Answers
Answer:
the repetition principle is important in scientific research, because the observational indexes are random variables,which requires a certain amount of samples to reveal their changing regularity
Related Questions
The homework question reads:
"A sample of gas in a cylinder of volume 3.42 L at 298 K
and 2.57 atm expands to 7.39 L by two different pathways.
Path A is an isothermal, reversible expansion. Path B has two
steps. In the fi rst step, the gas is cooled at constant volume to
1.19 atm. In the second step, the gas is heated and allowed to
expand against a constant external pressure of 1.19 atm until
the final volume is 7.39 L. Calculate the work for each path.
Answers
Answer:
Explanation:
this guy on brainly already did it:
Alleei
Virtuoso
4.8K answers
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Answer : The work done for path A and path B is -685.3 J and -478.1 J respectively.
Explanation :
To calculate the work done for path A :
First we have to calculate the moles of the gas.
where,
= initial pressure of gas = 2.57 atm
= initial volume of gas = 3.42 L
n = moles of gas = ?
R = gas constant = 0.0821 atm.L/mol.K
T = temperature of gas = 298 K
Now put all the given values in the above formula, we get:
According to the question, this is the case of isothermal reversible expansion of gas.
As per first law of thermodynamic,
where,
= internal energy
q = heat
w = work done
As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.
So, at constant temperature the internal energy is equal to zero.
The expression used for work done will be,
where,
w = work done on the system = ?
n = number of moles of gas = 0.359 mole
R = gas constant = 8.314 J/mole K
T = temperature of gas = 298 K
= initial volume of gas = 3.42 L
= final volume of gas = 7.39 L
Now put all the given values in the above formula, we get :
Thus, the work done of path A is, -685.3 J
To calculate the work done for path B :
The formula used for isothermally irreversible expansion is :
where,
w = work done
= external pressure = 1.19 atm
= initial volume of gas = 3.42 L
= final volume of gas = 7.39 L
Now put all the given values in the above formula, we get :
Thus, the work done of path B is, -478.1 J
1. Most of the chemicals included in your General Chemistry Lab kit can be discarded down a drain. Describe a situation in which you would need to neutralize a chemical before discarding down a drain.
Answers
Answer: Chemicals like acids and bases are harmful and must be neutralized before draining.
Explanation:
A strong acid or strong base is required to be diluted or neutralized before it is discarded in the drain as if is discarded without diluting and neutralization it can spill and splash from sink or drain and can harm people in chemistry lab, moreover the fumes of the discarded chemical on spilling can cause respiratory tract burning and can even cause fire hazard so it must be converted into less harmful form and then must be drained.
Inter-molecular forces determine the _______________ properties while intra-molecular forces determine the ________ properties of compounds.
Answers
Answer:
Physical
Chemical
Explanation:
Intermolecular forces are the forces that hold the molecules of a substance together in a particular state of matter. They decide the physical properties of a substance.
The intra molecular forces are the bond forces that hold atoms together in molecules. The nature of this bonding determines the chemical properties of substances.
You should set out support, like a cork ring or clamp, before removing the glassware from a glassware kit to place the glassware in and to stop it from _________. Thoroughly check that the glasswar is________ and that it does not have any _______before using it.
Answers
Answer:
(A) Slipping and breaking
(B) Clean and dry
(C) Cracks
Explanation:
This describes the process of unpacking a glassware for use.
You should set out support like a cork ring or clamp (these are simple machines that'll hold the glassware in place) before removing the glassware from a glassware kit; to place the glassware in and to stop it from slipping and breaking.
Thoroughly check that the glassware is clean and dry and that it does not have any cracks, before using it.
Classify each molecule as an alcohol, ketone, or aldehyde based on its name. Propanone (acetone) Choose... Ethanal Choose... 3-phenyl-2-propenal Choose... Butanone Choose... Ethanol Choose... 2-propanol Choose...
Answers
Answer:
1.) Propanone (ketone)
2.) Ethanal( aldehyde)
3.) 3-phenyl-2-propenal (aldehyde)
4.) Butanone (ketone)
5.) Ethanol ( alcohol)
6.) 2-propanol (alcohol)
Explanation:
In organic chemistry, ALCOHOL ( also known as alkanol) are compounds in which hydroxyl groups are linked to alkyl groups. They can be considered as being derived from the corresponding alkanes by replacing the hydrogen atoms with hydroxyl groups. The hydroxyl group is the functional group of the alcohol as it is responsible for their characteristic chemical properties. A typical example of alcohol is ethanol and 2-propanol.
Alkanals or ALDEHYDES have the general formula RCHO while alkanones or KETONES have the general formula RR'CO where R and R' may be alkyl or aryl groups. The main similarity between these two classes of compounds is the presence of the carbonyl group. In aldehydes, there is a hydrogen atom attached to the carbon In the carbonyl group while there is none on the ketones.
Some common examples of ketones are Propanone, Butanone while examples of aldehydes are Ethanal and 3-phenyl-2-propenal
What direction would equilibrium moves towards based on the following if we increased the volume of the container.
[tex]2A_{(g)} + 5B_{(g)} + 12C_{(g)}[/tex] ↔ [tex]14AC_{(g)} + 5B_{(s)}[/tex]
Answer choices:
a) reactants
b) no change
c) products
d) decrease in volume
Please help!
Answers
To answer this question, we will first find out the number of gaseous moles on each side of the equilibrium
on the left:
we have 2 moles of A, 5 moles of B and 12 moles of C
which gives us a grand total of 19 gaseous moles
on the right:
here, we have 14 moles of AC gas, we will not count the number of moles of B because it's a solid
giving us 14 gaseous moles on the right
Where does the reaction shift?
more gaseous moles means more space taken, because gas likes to fill all the space it can
if we have more volume, more gas can move around without colliding (reacting) with each other
Hence more volume favors the side with more gaseous moles
here, the left has more gaseous moles. So we can say that the reaction will shift towards the left, or the reactants side
Answer:
Explanation:
given reversible chemical reaction:
2A(g) + 5B(g) + 12C(g) ↔ 14AC(g) + 5B(s)
chemicals in solid form do not take up a lot of volume so change in container volume has no effect
look at chemicals in gas form only:
the total no. of moles of reactants in gas form = 2 + 5 + 12 = 19
the total no. of moles of products in gas form = 14
so an increase in volume of the container will favor the reaction direction with higher volume n high volume means higher no. of moles
the ans is the equilibrium will move towards a) reactants
For each of the following compounds, indicate the pH at which 50% of the compound will be in a form that possesses a charge and at which pH more than 99% of the compound will be in a form that possesses a charge.
ClCH2COOH (pKa = 2.86)
CH3CH2NH+3 (pKa = 10.7)
Express your answer using two decimal places
a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.
b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.
c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.
d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.
Answers
Answer:
a. 2..86 b. 4.86 c. 10.7 d. 8.7
Explanation:
a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.
Using the Henderson-Hasselbalch equation,
pH = pKa + log[A⁻]/[HA]
where [A⁻] = concentration of conjugate base (or charged form) and [HA] = concentration of acid.
At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1
So, pH = pKa + log[A⁻]/[HA]
pH = pKa + log1
pH = pKa = 2.86
b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.
Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.99x while the acidic concentration is remaining 1 % (1 - 0.99)x = 0.01x
Using the Henderson-Hasselbalch equation,
pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base (or charged form) = 0.99x and [HA] = concentration of acid = 0.01x.
pH = pKa + log0.99x/0.01x
pH = pKa + log0.99/0.01
pH = 2.86 + log99
pH = 2.86 + 1.996
pH = 4.856
pH ≅ 4.86
c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.
Using the Henderson-Hasselbalch equation,
pH = pKa + log[A⁻]/[HA]
where [A⁻] = concentration of conjugate base and [HA] = concentration of acid.
At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1
So, pH = pKa + log[A⁻]/[HA]
pH = pKa + log1
pH = pKa = 10.7
d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.
Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.01x while the acidic concentration is remaining 99 % (1 - 0.01)x = 0.99x (which possesses the charge).
Using the Henderson-Hasselbalch equation,
pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base = 0.01x and [HA] = concentration of acid = 0.99x.
pH = pKa + log0.01x/0.99x
pH = pKa + log1/99
pH = 10.7 - log99
pH = 10.7 - 1.996
pH = 8.704
pH ≅ 8.7
how many of the electrons in a molecule of ethane are not involved in bondind
Answers
Ethane consists of 6C−H bonds and 1C−C bond. Total number of bonds is 7. Each bond is made up of two electrons
#LETS STUDY#BRAINLEST LOVE❣️
Write a balanced half-reaction for the oxidation of liquid water to aqueous hydrogen peroxide in basic aqueous solution. Be sure to add physical state symbols where appropriate.
Answers
Answer:
O2(g) + 2H2O(l) --------> 2H2O2(aq) + 2e
Explanation:
An oxidation reaction reaction refers to a reaction in which electrons are lost. In this case, we are about to see the full balanced half-reaction for the oxidation of liquid water to aqueous hydrogen peroxide in basic aqueous solution.
The full equation is;
O2(g) + 2H2O(l) --------> 2H2O2(aq) + 2e
So, two electrons were lost in the process.
What is the energy change when 78.0 g of Hg melt at −38.8°C
Answers
Answer:
The correct answer is - 2.557 KJ
Explanation:
In this case, Hg is melting, the process is endothermic, so the energy change will have a positive sign.
we can calculate this energy by the following formula:
Q = met
where, m = mass,
e = specific heat
t = temperature
then,
Q = 78*0.14* (273-38.8)
here 0.14 = C(Hg)
= 2.557 Kj
Calculate the mass of sodium phosphate in aqueous solution to fully react with 37 g of chromium nitrate(III) an aqueous solution?(report answer in grams and only three Sigg figs do not put the unit)
Answers
Answer:
41 g
Explanation:
The equation of the reaction is;
Cr(NO3)3(aq)+Na3PO4(aq)=3NaNO3(s)+CrPO4(aq)
Number of moles of chromium nitrate = 37g/ 146.97 g/mol = 0.25 moles
1 mole of sodium phosphate reacts with 1 mole of chromium nitrate
x moles of sodium phosphate react as with 0.25 moles of chromium nitrate
x= 1 × 0.25/1
x= 0.25 moles
Mass of sodium phosphate = 0.25 moles × 163.94 g/mol
Mass of sodium phosphate = 41 g
A uniform plastic block floats in water with 50.0 % of its volume above the surface of the water. The block is placed in a second liquid and floats with 23.0 % of its volume above the surface of the liquid.
What is the density of the second liquid?
Express your answer with the appropriate units.
Answers
Answer:
density of second liquid = 650 kg/m³
Explanation:
Given that:
The volume of the plastic block submerged inside the water = 0.5 V
The force on the plastic block = [tex]\rho_1V_1g[/tex]
[tex]= 0.5p_1 V_g[/tex]
when the block is floating, the weight supporting the force (buoyancy force) is:
W [tex]= 0.5p_1 V_g[/tex]
[tex]\rho Vg = 0.5p_1 V_g[/tex]
[tex]\rho = 0.5 \rho _1[/tex]
where;
water density [tex]\rho _1[/tex] = 1000
[tex]\rho = 0.5 (1000)[/tex]
[tex]\rho = 500 kg/m^3[/tex]
In the second liquid, the volume of plastic block in the water = (100-23)%
= 77% = 0.7 V
The force on the plastic block is:
[tex]= 0.77p_2 V_g[/tex]
when the block is floating, the weight supporting the force (buoyancy force) is:
[tex]W = 0.77p_2 V_g[/tex]
[tex]\rho Vg = 0.77 \rho_2 V_g \\ \\ \rho = 0.77 \rho_2 \\ \\ 500 = 0.77 \rho_2 \\ \\ \rho_2 = 500/0.77[/tex]
[tex]\mathbf{ \rho_2 \simeq 650 \ kg/m^3}[/tex]
The compound IF5 contains Question 16 options: polar covalent bonds with partial negative charges on the F atoms. ionic bonds. polar covalent bonds with partial negative charges on the I atoms. nonpolar covalent bonds.
Answers
Answer:
See explanation
Explanation:
The molecule IF5 possesses five I-F polar bonds. However, the presence of polar bonds does not automatically imply that the molecule will be polar.
The geometry of the molecule is very important in determining the polarity of a compound. Since IF5 has a lone pair of electrons, the molecule is bent and as such there is a permanent dipole moment created in the molecule thereby making IF5 polar in nature.
Write a net ionic equation for the overall reaction that occurs when aqueous solutions of carbonic acid and sodium hydroxide are combined. Assume excess base.
Answers
Answer:
[tex]H_2CO_3(aq)+2OH^-(aq)\rightarrow (CO_3)^{2-}(aq)+2H_2O(l)[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to set up this net ionic equation, by firstly setting up the complete molecular equation as follows:
[tex]H_2CO_3(aq)+2NaOH(aq)\rightarrow Na_2CO_3(aq)+2H_2O(l)[/tex]
Thus, since carbonic acid is weak it merely ionizes whereas sodium hydroxides ionizes for the 100 % as it is strong; thus, we can write the complete ionic equation:
[tex]H_2CO_3(aq)+2Na^+(aq)+2OH^-(aq)\rightarrow 2Na^+(aq)+(CO_3)^{2-}(aq)+2H_2O(l)[/tex]
Whereas sodium ions act as the spectator ones to be cancelled out for us to obtain:
[tex]H_2CO_3(aq)+2OH^-(aq)\rightarrow (CO_3)^{2-}(aq)+2H_2O(l)[/tex]
Regards!
once formed, how are coordinate covalent bonds different from other covalent bonds?
Answers
Answer:
[tex]\boxed {\boxed {\sf {One \ atom \ donates \ both \ electrons \ in \ a \ pair}}}[/tex]
Explanation:
A covalent bond involves the sharing of electrons to make the atoms more stable, and so they satisfy the Octet Rule (8 valence electrons).
Typically each atom contributes an electron to form an electron pair. This is a single bond. There are also double bonds (two pairs of electrons), triple bonds (three pairs of electrons), and coordinate covalent bonds.
Sometimes, to satisfy the Octet Rule and achieve stability, one atom contributes both of the electrons in an electron pair. This is different from other covalent bonds because usually each of the 2 atoms contributes an electron to make a pair.
Gaseous BF3 and BCl3 are mixed in equal molar amounts. All B-F bonds have about the same bond enthalpy, as do all B-Cl bonds. Compare the numbers of microstates to explain why the mixture tends to react to form BF2Cl(g) and BCl2F(g
Answers
Solution :
[tex]$BF_3 (g) + BCl_3 (g) \rightarrow BF_2 Cl + BCl_F(g)$[/tex]
Explanation 1 :
Spontaneity of the reaction is based on two factors :
-- the tendency to acquire a state of minimum energy
-- the energy of a system to acquire a maximum randomness.
Now, since there isn't much difference in the bond enthalpies of B-F and B-Cl. So, we can say the major driving factor is tendency to acquire a state of maximum randomness.
Explanation 2 :
A system containing the [tex]\text{"chemically mixed"}[/tex] B halides has a [tex]\text{greater entropy}[/tex] than a system of [tex]$BCl_3$[/tex] and [tex]BF_3[/tex].
It has the same number of [tex]\text{gas phase molecules}[/tex], but more distinguishable kinds of [tex]\text{molecules}[/tex], hence, more microstates and higher entropy.
In the given range,at what temperature does oxy gen have the highest solubility?
Answers
Solubility of oxygen and oxygen compounds
Water solubility of oxygen at 25oC and pressure = 1 bar is at 40 mg/L water. In air with a normal composition the oxygen partial pressure is 0.2 atm. This results in dissolution of 40 . 0.2 = 8 mg O2/L in water that comes in contact with air.
What enzyme below is an exoenzyme?
A. Casease
B. Citrase
C. Catalase
D. Oxidase
Answers
Calculate the molarity of a 17.5% (by mass) aqueous solution of nitric acid. Select one: a. 2.74 m b. 4.33 m c. 0.274 m d. 3.04 m e. The density of the solution is needed to solve the problem.
Answers
Answer:
Option e.
Explanation:
Molarity is the concentration that indicates moles of solute in 1 L of solution.
We have another concentration, percent by mass.
Percent by mass indicates mass of solute in 100 g of solution.
Our solute is HNO₃, our solvent is water.
17.5 g of nitric acid is the mass of solute. We can convert them to moles:
17.5 g . 1mol / 63g = 0.278 moles
We do not have volume of solution. We assume the mass is 100 g because the percent by mass but we need density to state the volume.
Density = Mass / Volume
Mass / Density = Volume
Once we have the volume, we need to be sure the units is in L, to determine molarity
M = mol /L
any two functions of crystals
Answers
Answer:
1. Participating in calcium homeostatis storage of calcium.
2. High capacity calcium (Ca) regulation and protection against herbivory
[tex]\large \boxed{\sf 2 \: functions \: of \: crystals \: are :- } [/tex]
_________________
⟹
[tex] \sf \: \underline{ Calcium \: oxalate \: (CaOx) \: crystals} \: are \: distributed \: \\\sf among \: all \: taxonomic \: levels \\ \sf\: of \: photosynthetic \: organisms \: from \\ \sf \: small \: algae \: to \: angiosperms \: and \: giant \: gymnosperms .[/tex]
__________________
⟹
[tex]\sf Bone \: is \: mostly \: made \: of \: \underline{mineral \: crystals} \: \\ \sf and \: the \: protein \: collagen. \: The \: mineral \: crystals \: bone \\ \sf\: provide \: strength \: and \: rigidity \: for \: the \: matrix \: upon \: \\ \sf \: and \: within \: which \: they \: are \: deposited.[/tex]
A system receives 425 J of heat from and delivers 425 J of work to its surroundings. What is the change in internal energy of the system (in J)?
Answers
Answer:
0 J
Explanation:
Applying,
ΔE = q+w................ Equation 1
Where ΔE = change in internal energy of the system, q = Heat of the system, w = work of the system.
Note: q is positive, while w is negative
From the question,
Given: q = 425 J, w = -425 J
Substitute these values into equation 1
ΔE = 425-425
ΔE = 0 J
Hence the change in internal energy of the system is 0 J
Use the Ka values for weak acids to identify the best components for preparing buffer solutions with the given pH values.
Name Formula Ka
Phosphoric acid H3PO4 7.5 x 10^-3
Acetic acid CH3COOH 1.8 x 10^-5
Formic acid HCOOH 1.8 x 10^-4
pH 1.9 =_________
pH 5.0 = ________
pH 3.9= ________
Answers
Answer:
pH= 1.9 then [tex]H_{3} PO_{4}[/tex]
pH = 5.0 , [tex]CH_{3} COOH[/tex]
pH = 3.9 , HCOOH
As we know range left [tex]pH= pKa+/- 1[/tex]
A hypothetical A-B alloy of composition 53 wt% B-47 wt% A at some temperature is found to consist of mass fractions of 0.5 for both and phases. If the composition of the phase is 92 wt% B-8 wt% A, what is the composition of the phase
Answers
Answer:
the composition of the ∝ phase C∝ = 14 or [ 14 wt% B-86 wt% A ]
Explanation:
Given the data in the question;
Co = 53 or [ 53 wt% B-47 wt% A ]
W∝ = 0.5 = Wβ
Cβ = 92 or [ 92 wt% B-8 wt% A ]
Now, lets set up the Lever rule for W∝ as follows;
W∝ = [ Cβ - Co ] / [ Cβ - C∝ ]
so we substitute our given values into the expression;
0.5 = [ 92 - 53 ] / [ 92 - C∝ ]
0.5 = 39 / [ 92 - C∝ ]
0.5[ 92 - C∝ ] = 39
46 - 0.5C∝ = 39
0.5C∝ = 46 - 39
0.5C∝ = 7
C∝ = 7 / 0.5
C∝ = 14 or [ 14 wt% B-86 wt% A ]
Therefore, the composition of the ∝ phase C∝ = 14 or [ 14 wt% B-86 wt% A ]
What is "X" in the following reaction?
Answers
Sometimes in lab we collect the gas formed by a chemical reaction over water . This makes it easy to isolate and measure the amount of gas produced.
Suppose the CO, gas evolved by a certain chemical reaction taking place at 50.0°C is collected over water, using an apparatus something like that in the sketch, and the final volume of gas in the collection tube is measured to be 132. mL. Calculate the mass of CO, that is in the collection tube. Round your answer to 2 significant digits.
Answers
Answer:
0.17 g
Explanation:
Since the volume of gas collected is 132 mL, we need to find the number of moles of gas present in 132 mL.
So, number of moles, n = volume of gas, v/molar volume, V
n = v/V where v = 132 mL = 0.132 L and V = 22.4 L
So, substituting the values of the variables into the equation, we have
n = v/V
n = 0.132 L/22.4 L
n = 0.005893 mol
We then need to calculate the molar mass of CO, M = atomic mass of carbon + atomic mass of oxygen = 12 g/mol + 16 g/mol = 28 g/mol
Also, number of moles of gas, n = m/M where m = mass of CO and M = molar mass of CO
m = nM
m = 0.005893 mol × 28 g/mol
m = 0.165004 g
m ≅ 0.17 g to 2 significant digits
How can a Bose-Einstein condensate be formed? A. B super-heating a gas. B. By super-cooling certain types of solid. C. By super-cooling certain types of plasma. D. By super-heating a plasma
Answers
Answer:
C. By super-cooling certain types of plasma.
Explanation:
Bose-Einstein condensate is a state of matter whereby atoms or particles become cooled to a very low energy state leading to their condensation to give a single quantum state.
Note that plasma refers to atoms that have had some or even all of its electrons stripped away leaving only positively charged ions. Simply put, plasma is ionized matter.
When certain types of plasma are super cooled, Bose-Einstein condensate are formed.
42 Organic compound may have names ending in -ane, -ene, -ol or -oic acid. How many of these endings indicate the compounds contain double bonds in their molecules? * (1 Point)
Answers
Answer: Organic compounds ending with the name (-ene) indicate that the compounds contain double bonds in their molecules.
Explanation:
Organic compounds are those molecules that contains carbon atoms (as their main element), hydrogen and oxygen which are usually present. The presence of numerous organic compounds is due to the following properties of carbon:
--> the exceptional ability of carbon atoms to catenate, that is, to combine with one another to form straight chains, branched chains or ring compounds containing many carbon atoms.
--> The ease with which carbon combines with hydrogen, oxygen, Nitrogen and halogens
--> The ability of carbon atoms to form single, DOUBLE or triple bonds.
The organic compound that has the name ending with -ene are known as the alkenes. The members of the alkene series are formed from the alkanes by the removal of two hydrogen atoms and the introduction of a DOUBLE BOND in the carbon chain. They are named after the corresponding alkanes by changing the -ane ending to -ene.
Note: the systematic name of a compound is formed from the root hydrocarbon by adding a suffix and prefixes to denote the substitution of the hydrogen atoms.
Classify the processes as endothermic or exothermic.
a. Ice melting
b. Water condensing on surface
c. Baking a cake
d. The chemical reaction inside an instant cold pack.
e. A car using gasoline
Answers
endothermic absorbs heat
exothermic gives heat
a. endothermic
b. exothermic
c. endothermic
d. exothermic
a. Ice melting - endothermic
b. Water condensing on the surface - exothermic
c. Baking a cake - endothermic
d. The chemical reaction inside an instant cold pack - endothermic
e. A car using gasoline - exothermic
What is an exothermic and endothermic reaction?An exothermic reaction can be described as a thermodynamic chemical reaction that emits energy from the system to its surroundings usually in the form of light, heat, or sound.
While an endothermic reaction can be described as an opposite of an exothermic reaction where the energy gains in the form of heat. In exothermic chemical reactions, the bond energy is transformed into thermal energy.
In exothermic reactions, the reaction happens the form of the kinetic energy of molecules when the energy is released. The release of energy is due to the electronic transition of electrons from one energy level to another.
The burning of gasoline, and water condensation is also an exothermic reaction in which energy is released while ice melting and baking cake is an endothermic reaction.
Learn more about the exothermic process, here:
brainly.com/question/12321421
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A student prepares a aqueous solution of acetic acid . Calculate the fraction of acetic acid that is in the dissociated form in his solution. Express your answer as a percentage. You will probably find some useful data in the ALEKS Data resource.
Answers
Answer:
10.71%
Explanation:
The dissociation of acetic acid can be well expressed as follow:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:
Then:
The I.C.E Table is expressed as follows:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Initial 0.0014 0 0
Change - x +x +x
Equilibrium (0.0014 - x) x x
Recall that:
Ka for acetic acid CH₃COOH = 1.8×10⁻⁵
∴
[tex]K_a = \dfrac{[x][x]]}{[0.0014-x]}[/tex]
[tex]1.8*10^{-5} = \dfrac{[x][x]]}{[0.0014-x]}[/tex]
[tex]1.8*10^{-5} = \dfrac{[x]^2}{[0.0014-x]}[/tex]
[tex]1.8*10^{-5}(0.0014-x) = x^2[/tex]
[tex]2.52*10^{-8} -1.8*10^{-5}x = x^2[/tex]
[tex]2.52*10^{-8} -1.8*10^{-5}x - x^2 =0[/tex]
By rearrangement:
[tex]- x^2 -1.8*10^{-5}x +2.52*10^{-8}= 0[/tex]
Multiplying through by (-) and solving the quadratic equation:
[tex]x^2 +1.8*10^{-5}x-2.52*10^{-8}= 0[/tex]
[tex](-0.00015 + x) (0.000168 + x) =0[/tex]
x = 0.00015 or x = -0.000168
We will only consider the positive value;
so x=[CH₃COO⁻] = [H⁺] = 0.00015
CH₃COOH = (0.0014 - 0.00015) = 0.00125
However, the percentage fraction of the dissociated acetic acid is:
[tex]= \dfrac{ 0.00015}{0.0014}\times 100[/tex]
= 10.71%
A sample of oxygen occupies 1.00 L. If the temperature remains constant, and the pressure on the oxygen is decreased to one third the original pressure, what is the new volume
Answers
Answer:
3.00 L
Explanation:
P₁V₁ = P₂V₂
V₁ = 1.00 L
P₁ = (x) atm
P₂ = [tex]\frac{1}{3}[/tex] · (P₁) = [tex]\frac{x}{3}[/tex]
V₂ = unknown
(x atm)(1.00 L) = ( [tex]\frac{x}{3}[/tex] atm)(V₂)
divide both sides by ( [tex]\frac{x}{3}[/tex] atm)
( 1.00x )( [tex]\frac{3}{x}[/tex] ) = V₂
x cancels out
(1.00)(3) = V₂
V₂ = 3.00 L
