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Rhodium has an atomic radius of 0.1345 nm and density of 12.41 gm/cm3 . Determine whether it has an FCC or BCC crystal structure.

Answers

Answer 1

Answer:

Answer:

FCC.

Explanation:

Hello,

In this case, since the density is defined as:

$\rho =(n*M)/(Vc*N_A)$

Whereas n accounts for the number of atoms per units cell (2 for BCC and 4 for FCC), M the atomic mass of the element, Vc the volume of the cell and NA the Avogadro's number. Thus, for both BCC and FCC, the volume of the cell is:

$Vc_(BCC)=((4r)/(√(3) ) )^3=((4*0.1345x10^(-7)cm)/(√(3) ) )^3=2.997x10^(-23)cm^3\n\nVc_(FCC)=(2√(2)r)^(3) =(2√(2) *0.1345x10^(-7)cm)^3=5.506x10^(-23)cm^3$

Hence, we compute the density for each crystal structure:

$\rho _(BCC)=(n_(BCC)*M)/(Vc_(BCC)*N_A)=(2*102.9g/mol)/(2.337x10^(-23)cm^3*6.022x10^(23)/mol) =14.62g/cm^3\n\n\rho _(FCC)=(n_(FCC)*M)/(Vc_(FCC)*N_A)=(4*102.9g/mol)/(5.506x10^(-23)cm^3*6.022x10^(23)/mol) =12.41g/cm^3$

Therefore, since the density computed as a FCC crystal structure matches with the actual density, we conclude rhodium has a FCC crystal structure.

Regards.

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Which of the following combinations represents an element with a net charge of +1 with a mass number of 75?a) 35 (o), 35 (+), 34 (-)b) 40 (o), 40 (+), 39 (-)c) 40 (o), 35 (+), 34 (-)d) 37 (o), 38 (+), 34 (-)e) 40 (o), 35 (+), 35 (-)

You would like to make a 100 mL buffer solution at pH 8.00. Assuming you would like to accomplish this with a hypochlorous acid (HOCl) buffer (HOCl/NaOCl), Ka= 3.0 * 10-8. If the solution is 0.3 M in HOCl, what concentration of NaOCl would be necessary in the buffer solution to obtain a pH of 8.0?

Answers

Answer:

To obtain the pH of 8.0, the concentration of NaOCl needs to be 0.9 M in the 0.3 M HOCl solution

Explanation:

This problem can be solved by Henderson-Hasselbalch equation, which gives relation between the concentration of acid, its salt, pKa and the pH of the solution. This equation is given as,

$pH=-log(K_a)+log([NaOCl])/([HOCl])$

By placing the known variables in the above equation we get,

$8=-log(3*10^(-8))+log([NaOCl])/(0.3)$

$8-7.52=log([NaOCl])/([0.3])$

$10^(0.48)=([NaOCl])/(0.3)$

$[NaOCl]=10^(0.48)*{0.3}$

$[NaOCl]=0.9 M$

The above calculations show that the required concentration of NaOCl is 0.9 M.

Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V

An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:

Ni(s)VO2+(aq,0.083M)+2H+(aq,1.1M)+e−→→Ni2+(aq,2.5M)+2e−VO2+(aq,2.5M)+H2O(l)

Calculate the cell potential under these nonstandard concentrations.

Express the cell potential to two decimal places and include the appropriate units.

Answers

Answer:

Cell potential under non standard concentration is 4.09 v

Explanation:

To solve this problem we need to use Nernst Equation because concentrations of the components of the chemical reactions are differents to 1 M (normal conditions: 1 M , 1 atm).

Nernst equation at 25ºC is:

$E = E^(0) - [((0.0592)/(n) · log Q)]$

where

E: Cell potential (non standard conditions)

$E^(0)$ = Cell potential (standard conditions)

n: Number of electrons transfered in the redox reaction

Q: Reaction coefficient (we are going to get it from the balanced chemical reaction)

For example, consider the following general chemical reaction:

aA + bB --> cC + dD

where

a, b, c, d: coefficient of balanced chemical reaction

A,B,C,D: chemical compounds in the reaction.

Using the previous general reaction, expression of Q is:

$Q = (C^(c) * D^(d) )/(A^(a)*B^(b) )$

Previous information is basic to solve this problem. Let´s see the Nernst equation, we need to know: $E^(0)$ , n and Q

Let´s calcule potential in nomal conditions ( $E^(0)$ ):

1. We need to know half-reactions (oxidation and reduction), we take them from the chemical reaction given in this exercise:

Half-reactions: Eo (v):

$(VO_(2))^(2+)$ + e- --> $(VO_(2))^(+)$ -0.23

$Ni --> Ni^(2+)$ + 2 e- +0.99

Balancing each half-reaction, first we are going to balance mass and then we will balance charge in each half-reaction and then charge between half-reactions:

Half-reactions: Eo (v):

2 * [ $(VO_(2))^(2+)$ + e- --> $(VO_(2))^(+)$ ] -0.23

1 * [ $Ni --> Ni^(2+)$ + 2 e- ] +0.99

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

2 $(VO_(2))^(2+)$ + 2e- + Ni --> 2 $(VO_(2))^(+)$ + $Ni^(2+)$ + 2e- 0.76 v

Then global balanced chemical reaction is:

2 $(VO_(2))^(2+)$ + Ni --> 2 $(VO_(2))^(+)$ + $Ni^(2+)$

and the potential in nomal conditions is:

$E^(0)$ = 0.76 v

Also from the balanced reaction, we got number of electons transfered:

n = 2

2. Calculate Q:

Now using previous information, we can establish Q expression and we can calculate its value:

$Q = ([(VO_(2)+]^(2)* [Ni^(2+) )/([VO_(2+)]^(2) * Ni )]$

From the exercise we know:

$[VO_(2) ^(2+)] = 2.5 M$

$[VO_(2)+] = 0.083 M$

$[Ni^(2+)] = 2.5 M$

$[Ni] = 1 M, it is a pure solid, so its activity in Q is unit (1). It is also applied for pure liquids.$

$Q = ((2.5)^(2)* 2.5 )/((0.083)^(2) * 1) = 2,268.11$

3. Use Nernst equation:

Finally, we replace all these results in the Nernst equation:

$E = E^(0) - ((0.0592)/(n) - log Q)\n \nE = 0.76 - ((0.0592)/(2)-log (2,268.11) \nE = 0.76 - (0.0296 - 3.36)\nE = 4.09 v$

Cell potential under non standard concentration is 4.09 v

Final answer:

To calculate the cell potential under nonstandard conditions, we need to apply the Nernst Equation. This involves finding the reaction quotient (Q) from the given concentrations and then subtracting a value derived from Q and the number of electrons transferred, from the cell potential under standard conditions.

Explanation:

For calculating the cell potential under nonstandard conditions for an electrochemical cell, we need to use the Nernst equation. In this case, the Nernst Equation is Ecell = E∘cell - (0.0592/n) * logQ, where Q, the reaction quotient, is the ratio of the concentrations of the products to the reactants raised to their stoichiometric coefficients.

Given the half-cell reduction potentials, we can calculate the cell potential under standard conditions (E°cell) by subtracting the potential of the anode from the potential of the cathode (E°cell = Ecathode - Eanode = 0.99V - (-0.23V), resulting in E°cell = 1.22V.

Next, Q = [Ni2+]/([VO2+]×[H+]²), substituting the given concentrations, Q = (2.5)/(0.083×1.1²).

After calculating Q, we substitute all known values into the Nernst Equation and solve for Ecell. Hence, the cell potential under these nonstandard conditions can be calculated.

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In a 74.0-g aqueous solution of methanol, CH4O, the mole fraction of methanol is 0.140. What is the mass of each component?

Answers

Answer:

The correct answer is 16.61 grams methanol and 57.38 grams water.

Explanation:

The mole fraction (X) of methanol can be determined by using the formula,

X₁ = mole number of methanol (n₁) / Total mole number (n₁ + n₂)

X₁ = n₁/n₁ + n₂ = 0.14

n₁ / n₁ + n₂ = 0.14 ---------(i)

n₁ mole CH₃OH = n₁ mol × 32.042 gram/mol (The molecular mass of CH₃OH is 32.042 grams per mole)

n₁ mole CH₃OH = 32.042 n₁ g

n₂ mole H2O = n₂ mole × 18.015 g/mol

n₂ mole H2O = 18.015 n₂ g

Thus, total mole number is,

32.042 n₁ + 18.015 n₂ = 74 ------------(ii)

From equation (i)

n₁/n₁ + n₂ = 0.14

n₁ = 0.14 n₁ + 0.14 n₂

n₁ - 0.14 n₁ = 0.14 n₂

n₁ = 0.14 n₂ / 1-0.14

n₁ = 0.14 n₂/0.86 ----------(iii)

From eq (ii) and (iii) we get,

32.042 × 0.14/0.86 n₂ + 18.015 n₂ = 74

n₂ (32.042 × 0.14/0.86 + 18.015) = 74

n₂ = 74 / (32.042 × 0.14/0.86 + 18.0.15)

n₂ = 3.1854 mol

From equation (iii),

n₁ = 0.14/0.86 n₂

n₁ = 0.14/0.86 × 3.1854

n₁ = 0.5185 mol

Now, presence of water in the mixture is,

= 3.1854 mole × 18.015 gram per mole

= 57.38 grams

Methanol present in the mixture is,

= 0.5185 mol × 32.042 gram per mole

= 16.61 grams

Final answer:

In a 74.0 g aqueous solution of methanol with a mole fraction of 0.140, the mass of methanol is approximately 10.36 g and the mass of water is approximately 63.64 g.

Explanation:

The problem involves the calculation of the mass of the components of an aqueous solution of methanol (CH3OH). First, we need to know that the mole fraction is defined as the ratio of the number of moles of a component to the total number of moles of all components in the mixture.

Given that the mole fraction of methanol is 0.140, this means that the rest of the solution (i.e., water) is 1 - 0.140 = 0.860. To find the mass of each component, we need to consider the total mass of 74.0 g.

The mass of methanol can be calculated as 74.0 g * 0.140 = 10.36 g. And the mass of water would be 74.0 g * 0.860 = 63.64 g.

So, in this aqueous solution, you have approximately 10.36 g of methanol and 63.64 g of water.

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Balance the following equation and list the coefficients in order from left to right.SF4 + __ H2O → _H2SO3 + HE

Answers

Vascular tissue

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Cross section of celery stalk, showing vascular bundles, which include both phloem and xylem.

Detail of the vasculature of a bramble leaf.

Translocation in vascular plants

This article is about vascular tissue in plants. For transportation in animals, see Circulatory system.

Vascular tissue is a complex conducting tissue, formed of more than one cell type, found in vascular plants. The primary components of vascular tissue are the xylem and phloem. These two tissues transport fluid and nutrients internally. There are also two meristems associated with vascular tissue: the vascular cambium and the cork cambium. All the vascular tissues within a particular plant together constitute the vascular tissue system of that plant.

The cells in vascular tissue are typically long and slender. Since the xylem and phloem function in the conduction of water, minerals, and nutrients throughout the plant, it is not surprising that their form should be similar to pipes. The individual cells of phloem are connected end-to-end, just as the sections of a pipe might be. As the plant grows, new vascular tissue differentiates in the growing tips of the plant. The new tissue is aligned with existing vascular tissue, maintaining its connection throughout the plant. The vascular tissue in plants is arranged in long, discrete strands called vascular bundles. These bundles include both xylem and phloem, as well as supporting and protective cells. In stems and roots, the xylem typically lies closer to the interior of the stem with phloem towards the exterior of the stem. In the stems of some Asterales dicots, there may be phloem located inwardly from the xylem as well.

Between the xylem and phloem is a meristem called the vascular cambium. This tissue divides off cells that will become additional xylem and phloem. This growth increases the girth of the plant, rather than it

By pipet, 11.00 mL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask and diluted to the calibration mark. Determine the molarity of the resulting solution. A stock solution of potassium permanganate (KMnO4) was prepared by dissolving 13.0g KMnO4 with DI H2O in a 100.00-mL volumetric flask and diluting to the calibration mark. Determine the molarity of the solution Molarity= O.822 M

Answers

Answer:

1) 0.18106 M is the molarity of the resulting solution.

2) 0.823 Molar is the molarity of the solution.

Explanation:

1) Volume of stock solution = $V_1=11.00 mL$

Concentration of stock solution = $M_1=0.823 M$

Volume of stock solution after dilution = $V_2=50.00 mL$

Concentration of stock solution after dilution = $M_2=?$

$M_1V_1=M_2V_2$ ( dilution )

$M_2=(0.823 M* 11.00 mL)/(50 ,00 mL)=0.18106 M$

0.18106 M is the molarity of the resulting solution.

Molarity of the solution is the moles of compound in 1 Liter solutions.

$Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}* Volume (L)}$

Mass of potassium permanganate = 13.0 g

Molar mass of potassium permangante = 158 g/mol

Volume of the solution = 100.00 mL = 0.100 L ( 1 mL=0.001 L)

$Molarity=(13.0 g)/(158 g/mol* 0.100 L)=0.823 mol/L$

0.823 Molar is the molarity of the solution.

Final answer:

To determine the molarity of the resulting solution, we can use the formula M1V1 = M2V2. Plugging in the given values, we find that the molarity of the resulting solution is 0.180 MM.

Explanation:

To determine the molarity of the resulting solution, we need to use the formula:

M1V1 = M2V2

Where M1 is the molarity of the stock solution, V1 is the volume of the stock solution used, M2 is the molarity of the resulting solution, and V2 is the final volume of the resulting solution.

Using the given values, we have:

M1 = 0.823 MM

V1 = 11.00 mL

V2 = 50.00 mL

Substituting these values into the formula, we can find the molarity of the resulting solution.

M2 = (M1 * V1) / V2

Plugging in the values:

M2 = (0.823 MM * 11.00 mL) / 50.00 mL = 0.180 MM

The molarity of the resulting solution is 0.180 MM.

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Predict whether the compounds are soluble or insoluble in water

Answers

Answer:

The polar compounds are soluble in water while non polar are insoluble in water.

Explanation:

Solvent is the that part of solution which is present in large proportion and have ability to dissolve the solute. In simplest form it is something in which other substance get dissolve. The most widely used solvent is water, other examples are toluene, acetone, ethanol, chloroform etc.

Water is called universal solvent because of high polarity all polar substance are dissolve in it. Hydrogen is less electronegative while oxygen is more electronegative and because of difference in electronegativity hydrogen carry the partial positive charge while oxygen carry partial negative charge.

Water create electrostatic interaction with other polar molecules. The negative end of water attract the positive end of polar molecules and positive end of water attract negative end of polar substance and in this way polar substance get dissolve in it.

Example:

when we stir the sodium chloride into water the cation Na⁺ ions are surrounded by the negative end of water i.e oxygen and anion Cl⁻ is surrounded by the positive end of water i.e hydrogen and in this way all salt is get dissolved.

The chemicals that can dissolve in a certain solvent to create a homogenous mixture known as a solution are said to be soluble chemicals. The compounds that are soluble are: $KNO_3$ , $AgNO_3$ , and $CuBr_2$ .

As per this,

Nitrate ( $NO^{3-$ ) salts are often soluble in water. $KNO_3$ is potassium nitrate.
The majority of nitrate ( $NO^{3-$ ) salts, including silvernitrate, are soluble in water, including $AgNO_3$ .
The majority of bromide ( $Br^-$ ) salts, including copper(II) bromide, are water soluble.

Insoluble:

Lead(II) chloride, or $PbCl_2$ , is an exception and is regarded as being insoluble in water among the chloride ( $Cl^-$ ) salts.
Barium sulphate, also known as BaSO4, is an exception to the rule of most sulphate ( $SO^{4-$ ) compounds being soluble in water.

Thus, these are the classification of the compounds as per their solubility.

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Your question seems incomplete, the prpbable complete question is:

Predict whether the following compounds are soluble or insoluble in water. Soluble Insoluble PbCl2, BaSO4, KNO3, AgNO3, and CuBr2.

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