:

Through serial dilution principles, the minimum number of rinses needed to dilute the residual concentration of a solution in a flask to 0.000100M or below is approximately 7, considering a 1:10 dilution ratio for each rinse.

Explanation:

The question involves calculating the minimum number of rinse cycles required to reduce the residual concentration of a solution adhering to the walls of a flask below a target concentration, using serial dilution principles.

After each rinse, the concentration is diluted by the volume ratio, in this case, a 1:10 dilution per rinse because 1.00 ml of the original solution is mixed with 9.00 ml of solvent. The formula for the final concentration (Cf) after n rinses is given by Cf = Co × (Vo / Vt)n, where Co is the original concentration, Vo is the original volume adhering to the flask, Vt is the total volume after adding solvent, and n is the number of rinses.

To solve for n, we rearrange the equation to get n = log(Cf/Co) / log(Vo/Vt). Plugging in the given values (Co=0.900M, Cf=0.000100M, Vo=1ml, Vt=10ml), we find that the minimum number of rinses needed to reduce the concentration to 0.000100 M or below is calculated to be around 7. The exact number might slightly vary depending on rounding during calculation, but 7 rinses ensure meeting the target concentration threshold.

Answer:

True

Explanation:

Answer:

True

Explanation:

The half lives of unstable isotopes of elements vary from miliseconds to billions of years, for example Bismuth-209 has the longest half life known by the human and the scientists and it has still billions of years to this day, and there are elements that cannot withstand seconds before decaying into another element, so the statement is True.

Answer:

E(Z) > E(X)

Explanation:

X => 4.2 x 10¹¹J/50 Nucleons = 8.4 x 10⁹ J/Nu

Z => 8.4 x 10¹¹J/80 Nucleons = 1.1 x 10¹⁰ J/Nu

E(Z)1.1 x 10¹¹J/Nu > E(X)8.4 x 10⁹J/Nu

Answer:  

[tex]\boxed{\text{0.0301 g}}[/tex]

Step-by-step explanation:  

1. Calculate the initial moles of H₂S

We can use Avogadro's law: the number of moles of a gas is directly proportional to the volume if the pressure and temperature are constant.

[tex]\dfrac{n_{1} }{V_{1}} = \dfrac{n_{2}}{V_{2}}\\\\\dfrac{n_{1}}{\text{44.2 mL}} = \dfrac{1.97 \times 10^{-3} \text{ mol}}{\text{98.5 mL }}\\\\\dfrac{n_{1}}{44.2} = 2.00 \times 10^{-5} \text{ mol}\\\\n_{1} = 44.2 \times 2.00 \times 10^{-5} \text{ mol} = 8.84 \times 10^{-4} \text{ mol}[/tex]

2. Calculate the initial mass of H₂S

[tex]\text{Mass of H$_{2}$S} = 8.84 \times 10^{-4}\text{ mol H$_{2}$S} \times \dfrac{\text{34.08 g H$_{2}$S}}{\text{1 mol H$_{2}$S}} = \text{0.0301 g H$_{2}$S}\\\\\text{The container initially held }\boxed{\textbf{0.0301 g H$_{2}$S}}[/tex]

The overall reaction for the chemical equation in question can be obtained by balancing the equation. The initial equation C2H6 + 7/2O2 → 3H2O + 2CO2 is balanced by multiplying the coefficients by 2 to obtain: 2C2H6 + 7O2 → 6H2O + 4CO2. This represents the overall homogeneous reaction indicating the rate of consumption of reactants and formation of products.

In the realm of Chemistry, the overall reaction for the given chemicals is derived by multiplying each coefficient in the equation by 2, resulting in a balanced equation. Starting with the equation C₂H₆ + 7/2O₂ which leads to 3H₂O + 2CO2, multiplying the coefficients by 2 gives us: 2C₂H₆ + 7O₂ --> 6H₂O + 4CO2. This equation represents the overall reaction observed. It may also depict an elementary reaction showing first-order behavior. The overall reaction equation reflects the stoichiometry of this homogeneous reaction, signaling rates for the consumption of reactants and the formation of products.

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Answer: Option (B) is the correct answer.

Explanation:

According to Le Chatelier's principle, any disturbance causes in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.

For example, [tex]2NO_{2}(g) \rightleftharpoons N_{2}O_{4}(g)[/tex]

When we increase the temperature then the reaction will shift in a direction where there will be decrease in temperature.

This, means that the reaction will shift in the backward direction.

Thus, we can conclude that if the reaction is at equilibrium and the temperature increases, the equilibrium will shift so that there is more nitrogen dioxide.

Answer:

B. The equilibrium will shift so that there is more nitrogen dioxide.

Explanation:

Hello,

By considering that the mentioned reaction is exothermic, it means that the heat is like a product, if heat is added (increase the temperature) the equilibrium will shift leftwards, contributing to the inverse reaction; it is the formation of more nitrogen dioxide by recalling the Le Chatelier's principle.

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The magnesium ribbon, D. It forms a material to cast the tool mark.

When a magnesium ribbon is burnt in the presence of oxygen it gives out strong light and heat is produced. Apart from it, it leads to the production of substance called as magnesium oxide which is formed as the product due to the reaction of magnesium with the oxygen present in the air.

Tool marks are the mark which is created by tools while using them. In order to identify or locate them castes made up of magnesium oxide is utilized. When this is pasted on the suspected area, the tool mark of the suspected tool gets pasted on it.

The correct answer is option (C). Burning a magnesium ribbon highlights a toolmark is: It forms a fine, white powder within the mark.

When a magnesium ribbon is burned, it produces a very bright light, which is useful for illuminating scenes in low-light conditions, especially in forensic photography. However, the specific reason it is used to highlight toolmarks is due to the properties of the residue it leaves behind.

As the magnesium burns, it oxidizes and produces magnesium oxide, which is a fine, white powder. This powder adheres to the surfaces and edges of the toolmark, effectively filling in the grooves and ridges. When the excess powder is brushed away, the magnesium oxide remains within the indentations of the toolmark, thereby highlighting it.

This makes the toolmark more visible and easier to photograph and analyze. The white powder contrasts sharply with the surrounding material, which is particularly useful when the toolmark is on a dark or non-reflective surface.

The other options do not accurately describe the process:

A.) The shimmering light provides shadows that show the relief of the mark. - While the bright light from the burning magnesium can create shadows, the primary method by which it highlights a toolmark is by leaving behind the white powder, not just by casting shadows.

B.) It burns brightly to provide light for photography. - While this is true and is one of the reasons magnesium is used, it is not the direct method of highlighting the toolmark. The light helps in seeing the toolmark after the powder has been applied and excess removed.

D.) It forms a material to cast the toolmark. - This option is incorrect because magnesium does not form a material to cast the toolmark. Instead, it leaves behind a powder that fills the toolmark, making it more visible. Casting implies creating a three-dimensional copy, which is not the case here.

Answer:

12,742 km

Explanation:

Therefore, its radius is 6,371 km.

Answer:

12,742 km

Explanation:

radius is 6,371 km.

Answer:

Enthalpy => Heat Effects => changes in temperature

Explanation:

Rate of Rxn is affected by changes in 5 issues ...

C => Concentration

A => Surface Area

N => Nature (Chemical Structure)

T => Temperature (Enthalpy = Heat of Rxn)

C => Catalyst

The rate of a chemical reaction is generally more directly impacted by factors like concentration of reactants, temperature, presence of a catalyst, and surface area of reactants rather than enthalpy and entropy. While enthalpy and entropy can determine whether a reaction is feasible or not, they usually do not directly affect the rate of the reaction.

The rate of a chemical reaction can be affected by both enthalpy and entropy, but these factors usually do not impact the rate of the reaction directly. They are part of the factors that determine the feasibility of a reaction, not the speed. The factors that directly impact the rate of a reaction include concentration of reactants, temperature, presence of a catalyst, and surface area of reactants.

Enthalpy pertains to the heat content of the reaction, while entropy pertains to the degree of disorder or randomness in the system. However, while they can determine whether a reaction is spontaneous or not, they do not directly influence the rate of the reaction.

To conclude, the rate of a chemical reaction is typically affected more directly by factors other than enthalpy and entropy, such as the concentration of reactants, temperature, presence of a catalyst, and surface area of reactants.

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Answer:

Explanation:

1) Chemical equilibrium

2) Equilibrium constant, Kc:

Thus, the equation to use is:

3) Determine the concentration of O₂ (g)

4) Compute Kc

The water would need to B. absorb less energy during the plateau in order to melt.  

Salt when added to ice lowers the freezing point of the ice. So, if ice is added to the solid ice it doesn't let it to freeze rather the temperature of water may fall but it won't freeze.

For melting, it needs less energy after adding salt because salt itself absorbs the energy from the surroundings to help the phase transition of water from solid to liquid.

Answer: C. It would need to absorb more energy during the plateau in order to melt.

Explanation:

adding salt lowers the freezing point, meaning it needs to absorb more energy from its surrounding in order to melt.

Answer:

Explanation:

The specific question is how many moles of Ba(OH)₂ are present in 285 mL of 0.800 M Ba(OH)₂.

In a titration, the determination of the number of moles of the substance whose molarity is known, is the first step of the calculations.

Since you know the molar concentration, and the volume of the substance, in this case the base Ba(OH)₂, you can use the molarity formula to solve for the number of moles:

The name of the variables are:

Formula:

Solve for n:

Substitute the known values>

Answer: 0.288 moles of Ba(OH)₂.

Note that from the number of moles of one component (the base, in this case), you can calculate the molar concentration of the other component (the acid in this case),  because, at the equivalence point, both acid and base reactants are at the theoretical mole ratio, i.e. both are consumed.

Final answer:

To calculate the number of moles of Ba(OH)2 present in 285 mL of 0.800 M Ba(OH)2, multiply the volume (0.285 L) by the molarity (0.800 M), resulting in 0.228 moles of Ba(OH)2.

Explanation:

To determine the number of moles of Ba(OH)2 present in 285 mL of 0.800 M Ba(OH)2, we first convert the volume in milliliters to liters by dividing by 1000:

285 mL ÷ 1000 = 0.285 L

Now, by using the concentration of the Ba(OH)2 solution (Molarity, M), we can calculate the moles of Ba(OH)2:

Number of moles = Molarity (M) × Volume (L)

Number of moles = 0.800 M × 0.285 L

Number of moles = 0.228 moles

Therefore, there are 0.228 moles of Ba(OH)2 in the 285 mL of the solution provided by the chemist during the titration.