a common concentration unit used commercially is percent weight per volume (%w/v) which is the number of gram of solute in 100 ml of solution. a bottle of hydrogen peroxide contains 2.5 %w/v hydrogen peroxide (h2o2). what is the molar concentration of hydrogen peroxide in the solution?

Answer:

7,72Е24

Explanation:

if one mole is 6.022*10²³, then 6.41 moles are: 6.41*6.022*10²³*2=2*3.86*10²⁴=7.72E24 (atoms).

PS/ Formula of Br is Br₂.

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.

https://brainly.com/question/31519289

#SPJ2

Answer:

Explanation:

1) Data:

a) Solution of a dye:

b) Water of the pool, after mixing the dye:

2) Formulae:

a) Molarity: M = n / V in liter

b) Molar mass: MM = mass in grams / number of moles

3) Solution:

a) Number of moles of methylene blue in the prepared solution:

b) Number of moles of methylen blue in the pool = number of moles of methylen blue in the prepared solution

c) Volume of the pool:

And that is the answer, which has to be rounded to 3 significant figures, since the data are expressed with 3 significant figures.

Note: For all the effects, the 50.0 mL of the dye solution are neglictible in front of the volume of the pool.

Final answer:

To calculate the volume of water in an irregularly shaped pool using a known concentration of methylene blue dye, the dilution formula is applied with the dye's initial concentration and mass, resulting in the pool's volume being 83,689 liters.

Explanation:

To find the volume of water in the pool using the concentration of methylene blue dye, first, determine the amount of dye initially used to disperse in the pool water. Since 1.00 g of methylene blue was dissolved in 50.0 mL of water, and knowing the molar mass of methylene blue is 319.854 g/mol, we can calculate the molarity of the initial solution before it was diluted in the pool.

Moles of methylene blue = mass (g) / molar mass (g/mol) = 1.00 g / 319.854 g/mol = 0.00313 moles.

Thus, the initial concentration of the dye in the solution dispersed in the pool is 0.00313 moles / 0.050 L = 0.0626 M.

When this dye disperses evenly throughout the pool, its concentration decreases to 3.74 × 10-8 M. The total volume of the pool water can be calculated by applying the concept of dilution, where the moles of dye remain constant but the volume changes:

Therefore, the volume of water in the pool is 83,689 liters.

Answer:

An oxidation-reduction reaction  

Explanation:

[tex]\stackrel{\hbox{-2}}{\hbox{C}}\text{$_{6}$H$_{12}$O$_{6}$} + 6\stackrel{\hbox{0}}{\hbox{O}\text{$_{2}$}} \longrightarrow 6\stackrel{\hbox{+4}}{\hbox{C}}\text{O$_{2}$}} + 6\text{H$_{2}$}\stackrel{\hbox{-2}}{\hbox{O}}[/tex]

Look at the oxidation numbers of the atoms.

Each carbon atom in glucose loses four electrons (oxidation), and each oxygen atom in O₂ gains two electrons (reduction).

D: 3, 2, 3, 1 The Pb3(PO4) comes from having 3 Pb and 2 H3PO4. The H2 comes from the 2H3PO4

Answer:

The reaction will move to the left.

Explanation:

Ba(OH)₂ = Ba²⁺ + 2OH⁻,

Ba(OH)₂ is dissociated to Ba²⁺ and 2OH⁻.

H⁺ will combine with OH⁻ to form water.

So, the concentration of OH⁻ will decrease and the equilibrium is disturbed.

According to Le Châtelier's principle: when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.

So, the right choice is: the reaction will move to the left, is the choice that will not happen to the equilibrium.

Answer:

D.) The reaction will move to the left.

Explanation:

Answer:

For a substance to classify as a mineral, it must lie within certain parameters. It should be an inorganic solid, that is naturally occurring in nature (not synthesized), with an ordered internal structure and a definite chemical composition.

By definite chemical composition, geologists mean that the mineral must be have chemical constituents that have an unvarying chemical composition, or a chemical composition that oscillates withing a very limited and specific range.

An example is the mineral, halite. It has a chemical composition of one sodium atom and one chloride atom, represented as NaCl and is unchanging in this composition throughout nature.

Answer:

Explanation:

1) Data:

a) Mass of CuSO₄ ∙ XH₂O = 1.50 g

b) Mass of CuSO₄ = 0.96

c) X = ?

2) Additional needed data:

a) Molar Mass of CuSO₄ = 159,609 g/mol

b) Molar mass of H₂O = 18,01528 g/mol

3) Chemical principles and formulae used:

a) Law of conservation of mass

b) Molar mass = mass in grams / number of moles = m / n

4) Solution:

a) Law of conservation of mass:

b) Moles

c) Proportion:

Divide both mole amounts by the least of the two numbers, i.e. 0.0060

Then, the ratio of CuSO₄ to H₂O is 1 : 5 and the chemical formula is:

Hence, the value of X is 5.

Answer:

CuSO4*5H2O

X = 5

Explanation:

Step 1: Data given

Mass of the copper(II)sulfate hydrate = 1.50 grams

After dehydration they find that they are left with 0.96 g of the an-hydrate CuSO4.

Step 2: Calculate mass of water

Mass of water = mass of hydrate - mass of anhydrate

Mass water = 1.50 grams - 0.96 grams

Mass water = 0.54 grams

Step 3: Calculate moles of water

Moles H2O = mass / molar mass

Moles H2O = 0.54 grams / 18.02 g/mol

Moles H2O = 0.030 moles

Step 4: Calculate moles of CuSO4

Moles CuSO4 = 0.96 grams / 159.61 g/mol

Moles CuSO4 = 0.0060 moles

Step 5: Calculate mol ratio

We divide by the smallest amount of moles

H2O : 0.030 / 0.060 = 5

CuSO4: 0.0060 / 0.0060 = 1

For 1 mol CuSO4 we have 5 moles of H2O

CuSO4*5H2O

X = 5

Answer:

C. 1.62.

Explanation:

∵ pH = - log[H⁺],

[H⁺] = 0.024 M.

∴ pH = - log(0.024) = 1.62.

So, the right choice is: C. 1.62.

Answer: B got it correct

Explanation:

The  the mass of the cone using the triple beam balance will be 543.0 grams. correct option is A.

The triple beam balance is an instrument or apparatus  used to measure mass very accurately and  precisely and these  devices typically have a reading error or round off of  of ±0.05 grams.

The name of the instrument  refers to its three beams, in which  the middle beam is the largest and longest the far beam of medium size one average  and the front beam the smallest and lightest.

Therefore, the mass of the cone using the triple beam balance will be 543.0 grams. correct option is A.

Learn more about triple beam balance, here:

https://brainly.com/question/14422546

#SPJ2

Answer:

1:2:1:2

Explanation:

A molecule is made up of aggregates of small particles called atoms.

                   CH₄ + 2O₂ → CO₂ + 2H₂O

From the reaction above we have: 1 molecule of methane CH₄ reacting with 2 molecules of oxygen gas O₂ to produce 1 molecule of carbon dioxide gas CO₂ and 2 molecules of H₂O:

                   CH₄ + 2O₂ → CO₂ + 2H₂O

                      1    :     2   :     1      :     2    

Answer:

D. Zn → Zn²⁺ + 2e⁻, 2H⁺ + 2e⁻ → H₂.

Explanation:

Oxidation reaction:

Zn losses 2 electrons and is oxidized to Zn²⁺:

Zn → Zn²⁺ + 2e⁻.

Reduction reaction:

H⁺ gains 1 electron and is reduced to H:

2H⁺ + 2e⁻ → H₂.

So, the right choice is: D. Zn → Zn²⁺ + 2e⁻, 2H⁺ + 2e⁻ → H₂.

Answer:

D. Zn → Zn²⁺ + 2e⁻, 2H⁺ + 2e⁻ → H₂.

Explanation:

Answer via Educere/ Founder's Education

Answer:

[tex]\boxed{\text{positive deviation}}[/tex]

Explanation:

[tex]\text{If the molecules interact less favourably in the solution than in the individual}\\\text{liquids, they can break away more easily than they can in the separate liquids.}\\\text{There will be more molecules in the vapour phase than you would expect.}\\\text{The solution will show a } \boxed{\textbf{positive deviation}} \text{ from Raoult's Law.}[/tex]

Answer: K

Hope this helps you out!

Answer:

Potassium (K)

Explanation:

The reason that potassium is the more reactive metal is because it is in group 1, as opposed to group 11 for copper, and being in group 1 means that it has one valence electron, which it can give away by reacting with almost any other element, and in doing so, it becomes much more stable.

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]

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.

Best regards.

Organic molecules cause the least increase in conductivity when added to pure water, as they do not dissociate into ions, unlike ionic compounds which dissociate completely, and weak acids and bases which partially ionize.

The addition of organic molecules to pure water causes the least increase in conductivity. This is because organic molecules typically do not dissociate into ions when dissolved in water. On the other hand, ionic compounds dissociate almost completely in water, making them strong electrolytes and significantly increasing water's conductivity. Option D is correct .

Weak bases and acetic acid, a weak acid, only partially ionize in water, which would lead to a slight increase in conductivity compared to organic molecules, but much less than that caused by ionic compounds.

The weak bases mentioned do not refer to the dissociation of ionic compounds themselves but to the reactions of their polyatomic anions with water. Thus, when discussing weak bases in this context, it is important to consider these secondary reactions rather than primary dissociation.

Furthermore, acetic acid, being a weak acid, increases the hydronium ion concentration in water, but again not as much as a strong acid would. However, since both acetic acid and weak bases only partially ionize, they’re still considered weak electrolytes.

Answer:

48 c

Explanation:

its b on usa test prep

Answer:

Explanation:

1) Ionization equilibrium equation: given

2) Ionization equilibrium constant, at 25°C, Kw: given

3) Stoichiometric mole ratio:

As from the ionization equilibrium equation, as from the fact it is stated, the concentration of both ions, at 25°C, are equal:

4) A solution has a [OH⁻] = 3.4 × 10⁻⁵ M at 25 °C and you need to calculate what the [H₃O⁺(aq)] is.

Since the temperature is 25°, yet the value of Kw is the same, andy you can use these conditions:

Then you can substitute the known values and solve for the unknown:

As you see, the increase in the molar concentration of the ion [OH⁻] has caused the decrease in the molar concentration of the ion [H₃O⁺], to keep the equilibrium law valid.

Answer:

1. pH = -log[H^+]

2. Acidic: pH < 7.00, neutral: pH = 7.00, basic: pH > 7.00

3. pH + pOH = 14.00 at room temperature

Explanation:

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:

1.135 M.

Explanation:

The reaction is a second order reaction of HI,so the rate law of the reaction is: Rate = k[HI]².

1/[A] = kt + 1/[A₀],

where, k is the reate constant of the reaction (k = 1.57 x 10⁻⁵ M⁻¹s⁻¹),

t is the time of the reaction (t = 8 hours x 60 x 60 = 28800 s),

[A₀] is the initial concentration of HI ([A₀] = ?? M).

[A] is the remaining concentration of HI after hours ([A₀] = 0.75 M).

∵ 1/[A] = kt + 1/[A₀],

∴ 1/[A₀] = 1/[A] - kt

∴ 1/[A₀] = [1/(0.75 M)] - (1.57 x 10⁻⁵ M⁻¹s⁻¹)(28800 s) = 1.333 M⁻¹ - 0.4522 M⁻¹ = 0.8808 M⁻¹.

∴ [A₀] = 1/(0.0.8808 M⁻¹) = 1.135 M.

So, the concentration of HI 8 hours earlier = 1.135 M.

Final answer:

To find the concentration of HI 8 hours earlier in a second-order reaction, we utilize the integrated rate law specific for second-order reactions and solve for the initial concentration using the given rate constant and the time in seconds.

Explanation:

The reaction 2HI → H2 + I2 is second order in [HI] and second order overall, meaning the rate equation can be written as Rate = k[HI]2. Given the rate constant (k) of 1.57 × 10−5 M−1s−1 at 700°C and the current concentration of 0.75 M, we need to calculate the concentration 8 hours earlier. To solve this, we can use the integrated rate law for a second-order reaction:

[HI]−1 - [HI]0−1 = kt

Where [HI]0 is the initial concentration, [HI] is the concentration after time t, k is the rate constant, and t is the time in seconds. We have:

[HI] = 0.75 M

k = 1.57 × 10−5 M−1s−1

t = 8 hours × 3600 seconds/hour = 28800 seconds

Plugging these values into the integrated rate equation, we calculate [HI]0. Finally, we can find the concentration 8 hours earlier.

Answer:

I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)

Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)

a) E(Pt⁺²/Fe°) = - 1.668v

b) Process is Non-spontaneous if E(cell) < 0

Explanation:

Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔

Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)

As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.

E°(Fe⁺²) = -0.44v

E°(Pt⁺²) = +1.20v

E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)

= -0.44v - (+1.20v) = - 1.64v

[Fe⁺²] = 0.0066M

[Pt⁺²] = 0.057M

n = electrons transferred = 2

E(nonstd) = E°(std) - (0.0592/n)logQ);

Q = [Pt⁺²]/[Fe⁺²]

= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v

Also, if ΔG(cell) > 0 => indicates non-spontaneous process

ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)

The Nernst equation is used to obtain the cell potential under standard conditions from the cell potential under nonstandard conditions.

Pt(s) + Fe^2+(aq) -------> Pt^2+ (aq) + Fe(s)

We have the following information from the question;

[Fe2+] = 0.0066M,   [Pt2+] = 0.057M

The standard reaction potential is; 1.18 V - (-0.44V) = 1.62 V

Using Nernst equation;

Ecell = E°cell - 0.0592/n log Q

Where;

Q = [Pt2+] /[Fe2+] =  0.057M/ 0.0066M = 8.636

Ecell = 1.62 V - 0.0592/2 log(8.636)

Ecell = 1.59 V

Cu(s) + 2Ag+(aq) -------> Cu2+(aq) + 2Ag(s)

The standard reaction potential is; 0.80 V - 0.34 V = 0.46 V

Q = [Cu2+]/[Ag+]^2 = 0.013/0.013^2 = 76.92

Ecell =  0.46 V - 0.0592/2 log(76.92)

Ecell = 0.40 V

Co2+(aq) + Ti3+(aq) -------> Co3+(aq) + Ti2+(aq)

The standard reaction potential of the reaction is; 1.92 V  - (-0.37) = 2.29 V

Q = [Ti2+] [Co3+]/[Co2+] [Ti3+] = [0.0110] [0.030]/[0.050] [0.0055] = 0.00033/0.000275 = 1.2

Ecell = 2.29 V - 0.0592/1 log(1.2)

Ecell = 2.28 V

Mg(s) + Sn2+(aq) --------> Mg2+(aq) + Sn(s)

The standard reaction potential of the reaction is; (-0.14 V) - (-2.37 V) = 2.23 V

Q = [Mg^2+]/[Sn^2+] = [0.796 M]/[0.0170 M] = 46.8

Ecell = 2.23 V - 0.0592/2 log(46.8)

Ecell = 2.18 V

Each of the cell is spontaneous as written since Ecell in each case is positive.

Learn more:https://brainly.com/question/9743981

Answer:

D) The nuclei of unstable isotopes  

Explanation:

For example, the carbon-14 in carbon dating is radioactive.

When it decays, a neutron in the nucleus is converted into a proton, an electron, and an antineutrino.

[tex]_{0}^{1}\text{n} \longrightarrow \, _{1}^{1}\text{p} + \, _{-1}^{0}\text{e} + \overline{\nu}_{\text{e}}[/tex]

Answer:

Explanation:

Pair  2.50g of O₂ and 2.50g of  N₂

The atoms sample with the largest number of moles since the masses are the same would be the one with lowest molar mass according the the equation below:

Number of moles = [tex]\frac{mass }{molarmass}[/tex]

Atomic mass of O = 16g and N = 14g

Molar mass of O₂ = 16 x 2 = 32gmol⁻¹

Molar mass of N₂ = 14 x 2 = 28gmol⁻¹

Number of moles of O₂ = [tex]\frac{2.5}{32}[/tex] = 0.078mole

Number of moles of N₂ = [tex]\frac{2.5}{28}[/tex] =  0.089mole

We see that N₂ has the largest number of moles

Answer:

Pair A : 2.50 g N2    Pair B : 21.5 g n2    Pair C : 0.081 CO2

Explanation:

Answer:

Explanation:

1) Data:

a) Cylinder diameter: d = 17.0 cm

b) Cylinder height: H = 20.4 cm

c) p = 2.20 MPa

d) compound: BF₃ gas

e) mass, m = 0.0985 kg

2) Formulae:

a) Volume of a cylinder, V = π r² H

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

c) Ideal gas equation: pV = nRT

3) Solution:

a) Volume (V)

i) r = d / 2 = 17.0 cm / 2 = 8.50 cm

ii) V = V = π r² H = π (8.50 cm) ² (20.4 cm) = 4,630 cm³ = 4.630 liter

b) Number of moles (n)

i) molar mass BF₃: 67.82 g/mol

ii) n = mass in grams / molar mass = 98.5 g / 67.82 g/mol = 1.452 mol

c) Temperature

i) pressure conversion: 2.20 MPa = 22.21 atm

ii) pV = nRT ⇒ T = pV / (nR)

The result is given with 3 significant figures, since that is the number of significant figures used for the data.

To calculate the maximum safe operating temperature for the gas reaction vessel, use the ideal gas law to find the number of moles of boron trifluoride gas and then calculate the maximum safe operating temperature.

To calculate the maximum safe operating temperature for the gas reaction vessel, we need to consider the ideal gas law and the given quantities of gas. First, we convert the pressure from megapascals to pascals. Then, we use the ideal gas law equation to find the number of moles of boron trifluoride gas. Finally, we use the molar mass of boron trifluoride to convert the number of moles to grams, and then divide by the volume of the vessel to find the density. By rearranging the ideal gas law equation to solve for temperature, we can calculate the maximum safe operating temperature.

https://brainly.com/question/16694127

#SPJ3

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%.

Answer:

b) Gain or lose electrons

Explanation:

An ion is an electrically charged particle. For an atom to be charged, it must have gained or lost electron in the process and therefore, it becomes an ion.

The loss or gain of electrons is what makes an atom charged and eventually becomes an ion.

A positively charged ion is one that has lost an electron and it is called a cation. In such an ion, the number of electrons are lesser than those of protons. This is why they are cations

A negatively charged ion is one that has gained electrons. They are called anions. In such an ion, the number of electrons are greater than that of protons.

The empirical formula of the hydrocarbon is C1H3, calculated from the products of combustion. Using the molecular weight approximation of 132 amu, the molecular formula determined is C8H24.

To determine the empirical formula of the hydrocarbon, we analyze the products from its combustion. Since the sample produced 0.4931 g of CO2, we calculate the moles of carbon:

0.4931 g CO2 × (1 mol CO2/44.01 g CO2) = 0.01120 mol CO2
0.01120 mol CO2 corresponds to 0.01120 mol C (since there is 1 C in each CO2 molecule).

Similarly, given 0.2691 g of H2O:

0.2691 g H2O × (1 mol H2O/18.02 g H2O) = 0.01493 mol H2O
This gives us 0.02986 mol H (since there are 2 H in H2O).

The molar ratio of C to H can be simplified by dividing by the smaller number of moles:

Molar ratio C:H = 0.01120 mol C : 0.02986 mol H
We reduce this ratio to the simplest whole numbers to get the empirical formula:

(0.01120 / 0.01120) : (0.02986 / 0.01120) = 1:2.67
Approximating to whole numbers, we get C1H3 as the empirical formula.

To find the molecular formula, we use the given molecular weight (approximately 132 amu). Since the empirical formula weight of C1H3 is 12 (for C) + 3 (for H) = 15 amu, we calculate the multiplier:

Molecular weight / Empirical formula weight = 132 amu / 15 amu ≈ 8.8
This multiplier indicates the molecular formula is approximately 8 or 9 times the empirical formula. A whole number will result when using 8, thus we get the molecular formula C8H24.

The behavior of the cans of soft drinks in a malfunctioning refrigerator is explained by the properties of water's expansion on freezing and gas solubility changes under varying pressure and temperature, but none of the given options accurately explains why only diet soft drink cans ruptured.

The scenario with cans of soft drinks in a malfunctioning refrigerator can be explained by the behavior of carbonated beverages when exposed to changes in temperature and pressure. The key statements that describe this behavior are:

Therefore, while 'B. Water expands on freezing' is a correct statement, it does not fully explain why only diet soft drink cans ruptured. The differences in freezing point depression due to differing ingredient compositions between diet and regular soft drinks could provide a better explanation, albeit not directly addressed among the options given.

Answer:

If the planet has a lower orbital radius it means the planets proximity to the sun is farther away. The climate of the planet will be colder than a planet that would have a closer proximity to the sun.