Cite examples of how copper deposits occur. Choose one or more: A. as an agglomeration metal B. as a native metal C. in carbonate ore minerals D. in sulfide ore minerals

Answer:

8.68

Explanation:

pOH = 8.68

all you need is contained in the sheet

Answer:

Approximately [tex]8.68[/tex].

Explanation:

The [tex]\rm pOH[/tex] of a solution can be found from the hydroxide ion concentration [tex]\rm \left[OH^{-}\right][/tex] with the following equation:

[tex]\displaystyle \rm pOH = -\log_{10} \rm \left[OH^{-}\right][/tex].

On the other hand, the ion-product constant of water, [tex]K_{\text{w}}[/tex], relates the hydroxide ion concentration [tex]\rm \left[OH^{-}\right][/tex] of a solution to its hydronium ion concentration [tex]\rm \left[{H_3O}^{+}\right][/tex]:

[tex]K_\text{w} = \rm \left[{H_3O}^{+}\right] \cdot \rm \left[OH^{-}\right][/tex].

Hence the [tex]\rm \left[OH^{-}\right][/tex] of this solution:

[tex]\begin{aligned}\left[\mathrm{OH}^{-}\right] &= \frac{K_\text{w}}{\rm \left[{H_3O}^{+}\right]} \\ &= \frac{1.0 \times 10^{-14}}{4.8 \times 10^{-6}}\; \rm mol\cdot L^{-1} \approx 2.08333 \times 10^{-9}\; \rm mol\cdot L^{-1}\end{aligned}[/tex].

Therefore, the [tex]\rm pOH[/tex] of this solution would be:

[tex]\begin{aligned}\rm pOH &= -\log_{10} \rm \left[OH^{-}\right] \\ &\approx -\log_{10} \left(4.8 \times 10^{-6}\right) \approx 8.68\end{aligned}[/tex].

Note that by convention, the number of decimal places in [tex]\rm pOH[/tex] should be the same as the number of significant figures in [tex]\rm \left[OH^{-}\right][/tex].

For example, because the [tex]\rm \left[{H_3O}^{+}\right][/tex] from the question has two significant figures, the [tex]\rm \left[OH^{-}\right][/tex] here also has two significant figures. As a result, the [tex]\rm pOH[/tex] in the result should have two decimal places.

[tex]\\ \sf\longmapsto HC_2H_3O_2[/tex]

[tex]\\ \sf\longmapsto 1u+2(12u)+3(1u)+2(16u)[/tex]

[tex]\\ \sf\longmapsto 1u+24u+3u+48u[/tex]

[tex]\\ \sf\longmapsto 28u+48u[/tex]

[tex]\\ \sf\longmapsto 76u[/tex]

[tex]\\ \sf\longmapsto 76g/mol[/tex]

[tex]\\ \sf\longmapsto No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}[/tex]

[tex]\\ \sf\longmapsto 0.2=\dfrac{Given\:mass}{76}[/tex]

[tex]\\ \sf\longmapsto Given\:Mass=0.2\times 76[/tex]

[tex]\\ \sf\longmapsto Given\:Mass=1.52g[/tex]

Answer:

(a).

» E. SO₃, sulphur trioxide.

(b).

» A. CaO, Calcium oxide.

» D. K₂O, potassium oxide.

» E. BaO, barium oxide.

(c).

» B. Al₂O₃, Aluminium oxide.

» E. SnO₂, tin (IV) oxide.

(d).

» A. CO, carbon monoxide.

» B. NO, nitrogen monoxide.

Explanation:

[tex]{ \underline{ \sf{ \blue{christ \:† \: alone }}}}[/tex]

from

1moles=iatom

Mole=mass÷avogardos

Where

Avogadro's= 6.02×10²³

So moles = 65.0÷6.02×10²³

Atoms of zinc = 391.6 ×10²³

The number of atoms present in the given mass of Zinc that is 65.0gm is [tex]5.99\times10^{ 23}[/tex].

Atoms are the basic building blocks of matter. They are the smallest units of an element that retain the chemical properties of that element.

Now, to determine the number of atoms in a given number of moles, we can use Avogadro's number, which is approximately  [tex]6.022 \times10^{23}[/tex]atoms per mole.

First, we calculate the number of moles of zinc in 65.0g by dividing the given mass by the molar mass of zinc. The molar mass of zinc (Zn) is 65.38 g/mol.

Number of moles = Mass / Molar mass

Number of moles = 65.0g / 65.38 g/mol ≈ 0.9942 mol

Next, multiply the number of moles by Avogadro's number to find the number of atoms.

Number of atoms =[tex]Number of moles \times Avogadro's number[/tex]

Number of atoms = [tex]0.9942[/tex]mol × [tex]6.022 \times10^{23}[/tex] atoms/mol

Therefore, approximately [tex]5.99\times10^{ 23}[/tex] atoms are present in 65.0g of zinc.

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The third quantum number (m) describes the specific orbital within a sublevel. The correct answer is option D.

Quantum numbers are a set of four numbers that describe the properties of an electron in an atom. They specify the energy, position, and orientation of an electron in an atom.

The third quantum number (m) is also called the magnetic quantum number. It describes the orientation of the orbital in space. Each orbital within a sublevel has a different orientation in space, and the magnetic quantum number tells us which orbital we are talking about.

The value of m can range from -l to +l, where l is the second quantum number (the angular momentum quantum number). The value of l determines the number of orbitals in a sublevel, and the value of m determines the specific orbital within that sublevel.

In conclusion, the third quantum number (m) describes the specific orbital within a sublevel. It tells us how the orbital is oriented in space, and it can have values ranging from[tex]\rm -l \ to +l[/tex], where l is the second quantum number.

Option D is the correct answer.

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Answer:

5.90

Explanation:

Initial moles of CH3COO- = 10.0/1000 x 0.75 = 0.0075 mol

Moles of HCl added = 5.0/1000 x 0.10 = 0.0005 mol

CH3COO- + HCl => CH3COOH + Cl-

Moles of CH3COO- left = 0.0075 - 0.0005 = 0.007 mol

Moles of CH3COOH formed = moles of HCl added = 0.0005 mol

pH = pKa + log([CH3COO-]/[CH3COOH])

= -log Ka + log(moles of CH3COO-/moles of CH3COOH)

= -log(1.78 x 10^(-5)) + log(0.007/0.0005)

= 5.90

Answer:

The correct answer is 5.895.

Explanation:

The reaction will be,

CHCOO⁻ + H+ ⇔ CH₃COOH

Both the HCl and the acetate are having one n factor.

The millimoles of CH₃COO⁻ is,

= Volume in ml × molarity = 10 × 0.75 = 7.5

The millimoles of HCl = Volume in ml × molarity = 5 × 0.1 = 0.5

Therefore, 0.5 will be the millimoles of CH₃COOH formed, now the millimoles of the CH₃COO⁻ left will be, 7.5-0.5 = 7.0

The volume of the solution is, 10+5 = 15 ml

The molarity of CH₃COO⁻ is, millimoles / volume in ml = 7/15

The molarity of CH₃COOH is 0.5/15

pH = pKa + log[CH₃COO⁻]/[CH₃COOH]

= 4.74957 + 1.146

= 5.895

Answer:

pH=0.22.

[tex][OH^-]=1.66x10^{-14}M[/tex]

Explanation:

Hello,

In this case, since the pH is directly computed from the given concentration of hydronium ions:

[tex]pH=-log([H_3O^+])=-log(0.6M)=0.22[/tex]

It is widely known that the pH and POH are directly related via:

[tex]pH+pOH=14[/tex]

Therefore, the pOH is:

[tex]pOH=14-pH\\\\pOH=14-0.22=13.78[/tex]

Thus, the concentration of hydroxyl ions are computed from the pOH:

[tex]pOH=-log([OH^-]}\\\\[/tex]

[tex][OH^-]=10^{-pOH]=10^{-13.78}[/tex]

[tex][OH^-]=1.66x10^{-14}M[/tex]

Regards.

Answer:

[tex]\boxed {\boxed {\sf 0.0769 \ M}}[/tex]

Explanation:

We are asked to find the molarity of an acid given the details of a titration experiment. The formula for titration is as follows:

[tex]M_AV_A= M_B V_B[/tex]

In this formula, M is the molarity of the acid or base and V is the volume of the acid or base. The molarity of the hydrochloric acid (HCl) is unknown and the volume is 25.0 milliliters.

[tex]M_A * 25.0 \ mL = M_BV_B[/tex]

The molarity of the sodium hydroxide (NaOH) is 0.108 molar and the volume is 17.80 milliliters.

[tex]M_A * 25.0 \ mL = 0.108 \ M * 17.80 \ mL[/tex]

We are solving for the molarity of the acid and we must isolate the variable [tex]M_A[/tex]. It is being multiplied by 25.0 milliliters. The inverse operation of multiplication is division, so we divide both sides of the equation by 25.0 mL.

[tex]\frac {M_A * 25.0 \ mL }{25.0 \ mL}= \frac{0.108 \ M * 17.80 \ mL }{25.0 \ mL}[/tex]

[tex]M_A= \frac{0.108 \ M * 17.80 \ mL }{25.0 \ mL}[/tex]

The units of milliliters cancel.

[tex]M_A= \frac{0.108 \ M * 17.80 }{25.0 }[/tex]

[tex]M_A= \frac{1.9224}{25.0 } \ M[/tex]

[tex]M_A= 0.076896 \ M[/tex]

The original measurements have 3 and 4 significant figures. We must round our answer to the least number of sig figs, which is 3. For the number we calculated, that is the ten-thousandth place. The 9 to the right of this place tells us to round the 8 up to a 9.

[tex]M_A \approx 0.0769 \ M[/tex]

The molarity of the hydrochloric acid is 0.0769 Molar.

Answer:1 )T2=134°C   2) T2=339.48°C. 3)

P=817.59 mmHg.

Explanation:

1.Given ;

pressure, P1 of neon gas = 0.646 atm

temperature, T1 =242oC + 273=515oC

Volume, V1 =515ml

Volume V2= 407ml

temperature , T 2= ?

Solution;

And at constant pressure, the volume cools at V2=407 mL at T2=?

From ideal gas equation, PV=nRT

V/T=constant

therefore

V1/V2=T1/T2 = T2=(V2 xT1)/V1

T2=(407 mL x 515 K)/515 mL= 407K.

T2= 407K -273= 134°C.   recall 0°C=273 K)

2..Given ;

pressure, P1 of neon gas = 0.633 atm

temperature, T1 =261oC + 273=534oC

Volume, V1 =694ml

Volume V2= 796ml

temperature , T 2= ?

Solution;

And at constant pressure, the volume expands  at V2=796mL at T2=?

From ideal gas equation, PV=nRT

V/T=constant

therefore

V1/V2=T1/T2 = T2=(V2 xT1)/V1

T2=(796 mL x 534 K)/694mL= 612.48K.

T2= 612.48K -273= 339.48°C. recall 0°C=273 K

3

Given;

moles of CO2= n=0.962 mol,

temperature T=20°C=20+273 K =293 K,

volume V=21.5 L,

gas constant R at L·mmHg/mol·K= 62.3637 L mmHg mol^-1 K^-1

Using  ideal gas equation PV=nRT

P=nRT/V

P=(0.962 mol)x(62.3637mmHg mol^-1 K^-1)x(293 K)/(21.5L)

P=817.59 mmHg.

Answer:

0.100 moles of barium sulfate are produced from 0.100 moles of barium chloride.

Explanation:

Barium chloride and sodium sulfate react according to the following balanced reaction:

BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagent and products participate in the reaction:

Then you can apply the following rule of three: if 1 mole of BaCl₂ produces 1 mole of BaSO₄, 0.100 mole of BaCl₂ how many moles of BaSO₄ does it produce?

[tex]amount of moles of BaSO_{4} =\frac{0.100 mole of BaCl_{2}* 1 mole of BaSO_{4} }{1 mole of BaCl_{2}}[/tex]

amount of moles of BaSO₄= 0.100

0.100 moles of barium sulfate are produced from 0.100 moles of barium chloride.

Answer:

5.94×10¯³

Explanation:

The following data were obtained from the question:

Concentration of N2, [N2] = 0.036 M

Concentration of H2, [H2] = 0.15 M

Equilibrium constant (Kc) = 0.29 M

Concentration of NH3, [NH3] =.…?

The equation for the reaction is given below:

N2(g) + 3H2(g) <=> 2NH3(g)

Thus, we can determine the concentration of NH3 by using the equilibrium expression for the reaction.

This is illustrated below:

The equilibrium constant for a reaction is simply defined as the ratio of the concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Kc = [NH3]² / [N2] [H2]³

Inputting the value of Kc, [N2], and [H2] the value of [NH3]can be obtained as follow:

Kc = [NH3]² / [N2] [H2]³

0.29 = [NH3]²/ 0.036 × 0.15³

Cross multiply

[NH3]² = 0.29 × 0.036 × 0.15³

[NH3]² = 3.5235×10¯⁵

Take the square root of both side

[NH3] = √(3.5235×10¯⁵)

[NH3] = 5.94×10¯³

Therefore, the concentration of NH3, [NH3] is 5.94×10¯³ M.

Answer:

[Cu²⁺] = 2.01x10⁻²⁶

Explanation:

The equilibrium of Cu(CN)₄²⁻ is:

Cu²⁺ + 4CN⁻ ⇄ Cu(CN)₄²⁻

And Kf is defined as:

Kf = 1.0x10²⁵ = [Cu(CN)₄²⁻] / [Cu²⁺] [CN⁻]⁴

As Kf is too high you can assume all Cu²⁺ is converted in Cu(CN)₄²⁻ -Cu²⁺ is limiting reactant-, the new concentrations will be:

[Cu²⁺] = 0

[CN⁻] = 0.33M - 4×2.2x10⁻³ = 0.3212M

[Cu(CN)₄²⁻] = 2.2x10⁻³

Some [Cu²⁺] will be formed and equilibrium concentrations will be:

[Cu²⁺] = X

[CN⁻] = 0.3212M + 4X

[Cu(CN)₄²⁻] = 2.2x10⁻³ - X

Where X is reaction coordinate

Replacing in Kf equation:

1.0x10²⁵ = [2.2x10⁻³ - X] / [X] [0.3212M +4X]⁴

1.0x10²⁵ = [2.2x10⁻³ - X] / 0.0104858X + 0.524288 X² + 9.8304 X³ + 81.92 X⁴ + 256 X⁵

1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ = 2.2x10⁻³ - X

1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ - 2.2x10⁻³ = 0

Solving for X:

X = 2.01x10⁻²⁶

As

[Cu²⁺] = X

Answer:

A clinical thermometer is a thermometer used to measure human body temperature. Most made in the 20th century are mercury-in-glass thermometers. They are accurate and sensitive, having a narrow place where the mercury level rises very fast. A kink in the tube stops the mercury level from falling on its own.

Answer:

83914.52 grams

Explanation:

Given that,

Weight of a man is 185 lb

We need to find his weight in grams

For this, we must know the relation between lb and grams.

We know that,

1 lb = 453.592 grams

To find the mass of man in grams, the step is :

185 lb = (453.592 × 185) grams

= 83914.52 grams

So, the mass of a man is 83914.52 grams.

Answer:

226Ra88→222Rn86+4He2

Explanation:

An α-particle usually consists of a helium nucleus which indicates the type of decay that was undergone in this radioactive process.

During α-decay(alpha decay), an atomic nucleus emits an alpha particle.

Answer:

21 KJ/mol

Explanation:

For this question, we have to start with the linear structure of 2-methylbutane. With the linear structure, we can start to propose all the Newman projections keep it in mind that the point of view is between carbons 2 and 3 (see figure 1).

Additionally, we have several energy values for each interaction present in the Newman structures:

-) Methyl-methyl gauche: 3.8 KJ/mol

-) Methyl-H eclipse: 6.0 KJ/mol

-) Methyl-methyl eclipse: 11.0 KJ/mol

-) H-H eclipse: 4.0 KJ/mol

Now, we can calculate the energy for each molecule.

Molecule A

In this molecule, we have 2 Methyl-methyl gauche interactions only, so:

(3.8x2) = 7.6 KJ/mol

Molecule B

In this molecule, we have a Methyl-methyl eclipse interaction a Methyl-H eclipse interaction and an H-H eclipse interaction, so:

(11)+(6)+(4) = 21 KJ/mol

Molecule C

In this molecule, we have 1 Methyl-methyl gauche interaction only, so:

3.8 KJ/mol

Molecule D

In this molecule, we have three Methyl-H eclipse interaction, so:

(6*3) = 18 KJ/mol

Molecule E

In this molecule, we have 1 Methyl-methyl gauche interaction only, so:

3.8 KJ/mol

Molecule F

In this molecule, we have a Methyl-methyl eclipse interaction a Methyl-H eclipse interaction and an H-H eclipse interaction, so:

(11)+(6)+(4) = 21 KJ/mol

The structures with higher energies would be less stable. In this case, structures B and F with an energy value of 21 KJ/mol (see figure 2).

I hope it helps!

Answer:

Sodium, magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon are the elements of third period.

Explanation:

There are three energy levels in sodium atom. It has 11 electrons revolving around the nucleus. the atomic radius of sodium atom is 227 ppm. Magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon has also three energy levels like sodium because all these elements belongs to third period. There are 12 electrons present in magnesium, 13 in aluminium, 14 in silicon, 15 in phosphorus, 16 in sulfur, 17 in chlorine, and 18 electrons in argon. The atomic radius of magnesium atom is 173 ppm.  The atomic radius of aluminium atom is 143 ppm.  The atomic radius of silicon atom is 111 ppm.  The atomic radius of phosphorus atom is 98 ppm.  The atomic radius of sulfur atom is 87 ppm. The atomic radius of chlorine atom is 79 ppm and the atomic radius of argon atom is 71 ppm.

Answer:

-434.14 kJ

Explanation:

Step 1: Write the balanced equation

4 NH₃(g) + 5 O₂(g) ⇒ 4 NO(g) + 6 H₂O(g)

Step 2: Calculate the standard free energy change (ΔG°r) for the reaction

We will use the following expression.

ΔG°r = 4 mol × ΔG°f(NO(g)) + 6 mol × ΔG°f(H₂O(g)) - 4 mol × ΔG°f(NH₃(g)) - 5 mol × ΔG°f(O₂(g))

ΔG°r = 4 mol × (86.55 kJ/mol) + 6 mol × (-228.57 kJ/mol) - 4 mol × (-16.45 kJ/mol) - 5 mol × (0 kJ/mol)

ΔG°r = -959.42 kJ

Step 3: Calculate the standard free energy change for 1.81 moles of NH₃

959.42 kJ are released per 4 moles of NH₃.

[tex]\frac{-959.42 kJ}{4mol} \times 1.81mol = -434.14 kJ[/tex]

Answer:

the acid strength is the order of [tex]\mathsf{HF _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5OH _{(aq)} }[/tex]

Explanation:

Given that :

a . [tex]\mathsf{C_6H_5COO^- _{(aq)} + C_6H_5OH _{(aq)} \to C_6H_5COOH _{(aq)} + C_6H_5O^- _{(aq)}}[/tex]

b.  [tex]\mathsf{ F^- _{(aq)} + C_6H_5OH _{(aq)} \to C_6H_5O^- _{(aq)} + HF _{(aq)} }[/tex]

c.  [tex]\mathsf{C_6H_5COOH _{(aq)} + F^- _{(aq)} \to HF _{(aq)} + C_6H_5COO^- _{(aq)} }[/tex]

Acid strength is the ability of an acid  to dissociate into a proton and an anion. Take for instance.

HA  ↔ H⁺ + A⁻

The  acid strength of the following compounds above are:

[tex]\mathsf{C_6H_5OH _{(aq)} }[/tex] = 1.00 × 10⁻¹⁰

[tex]\mathsf{HF _{(aq)} }[/tex] = 6.6 × 10⁻⁴

[tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] = 6.3 × 10⁻⁵

As the acid dissociation constant increases the relative acid strength also increases.

From above, the acid strength is the order of [tex]\mathsf{HF _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5OH _{(aq)} }[/tex]

[tex]\mathsf{C_6H_5COO^- }[/tex], [tex]\mathsf{C_6H_5O^- _{(aq)}}[/tex] and F⁻ are Bronsted- Lowry acid

Bronsted- Lowry acid are molecule or ion that have the ability to donate a proton.

Answer:

Explanation:

The moles of H2 and N2 are as follows respectively, 0.3915mol of H2 and 0.1305 mol of N2.

0.2175,0.0725 moles of [tex]H_2[/tex] and [tex]N_2[/tex] can be formed by the decomposition of 0.145 mol of ammonia, [tex]NH_3[/tex]

The reaction for the decomposition of ammonia is as follows:-

[tex]N_2+3H_2 \rightarrow 2NH_3[/tex]

Calculate the mole  of [tex]H_2[/tex] and [tex]N_2[/tex] as follows:-

[tex]Mole\ of \ H_2=0.145\ mol\ NH_3\times\frac{3\ mol\ H_2}{2\ mol\ NH_3} \\\\=0.2175\ mol\ H_2[/tex]

[tex]Mole\ of \ N_2=0.145\ mol\ NH_3\times\frac{1\ mol\ N_2}{2\ mol\ NH_3} \\\\=0.0725\ mol\ H_2[/tex]

Hence, the number of moles of [tex]H_2[/tex] and [tex]N_2[/tex]  are 0.2175 mol, and 0.0725 mol.

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That is because one of the Hydrogens of Formic acid is a part of the functional group (COOH). so to make sure the reader understands that, they second H is written separately

Answer:

D) these all have the same boiling point

Answer:

The rate would have doubled

Explanation:

Answer:

A) CH3CH2CH2CH2CH2CH2OH

Explanation:

For this question, we have the following answer options:

A) CH3CH2CH2CH2CH2CH2OH

B) (CH3CH2)2CH(OH)CH2CH3

C) (CH3CH2)2CHOHCH3

D) (CH3CH2)3COH

E) (CH3CH2)2C(CH3)OH

We have to remember the reaction mechanism of the substitution reaction with [tex]PBr_3[/tex]. The idea is to generate a better leaving group in order to add a "Br" atom.

The [tex]PBr_3[/tex] attacks the "OH" generation new a bond to P (O-P bonds are very strong), due to this new bond we will have a better leaving group that can remove the oxygen an allow the attack of the Br atom to generating a new C-Br bond. This is made by an Sn2 reaction. Therefore we will have a faster reaction with primary substrates. In this case, the only primary substrate is molecule A. So, "CH3CH2CH2CH2CH2CH2OH" will react faster.

See figure 1

I hope it helps!

Ortho and para hydrogen are nuclei forms

Answer:

the answer to your question is ar 126

Answer:

2-methyl-2-pentyl-1,3-dioxolane

Explanation:

In this case, we have two reactions:

First reaction:

1-heptyne + mercuric acetate -------> Compound A

Second reaction:

Compound A + HOCH2CH2OH -------> Compound C

First reaction

In the first reaction, we have as a main functional group a triple bond. We have to remember that mercuric acetate in sulfuric acid will produce a ketone. The carbonyl group (C=O) would be placed in the most substituted carbon of the triplet bond (in this case, carbon 2). With this in mind, we will have as a product: heptan-2-one. (See figure 1).

Second reaction

In this reaction, we have as reagents:

-) Heptan-2-one

-) Ethylene-glycol [tex]HOCH_2CH_2OH[/tex]

-) Sulfuric acid [tex]H_2SO_4[/tex]

When we put ethylene-glycol with a ketone or an aldehyde we will form a cyclic acetal. In this case, this structure would be formed on carbon 2 forming 2-methyl-2-pentyl-1,3-dioxolane. (See figure 2).

I hope it helps!

Answer:

see the photo

Explanation:

it was the answer

Answer:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases.

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases.

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases.

Explanation:

Hello.

In this case, we can understand a higher entropy when more disorder is present and a lower entropy when less disorder is present, thus:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases since iodine has the greatest molar mass (254 g/mol) and fluorine the least molar mass (38 g/mol).

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases since hydrogen peroxide weights 34 g/mol as well as hydrogen sulfide but the peroxide has more bonds (more complex, higher entropy).

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases since diamond has a well-ordered structure and amorphous carbon has a very disordered one.

Best regards.

Acetaminophen is used to treat patients with aches and fever. It is safe to give a dose of 4 grams in 24 hours period.

It is an effective medicine which does not require much time to provide relief to the patient.

Usually tylenol is the best medicine which consists of Acetaminophen in appropriate quantity to provide relief from pain and fever.

The reporting for the doses introduced to the patient must be routine since it is not an emergency case.

The correct answer is a. Routine

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