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.
Methylamine is a base because it can bond to H+. Draw Lewis structures to show how methylamine reacts with water to form a hydroxide ion.
Attached are the lewis structures,
The proton released by water bonds to the nitrogen, because it can hold charges relatively better than carbon because of it's bigger size
Why can long chain fatty acids can form micelles in solutions with pH > 7 but are insoluble in pH < 5
Answer:
In basic conditions that is ( pH > 7 ), the equilibrium shifts towards right and produces a lot of (-ve) negatively charged fatty acids which are polar, In water, since they have long hydrophobic hydrocarbon part, this form micelles where the hydrocarbon part remain inside the sphere and -coo- group remain outside the sphere due to H-bonding interaction with water.
At ( pH < 5 ) I.e acidic conditions, the equilibrium shift to the left giving neutral molecules which can not have stronger H-bonding interaction .
So micelles cant form as they become insoluble.
How many atoms are in 65.0g of zinc?
from
1moles=iatom
Mole=mass÷avogardos
Where
Avogadro's= 6.02×10²³
So moles = 65.0÷6.02×10²³
Atoms of zinc = 391.6 ×10²³
The number of atoms present in the given mass of Zinc that is 65.0gm is [tex]5.99\times10^{ 23}[/tex].
Atoms are the basic building blocks of matter. They are the smallest units of an element that retain the chemical properties of that element.
Now, to determine the number of atoms in a given number of moles, we can use Avogadro's number, which is approximately [tex]6.022 \times10^{23}[/tex]atoms per mole.
First, we calculate the number of moles of zinc in 65.0g by dividing the given mass by the molar mass of zinc. The molar mass of zinc (Zn) is 65.38 g/mol.
Number of moles = Mass / Molar mass
Number of moles = 65.0g / 65.38 g/mol ≈ 0.9942 mol
Next, multiply the number of moles by Avogadro's number to find the number of atoms.
Number of atoms =[tex]Number of moles \times Avogadro's number[/tex]
Number of atoms = [tex]0.9942[/tex]mol × [tex]6.022 \times10^{23}[/tex] atoms/mol
Therefore, approximately [tex]5.99\times10^{ 23}[/tex] atoms are present in 65.0g of zinc.
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A battery is an example of a(n) _________. A. anode B. voltaic cell C. cathode D. electrolytic cell
Answer:
The answer is D) Electrolytic cell
Explanation:
An electrolytic cell is a device used for the decomposition by the electrical current of ionized substances called electrolytes.
When the two electrodes are connected by a wire, electrical energy is produced, and a flow of electrons takes place from the electrode.
These cells are the closest thing to a galvanic battery.
Answer:
b. voltaic cell
Explanation:
Founders Education answer. had to take this quiz 4 times
A sample of argon gas (molar mass 40 g) is at four times the absolute temperature of a sample of hydrogen gas (molar mass 2 g). Find the ratio of the rms speed of the argon molecules to that of the hydrogen. Assume hydrogen molecule has only translational degree of freedom.
Answer:
Ratio of Vrms of argon to Vrms of hydrogen = 0.316 : 1
Explanation:
The root-mean-square speed measures the average speed of particles in a gas, and is given by the following formula:
Vrms = [tex]\sqrt{3RT/M}[/tex]
where R is molar gas constant = 8.3145 J/K.mol, T is temperature in kelvin, M is molar mass of gas in Kg/mol
For argon, M = 40/1000 Kg/mol = 0.04 Kg/mol, T = 4T , R = R
Vrms = √(3 * R *4T)/0.04 = √300RT
For hydrogen; M = 1/1000 Kg/mol = 0.001 Kg/mol, T = T, R = R
Vrms = √(3 * R *T)/0.001 = √3000RT
Ratio of Vrms of argon to that of hydrogen = √300RT / √3000RT = 0.316
Ratio of Vrms of argon to that of hydrogen = 0.316 : 1
write the balanced nuclear equation for the radioactive decay of radium-226 to give radon-222, and determine the type of decay
Answer:
226Ra88→222Rn86+4He2
Explanation:
An α-particle usually consists of a helium nucleus which indicates the type of decay that was undergone in this radioactive process.
During α-decay(alpha decay), an atomic nucleus emits an alpha particle.
What are the conditions that are favorable for extensive solid solubility of one element in another (Hume-rothery rules)
Answer:
Atomic radius less than 15%, similar structure and same valency.
Explanation:
The conditions that are favorable for extensive solid solubility of one element in another are the following.
The atomic radius of the solute and solvent atoms must be less than 15%. The structure of both solute and solvent are similar. Solubility completes when both have same valency. Valency means number of electrons in the outermost shell. If both solute and solvent has same number of electrons so it will be completely soluble in each other.
The conditions that are favorable for extensive solid solubility of one element in another is the same size, electrongativity and valency.
What is Hume - Rothery rules?Hume - Rothery rules are the sets of some important rules which gives idea about the desired condition for the formation of solid solution.
Following main points are described in this rule:
Difference between the size of the solute and the solvent should be less than 15%.Electronegativity difference between the solute and solvents should be small.And they both should have same valency, means same no. of electrons in the outermost shell.Hence size, electronegativity and valency are the conditions.
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The solubility of calcium oxalate, CaC2O4, in pure water is 4.8 × 10‑5 moles per liter. Calculate the value of Ksp for silver carbonate from this data.
Answer:
2.3 × 10⁻⁹
Explanation:
Step 1: Write the reaction for the solution of calcium oxalate
CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)
Step 2: Make an ICE chart
We can relate the molar solubility (S) with the solubility product constant (Ksp) through an ICE chart.
CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)
I 0 0
C +S +S
E S S
The solubility product constant is:
Ksp = [Ca²⁺] × [C₂O₄²⁻] = S² = (4.8 × 10⁻⁵)² = 2.3 × 10⁻⁹
Both formic acid and carbonic acid contain two hydrogen atoms. Why is the chemical formula of formic acid written HCHO₂ (with the two hydrogen atoms listed separately), while the chemical formula of carbonic acid is written H₂CO₃ (with the two hydrogen atoms written together)?
Please explain!
That is because one of the Hydrogens of Formic acid is a part of the functional group (COOH). so to make sure the reader understands that, they second H is written separately
Ortho and para hydrogen are....... a). molecular form. b). Nuclear form. c) allotropic form. d). All
Ortho and para hydrogen are nuclei forms
Complete the following equation of nuclear transmutation.
23892U + 126C → 24498Cf + 6 ______
Complete the following equation of nuclear transmutation.
U + C → Cf + 6 ______
A) 1n
B) 0 e
C) 0 e
D) 1H
E) 0g 0 -1 +1 1 0
Answer:
Option A. 1 0n
Explanation:
Details on how to balanced the equation for the reaction given in the question above can be found in the attached photo.
The missing part of the transmutation equation as it has been shown is 1/o n. Option A
What is nuclear transmutation?Nuclear transmutation is the process of shifting the number of protons in an atom's nucleus to change one element into another. Nuclear processes that change one atomic nucleus into another with a different atomic number are involved.
The production of nuclear energy, radioactive decay, and the creation of new isotopes for use in science and industry all depend on nuclear transmutation, a fundamental idea in nuclear physics.
We have the equation as;
238/92 U + 12/6 C ----> 244/98 Cf + 6 1/0 n
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The following reactions all have K < 1. 1) a. C6H5COO- (aq) + C6H5OH (aq) → C6H5COOH (aq) + C6H5O- (aq) b. F- (aq) + C6H5OH (aq) → C6H5O- (aq) + HF (aq) c. C6H5COOH (aq) + F- (aq) → HF (aq) + C6H5COO- (aq) Arrange the substances based on their relative acid strength.
Answer:
the acid strength is the order of [tex]\mathsf{HF _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5OH _{(aq)} }[/tex]
Explanation:
Given that :
a . [tex]\mathsf{C_6H_5COO^- _{(aq)} + C_6H_5OH _{(aq)} \to C_6H_5COOH _{(aq)} + C_6H_5O^- _{(aq)}}[/tex]
b. [tex]\mathsf{ F^- _{(aq)} + C_6H_5OH _{(aq)} \to C_6H_5O^- _{(aq)} + HF _{(aq)} }[/tex]
c. [tex]\mathsf{C_6H_5COOH _{(aq)} + F^- _{(aq)} \to HF _{(aq)} + C_6H_5COO^- _{(aq)} }[/tex]
Acid strength is the ability of an acid to dissociate into a proton and an anion. Take for instance.
HA ↔ H⁺ + A⁻
The acid strength of the following compounds above are:
[tex]\mathsf{C_6H_5OH _{(aq)} }[/tex] = 1.00 × 10⁻¹⁰
[tex]\mathsf{HF _{(aq)} }[/tex] = 6.6 × 10⁻⁴
[tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] = 6.3 × 10⁻⁵
As the acid dissociation constant increases the relative acid strength also increases.
From above, the acid strength is the order of [tex]\mathsf{HF _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5COOH _{(aq)} }[/tex] > [tex]\mathsf{C_6H_5OH _{(aq)} }[/tex]
[tex]\mathsf{C_6H_5COO^- }[/tex], [tex]\mathsf{C_6H_5O^- _{(aq)}}[/tex] and F⁻ are Bronsted- Lowry acid
Bronsted- Lowry acid are molecule or ion that have the ability to donate a proton.
A buffer is prepared such that [H2PO4-] = 0.095M and [HPO42-] = 0.125M? What is the pH of this buffer solution? (pKa = 7.21 for H2PO4-)
Answer:
pH of the buffer is 7.33
Explanation:
The mixture of the ions H₂PO₄⁻ and HPO₄²⁻ produce a buffer (The mixture of a weak acid, H₂PO₄⁻, with its conjugate base, HPO₄²⁻).
To find pH of a buffer we use H-H equation:
pH = pka + log [A⁻] / [HA]
Where A⁻ is conjugate base and HA weak acid.
For the H₂PO₄⁻ and HPO₄²⁻ buffer:
pH = pka + log [HPO₄²⁻] / [H₂PO₄⁻]
Computing values of the problem:
pH =7.21 + log [0.125M] / [0.095M]
pH = 7.33
pH of the buffer is 7.33
How many moles of NaF must be dissolved in 1.00 liter of a saturated solution of PbF 2 at 25°C to reduce the [Pb 2+] to 1.0 × 10 –6 M? The K sp for PbF 2 at 25 °C is 4.0 × 10 –8.
Answer:
0.1957 moles of NaF
Explanation:
The Pb²⁺ and F⁻ are in equilibrium with PbF₂ as follows:
PbF₂(s) ⇄ Pb²⁺(aq) + 2F⁻(aq)
Where Ksp expression is:
Ksp = 4.0x10⁻⁸ = [Pb²⁺] [F⁻]²
A saturated solution contains the maximum possible amount of Pb²⁺ and F⁻. That is:
PbF₂(s) ⇄ Pb²⁺(aq) + 2F⁻(aq)
PbF₂(s) ⇄ X + 2X
Where X is amount of ions presents in solution
4.0x10⁻⁸ = [Pb²⁺] [F⁻]²
4.0x10⁻⁸ = [X] [2X]²
4.0x10⁻⁸ = 4X³
4.0x10⁻⁸/4 = X³
1.0x10⁻⁸ = X³
2.15x10⁻³M = X
That means initial concentration of Pb²⁺ is = X = 2.15x10⁻³M and [F⁻] = 2X = 4.30x10⁻³M
Now, using again Ksp, if you want a [Pb²⁺] = 1.0x10⁻⁶M, the [F⁻] you need is:
4.0x10⁻⁸ = [Pb²⁺] [F⁻]²
4.0x10⁻⁸ = [1.0x10⁻⁶M] [F⁻]²
0.04M = [F⁻]²
0.2M = [F⁻]
You need a final concentration of 0.2M of F⁻. As initial concentration was 4.30x10⁻³M and volume of the buffer is 1.00L, the moles of F⁻ = moles of NaF you must add are:
0.2M - 4.30x10⁻³M =
0.1957 moles of NaF4NH3(g) 5O2(g)4NO(g) 6H2O(g) Using standard thermodynamic data at 298K, calculate the free energy change when 1.81 moles of NH3(g) react at standard conditions.
Answer:
-434.14 kJ
Explanation:
Step 1: Write the balanced equation
4 NH₃(g) + 5 O₂(g) ⇒ 4 NO(g) + 6 H₂O(g)
Step 2: Calculate the standard free energy change (ΔG°r) for the reaction
We will use the following expression.
ΔG°r = 4 mol × ΔG°f(NO(g)) + 6 mol × ΔG°f(H₂O(g)) - 4 mol × ΔG°f(NH₃(g)) - 5 mol × ΔG°f(O₂(g))
ΔG°r = 4 mol × (86.55 kJ/mol) + 6 mol × (-228.57 kJ/mol) - 4 mol × (-16.45 kJ/mol) - 5 mol × (0 kJ/mol)
ΔG°r = -959.42 kJ
Step 3: Calculate the standard free energy change for 1.81 moles of NH₃
959.42 kJ are released per 4 moles of NH₃.
[tex]\frac{-959.42 kJ}{4mol} \times 1.81mol = -434.14 kJ[/tex]
Could someone help with this? Much appreciated!
Answer:
The 3rd answer down.
Na²O (sodium oxide) will be a base when exposed to water H²O
Explanation:
Sodium Oxide Na²O, will become Sodium Hydroxide after being exposed to water (at 80% I believe).
The oxygen ion in Na²O has 2 extra electrons which makes it highly charged and very attractive to hydrogen ions. The attraction is so strong that when Na²O comes in contact with H²O, the O(-2) strips off a hydrogen from water, forming 2 x OH ions which of course are still strongly basic.
In a reversible reaction, the endothermic reaction absorbs ____________ the exothermic reaction releases. A. less energy than B. None of these, endothermic reactions release energy C. the same amount of energy as D. more energy than
Answer: C. the same amount of energy as
Explanation:
A reversible reaction is a chemical reaction where the reactants form products that, in turn, react together to give the reactants back.
Reversible reactions will reach an equilibrium point where the concentrations of the reactants and products will no longer change.
[tex]A+B\rightleftharpoons C+D[/tex]
Thus if forward reaction is exothermic i.e. the heat is released , the backward reaction will be endothermic i.e. the heat is absorbed and in same amount.
The amount of energy released will be equal and opposite in sign to the energy absorbed in that reaction.
Answer:
C.) the same amount of energy as
Explanation:
I got it correct on founders edtell
5. Calcule las concentraciones cuando se alcanza el equilibrio si partimos de unas concentraciones iniciales [A]=[B]=1M ; [C]=[D]=0M y una constante de equilibrio de 5.
Las concentraciones en el equilibrio para la reacción química presentada son:
[tex][A] = [B] = 1-x = 1-0.69 = 0.31 M\\[C] = [D] = x = 0.69 M[/tex]
Consideremos la siguiente reacción química genérica:
A + B ⇄ C + D
Para calcular las concentraciones en el equilibrio, debemos construir una Tabla ICE. Cada fila representa una instancia (Inicial, Cambio, Equilibrio) y la completamos con la concentración o cambio de concentración ("x" para concentraciones desconocidas). Como inicialmente no hay productos, la reacción se desplazará hacia la derecha para alcanzar el equilibrio.
A + B ⇄ C + D
I 1 1 0 0
C -x -x +x +x
E 1-x 1-x x x
La constante de equilibrio, Kc, es:
[tex]Kc = 5 = \frac{[C][D]}{[A][B]} = \frac{x^{2} }{(1-x)^{2} } \\\sqrt{5} = x/1-x\\x = 0.69[/tex]
Las concentraciones en el equilibrio son:
[tex][A] = [B] = 1-x = 1-0.69 = 0.31 M\\[C] = [D] = x = 0.69 M[/tex]
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If a substance has a half-life of 55.6 s, and if 230.0 g of the substance are present initially, how many grams will remain after 10.0 minutes?
Answer:
[tex]m=0.127g[/tex]
Explanation:
Hello,
In this case, for a first-order reaction, we can firstly compute the rate constant from the given half-life:
[tex]k=\frac{ln(2)}{t_{1/2}} =\frac{ln(2)}{55.6s}=0.0125s^{-1}[/tex]
In such a way, the integrated first-order law, allows us to compute the final mass of the substance once 10.0 minutes (600 seconds) have passed:
[tex]m=m_0*exp(-kt)=230.0g*exp(-0.0125s^{-1}*600s)\\\\m=0.127g[/tex]
Best regards.
A 0.100 M solution of NaOH is used to titrate an HCl solution of unknown concentration. To neutralize the solution, an average volume of the titrant was 38.2 mL. The starting volume of the HCl solution was 20 mL. What's the concentration of the HCl? answer options: A) 0.788 M B) 0.284 M C) 3.34 M D) 0.191 M
Answer: it is A
Explanation: I am sure
Answer:
0.191 M
Explanation:
i took the test.
What is the molarity of an HCl solution if 25.0 mL of this solution required 17.80 mL of 0.108 M NaOH to reach the end point in a titration?
Answer:
[tex]\boxed {\boxed {\sf 0.0769 \ M}}[/tex]
Explanation:
We are asked to find the molarity of an acid given the details of a titration experiment. The formula for titration is as follows:
[tex]M_AV_A= M_B V_B[/tex]
In this formula, M is the molarity of the acid or base and V is the volume of the acid or base. The molarity of the hydrochloric acid (HCl) is unknown and the volume is 25.0 milliliters.
[tex]M_A * 25.0 \ mL = M_BV_B[/tex]
The molarity of the sodium hydroxide (NaOH) is 0.108 molar and the volume is 17.80 milliliters.
[tex]M_A * 25.0 \ mL = 0.108 \ M * 17.80 \ mL[/tex]
We are solving for the molarity of the acid and we must isolate the variable [tex]M_A[/tex]. It is being multiplied by 25.0 milliliters. The inverse operation of multiplication is division, so we divide both sides of the equation by 25.0 mL.
[tex]\frac {M_A * 25.0 \ mL }{25.0 \ mL}= \frac{0.108 \ M * 17.80 \ mL }{25.0 \ mL}[/tex]
[tex]M_A= \frac{0.108 \ M * 17.80 \ mL }{25.0 \ mL}[/tex]
The units of milliliters cancel.
[tex]M_A= \frac{0.108 \ M * 17.80 }{25.0 }[/tex]
[tex]M_A= \frac{1.9224}{25.0 } \ M[/tex]
[tex]M_A= 0.076896 \ M[/tex]
The original measurements have 3 and 4 significant figures. We must round our answer to the least number of sig figs, which is 3. For the number we calculated, that is the ten-thousandth place. The 9 to the right of this place tells us to round the 8 up to a 9.
[tex]M_A \approx 0.0769 \ M[/tex]
The molarity of the hydrochloric acid is 0.0769 Molar.
write any four characties of vertebratas?
Answer:
1. to bend
2. to sit
3. to walk
4. To stand
Explanation:
As a reaction proceeds, the ratio between the rate of consumption of reactant and the rate of formation of product:
Answer:
Depends on the reaction.
Explanation:
Hello,
In this case, the answer is depends on the reaction since the ratios between the rates of both consumption and formation depend upon the stoichiometric coefficients in the chemical reaction. For instance, for the reaction:
A -> 2B
The relationship is:
[tex]\frac{1}{-1}r_A =\frac{1}{2} r_B[/tex]
Therefore, we can see that the rate of consumption of A half the rate of formation of B, but is we consider the following chemical reaction:
2A -> B
The relationship is:
[tex]\frac{1}{-2}r_A =\frac{1}{1} r_B[/tex]
Therefore we can see that the rate of consumption of A doubles the rate of consumption of B.
Best regards.
Consider the acid H3PO4. This acid will react with water by the following equation. H3PO4+H2O↽−−⇀H2PO−4+H3O+ What will be true of the resulting conjugate base H2PO−4? Select the correct answer below: H2PO−4 can act as an acid.
Answer:
H+/PO-4^-2
Explanation:
hydrogen has dissolved completely
In the given reaction conjugate base is H₂PO₄⁻, it also behave as a weak acid.
What is acid - conjugate base pair?
An acid and conjugate base pairs are those pairs in which they are differentiated by the one atom of hydrogen atom.
Given chemical reaction is:
H₃PO₄ + H₂O → H₂PO₄⁻ + H₃O⁺
In the above reaction H₃PO₄ is an acid as it gives H⁺ ion to the solution and formed H₂PO₄⁻, which is a conjugate base of H₃PO₄ acid. H₂PO₄⁻ will also behave as an acid because it have H⁺ ion to gives in the solution but nature of this acid is weak as they not readily dissociates.
Hence, H₂PO₄⁻ is a conjugate base.
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Determine which set of properties correctly describes copper (Cu)?
A. Giant structure, conducts electricity, high melting point, soluble in water, malleable
B. Malleable, brittle, soluble in oil or gasoline, high melting point, simple structure
C. Ionic lattice, conducts electricity, soluble in oil or gasoline, low melting point, ductile
D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice
Answer:
D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice
Explanation:
Copper is a metal with an atomic number of 29. This metal is soft and reddish in color which explains why it is very malleable(beaten to form various shapes without breaking).
All metals are good conductors of electricity including copper which is also a metal. Metals generally are insoluble in water. Copper also has a high melting point which is a characteristic of metals due to their giant structure and metallic lattice which makes it difficult to be broken down.
Of the following two gases, which would you predict to diffuse more rapidly? PLZZ HELPP PLZ PLZ PLZ
Answer:
CO2 will diffuse more rapidly.
Explanation:
From Graham's law of diffusion, we understood that the rate of diffusion of a gas is inversely proportional to the square root of its density as shown below:
Rate (R) & 1/√Density (d)
R & 1/√d
But, the density of a gas is directly proportional to the relative molecular mass (M) of the gas.
Thus, we can say that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. This can be represented mathematically as:
Rate (R) & 1/√Molar mass (M)
R & 1/√M
From the above illustration, we can say that the lighter the gas, the faster the rate of diffusion and the heavier the gas, the slower the rate of diffusion.
Now, to answer the question given above,let us determine the molar mass of Cl2 and CO2.
This is illustrated below:
Molar mass of Cl2 = 2 x 35.5 = 71 g/mol
Molar mass of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol
Summary
Gas >>>>>> Molar mass
Cl2 >>>>>> 71 g/mol
CO2 >>>>> 44 g/mol
From the illustration above, we can see that CO2 is lighter than Cl2.
Therefore, CO2 will diffuse more rapidly.
Answer: CO2
Explanation:
What 2 variables are needed to calculate the percent yield?
Answer:
isn’t it reactants and products?
Explanation:
Answer: reactants and products
Explanation:
If an experiment calls for 0.200mole acetic acid (Hc2H3O2)how many grams of glacial acetic acid do we need?
[tex]\\ \sf\longmapsto HC_2H_3O_2[/tex]
[tex]\\ \sf\longmapsto 1u+2(12u)+3(1u)+2(16u)[/tex]
[tex]\\ \sf\longmapsto 1u+24u+3u+48u[/tex]
[tex]\\ \sf\longmapsto 28u+48u[/tex]
[tex]\\ \sf\longmapsto 76u[/tex]
[tex]\\ \sf\longmapsto 76g/mol[/tex]
No of moles=0.2molGiven mass=?[tex]\\ \sf\longmapsto No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}[/tex]
[tex]\\ \sf\longmapsto 0.2=\dfrac{Given\:mass}{76}[/tex]
[tex]\\ \sf\longmapsto Given\:Mass=0.2\times 76[/tex]
[tex]\\ \sf\longmapsto Given\:Mass=1.52g[/tex]
An individual was injected with 80 mg of inulin and 960,000 counts per min (cpm) of tritium-labeled water (3H20) to determine the volume of various body fluid compartments. After equilibration a blood sample was obtained and the plasma inulin concentration was 0.5 mg% and the plasma activity (concentration) of tritium was 20 cpm/ml. The volumes of which body compartments can be determined?
The measurement of body fluid compartments can be achieved by the dilution of chemical compounds that only circulate and disperse in the region of selected areas in the body. The dilution process is dependent on how the concentration is defined.
Given that:
the concentration of plasma insulin after equilibrium = 0.5 mg %∴
Concentration C = 0.5 mg/100
Concentration C = 0.005 mg/ml
The mass of insulin = 80 mgSince the mass amount of the chemical compound(i.e. insulin) and the concentration is known.
The volume of the body fluid compartment can be calculated as:
[tex]\mathbf{volume = \dfrac{\text{mass of the marker }}{concentration }}[/tex]
[tex]Volume = \dfrac{80 \ mg}{0.005 \ mg/ml}[/tex]
Volume = 16000 ml
Thus, it is known that insulin is generally utilized for the measurement of the extracellular fluid volume and serves as a cell impermeant marker.
As a result;
The volume of the extracellular fluid compartment is 16000 ml.
However, the tritium-labeled water is a good marker for the entire body fluid compartment due to the fact that:
its diffusion occurs throughout the entire body,it is identical to water and;the equilibrium concentration is typically easy to measure due to the radioactive characteristics of tritium.Given that:
plasma activity of tritium = 20 cpm/ml
i.e.
In 1 ml of plasma, 20 cpm of tritium is present.
As such, in 960,000 counts per min (cpm) of tritium-labeled water, the volume of the whole body compartment is:
[tex]\mathbf{= \dfrac{960000}{20} ml \plasma}[/tex]
= 48000 ml of plasma
Therefore, we can conclude that the volumes of the body compartment that can be determined are:
The volume of the extracellular fluid compartment, which is 16000 ml.The volume of the whole body compartment, which is 48000 mlLearn more about body fluid compartments here:
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A sample of ammonia gas was allowed to come to equilibrium at 400 K. 2NH3(g) <----> N2(g) 3H2(g) At equilibrium, it was found that the concentration of H2 was 0.0484 M, the concentration of N2 was 0.0161 M, and the concentration of NH3 was 0.295 M. What was the initial concentration of ammonia
Answer:
0.327 M
Explanation:
Step 1: Write the balanced equation
2 NH₃(g) ⇄ N₂(g) + 3H₂(g)
Step 2: Make an ICE chart
2 NH₃(g) ⇄ N₂(g) + 3 H₂(g)
I x 0 0
C -2y +y +3y
E x-2y y 3y
Step 3: Find the value of y
The concentration of N₂ at equilibrium is 0.0161 M. Then,
y = 0.0161
Step 4: Find the value of x
The concentration of NH₃ at equilibrium is 0.295 M. Then,
x-2y = 0.295
x-2(0.0161) = 0.295
x = 0.327