Answer:
0.93 M
Step-by-step Explanation:
First, we have to calculate the molar mass (MM) of ZnSO₄ by using the molar mass of each chemical element:
MM(ZnSO₄) = 65.4 g/mol Zn + 32 g/mol S + (16 g/mol x 4) = 161.4 g/mol
Then, we divide the mass of ZnSO₄ into its molar mass to obtain the number of moles:
moles ZnSO₄ = mass/MM = 150 g/(161.4 g/mol)= 0.93 mol
Since the molarity of a solution expresses the number of moles of solute per liter of solution, we calculate the molarity (M) as follows:
M = moles ZnSO₄/1 L = 0.93 mol/1 L = 0.93 M
For the following list of acids, rank the acids in strength from weakest acid to strongest acid.
a. FCH2OH
b. F2CHOH
c. CH3OH
d. F3COH
Answer:
CH3OH < FCH2OH < F2CHOH < F3COH
Explanation:
Let us recall that, for a carboxylic acid, the dissociation of the acid yields;
RCOOH ⇄RCOO^- + H^+
The ease of dissociation and release of the hydrogen ion depends on the nature of the group designated R.
When R is is a highly electronegative element, the -I inductive effect causes the hydrogen to become less tightly held by the C-Cl bond.
As the number of electron withdrawing substituents increaseses, the acid ionizes much more and becomes stronger.CH3OH < FCH2OH < F2CHOH < F3COH
Hence, the order of decreasing acid strength is;
Each set of quantum numbers to the correct sub shell description
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.
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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4) In water, Vanillin, C8H8O3, has a solubility of 0.070 moles of vanillin per liter of solution at 25C. What will be produced if 5.00 g of vanillin are added to 1 L of water at 25 C
Answer:
An unsaturated solution.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to firstly realize we need to calculate the grams of vanillin in 0.070 moles by using its molar mass of 152.15 g/mol:
[tex]m=0.070mol*\frac{152.15 g}{1mol} =10.65g[/tex]
Thus, since the solubility is 10.65 g per 1 L of solution, we can notice 5.00 g will complete dissolve and produce an unsaturated solution.
Best regards!
In the given range,at what temperature does oxy gen have the highest solubility?
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.
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
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.
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
Please help me order these bonds urgent
Answer:
From least polar covalent to most polar covalent;
S-I< Br-Cl < N-H< Te-O
From most ionic to least ionic
Cs-F> Sr-Cl> Li- N> Al-O
Explanation:
Electro negativity refers to the ability of an atom in a bond to attract the shared electrons of the bond towards itself.
Electro negativity difference between two atoms is a key player in the nature of bond that exists between any two atoms. A large difference in electron negativity leads to an ionic bond while an intermediate difference in electro negativity leads to a polar covalent bond.
Based on electro negativity differences, the bonds in the answer have been arranged in order of increasing polar covalent nature or decreasing ionic nature.
Elements that have the same number of electron rings are ?
Answer:
are in the same orbital
Explanation:
Answer:
are in the same orbit
Explanation:
Write the separation scheme for the isolation of triphenylmethanol from the reaction mixture once the reaction is complete. The separation begins after the addition of HCl and water to the reaction and includes the column chromatography procedure to further purify crude triphenylmethanol isolated in the day 1 procedure.
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A covalent bond is formed by the following process
Answer:
Covalent bonding occurs when pairs of electrons are shared by atoms.
Explanation:
Atoms will covalently bond with other atoms in order to gain more stability, which is gained by forming a full electron shell. By sharing their outer most (valence) electrons, atoms can fill up their outer electron shell and gain stability.
5.60g of glyceraldehydes was dissolved in 10ml of a solvent and placed in a 50mm cell if the rotation is 1.74 calculate the specific rotation?
Answer:
6.214 degrees-mL/gdm
Explanation:
The specific rotation α' = α/LC where α = observed rotation, L = length of tube and C = concentration of solution.
Given that α = 1.74, L = length of cell = 50 mm = 0.50 dm and C = m/V where m = mass of glyceraldehyde = 5.60 g and V = volume = 10 ml
So, C = m/V = 5.60 g/10 ml = 0.560 g/ml
Since α' = α/LC
substituting the values of the variables into the equation, we have
α' = α/LC
α' = 1.74/(0.50 dm × 0.560 g/ml)
α' = 1.74/(0.28 gdm/l)
α' = 0.006214 °mL/gdm
α' = 6.214 °mL/gdm
α' = 6.214 degrees-mL/gdm
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
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.
What enzyme below is an exoenzyme?
A. Casease
B. Citrase
C. Catalase
D. Oxidase
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.
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 chemical reaction in a bomb calorimeter evolves 3.86 kJ of energy in the form of heat. If the temperature of the bomb calorimeter increases by 4.17 K, what is the heat capacity of the calorimeter?
Answer:
925.66 J/K
Explanation:
Applying,
Q = CΔt............. Equation 1
Where Q = amount of heat, C = heat capacity of the calorimeter, Δt = rise in temperature.
make C the subject of the equation
C = Q/Δt.............. Equation 2
From the question,
Given: Q = 3.86 kJ = 3860 J, Δt = 4.17K
Substitute into equation 2
C = 3860/4.17
C = 925.66 J/K
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
Determine the molecular formula of a compound if it is composed of 40.92% carbon, 4.58% hydrogen, and 54.50% oxygen. The molar mass is 176.12 g/mol.
A) C3H8O3.
B) CH2O.
C) C2H3O2.
D) C3H4O3.
Answer:
No correct answer listed. See explanation for defense.
Explanation:
Given
C: 40.92% => 40.92g/100wt => (40.92/12)moles C = 3.41 moles O
H: 4.58% => 4.58g/100wt => (4.58/1)moles H = 4.58 moles H
O: 54.50% => 54.5g/100wt => (54.5/16)moles O = 3.41 moles O
Empirical ratio => C : H: O => (3.41/3.41) : (4.58/3.41) : (3.41/3.41) => 1 : 1.34 : 1
=> C : H : O => 3(1 : 1.34 : 1) => 3 : 4 : 3 => Empirical Formula C₃H₄O₃
Molecular Weight = Empirical Formula Wt x N
176.12 = 88 x N
N = whole number multiple of empirical formula = 176.12/88 = 2
∴ Molecular Formula => (C₃H₄O₃)₂ => C₆H₈O₆
Note => Only ionic compounds (salts) have subscripts reduced to lowest whole number ratios. Molecular compounds as C₆H₈O₆ are not reduced to lowest whole number ratios. Therefore, there is no correct answer in the answer choice list for the 'Molecular Formula'. Doc :-)
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...
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
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
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]
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]
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
Water put into a freezer compartment in the same refrigerator goes into a state of less molecular disorder when it freezes. Is this an exception to the entropy principle
Answer:
No it is not an exception to this principle
Explanation:
Work was carried out by this compressor to reduce the entropy of ice. What this means is that the ice gave out heat which is as a result of the work that the compressor was putting in. there are violations of this principle
the entropy principle has that the entropy of the universe is always going to be more than 0 (system + surrounding). in this question, the that of the system is negative while that of the surrounding is positive. As the refrigerator was cooling the water, the air outside was getting heated. Outside this refrigerator, the gain in entropy is more than the entropy that was lost in the water.
the entropy of the universe once again is more than 0.
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
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 ]
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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Hazmat products warnings or labels allowed in fc
Answer:
The Hazmat products warnings or labels allowed in fc include:
1. Fully Regulated Aerosol Placard
2. Fully Regulated Flammable Solid Placard
3. Fully Regulated Flammable
4. Lithium-Ion/Metal Battery label
Explanation:
Hazmat products (including explosives, flammable liquids and solids, and gases, etc.) are classified as dangerous substances and materials that pose a risk to people during their storage, handling, or transportation. The requirement for this Hazmat classification is to show that the identified products require diligence, carefulness, and alertness in handling, transporting, and storing them. The reason for this is that mishaps can occur. Some of them can also cause fire outbreaks.
