1.) A sample of neon gas at a pressure of 0.646 atm and a temperature of 242 °C, occupies a volume of 515 mL. If the gas is cooled at constant pressure until its volume is 407 mL, the temperature of the gas sample will be ________°C.
2.) A sample of argon gas at a pressure of 0.633 atm and a temperature of 261 °C, occupies a volume of 694 mL. If the gas is heated at constant pressure until its volume is 796 mL, the temperature of the gas sample will be___________°C.
3.) 0.962 mol sample of carbon dioxide gas at a temperature of 20.0 °C is found to occupy a volume of 21.5 liters. The pressure of this gas sample ismm ____________ Hg.

Answers

Answer 1

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.


Related Questions

Ammonia, methane, and phosphorus trihydride are three different compounds with three different boiling points. Rank their boiling points in order from lowest to highest.
A. CH4< NH3 < PH3
B. NH3 < PH3< CH4
C. CH4 < PH3 < NH3
D. NH3 < CH4< PH3
E. PH3< NH3 < CH4

Answers

Answer:

B. NH3 < PH3< CH4

Explanation:

Hello,

In this case, taking into account that the boiling point of ammonia, methane and phosphorous trihydrate are -33.34 °C , -161.5 °C  and -87.7 °C , clearly, methane has the lowest boiling point (most negative) and ammonia the greatest boiling point (least negative), therefore, ranking is:

B. NH3 < PH3< CH4

Best regards.

g Which ONE of the following pairs of organic compounds are NOT pairs of isomers? A) butanol ( CH3-CH2-CH2-CH2-OH ) and diethyl ether ( CH3–CH2–O–CH2–CH3 ) B) isopentane ( (CH3)2-CH-CH2-CH3 ) and neopentane ( (CH3)4C ) C) ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 ) D) acrylic acid ( CH2=CH-COOH ) and propanedial ( OHC–CH2–CHO ) E) trimethylamine ( (CH3)3N ) and propylamine ( CH3-CH2-CH2-NH2 )

Answers

Answer:

ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 )

Explanation:

Isomers are compounds that have the same molecular formula but different structural formulas. Hence any pair of compounds that can be represented by exactly the same molecular formula are isomers of each other.

If we look at the pair of compounds; ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 ), one compound has molecular formula, C2H7ON while the other has a molecular formula, C2H5ON, hence they are not isomers of each other.

Solid cesium bromide has the same kind of crystal structure as CsCl which is pictured below: If the edge length of the unit cell is 428.7 pm, what is the density of CsBr in g/cm3.

Answers

Answer:

[tex]\mathbf {density \ d =4.4845 \ g/cm^3}[/tex]

Explanation:

Let recall the crystal structure of CsBr obtains a BCC structure. In a BCC structure, there exist only two atom per cell.

The density d of CsBr in g/cm³ can be calculated by using the formula:

[tex]\mathtt{ density \ d = \dfrac{z \times molar\ mass \ (M)}{ edge \ length \ (a) \ \times avogadro's \ number \ (N)}}[/tex]

where;

z = 1 mole of CsBr

edge length = 428.7 pm = (4.287 × 10⁻⁸)³ cm

molar mass of CsBr = 212.81 g/mol

avogadro's number = 6.023 × 10²³

[tex]\mathtt{ density \ d = \dfrac{1 \times 212.81}{(4.287 \times 10^{-8})^3 \times 6.023 \times 10^{23}}}[/tex]

[tex]\mathtt{ density \ d = \dfrac{ 212.81}{47.4540533}}[/tex]

[tex]\mathbf {density \ d =4.4845 \ g/cm^3}[/tex]

Which of the following contains a nonpolar covalent bond?
O A. Co
B. NaCl
O C. 02
O D. HE

Answers

I think the answer is C. 02

Answer:

The answer is o2

Explanation:

I took the test

What is the balanced equation for the reaction of aqueous cesium sulfate and aqueous barium perchlorate?

Answers

Answer:

The balanced chemical reaction is given as:

[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]

Explanation:

When aqueous cesium sulfate and aqueous barium perchlorate are mixed together it gives white precipitate barium sulfate and aqueous solution od cesium perchlorate.

The balanced chemical reaction is given as:

[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]

According to reaction, 1 mole of cesium sulfate reacts with 1 mole of barium perchlorate to give 1 mole of a white precipitate of barium sulfate and 2 moles of cesium perchlorate.

Write the equations that represent the first and second ionization steps for sulfuric acid (H2SO4) in water.

Answers

Answer:

[tex]H_2SO_4(aq)\rightarrow H^+(aq)+HSO_4^-\\\\HSO_4^-(aq)\rightarrow H^+(aq)+SO_4^{2-}rightarrow[/tex]

Explanation:

Hello,

In this case, given that the sulfuric acid is a diprotic acid (two hydrogen ions) we can identify two ionization reactions, the first one, showing up the dissociation of the first hydrogen to yield hydrogen sulfate ions and the second one, showing up the dissociation of the hydrogen sulfate ions to hydrogen ions and sulfate ions by separated as shown below:

[tex]H_2SO_4(aq)\rightarrow H^+(aq)+HSO_4^-\\\\HSO_4^-(aq)\rightarrow H^+(aq)+SO_4^{2-}[/tex]

They are have one-sensed arrow, since sulfuric acid is a strong acid.

Regards.

The equations that represent the first and second ionization steps for sulfuric acid in water are H₂SO₄→HSO₄+H⁺ & HSO₄⁻→SO₄⁻+H⁺ respectively.

What is ionization reaction?

Ionization reactions are those reactions in which atom or molecule will convert into ion by bearing a positive or negative charge on itself.

In water in the following way ionization of sulphuric acid takes place:

In the first ionization step one hydrogen atom (H⁺) will loose from the sulphuric acid molecule as:

        H₂SO₄ → HSO₄⁻ + H⁺

In the second ionization step another hydrogen atom will also loose and we get the sulphate ion (SO₄⁻) and one proton (H⁺) as:

        HSO₄⁻ → SO₄⁻ + H⁺

Hence, two steps are shown above.

To know more about ionization reaction, visit the below link:
https://brainly.com/question/1445179

When titrating a strong acid with a strong base, after the equivalence point is reached, the pH will be determined exclusively by: Select the correct answer below:
A) hydronium concentration
B) hydroxide concentration
C) conjugate base concentration
D) conjugate acid concentration

Answers

Answer:

B) hydroxide concentration

Explanation:

Hello,

In this case, since we are talking about strong both base and acid, since the base is the titrant and the acid the analyte, once the equivalence point has been reached, some additional base could be added before the experimenter realizes about it, therefore, since the titrant is a strong base, it completely dissociates in hydroxide ions and metallic ions which allows us to compute the pOH of the solution by known the hydroxide ions concentration.

After that, due to the fact that the pH is related with the pOH as shown below:

pH=14-pOH

We can directly compute the pH.

Best regards.

What is the ph of 0.36M HNO3 ?

Answers

Answer:

0.44

Explanation:

We know that the pH of any acid solution is given by the negative logarithm of its hydrogen ion concentration. Hence, if I can obtain the hydrogen ion concentration of any acid, I can obtain its pH.

For the acid, HNO3, [H^+] = [NO3^-]= 0.36 M

pH= -log [H^+]

pH= - log[0.36]

pH= 0.44

Calculate the molarity of a solution containing 29g of glucose (C 6 H 12 O 6 ) dissolved in 24.0g of water. Assume the density of water is 1.00g/mL.

Answers

Answer:

whats the ph  ofpoh=9.78

Explanation:

A piece of plastic sinks in oil but floats in water. Place these three substances in order from lowest density to greatest density.

Answers

Answer:

[tex]\rho _{oil}<\rho _{plastic}<\rho _{water}[/tex]

Explanation:

Hello,

In this case, since water and oil are immiscible due to the oil's nonpolarity and water's polarity, when mixed, the oil remains on the water since it is less dense than water. In such a way, for a plastic sunk in the oil and floating on the water (in middle of them) we can conclude that the plastic have a mid density, therefore, the required organization is:

[tex]\rho _{oil}<\rho _{plastic}<\rho _{water}[/tex]

Best regards.

Im really confused and select all that apply questions scare me.

Answers

Answer:

The 3rd one

Explanation:

15. Calculate the critical angle of glass and water combination. Show your calculation. 16. What is the critical angle for the interface between Mystery A and glass

Answers

Answer:

15. Critical angle of glass and water combination, θ = 62.45°

16. Critical angle for the interface between Mystery A and glass, θ = 37.93°

Note; The question is incomplete. The complete question is as follows:

Medium Air Water Glass Mystery A Mystery B Table-2 Speed (m/s) 1.00 C 0.75 c 0.67 0.41 c 0.71 c n 1.00 1.33 1.50 Index of Refraction n of a given medium is defined as the ratio of speed of light in vacuum, c to the speed of light in a medium, v. n = c/v

Table-4: Incident Angle (degrees) Reflected Angle Refracted angle (degrees) (degrees) % Intensity of reflected ray 0 10 20 30 40 50 N/A N/A N/A 30 40 50 0 11.3 22.7 34.2 46.3 59.5 N/A N/A N/A 0.67 1.22 3.08 % Intensity of refracted ray 100 100 100 99.33 98.78 96.92

When rays travel from a denser medium to a less dense medium, we can define a critical angle of incidence θ such that refracted angle θ₂ = 90°. Applying Snell's law: Critical angle θ = sin-1(n₂/n₁).

When the angle of incidence is greater than the critical angle, 100% of the light intensity is reflected. This is called total internal reflection because all the light is reflected.

15. Calculate the critical angle of glass and water combination. Show your calculation.

16. What is the critical angle for the interface between Mystery A and glass?

Explanation:

15.  Applying Snell's law; Critical angle θ = sin-1(n₂/n₁).

where n₂,refractive index of water = 1.33, n₁, refractive index of glass = 1.50 since glass is denser than water

θ = sin-1(1.33/1.50)

θ = 62.45°

Critical angle of glass and water combination, θ = 62.45°

16.  Refractive index of mystery A , n = c/v

where v = 0.41 c

therefore, n = c / 0.41 c = 2.44

Critical angle for the interface between Mystery A and glass, θ = sin-1(n₂/n₁).

where n₂,refractive index of glass = 1.50, n₁, refractive index of mystery A = 2.44 since mystery A is denser than glass as seen from its refractive index

θ = sin-1(1.50/2.44)

θ = 37.93°

Critical angle for the interface between Mystery A and glass, θ = 37.93°

If the theoretical yield of a reaction is 332.5 g and the percent yield for the reaction is 38 percent, what's the actual yield of product in grams? \

A. 8.74 g
B. 12616 g
C. 116.3 g
D. 126.4 g

Answers

Answer: D - 126.4g

Explanation:  

% Yield = Actual Yield/Theoretical Yield

38% = Actual Yield/332.5

38/100 = Actual Yield/332.5

(.38)(332.5) = 126.35 g = 126.4 g Actual Yield

Answer:

is D. the correct answer

Explanation:

I'm not sure if it is. Please let me know if I'm mistaking.

3,3-dibromo-4-methylhex-1-yne​

Answers

Explanation:

see the attachment. hope it will help you...

Heterocyclic aromatic compounds undergo electrophilic aromatic substitution in a similar fashion to that undergone by benzene with the formation of a resonance-stabilized intermediate. Draw all of the resonance contributors expected when the above compound undergoes bromination

Answers

Answer:

See explanation

Explanation:

When we talk about electrophilic substitution, we are talking about a substitution reaction in which the attacking agent is an electrophile. The electrophile attacks an electron rich area of a compound during the reaction.

The five membered furan ring is aromatic just as benzene. This aromatic structure is maintained during electrophilic substitution reaction. The attack of the electrophile generates a resonance stabilized intermediate whose canonical structures have been shown in the image attached.

What creation would you make if you had access to any of the chemical elements? can someone answer this for me please.​

Answers

Answer:

Depends on the chemicals.

Explanation:

The creation of something depends on the chemical we have. If I have chemicals such as acid and base so I can produce salt and water by mixing acid and base. If I have ethanol, vegetable oil and sodium hydroxide which is  a catalyst, I can produce biodiesel which can be used in vehicles instead of fossil fuel. If I have sodium element and chlorine gas, I can produce sodium chloride also known as table salt.

Which of the following provides a characteristic of
MgO(s) with a correct explanation?
Choose 1 answer:
А
It is hard because its ions are held together by strong
electrostatic attractions.
B
It is malleable because its atoms can easily move past
one another without disrupting the bonding.
It is a poor conductor of electricity because its
electrons are tightly held within covalent bonds and
lone pairs.
It has a high melting point because its molecules
interact through strong intermolecular forces.

Answers

Answer:

А It is hard because its ions are held together by strong electrostatic attractions.

B It is malleable because its atoms can easily move past one another without disrupting the bonding.  

Explanation:

These are correct explanations of the properties of magnesium.

C is wrong. Mg is a good conductor of electricity and it has metallic bonds.

D is wrong. Mg has no molecules. It has no intermolecular forces.

9
What might happen if acidic chemicals were emitted into
the air by factories? Choose the best answer.
A
The acid would destroy metallic elements in the air
B
The acid would be neutralized by bases within clouds
C
Acid rain might destroy ecosystems and farmland
D
Violent chemical reactions would take place within the
atmosphere
co search
O
BI

Answers

Your answer is B. Acid rain might destroy ecosystems and farmland

Select the true statement concerning voltaic and electrolytic cells. Select one: a. Voltaic cells involve oxidation-reduction reactions while electrolytic cells involve decomposition reactions. b. Voltaic cells require applied electrical current while electrolytic cells do not. . c. all electrochemical cells, voltaic and electrolytic, must have spontaneous reactions. d. Electrical current drives nonspontaneous reactions in electrolytic cells.

Answers

Answer:

Electrical current drives nonspontaneous reactions in electrolytic cells.

Explanation:

Electrochemical cells are cells that produce electrical energy from chemical energy.

There are two types of electrochemical cells; voltaic cells and electrolytic cells.

A voltaic cell is an electrochemical cell in which electrical energy is produced from spontaneous chemical process while an electrolytic cell is an electrochemical cell where electrical energy is produced from nonspontaneous chemical processes. Current is needed to drive these nonspontaneous chemical processes in an electrolytic cell.

Answer:

electrolytic cells generate electricity through a non-spontaneous reaction while voltaic cells absorb electricity to drive a spontaneous reaction.

Explanation:

Answer via Educere/ Founder's Education

(a) Identify the name of the method used below for the separation.
(b) Give one more application of this method of separation.
(c) What is the name for the line at position B ?
(d) what conclusions can you draw about the colours present in sweets C and D ?

Answers

Answer:

(a) Chromatography

(b) DNA fingerprinting

(c) Origin

(d) Sweet C consists of more colours than sweet D.

  ii. The speed of colours in sweet C are proportional to one another, while that of colours in D is not.

Explanation:

Chromatography is one of the physical method of separating mixtures. This process composed of the ability of the constituents in a mixture to separate by virtue of rate of movement through a medium, thus separates into constituents.

It can be used to determine the soluble constituents of a given mixture. And for purification purpose.

What is the final volume V2 in milliliters when 0.551 L of a 50.0 % (m/v) solution is diluted to 23.5 % (m/v)?

Answers

Answer:

[tex]V_2=1.17L[/tex]

Explanation:

Hello,

In this case, for dilution processes, we must remember that the amount of solute remains the same, therefore, we can write:

[tex]V_1C_1=V_2C_2[/tex]

Whereas V accounts for volume and C for concentration that in this case is %(m/v). In such a way, the final volume V2 turns out:

[tex]V_2=\frac{V_1C_1}{C_2}= \frac{0.551L*50.0\%}{23.5\%}\\ \\V_2=1.17L[/tex]

Best regards.

place the following substances in Order of decreasing boiling point H20 N2 CO

Answers

Answer:

-195.8º < -191.5º < 100º

Explanation:

Water, or H20, starts boiling at 100ºC.

Nitrogen, or N2, starts boiling at -195.8ºC.

Carbon monoxide, or C0, starts boiling at -191.5ºC.

When we place these in order from decreasing boiling point:

-195.8º goes first, then -191.5º, and 100º goes last.

Answer:

therefore, N2, CO, H20

Decreasing boiling point

Explanation:

the bond existing in H2O is hydrogen bond

bond existing in N2 is covalent bond, force existing is dipole-dipole-interaction

bond existing in CO is covalent bond , force existing between is induced -dipole- induced dipole-interaction

hydrogen bond is the strongest , followed by dipole-dipole-interaction and induced -dipole- induced dipole-interaction

the stronger the bond , the higher the boiling point

therefore, N2, CO, H20

-------------------------------------->

Decreasing boiling point

Consider these metal ion/metal standard reduction potentials Cd2+(aq)|Cd(s) Zn2+(aq)|Zn(s) Ni2+(aq)|Ni(s) Cu2+(aq)|Cu(s) Ag+(aq)|Ag(s) -0.40 V -0.76 V ‑0.25 V +0.34 V +0.80 V Based on the data above, which species is the best reducing agent?

Answers

Answer:

The best reducing agent is Zn(s)

Explanation:

A reducing agent must to be able to reduce another compound, by oxidizing itself. Consequently, the oxidation potential must be high. The oxidation potential of a compound is the reduction potential of the same compound  with the opposite charge. Given the reduction potentials, the best reducing agent will be the compound with the most negative reduction potential. Among the following reduction potentials:

Cd₂⁺(aq)|Cd(s) ⇒ -0.40 V

Zn²⁺(aq)|Zn(s) ⇒ -0.76 V

Ni²⁺(aq)|Ni(s) ⇒‑0.25 V

Cu²⁺(aq)|Cu(s) ⇒ +0.34 V

Ag⁺(aq)|Ag(s) ⇒ +0.80 V

The most negative is Zn²⁺(aq)|Zn(s) ⇒ -0.76 V

From this, the most reducing agent is Zn. Zn(s) is oxidized to Zn²⁺ ions with the highest oxidation potential (0.76 V).

Calculate the amount of heat that must be absorbed by 10.0 g of ice at –20°C to convert it to liquid water at 60.0°C. Given: specific heat (ice) = 2.1 J/g·°C; specific heat (water) = 4.18 J/g·°C; ΔH fus = 6.0 kJ/mol.

Answers

Answer:

The amount of heat to absorb is 6,261 J

Explanation:

Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.

The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.

So:

Heat required to raise the temperature of ice from -20 °C to 0 °C

Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).

In this case, m= 10 g, specific heat of the ice= 2.1 [tex]\frac{J}{g*C}[/tex] and ΔT=0 C - (-20 C)= 20 C

Replacing: Q= 10 g*2.1 [tex]\frac{J}{g*C}[/tex] *20 C and solving: Q=420 J

Heat required to convert 0 °C ice to 0 °C water

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

In this case, being 1 mol of water= 18 grams: Q= 10 g*[tex]6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}[/tex]= 3.333 kJ= 3,333 J (being kJ=1,000 J)

Heat required to raise the temperature of water from 0 °C to 60 °C

In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 [tex]\frac{J}{g*C}[/tex] and ΔT=60 C - (0 C)= 60 C

Replacing: Q= 10 g*4.18 [tex]\frac{J}{g*C}[/tex] *60 C and solving: Q=2,508 J

Finally, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

The amount of heat to absorb is 6,261 J

The amount of heat to absorb is 6,261 J.

Calculation for heat:

Heat required to raise the temperature of ice from -20 °C to 0 °C.

The formula for specific heat is used to calculate the amount of heat

Q = c * m * ΔT

Where,

Q =heat exchanged by a body,

m= mass of the body

c= specific heat

ΔT= change in temperature

Given:

m= 10 g,

specific heat of the ice= 2.1

ΔT=0 C - (-20 C)= 20 C

On substituting the values:

Q= 10 g*2.1  *20 C

Q=420 J

Heat required to convert 0 °C ice to 0 °C water.

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

Heat required to raise the temperature of water from 0 °C to 60 °C

m= 10 g,

Specific heat of the water= 4.18  

ΔT=60 C - (0 C)= 60 C

On substituting:

Q= 10 g*4.18  *60 C

Q=2,508 J

Thus, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

The amount of heat to absorb is 6,261 J

Find more information about Specific heat here:

brainly.com/question/13439286

When we react a weak acid with a strong base of equal amounts and concentration, the component of the reaction that will have the greatest effect on the pH of the solution is:______.
a. the acid.
b. the base.
c. the conjugate acid.
d. the conjugate base.

Answers

Answer:

d. the conjugate base.

Explanation:

The general reaction of a weak acid, HA, with a strong base YOH, is:

HA + YOH → A⁻ + H₂O + Y⁻

Where A⁻ is the conjugate base of the weak acid and Y⁻ usually is a strong electrolyte.

That means after he complete reaction you don't have weak acid nor strong base, just conjugate base that will be in equilibrium with water, thus (Strong electrolyte doesn't change pH:

A⁻ + H₂O ⇄ HA + OH⁻

As the equilibrium is producing OH⁻, the pH of the solution is being affected for the conjugate base

Right option:

d. the conjugate base.

Write a balanced chemical equation for the base hydrolysis of methyl butanoate with NaOH. (Use either molecular formulas or condensed structural formulas, but be consistent in your equation.)

Answers

Explanation:

C5H10O2 + NaOH = C2H5COONa + C2H5OH

your result are : sodium propanoate and ethanol

A balanced chemical equation represents atoms and their numbers with their charge. The balanced equation for base hydrolysis is C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH.

What is hydrolysis?

Base hydrolysis is the splitting of the ester linkage by the basic molecule. As the result the acidic ester portion makes the salt, and also alcohol is produced as the by-product.

The base hydrolysis of methyl butanoate is shown as,

C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH

Here, sodium propanoate and ethanol are produced by the splitting of methyl butanoate in the presence of the base (NaOH).

Therefore, C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH is balanced reaction.

Learn more about hydrolysis here:

https://brainly.com/question/22078321

#SPJ2

Refer to the figure.
30. How many planes are shown in the figure?
31. How many planes contain points B, C, and E?
32. Name three collinear points.
3. Where could you add point G on plane N
so that A, B, and G would be collinear?
4. Name a point that is not coplanar with
A, B, and C.
5. Name four points that are coplanar.
BN

Answers

Answer:

  30. 5 planes are shown

  31. 1 plane

  32. CEF

  33. on line AB

  34. E or F

   35. ABCD or BCEF or CDEF or ACEF

Explanation:

30. Each of the surfaces of the rectangular pyramid is a plane. There are 5 planes.

__

31. 3 points define one plane only.

__

32. The only points shown on the same line segment are points E, F, and C.

__

33. If G is to be collinear with A and B, it must lie on line AB.

__

34. The only points shown that are not on plane N are points E and F. Either of those will do.

__

35. There are three planes that have 4 points shown on them. The four points that are on the same plane are any of ...

ABCDBCEFCDEF

Plane ACEF is not shown on the diagram, but we know that those 4 points are also coplanar. (Any point not on line CE, together with the three points on that line, will define a plane with 4 coplanar points.)

A hypothetical metal crystallizes with the face-centered cubic unit cell. The radius of the metal atom is 198 picometers and its molar mass is 195.08 g/mol. Calculate the density of the metal in g/cm3.

Answers

Answer:

7.38 g/cm³ is the density of the metal

Explanation:

In a Face-centered cubic unit cell you have 4 atoms. Also, the edge length is √8×r (r is radius of the atom).

To solve this problem, we need first to calculate the volume of the unit cell and then, with molar mass calculate the mass of 4 atoms. As density is the ratio between mass and volume we can obtain this value.

Volume of the unit cell

Volume = a³

a = √8×r

(r = 198x10⁻¹²m)

a = 5.6x10⁻¹⁰ m

Volume = 1.756x10⁻²⁸ m³

1m = 100cm → 1m³ = (100cm)³:

1.756x10⁻²⁸ m³× ((100cm)³ / 1m³) =

1.756x10⁻²² cm³ → Volume of the unit cell in cm³Mass of the unit cell:

There are 4 atoms of gold:

4 atoms × (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles of gold

As 1 mole weighs 195.08g:

6.64x10⁻²⁴ moles of gold × (195.08g / mol) =

1.296x10⁻²¹g is the mass of the unit cellDensity of the metal:

1.296x10⁻²¹g / 1.756x10⁻²² cm³ =

7.38 g/cm³ is the density of the metal

The density of the metal is 7.40 g/cm³

In cubic crystal system, face-centered cubic FFC is the name given to sort of atom arrangement observed in which structure is made up of atoms organized in a cube with a portion of an atom in each corner and six extra atoms in the center of each cube face.

It is expressed by using the formula:

[tex]\mathbf{\rho = \dfrac{Z \times M}{N_A\times a^}}[/tex]

where;

[tex]\rho[/tex] = density of the metalZ = atoms coordination no = 4 (for FCC)Molar mass (M) = 195.8 g/molAvogadro's constant (NA) = 6.022 × 10²³ /mola = edge length

For face-centered cubic FFC;

The edge length  [tex]\mathbf{a =2 \sqrt{2}\times r }[/tex]

[tex]\mathbf{a =2 \sqrt{2}\times 198 \ pm }[/tex]

[tex]\mathbf{a =560.0285 \ pm }[/tex]

a = 5.60 × 10⁻⁸ cm

Replacing it into the previous equation, we have:

[tex]\mathbf{\rho = \dfrac{4 \times 195.8}{6.022 \times 10^{23} \times( 5.60 \times 10^{-8} )^3}}[/tex]

[tex]\mathbf{\rho = 7.40\ g/cm^3 }[/tex]

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The half-life of radium-226 is 1620 years. What percentage of a given amount of the radium will remain after 900 years

Answers

Answer:

68%

Explanation:

Since we need a percentage we can use any number we want for our initial value.

5(1/2)^900/1620 = 3.40

(3.40 / 5)*100 = 68%

To make sure lets use a different initial amount

1(1/2)^900/1620 = 0.68

(0.68/1) * 100 = 68%

The percentage of radium that will remain after 900 years is 68%.

To solve this question, we'll assume the initial amount of radium-226 to be 1.

Now, we shall proceed to obtaining the percentage of radium-226 that will after 900 years. This can be obtained as illustrated below:

Step 1

Determination of the number of half-lives that has elapsed.

Half-life (t½) = 1620 years

Time (t) = 900 years

Number of half-lives (n) =?

[tex]n = \frac{t}{t_{1/2}}\\\\n = \frac{900}{1620}\\\\n = \frac{5}{9}[/tex]

Step 2:

Determination of the amount remaining

Initial amount (N₀) = 1

Number of half-lives (n) = 5/9

Amount remaining (N) =?

[tex]N = \frac{N_{0} }{2^{n}}\\\\N = \frac{1}{2^{5/9}}[/tex]

N = 0.68

Step 3

Determination of the percentage remaining.

Initial amount (N₀) = 1

Amount remaining (N) = 0.68

Percentage remaining =?

Percentage remaining = N/N₀ × 100

Percentage remaining = 0.68/1 × 100

Percentage remaining = 68%

Therefore, the percentage amount of radium-226 that remains after 900 years is 68%

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Find the pH of these buffer solutions using the information provided: 1L solution containing 80g of lactic acid (MW

Answers

Answer:

pH of the solution is 2.0

Explanation:

The lactic acid is a weak acid that is in equilibrium with water as follows:

Lactic acid + H2O ⇄ Lactate + H₃O⁺

And Ka for lactic acid: 1.38x10⁻⁴

Ka = 1.38x10⁻⁴ = [Lactate] [H₃O⁺] / [Lactic acid]

Initial concentration of lactic acid is (MW: 112.06g/mol):

80g * (1mol / 112.06g) / 1L = 0.714M

The equilibrium concentration of the species in the equilibrium are:

[Lactate] = X

[H₃O⁺] = X

[Lactic acid] = 0.714-X

Replacing in Ka expression:

1.38x10⁻⁴ = [X] [X] / [0.714-X]

9.8532x10⁻⁵ -  1.38x10⁻⁴X = X²

9.8532x10⁻⁵ -  1.38x10⁻⁴X - X² = 0

Solving for X:

X = -1.0x10⁻². False solution, there is no negative concentrations

X = 9.86x10⁻³M. Right solution.

As [H₃O⁺] = X

[H₃O⁺] = 9.86x10⁻³M

and pH = -log [H₃O⁺] = -log 9.86x10⁻³M

pH = 2.0

pH of the solution is 2.0
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