g Calculate the concentration of sulfate ion when BaSO4 just begins to precipitate from a solution that is 0.0758 M in Ba2+.

Answers

Answer 1

Answer:

1.42 × 10⁻⁹ M

Explanation:

Step 1: Given data

Concentration of Ba²⁺ ([Ba²⁺]): 0.0758 MSolubility product constant of BaSO₄ at 25°C: 1.08 × 10⁻¹⁰

Step 2: Write the reaction for the solution of BaSO₄

BaSO₄(s) ⇄ Ba²⁺(aq) + SO₄²⁻(aq)

Step 3: Calculate the concentration of SO₄²⁻

We will use the following expression.

Ksp = 1.08 × 10⁻¹⁰ = [Ba²⁺] × [SO₄²⁻]

[SO₄²⁻] = 1.08 × 10⁻¹⁰ / [Ba²⁺] = 1.08 × 10⁻¹⁰ / 0.0758 = 1.42 × 10⁻⁹ M


Related Questions

Qualitatively estimate the relative melting points for each of the solids, and rank them in decreasing order.
Rank from highest to lowest melting point. To rank items as equivalent, overlap them.
sodium chloride
graphite
solid ammonia

Answers

Answer:

Graphite> sodium chloride> solid ammonia

Explanation:

Melting points of solids has a lot to do with the nature of intermolecular forces in the solid. A substance melts when the intermolecular forces holding the crystal lattice has been overcome such that that the crystal structure of the solid just collapses.

Graphite consists of covalently bonded layers of carbon atom which form a giant lattice. The melting point of graphite is very high because of the fact that the strong covalent bonds that hold the carbon atoms together in the layers require a lot of heat energy to break. Grapoghite melts at about 3600°C

Sodium chloride is an ionic compound that melts at about 801°C. The lattice is composed of alternate sodium and chloride ions.

Solid ammonia is held together by much weaker intermolecular interaction hence it has a melting point of about −77.73 °C.

Determine the volume occupied by 10 mol of helium at 27 ° C and 82 atm

please.

Answers

Answer:

3.00 L

Explanation:

Convert the pressure to Pascals.

P = 82 atm × (101325 Pa/atm)

P = 8,308,650 Pa

Convert temperature to Kelvins.

T = 27°C + 273

T = 300 K

Use ideal gas law:

PV = nRT

(8,308,650 Pa) V = (10 mol) (8.314 J/mol/K) (300 K)

V = 0.00300 m³

If desired, convert to liters.

V = (0.00300 m³) (1000 L/m³)

V = 3.00 L

Answer:

[tex]\large \boxed{\text{3.0 L}}[/tex]

Explanation:

[tex]\begin{array}{rcl}pV &=& nRT\\\text{82 atm} \times V & = & \text{10 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{300.15 K}\\82V & = & \text{246 L}\\V & = & \textbf{3.0 L} \\\end{array}\\\text{The volume of the balloon is $\large \boxed{\textbf{3.0 L}}$}[/tex]

what are the similarities between amorphous solid and crystalline solid

Answers

Answer:

solid dont know

Explanation:

so sorry ask another

It takes 242. kJ/mol to break a chlorine-chlorine single bond. Calculate the maximum wavelength of light for which a chlorine-chlorine single bond could be broken by absorbing a single photon. Round your answer to 3 significant digits. single by absorbing a significant digit.

Answers

Answer:

495nm

Explanation:

The energy of a photon could be obtained by using:

E = hc / λ

Where E is energy of a photon, h is Planck's constant (6.626x10⁻³⁴Js), c is speed of the light (3x10⁸ms⁻¹) and λ is wavelength.

The energy to break 1 mole of Cl-Cl bonds is 242kJ = 242000J. The energy yo break a single bond is:

242000J/mol ₓ (1mol / 6.022x10²³bonds) = 4.0186x10⁻¹⁹J/bond.

Replacing in the equation:

E = hc / λ

4.0186x10⁻¹⁹J = 3x10⁸ms⁻¹ₓ6.626x10⁻³⁴Js / λ

λ = 4.946x10⁻⁷m

Is maximum wavelength  of light that could break a Cl-Cl bond.

Usually, wavelength is given in nm (1x10⁻⁹m / 1nm). The wavelength in nm is:

4.946x10⁻⁷m ₓ (1nm / 1x10⁻⁹m) =

495nm

An aqueous solution of cobalt(II) fluoride, , is made by dissolving 6.04 grams of cobalt(II) fluoride in sufficient water in a 200. mL volumetric flask, and then adding enough water to fill the flask to the mark. What is the weight/volume percentage of cobalt(II) fluoride in the solution

Answers

Answer:

[tex]w/v\%=3.02\frac{g}{mL} \%[/tex]

Explanation:

Hello,

In this case, we first define the formula for the calculation of weight/volume percentage considering cobalt (II) fluoride as the solute, water the solvent and the both of them as the solution:

[tex]w/v\%=\frac{mass_{solute}}{V_{solution}}*100\%[/tex]

In such a way, since the mass of the solute is given as 6.04 g and the final volume of the solution 200 mL, the weight/volume percentage turns out:

[tex]w/v\%=\frac{6.04g}{200mL}*100\%\\\\w/v\%=3.02\frac{g}{mL} \%[/tex]

Regards.

For each reaction, write the chemical formulae of the oxidized reactants in the space provided. Write the chemical formulae of the reduced reactants.
reactants oxidized _____
reactants reduced _____
a. 2Fe(s)+3Pb(NO3)2(aq)→3Pb(s)+2Fe(NO3)3(aq)
b. AgNO3(aq)+Cu(s)→2Ag(s)+CuNO)2(a)
c. 3AgNO(aq)+Al()→3Ags)+Al(NO3)3(aq)

Answers

Answer:

a. Oxidized: Fe(s)

Reduced: Pb(NO3)2

b.Oxidized: Cu(s)

Reduced: AgNO3

c. Oxidized: Al(s)

Reduced: AgNO3

Explanation:

In a redox reaction, one reactant is been oxidized whereas the other is reduced.. The reduced reactant is the one that is gaining electrons and the oxidized one is loosing electrons.

In the reactions:

a. 2Fe(s)+3Pb(NO3)2(aq)→3Pb(s)+2Fe(NO3)3(aq)

The Fe is as reactant as Fe(s) (Oxidation state 0) and the product is +3 (Because NO3, nitrate ion, is always -1). That means Fe is oxidized. The Pb as reactant is +2 and as product 0 (Gaining 2 electrons). Pb(NO3)2 is reduced

b. 2AgNO3(aq)+Cu(s)→2Ag(s)+Cu(NO3)2(a)

AgNO3 is +1 and Ag(s) is 0. AgNO3 is reduced. Cu(s) is 0 as reactant and +2 as product. Cu(s) is been oxidized

c. 3AgNO3(aq)+Al(s)→3Ag(s)+Al(NO3)3(aq)

Here, in the same way, AgNO3 is +1 as reactant and 0 as product. AgNO3 is reduced. And Al(s) is 0 as reactant but + 3 as product. Al(s) is oxidized.

What are the concentrations of [K+], [OH-], [CO32-] and [H+], in a 1.2 M solution of K2CO3 ? (Note: Question is asking for concentrations and not pH) g

Answers

Answer:

The concentrations are: [K⁺] = 1.2 M, [OH⁻] = 0.016 M, [CO₃²⁻] = 1.18 M and [H⁺] = 6.25x10⁻¹³ M.

Explanation:

The dissociation equation of K₂CO₃ in water is:

K₂CO₃(aq) ⇄  K⁺(aq) + CO₃²⁻(aq)     (1)

Also, the CO₃²⁻ will react with water as follows:

CO₃²⁻(aq) + H₂O(l) ⇄ HCO₃⁻(aq) + OH⁻(aq)    (2)

                         

The constant of the reaction (2) is:    

[tex] Kb = \frac{[OH^{-}][HCO_{3}^{-}]}{[CO_{3}^{-2}]} = 2.08 \cdot 10^{-4} [/tex]

The solution of K₂CO₃ is 1.2 M, and since the mole ratio of K₂CO₃ with K⁺ and CO₃²⁻ is 1:1, then we have:                      

[tex] [K_{2}CO_{3}] = [K^{+}] = [CO_{3}^{-2}] = 1.2 M [/tex]

Now, from equation (2) we have:

CO₃²⁻(aq) + H₂O(l) ⇄ HCO₃⁻(aq) + OH⁻(aq)    (3)

1.2 - x                                x               x

[tex] 2.08 \cdot 10^{-4} = \frac{[OH^{-}][HCO_{3}^{-}]}{[CO_{3}^{-2}]} [/tex]

[tex] 2.08 \cdot 10^{-4} = \frac{x^{2}}{1.2 - x} [/tex]

[tex] 2.08 \cdot 10^{-4}*(1.2 - x) - x^{2} = 0 [/tex]    (4)  

By solving equation (4) for x we have:

x = 0.016 M = [HCO₃⁻] = [OH⁻]        

Hence, the CO₃²⁻ concentration is:                        

[CO₃²⁻] = 1.2 M - 0.016 M = 1.18 M

Finally, the concentration of [H⁺] is:

[tex] [H^{+}][OH^{-}] = 10^{-14} [/tex]

[tex][H^{+}] = \frac{10^{-14}}{[OH^{-}]} = \frac{10^{-14}}{0.016} = 6.25 \cdot 10^{-13} M[/tex]      

Therefore, the concentrations are: [K⁺] = 1.2 M, [OH⁻] = 0.016 M, [CO₃²⁻] = 1.18 M and [H⁺] = 6.25x10⁻¹³ M.

I hope it helps you!


When the owners of some wells in Pallerla started using high-powered motors to
draw water from the wells, the owners of other wells noticed that their wells were
drying up. Discuss the possible solution to the problem solutions to the problem​

Answers

Answer:

The possible solution is to balance the rate of water removal from the well to the rate of natural recharge of the well from its underground aquifer.

Explanation:

A well is an excavation in the earth, made with the aim of extracting water from the aquifers. The water from a well can be drawn up by the means of a pump, containers, such as buckets, or by hand. Aquifers can also be recharged through a well.

Well draw down occurs when water from the well is drained faster than it is naturally recharged from the aquifer. This can be as a result of over pumping, extended drought, among other factors. The use of the high-powered motor in this case, for pumping, might be the possible cause of the well drying up. The situation might have resulted from the pump drawing out water from the well at a rate tat exceeds the rate at which it is recharged naturally, causing the well water to start drying up. There's also a possibility that the well is pumped indiscriminately, possibly leading to wastage of water.

The solution to this problem is to give the well a time duration for it to recharge itself. Then, the rate of recharges should be calculated and determined by an hydrologist. When all these is done, a pump with a motor power that does not exceed the calculated recharge rate should be used in place of the high-powered motor. Also, water usage should be brought to the minimum level to prevent unnecessary pumping due to excessive, wasteful use of water.

If a radioactive isotope of thorium (atomic number 90, mass number 232) emits 6 alpha particles and 4 beta particles during the course of radioactive decay, what is the mass number of the stable daughter product?

Answers

Answer:

The mass number of the stable daughter product is 208

Explanation:

First thing's first, we have to write out the equation of the reaction. This is given as;

²³²₉₀Th → 6 ⁴₂α  +  4 ⁰₋₁ β + X

In order to obtain the identity of X, we have to obtain it's mass numbers and atomic number.

There is conservation of matter so we expect the mass number to remain the same in both the reactant and products.

Mass Number

Reactant = 232

Product = (6* 4 = 24) + (4 * 0 = 0) + x = 24 + x

since reactant = product

232 = 24 + x

x = 232 - 24 = 208

Atomic Number

Reactant = 90

Product = (6* 2 = 12) + (4 * -1 = -4) + x = 8 + x

since reactant = product

90 = 8 + x

x = 90 - 8 = 82

A buffered solution containing dissolved aniline, C6H5NH2, and aniline hydrochloride, C6H5NH3Cl, has a pH of 5.57 . A. Determine the concentration of C6H5NH+3 in the solution if the concentration of C6H5NH2 is 0.200 M. The pKb of aniline is 9.13. g

Answers

Answer:

[C₆H₅NH₃⁺] = 0.0399 M

Explanation:

This excersise can be easily solved by the Henderson Hasselbach equation

C₆H₅NH₃Cl → C₆H₅NH₃⁺  + Cl⁻

pOH = pKb + log (salt/base)

As we have value of pH, we need to determine the pOH

14 - pH = pOH

pOH = 8.43  (14 - 5.57)

Now we replace data:

pOH = pKb + log ( C₆H₅NH₃⁺/  C₆H₅NH₂ )

8.43 = 9.13 + log (  C₆H₅NH₃⁺ / 0.2 )

-0.7 = log (  C₆H₅NH₃⁺ / 0.2 )

10⁻⁰'⁷ = C₆H₅NH₃⁺ / 0.2

0.19952 = C₆H₅NH₃⁺ / 0.2

C₆H₅NH₃⁺ = 0.19952 . 0.2  = 0.0399 M

A solution of malonic acid, H2C3H2O4, was standardized by titration with 0.0990 M NaOH solution. If 20.52 mL mL of the NaOH solution is required to neutralize completely 11.13 mL of the malonic acid solution, what is the molarity of the malonic acid solution

Answers

Answer:

0.0913 M

Explanation:

We'll begin by writing the balanced equation for the reaction.

This is given below:

H2C3H2O4 + 2NaOH —> C3H2Na2O4 + 2H2O

From the balanced equation above, we obtained the following:

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 2

Data obtained from the question include:

Molarity of base, NaOH (Mb) = 0.0990 M

Volume of base, NaOH (Vb) = 20.52 mL

Volume of acid, H2C3H2O4 (Va) = 11.13 mL

Molarity of acid, H2C3H2O4 (Ma) =..?

The molarity of the acid, H2C3H2O4 can be obtained as follow:

MaVa/MbVb = nA/nB

Ma x 11.13 / 0.0990 x 20.52 = 1/2

Cross multiply

Ma x 11.13 x 2 = 0.0990 x 20.52 x 1

Divide both side by 11.13 x 2

Ma = (0.0990 x 20.52)/ (11.13 x 2)

Ma = 0.0913 M

Therefore, the molarity of malonic acid, H2C3H2O4 solution is 0.0913 M

According to the collision theory, all collisions do not lead to reaction. Which choice gives both reasons why not all collisions between reactant molecules lead to reaction? 1. The total energy of two colliding molecules is less than some minimum amount of energy. 2. Molecules cannot react with each other unless a catalyst is present. 3. Molecules that are improperly oriented during collision will not react. 4. Solids cannot react with gases.

Answers

Answer: 3. Molecules that are improperly oriented during collision will not react.

Explanation:

According to the collision theory , the number of collisions that take place per unit volume of the reaction mixture is called collision frequency. The effective collisions are ones which result into the formation of products.

It depends on two factors:-

1. Energy factor:  For collision to be effective,  the colliding molecules must have energy more than a particular value called as threshold energy.

2. Orientation factor: Also the colliding molecules must have proper orientation at the time of collision to result into formation of products.

Thus not all collisions between reactant molecules lead to reaction because molecules that are improperly oriented during collision will not react.

What are the solutions to the quadratic equation 2x2 + 10x - 48 = 0?

Answers

Answer:

x = 3 , x= -8

Explanation:

[tex]2x^2+10x-48\\=2\left(x^2+5x-24\right)\\x^2+5x-24\\=\left(x^2-3x\right)+\left(8x-24\right)\\=x\left(x-3\right)+8\left(x-3\right)\\=\left(x-3\right)\left(x+8\right)\\=2\left(x-3\right)\left(x+8\right)\\2\left(x-3\right)\left(x+8\right)=0\\x-3=0\\x = 0+3\\x = 3\\x+8=0\\x+8-8=0-8\\x=-8\\x=3,\:x=-8[/tex]

A certain radioactive nuclide has a half life of 1.00 hour(s). Calculate the rate constant for this nuclide. s-1 Calculate the decay rate for 1.000 mole of this nuclide. decays s-1

Answers

Answer:

k= 1.925×10^-4 s^-1

1.2 ×10^20 atoms/s

Explanation:

From the information provided;

t1/2=Half life= 1.00 hour or 3600 seconds

Then;

t1/2= 0.693/k

Where k= rate constant

k= 0.693/t1/2 = 0.693/3600

k= 1.925×10^-4 s^-1

Since 1 mole of the nuclide contains 6.02×10^23 atoms

Rate of decay= rate constant × number of atoms

Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms

Rate of decay= 1.2 ×10^20 atoms/s

Determine the volumes of 0.10 M CH3COOH and 0.10 M CH3COONa required to prepare 10 mL of the following pH buffers: pH 4.7, pH 5.7. (Note: the pKa of CH3COOH

Answers

Answer:

pH 4.7: 5mL of 0.10 M CH3COOH and 5mL 0.10 M CH3COONa

pH 5.7: 0.91mL of 0.10 M CH3COOH and 9.09mL 0.10 M CH3COONa

Explanation:

pKa acetic acid, CH3COOH = 4.7

It is possible to determine pH of a buffer using H-H equation:

pH = pka + log [A⁻] / [HA]

For the acetic buffer,

pH = 4.7 + log [CH3COONa] / [CH3COOH]

As you want a pH 4.7 buffer:

4.7 = 4.7 + log [CH3COONa] / [CH3COOH]

1 = [CH3COONa] / [CH3COOH]

That means you need the same amount of both species of the buffer to make the pH 4.7 buffer. That is:

5mL of 0.10 M CH3COOH and 5mL 0.10 M CH3COONa

For pH 5.7:

5.7 = 4.7 + log [CH3COONa] / [CH3COOH]

1 = log [CH3COONa] / [CH3COOH]

10 = [CH3COONa] / [CH3COOH] (1)

That means you need 10 times [CH3COONa] over [CH3COOH]

And as you know:

10mL=  [CH3COONa] + [CH3COOH] (2)

Replacing (1) in (2):

10 = 10mL + [CH3COOH] / [CH3COOH]

10[CH3COOH] = 10mL + [CH3COOH]

11[CH3COOH] = 10mL

[CH3COOH] = 0.91mL

And [CH3COONa] = 10mL - 0.91mL =

[CH3COONa] = 9.09mL

That is:

0.91mL of 0.10 M CH3COOH and 9.09mL 0.10 M CH3COONa

The volumes according to the pH are as follows:

(i) 5mL of 0.10 M CHCOOH and 5mL 0.10 M CHCOONa for pH 4.7

(ii) 0.91mL of 0.10 M CHCOOH and 9.09mL 0.10 M CHCOONa pH 5.7

Calculating the volume of chemicals needed:

Given that pKa of acetic acid, CH₃COOH = 4.7

The pH of a buffer using the H-H equation is given by:

pH = pKa + log [A⁻] / [HA]

For the acetic buffer,

pH = 4.7 + log [CH₃COONa] / [ CH₃COOH]

4.7 = 4.7 + log [CH₃COONa] / [ CH₃COOH]

0 = log [CH₃COONa] / [ CH₃COOH]

takin antilog on both sides of the equation we get:

1 = [CHCOONa] / [CHCOOH]

It implies that the same amount of both species is needed to make the pH 4.7 buffer.

So,

5mL of 0.10 M CHCOOH and 5mL 0.10 M CHCOONa makes a buffer of pH 4.7

Similarly:

5.7 = 4.7 + log [CH₃COONa] / [CH₃COOH]

1 = log [CH₃COONa] / [CH₃COOH]

takin antilog on both sides of the equation we get:

10 = [CHCOONa] / [CHCOOH]

10[CHCOOH] = [CHCOONa]

It implies that we need 10 times [CH₃COONa] as much of [CH₃COOH]

We have to prepare 10 mL of buffer, so:

10mL=  [CH₃COONa] + [CH₃COOH]

10mL = 11[CH₃COOH]

[CH₃COOH] = 0.91mL

So, [CH₃COONa] = 10mL - 0.91mL

[CHCOONa] = 9.09mL

Therefore,

0.91mL of 0.10 M CH3COOH and 9.09mL 0.10 M CH3COONa is required to make a buffer of pH 5.7

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The overall process that uptakes energy-poor molecules (CO2 and H2O) from their reservoirs in nature and converts them into energy-rich molecules is

Answers

Answer:

Photosynthesis

Explanation:

Photosynthesis is a process by  which photosynthetic organisms use the energy of captured sunlight to convert energy-poor molecules such as carbon (iv), CO₂ and water (H₂O) into energy-rich organic molecules sch as carbohydrates e.g. glucose.

Photosynthesis occurs in a variety of bacteria  and algae as well in vascular plants. The overall equation for the reaction of photosynthesis is as follows:

CO₂ + H₂O + light------->  (CH₂O) + O₂

It is a redox reaction in which  water donates electrons (as hydrogen) for the reduction of  CO₂ to carbohydrate, (CH₂O).

Carbohydrates are energy-rich molecules which serves as energy sources for many living organisms.

The diagram shows two waves.
How do the frequencies of the waves compare?
Wave A has a lower frequency because it has a
smaller amplitude.
Wave A has a higher frequency because it has a
shorter wavelength.
The waves have the same frequency because they
have the same wavelength.
The waves have the same frequency because they
have the same amplitude.

Answers

Answer:

Wave A has a higher frequency because it has a shorter wavelength.

Explanation:

The frequency of a wave and the wave length are related by the following equation:

Velocity (v) = wave length (λ) x frequency (f)

v = λf

If we make frequency (f) the subject of the above equation, we will have:

f = v/λ

Let the velocity (v) be constant.

f = v/λ

f & 1/λ

From the equation above,

We can see that the frequency (f) is inversely proportional to the wavelength (λ).

This implies that a wave with a high frequency, will have a short wavelength and a wave with a short frequency will have a longer wavelength.

Now considering wave A and B in the diagram above,

Wave A will have a higher frequency because it has a shorter wavelength as explained above.

Answer:

it is the second option

Explanation:

Match each property of a liquid to what it indicates about the relative strength of the intermolecular forces in that liquid.

Strong intermolecular forces

Weak intermolecular forces

Answers

Answer:

Strong intermolecular forces:  an increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point.

Weak intermolecular forces: a decrease in viscosity, a decrease in surface tension, an increase in vapor pressure and an increase in boiling point.

Explanation:

Intermolecular forces are forces of attraction or repulsion between neighboring molecules in a substance. These intermolecular forces inclde dispersion forces, dipole-dipole interactions, hydrogen bonding, and ion-dipole forces.

The strength of the intermolecular forces in a liquid usually affects the various properties of the liquid such as viscosity, surface tension, vapour pressure and boiling point.

Strong intermolecular forces in a liquid results in the following; an increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point of the liquid.

Weak intermolecular forces in a liquid results in the following; a decrease in viscosity, a decrease in surface tension, an increase in vapor pressure and an increase in boiling point of that liquid.

Strong intermolecular force is defined as the increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point while  weak intermolecular forces define as the decrease in viscosity, a decrease in surface tension, an increase in vapor pressure, and an increase in boiling point.

Intermolecular forces are forces of attraction or repulsion between neighboring molecules in a substance. These intermolecular forces include as follows:-

Dispersion forcesDipole-dipole interactionsHydrogen bondingion-dipole forces.

Strong intermolecular forces in a liquid result in the following; an increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point of the liquid.

Weak intermolecular forces in a liquid result in the following; a decrease in viscosity, a decrease in surface tension, an increase in vapor pressure, and an increase in the boiling point of that liquid.

For more information, refer to the link:-

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What word or two-word phrase best describes the shape of the water ( H2O ) molecule?

Answers

Answer:

Water (H2O) is an inorganic chemical compound formed by two hydrogen (H) and one oxygen (O) atoms. 3 This molecule is essential in the life of living beings, serving as a medium for the metabolism of biomolecules, is found in nature in its three states and was key to its formation. It is necessary to distinguish between drinking water and pure water, since the first is a mixture that also contains salts in solution; this is why in the laboratory and in other areas distilled water is used.

Explanation:

I hope I've helped

According to the molecular geometry, the V-shape or bent structure best describes the shape of water molecule.

What is molecular geometry?

Molecular geometry can be defined as a three -dimensional arrangement of atoms which constitute the molecule.It includes parameters like bond length,bond angle and torsional angles.

It influences many properties of molecules like reactivity,polarity color,magnetism .The molecular geometry can be determined by various spectroscopic methods and diffraction methods , some of which are infrared,microwave and Raman spectroscopy.

They provide information about geometry by taking into considerations the vibrational and rotational absorbance of a substance.Neutron and electron diffraction techniques provide information about the distance between nuclei and electron density.

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A vehicle travels 2345 meter in 35 second toward the evening sun in the West. What is its speed? A. 47 m/s West

Answers

Explanation:

Speed = 2345 ÷ 35 = 67m/s

The amount of space an object takes up is called _____. gravity weight mass volume

Answers

Volume is the amount of space an object takes up

Which of the following best describes hydrocarbons? a. Alkanes in which a hydrogen atom is replaced by a hydroxyl group b. Binary compounds of carbon and hydrogen c. Organic compounds containing water and carbon d. Covalently bonded carbon compounds which have intermolecular force attractions to hydrogen compounds e. Compounds which are formed by the reaction of a naturally occurring carbon-containing substance and water

Answers

Answer:

b. Binary compounds of carbon and hydrogen

Explanation:

Before proceeding, Hydrocarbons refers to organic chemical compounds composed exclusively of hydrogen and carbon atoms. This means the only elements present in an hydrocarbon are;

- Carbon

- Hydrogen

Looking through the options;

- Option A: This is wrong because the hydroxyl group contains oxygen and hydrocarbons contain only hydrogen and carbon.

- option B: This is correct. Binary compounds refers to compounds with just two elements.

- option C: This is wrong because water contains oxygen and hydrocarbons contain only hydrogen and carbon.

- option D: Carbon atoms can contain other elements so this option is wrong.

- option E: This also wrong because we had already gotten the correct option.

Choose the incorrect statement A. Replacing 3 of the H of LiALH4 with OR groups make it a more reactive reducing agents than NaBH4. B. Ester functional group has a higher priority than aldehyde in nomenclature. C. Ketones are more reactive than esters. D. Aldehydes are less reactive than acyl halides. E. Reactions of aldehydes with (1) NaBH4 and (2) H3O+ form primary alcohols.

Answers

Answer:

Replacing 3 of the H of LiALH4 with OR groups make it a more reactive reducing agents than NaBH4

Explanation:

LiAlH4 and NaBH4 are two well known reducing agents in organic chemistry. These two reducing agents function by transfer of hydrogen to the substrate.

If the hydrogen atoms in LiAlH4 are replaced by the -OR moiety, the new compound will be far less reducing than NaBH4 because the hydrogen atoms necessary to effect the reduction has been removed. Thus, the new compound containing -OR moiety can never be more reducing than NaBH4. This implies that the statement written in the answer is false as written.

What amounts of sodium benzoate would be required to prepare 2.5L of 0.35M benzoic buffer solution with a pH of 6.10? Ka of benzoic acid = 6.5 x 10-5 MW benzoic acid, HC7H5O2, is 122.01 MW sodium benzoate, NaC7H5O2, is 144.01

Answers

Answer:

Benzoic acid: 1.288g

Sodium benzoate: 124.48g

Explanation:

Benzoic acid, HC7H5O2 is in equilibrium with its conjugate base, C7H5O2⁻ producing a buffer. The pH of the buffer can be determined following H-H equation:

pH = pKa + log [C7H5O2⁻] / [HC7H5O2] (1)

Where pH is desire pH = 6.10 pKa is -log Ka = 4.187 and [] are molar concentrations of the buffer.

As you want to prepare 2.5L of a 0.35M of buffer, moles of buffer are:

2.5L ₓ (0.35mol / L) = 0.875moles of buffer.

And you can write:

0.875 moles = [C7H5O2⁻] + [HC7H5O2] (2)

Replacing (2) in (1)

pH = pKa + log [C7H5O2⁻] / [HC7H5O2]

6.10 = 4.187 + log [C7H5O2⁻] / [HC7H5O2]

1.913 =  log [C7H5O2⁻] / [HC7H5O2]

81.846 = 0.875mol - [HC7H5O2] / [HC7H5O2]

81.846 [HC7H5O2] = 0.875mol - [HC7H5O2]

82.846 [HC7H5O2] = 0.875mol

[HC7H5O2] = 0.01056 moles

And moles of the benzoate, [C7H5O2⁻]:

[C7H5O2⁻] = 0.875mol - 0.01056mol =

[C7H5O2⁻] = 0.8644mol

Using molar mass of benzoic acid and sodium benzoate, amount of each compound you must add to prepare 2.5L of the buffer are:

Benzoic acid: 0.01056mol ₓ (122.01g/mol) = 1.288g

Sodium benzoate: 0.8644mol ₓ (144.01g/mol) = 124.482g

Write the net ionic equation for any precipitation reaction that may be predicted when aqueous solutions of manganese(II) nitrate and sodium hydroxide are combined.

Answers

Answer:

Explanation:

Mn( NO₃ )₂ + 2Na OH = Mn( OH)₂ (s) ↓ +  2Na NO₃

Converting into ions

Mn⁺ + 2 NO₃⁻ + 2 Na⁺ + 2 OH⁻ = Mn( OH)₂ + 2 Na⁻ + 2 NO₃⁻

Cancelling out common terms

Mn⁺ + 2 OH⁻ = Mn( OH)₂

this is net ionic equation required.

A student mixes 43.8 mL of acetone (58.08 g/mol, 0.791 g/mL) with excess benzaldehyde and NaOH to produce 79.4 g of (1E,4E)-1,5-diphenylpenta-1,4-dien-3-one (234.29 g/mol). What is the percent yield of this student's experiment

Answers

Answer:

% yield of the student's experiment is

[tex]\frac{0.34}{0.60}[/tex] ˣ 100 = 56.67%

Explanation:

given

volume of acetone= 43.8 mL

molar weight of acetone = 58.08 g/mol

density of acetone = 0.791 g/mL

A student mixes 43.8 mL of acetone (58.08 g/mol, 0.791 g/mL)

43.8 mL = 43.8mL × 0.791g/mL

= 34.6458g ≈34.65g

1 mole of acetone = 58.08g

∴34.65g = 34.65g/58.08g

= 0.60mol

molecular weight of the product 1,5-diphenylpenta-1,4-dien-3-one = 234.29 g/mol

mole = mass/ molar weight

mole = 79.4g/ 234.29g/mol

mole(n) = 0.3389mol ≈ 0.34mol

1 mole of acetone will produce 1 mole of the product

∴0.60mol of acetone will produce 0.60mol of the product

but we get 0.34mol of the product

∴ % yield of the student's experiment is

[tex]\frac{0.34}{0.60}[/tex] ˣ 100 = 56.67%

A researcher places a reactant for decomposition in an expandable reaction chamber and purges the air from the vessel with nitrogen gas. The 500mL reaction vessel is sealed at a pressure of 1.00atm and 390K. If the decomposition reaction was triggered by an electrical shock, producing 3.1g of oxygen gas, what would the volume (L) of the reaction vessel be if the temperature and pressure were kept constant

Answers

Answer:

3.1 L

Explanation:

Step 1: Given data

Pressure (P): 1.00 atmTemperature (T): 390 KMass of oxygen (m): 3.1 gVolume (V): ?

Step 2: Calculate the moles of oxygen

The molar mass of oxygen is 32.00 g/mol.

[tex]3.1g \times \frac{1mol}{32.00g} = 0.097mol[/tex]

Step 3: Calculate the volume of the container

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T / P

V = 0.097 mol × (0.0821 atm.L/mol.K) × 390 K / 1.00 atm

V = 3.1 L

a certain compound was found to contain 54.0 g of carbon and 10.5 grams of hydrogen. its relative molecular mass is 86.0. find the empirical and molecular formulas

Answers

Answer:

empirical formula = C3H7

molecular formula = C6H14

g Increasing the number of unsaturations in a fatty acid ____________ the melting temperature of the fatty acid.

Answers

Answer:

Decreases

Explanation:

Fatty acid which have the double bond or triple bond are called unsaturated fatty acids. Because of the double or triple bond, unsaturated fatty acids are loosely packed and form some distance among molecules which lowers the melting point of unsaturated fatty acids.

So, if the unsaturation of fatty acid will increase, it leads to more branched and loosely packed molecules and decreases the melting temperature accordingly.

What mass of aluminum metal can be produced per hour in the electrolysis of a molten aluminum salt by a current of 21 A? Express your answer using two significant figures.

Answers

Answer

mass of aluminum metal= 7 .0497g of Al

Explanation:

current = 21 A

time = 1 hour = 60 X 60 = 3600 s

quantity of electricity passed = current X time = 21X 3600 = 75600 C

Following the electrolysis the below reaction will occur :

Al3+ + 3e- --------> Al

therefore, 3F i.e. 3 X 96500 C = 289500 C gives 1 mole of Al

so 1 C will produce 1/289500 moles of Al

so 108000 C will produce 1/289500 X 75600 = 0.2611 moles of Al

now 1 mole of aluminium weighs = 27 g/mole

so 0.2611 moles of Al = 0.2611 X 27 = 7 .0497 g

mass of aluminum metal= 7 .0497 g of Al

The mass of aluminum metal can be produced per hour in the electrolysis of a molten aluminum salt by a current of 21 A is 7.05 g

We'll begin by calculating the the quantity of electricity used. This can be obtained as follow:

Current (I) = 21 A

Time(t) = 1 h = 60 × 60 = 3600 s

Quantity of electricity (Q) =?

Q = it

Q = 21 × 3600

Q = 75600 C

Finally, we shall determine the mass of the aluminum metal produced.

Al³⁺ + 3e —> Al

Recall:

1 mole of Al = 27 g

1 electron (e) = 96500 C

Thus,

3 electrons = 3 × 96500 = 289500 C

From the balanced equation above,

289500 C of electricity produced 27 g of Al.

Therefore,

75600 C of electricity will produce = (75600 × 27) / 289500 = 7.05 g of Al

Thus, the mass of the aluminum metal obtained is 7.05 g

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