Answer:
25.5mL of 0.100M NaOH are needed to reach buffer capacity.
Explanation:
The buffer capacity is reached when the ratio between moles of conjugate base (Sodium acetate) and moles of weak acid (Acetic acid) is 10:
Moles sodium acetate / Moles Acetic acid = 10
The reaction of acetic acid, HA, with NaOH, to produce sodium acetate, NaA is:
HA + NaOH → H2O + NaA
That means the moles of NaOH added = Moles of HA that are being subtracted and moles of NaA that are been produced.
The initial moles of each species is:
Acetic acid:
23.34mL = 0.02334L * (0.147mol / L) = 0.00343 moles Acetic Acid
Sodium Acetate:
33.66mL = 0.03366L * (0.185mol / L) = 0.00623 moles Sodium Acetate
We can write the moles of each species when NaOH is added as:
Moles sodium acetate / Moles Acetic acid = 10
0.00623 moles + X / 0.00343 moles - X = 10
Where X are moles of NaOH added
Solving for X:
0.00623 moles + X = 0.0343 moles - 10X
11X = 0.0281
X = 0.00255 moles of NaOH are needed
In Liters:
0.0255mol NaOH * (1L / 0.100mol) = 0.0255L of 0.100M NaOH are needed =
25.5mL of 0.100M NaOH are needed to reach buffer capacity
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.
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]
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.
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.
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.
Answer:
Explanation:
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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
If a 0.320 mM solution of MnO41- has an absorbance of 0.480 at 525 nm in a 1.000 cm cell. What is the concentration of a MnO41- solution that has absorbance of 0.490 in the same cell at that wavelength?
Answer:
Hence the concentration of a MnO41- solution that has absorbance of 0.490 in the same cell at that wavelength is 0.3266.
Explanation:
Now A = el, el=const
Then,
[tex]A2 / A1 = C2/ C1\\\\0.49/ 0.48 = C2 / 0.32\\\\C2 = 0.3266[/tex]
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.
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
It is found that, when a dilute gas expands quasistatically from 0.40 to 5.0 L, it does 210 J of work. Assuming that the gas temperature remains constant at 300 K, how many moles of gas are present
Answer:
[tex]n=0.033mole[/tex]
Explanation:
From the question we are told that:
Initial volume [tex]V_1=0.40L[/tex]
Final Volume[tex]V_2=5.0L[/tex]
Work [tex]W=210J[/tex]
Temperature [tex]T=300k[/tex]
Generally the equation for Ideal gas is mathematically given by
[tex]W=nRTIn\frac{V_2}{V_1}[/tex]
[tex]n=\frac{W}{RTIn\frac{V_2}{V_1}}[/tex]
[tex]n=\frac{210}{8.32*300In\frac{5.0}{0.4}}[/tex]
[tex]n=0.033mole[/tex]
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
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:
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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.
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 many of the electrons in a molecule of ethane are not involved in bondind
Ethane consists of 6C−H bonds and 1C−C bond. Total number of bonds is 7. Each bond is made up of two electrons
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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)
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.
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)?
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
How much heat capacity, in joules and in calories, must be added to a 75.0-g iron block with a specific heat of 0.499J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C?
Answer:
56511.75 J
13506.3 Calories
Explanation:
Applying,
Q = cm(t₂-t₁).................. Equation 1
Where Q = amount of heat, m = mass of the iron, c = specific heat capacity of the iron, t₁ = initial temperature, t₂ = final temperature.
From the question,
Given: m = 75 g, c = 0.499 J/g.°C, t₂ = 1535°C, t₁ = 25°C
Substitute these values into equation 1
Q = 75(0.499)(1535-25)
Q = 75(0.499)(1510)
Q = 56511.75 J
Q in Calories is
Q = (56511.75×0.239)
Q = 13506.3 Calories
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)
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
What is true of all matter?
A. It pushes or pulls on objects.
B. You can see it.
C. It gives off heat energy.
D. It has mass.
What is "X" in the following reaction?
Read the chemical equation.
N2 + 3H2 - 2NH3
Using the volume ratio, determine how many liters of NH3 is produced if 4.2 liters of H2 reacts with an excess of N2
if all measurements are taken at the same temperature and pressure?
A 2.8 liters
B 3.2 liters
C 5.4 liters
D 6.3 liters
Answer:
A 2.8 liters
Explanation:
Step 1: Write the balanced equation
N₂ + 3 H₂ ⇄ 2 NH₃
Step 2: Establish the appropriate volume ratio
At the same temperature and pressure, the volume ratio of H₂ to NH₃ is 3:2.
Step 3: Calculate the volume of ammonia produced from 4.2 L of hydrogen
4.2 L H₂ × 2 L NH₃/3 L H₂ = 2.8 L
any two functions of crystals
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]
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.
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.
A one electron species, X m, where m is the charge of the one electron species and X is the element symbol, loses its one electron from its ground state when it absorbs 3.49 x 10-17 J of energy. Using the prior information, the charge of the one electron species is:_____________
a. +8
b. +2
c. +3
d. +1
e. +4
Answer:
Option C
Explanation:
From the question we are told that:
Difference in energy [tex]\delta E =3.49 * 10^{-17} J[/tex]
The Ground state Difference in energy at n=1
[tex]\delta E_g = 2.18 * 10^{-18} × Z^2[/tex]
Generally the equation for Difference in energy is mathematically given by
[tex]\delta E=\delta E_g[/tex]
Therefore
[tex]3.49 * 10^{-17} = 2.18 * 10^{-18} * Z^2[/tex]
[tex]Z^2=16[/tex]
[tex]Z=4[/tex]
Therefore
Charge on element Z Q_Z
[tex]Q_Z= Atomic\ no\. of\ element - No.\ of\ electrons\ of\ element[/tex]
[tex]Q_Z =4-1[/tex]
[tex]Q_Z=+3[/tex]
Option C
The decomposition of ethyl amine, C2H5NH2, occurs according to the reaction: C2H5NH2(g)⟶C2H4(g)+NH3(g) At 85∘C, the rate constant for the reaction is 2.5 x 10-1 s-1. What is the half-life (in sec) of this reaction?
Answer:
2.772 seconds
Explanation:
Given that;
t1/2 = 0.693/k
Where;
t1/2 = half life of the reaction
k= rate constant
Note that decomposition is a first order reaction since the rate of reaction depends on the concentration of one reactant
t1/2 = 0.693/2.5 x 10-1 s-1
t1/2= 2.772 seconds
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= ________
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]
Viết các đồng phân cấu tạo mạch hở của C4H6O2 cùng nhóm chức axit
Answer:
+ axit
CH2=CH-CH2-COOH,
CH3-CH=CH-COOH (tính cả đồng phân hình học)
CH2=C(CH3)-COOH.
+ este
HCOOCH=CH-CH3 (tính cả đồng phân hình học)
HCOO-CH2-CH=CH2,
HCOOC(CH3)=CH2.
CH3COOCH=CH2
CH2=CH-COOCH3
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.
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!
Inter-molecular forces determine the _______________ properties while intra-molecular forces determine the ________ properties of compounds.
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.
a chemist combines 4.9 g of nitrogen gas with 9.4 grams of nitrogen gas to form 11.4 g of ammonia 2.9 g of nitrogen is remaining
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
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
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.
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.
Determine the equilibrium constant, Keq, at 25°C for the reaction
2Br- (aq) + I2(s) <--> Br2(l) + 2I- (aq)
Eocell = (0.0257/n) lnKeq, Calculate Eocell from Use this equation to calculate K value.
Eo (I2/I-) = +0.53, Eo (Br2/Br-) = +1.07,
Explanation:
The given chemical reaction is:
[tex]2Br^- (aq) + I_2(s) <-> Br_2(l) + 2I^- (aq)[/tex]
[tex]E^ocell=oxidation potential of anode + reduction potential of cathode\\[/tex]
The relation between Eo cell and Keq is shown below:
[tex]deltaG=-RTlnK_e_q\\delta=-nFE^o cell\\=>nFE^o cell=RTlnK_e_q\\lnK_e_q=\frac{nF}{RT} E^o cell[/tex]
The value of Eo cell is:
Br- undergoes oxidation and I2 undergoes reduction.
Reduction takes place at cathode.
Oxidation takes place at anode.
Hence,
[tex]E^ocell= (-1.07+0.53)V\\=-0.54V[/tex]
F=96485 C/mol
n=2 mol
R=8.314 J.K-1.mol-1
T=298K
Substitute all these values in the above formula:
[tex]ln K_e_q=\frac{2mol* 96485 C/mol}{8.314 J.K^-^1.mol^-^1x298K} \\\\lnK_e_q=77.8\\K_e_q=e^7^7^.^8\\=>K_e_q=6.13x10^3^3[/tex]
Answer:
Keq=6.13x10^33
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.
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.
What enzyme below is an exoenzyme?
A. Casease
B. Citrase
C. Catalase
D. Oxidase
