Answer:
[tex]u/n = \frac{-e^2}{4\pi \epsilon r} (ln2)[/tex]
Explanation:
given data
we will take here
sodium ions = positive charge
chloride ions = negative cgarge
solution
as when we take Na positive charge so Number of origin is
d = 0
and here pair of ions with negative charge at d = - r
and d = +r
therefore
[tex]u = \frac{-2e^2}{4\pi \epsilon r} \times \frac{1}{r} \times \sum _n {\frac{1}{n}}(-1)^{n-1}[/tex]
we will use here Taylor series approx method
[tex]u = \frac{-e^2}{2\pi \epsilon r} (ln2)[/tex]
and N/2 pair will contribute here
so
[tex]u = \frac{N}{2} \frac{-e^2}{2\pi \epsilon r} (ln2)[/tex]
so energy per ion will be here
[tex]u/n = \frac{-e^2}{4\pi \epsilon r} (ln2)[/tex]
The average molecular speed in a sample of Ar gas at a certain temperature is 391 m/s. The average molecular speed in a sample of Ne gas is ______ m/s at the same temperature.
Answer:
550 m/s
Explanation:
The average molecular speed (v) is the speed associated with a group of molecules on average. We can calculate it using the following expression.
[tex]v = \sqrt{\frac{3 \times R \times T}{M} }[/tex]
where,
R: ideal gas constantT: absolute temperatureM: molar mass of the gasWe can use the info of argon to calculate the temperature for both samples.
[tex]T = \frac{v^{2} \times M}{3 \times R} = \frac{(391m/s)^{2} \times 39.95g/mol}{3 \times 8.314J/k.mol} = 2.45 \times 10^{5} K[/tex]
Now, we can use the same expression to find the average molecular speed in a sample of Ne gas.
[tex]v = \sqrt{\frac{3 \times R \times T}{M} } = \sqrt{\frac{3 \times (8.314J/k.mol) \times 2.45 \times 10^{5}K }{20.18g/mol} } = 550 m/s[/tex]
Calculate the height of a column of water at 25 °C that corresponds to normal atmospheric pressure. The density of water at this temperature is 1.0 g/
Answer:
10.328 m
Explanation:
normal atmospheric pressure = 101325 Pa
density of water at 25 °C = 1.0 g/cm^3 = 1000 kg/m^3
pressure = pgh
where p = density
g = acceleration due to gravity = 9.81 m/s^2
h = height of column
imputing values, we have
101325 = 1000 x 9.81 x h
height of column h = 101325/9810 = 10.328 m
Determine the radius of an Al atom (in pm) if the density of aluminum is 2.71 g/cm3 . Aluminum crystallizes in a face centered cubic structure with an edge leng
Answer:
143pm is the radius of an Al atom
Explanation:
In a face centered cubic structure, FCC, there are 4 atoms per unit cell.
First, you need to obtain the mass of an unit cell using molar mass of Aluminium and thus, obtain edge length and knowing Edge = √8R you can find the radius, R, of an Al atom.
Mass of an unit cell
As 1 mole of Al weighs 26.98g. 4 atoms of Al weigh:
4 atoms × (1mole / 6.022x10²³atoms) × (26.98g / mole) = 1.792x10⁻²²g
Edge length
As density of aluminium is 2.71g/cm³, the volume of an unit cell is:
1.792x10⁻²²g × (1cm³ / 2.71g) = 6.613x10⁻²³cm³
And the length of an edge of the cell is:
∛6.613x10⁻²³cm³ = 4.044x10⁻⁸cm = 4.044x10⁻¹⁰m
Radius:
As in FCC structure, Edge = √8 R, radius of an atom of Al is:
4.044x10⁻¹⁰m = √8 R
1.430x10⁻¹⁰m = R.
In pm:
1.430x10⁻¹⁰m ₓ (1x10¹²pm / 1m) =
143pm is the radius of an Al atomThe radius of the atom of Al in the FCC structure has been 143 pm.
The FCC lattice has been contributed with atoms at the edge of the cubic structure.
The FCC has consisted of 4 atoms in a lattice.
The mass of the unit cell of Al can be calculated as:[tex]\rm 6.023\;\times\;10^2^3[/tex] atoms = 1 mole
4 atoms = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles
The mass of 1 mole Al has been 26.98 g/mol.
The mass of [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles × 26.98 g
The mass of 1 unit cell of Al has been = 1.792 [tex]\rm \bold{\times\;10^-^2^2}[/tex] g.
The volume of the Al cell can be calculated as:Density = [tex]\rm \dfrac{mass}{volume}[/tex]
Volume = Density × Mass
The volume of Al unit cell = 2.71 g/[tex]\rm cm^3[/tex] × 1.792 [tex]\rm \times\;10^-^2^2[/tex] g
The volume of Al cell = 6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex]
The volume of the cube has been given as:Volume = [tex]\rm edge\;length^3[/tex]
6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex] = [tex]\rm edge\;length^3[/tex]
Edge length = [tex]\rm \sqrt[3]{6.613\;\times\;10^-^2^3}[/tex] cm
Edge length = 4.044 [tex]\rm \times\;10^-^8[/tex] cm
Edge length = 4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.
In an FCC lattice structure, the radius of the atom can be given by:Edge length = [tex]\rm \sqrt{8\;\times\;radius}[/tex]
4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = [tex]\rm \sqrt{8\;\times\;radius}[/tex]
Radius = 1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.
1 m = [tex]\rm 10^1^2[/tex] pm
1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = 143 pm.
The radius of the atom of Al in the FCC structure has been 143 pm.
For more information about the FCC structure, refer to the link:
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Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is
Answer:
4.26 %
Explanation:
There is some info missing. I think this is the original question.
Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10 ⁻⁴.
Step 1: Given data
Initial concentration of the acid (Ca): 0.249 M
Acid dissociation constant (Ka): 4.50 × 10 ⁻⁴
Step 2: Write the ionization reaction for nitrous acid
HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)
Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])
We will use the following expression.
[tex][A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M[/tex]
Step 4: Calculate the percent ionization of nitrous acid
We will use the following expression.
[tex]\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%[/tex]
Consider the following precipitation reaction occurring in aqueous solution:
3 SrCl2(aq)+2 Li3PO4(aq) →Sr3(PO4)2(s)+6 LiCl(aq)
Write the complete ionic equation and the net ionic equation for this reaction.
Answer:
[tex]3Sr^{+2}+6Cl^{-}+6Li^{+}+2PO_{4}^{3-}-->Sr_{3}(PO_{4})_{2}+6Li^{+}+6Cl^{-}\\\\3Sr^{+2}+2PO_{4}^{3-} --->Sr_{3}(PO_{4})_{2}[/tex]
First equation is the complete ionic equation.
Second equation is the net ionic equation.
Name of th molecule
1. CH3CH2CHClCHBrCH3
2.C=C-CH3
CH3CH=CHCH2
Answer:
1: 2-bromo-3-chloropentane
Explanation:
find longest carbon chain =5
place the Br and Cl on the carbon chain
follow naming rules I guess
If an electron has a principal quantum number (n) of 7 and an angular momentum quantum number (l) of 1, the subshell designation is ________
Answer:
7p
Explanation:
principal quantum number is 7
n=7( principle shell)
angular momentum quantum number gives sub shell
l = 1 means it is p orbital
so answer is 7p orbital
How are scientific questions answered?
A. Through observing and measuring the physical world
B. Through testing a theory about the physical world
c. Through forming a hypothesis about the question
D. Through predicting a solution about the question
SUBM
Answer:
Option B
Explanation:
Scientific question are answered through experimentation, through testing the theory about the physical world.
Answer: its A
through observing and measuring the physical world
Explanation:
What is the final pH of a solution with an initial concentration of 2.5mM Ascorbic acid (H2C6H6O6) which has the following Kas: 7.9x10-5 and 1.6x10-12
Answer:
pH = 3.39
Explanation:
The equilibrium in water of ascorbic acid (With its conjugate base) is:
H₂C₆H₆O₆(aq) + H₂O(l) ⇄ HC₆H₆O₆⁻(aq) + H₃O⁺(aq)
Where the acidic dissociation constant is written as:
Ka = 7.9x10⁻⁵ = [HC₆H₆O₆⁻] [H₃O⁺] / [H₂C₆H₆O₆]
H₂O is not taken in the Ka expression because is a pure liquid.
As initial concentration of H₂C₆H₆O₆ is 2.5x10⁻³M, the equilibrium concentration of each species in the equilibrium is:
[H₂C₆H₆O₆] = 2.5x10⁻³M - X
[HC₆H₆O₆⁻] = X
[H₃O⁺] = X
Replacing in the Ka expression:
7.9x10⁻⁵ = [X] [X] / [2.5x10⁻³M - X]
1.975x10⁻⁷ - 7.9x10⁻⁵X = X²
0 = X² + 7.9x10⁻⁵X - 1.975x10⁻⁷
Solving for X:
X = -0.00048566→ False solution, there is no negative concentrations
X = 0.00040666 → Right solution
As [H₃O⁺] = X, [H₃O⁺] = 0.00040666
pH is defined as -log [H₃O⁺];
pH = -log 0.00040666,
pH = 3.39determine the rate of reaction that follows the rate= k[A]^m[B]^n
rate=0.2*3^1*3^2=0.2*3*9=5.4(mol/L)s so the correct answer is C.
Identify the correctly written chemical reaction
A. Reactant + Reactant = Product
B. Reactant + Reactant → Product + Product
C. Reactant + Product → Reactant + Product
D. Product + Product Reactant + Reactant
Answer:
B. Reactant + Reactant -> Product + Product
Explanation:
Reactants are substances that- as the name suggests- reacts with other substances at the beginning of a reaction
Products are substances that are produced as a result of the reaction
Typically, when writing a chemical reaction, an arrow is used to show the direction the reaction is moving. In this case, the arrows in options B and C suggest that the reaction only moves in one direction- forwards
And as mentioned above, reactants are the substances at the start of the reaction, they're what mixes together to form a new product.
To keep things simple:
Products can't be at the beginning of a reaction since they weren't formed yet.
Similarly, reactants can't be part of the products since they already existed and didn't need to be made. In a lot cases, the reactants would be completely used up to make the products
As such, only one possible chemical reaction would follow that reasoning:
Reactant + Reactant -> Product + Product
Reactant + Reactant → Product + Product is the correctly written chemical reaction. Hence, option B is correct.
What is a chemical equation?A chemical equation is a mathematical expression of the chemical reaction which represents the product formation from the reactants.
In an equation, the reactants are written on the left-hand side and the products are written on the right-hand side demonstrated by one-headed or two-headed arrows.
Hence, option B is correct.
Learn more about the chemical equation here:
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A 1.555-g sample of baking soda decomposes with heat to produce 0.991 g Na2CO3. Refer to Example Exercise 14.l and show the calculation for the theoretical yield of Na2CO3.
What is the percent yield of sodium carbonate, Na2CO3?
6. A 1473-g unknown mixture with baking soda is heated and has a mass loss of 0.325 g. Refer to Example Exercise 14.2 and show the calculation for the percentage NaHCOs in the mixture.
Answer:
a) 101%
b)59.7%
Explanation:
The equation for the thermal decomposition of baking soda is shown;
2NaHCO3 → Na2CO3 + H2O + CO2
Number of moles of baking soda= mass/molar mass= 1.555g/84.007 g/mol = 0.0185 moles
From the reaction equation;
2 moles of baking soda yields 1 mole of sodium carbonate
0.0185 moles of baking soda will yield = 0.0185 moles ×1 /2 = 9.25 ×10^-3 moles of sodium carbonate.
Therefore, mass of sodium carbonate= 9.25 ×10^-3 moles × 106gmol-1= 0.9805 g of sodium carbonate. This is the theoretical yield of sodium carbonate.
%yield = actual yield/theoretical yield ×100
% yield = 0.991/0.9805 ×100
%yield = 101%
Since ;
2NaHCO3 → Na2CO3 + H2O + CO2
And H2O + CO2 ---> H2CO3
Hence I can write, 2NaHCO3 → Na2CO3 + H2CO3
Molar mass of H2CO3= 62.03 gmol-1
Molar mass of baking soda= 84 gmol-1
Therefore, mass of baking soda=
0.325/62.03 × 2 × 84 = 0.88 g of NaHCO3
% of NaHCO3= 0.88/1.473 × 100 = 59.7%
The decomposition reaction of baking soda is a reaction in which water and carbon dioxide ae given off as gaseous products.
5. The theoretical yield of Na₂CO₃ is approximately 0.9809 gramsThe percentage yield of sodium carbonate is approximately 101.02%.6. Percentage of NaHCO₃ in the mixture is approximately 59.76%.Reasons:
Mass of baking soda = 1.555 g
Mass of Na₂CO₃ produced = 0.991 g
Required:
Calculation for the theoretical yield
Solution:
Theoretical yield (mass) of Na₂CO₃ produced is found as follows;
Molar mass of Na₂CO₃ = 105.9888 g/mol
Molar mass of NaHCO₃ = 84.007 g/mol
[tex]\displaystyle 1.555 \, g \, NaHCO_3 \times \frac{1 \, mol \, NaHCO_3}{84.007 \, g \, NaHCO_3} \times \frac{1 \, mol \, Na_2CO_3}{2 \, mol \, NaHCO_3} \times 105.9888 \ g \approx 0.9809 \, g \, Na_2CO_3[/tex]
The theoretical yield of Na₂CO₃ ≈ 0.9809 grams.
The percentage yield is given as follows;
[tex]\displaystyle Percentage \ yield = \mathbf{\frac{Actual \, Yield}{Theorectical \, Yield} \times 100 \%}[/tex]
The percentage yield of Na₂CO₃ is therefore;
[tex]\displaystyle Percentage \ yield \ of \ Na_2CO_3= \frac{0.991}{0.9809} \times 100 \% \approx \underline{ 101.02 \%}[/tex]
(Some baking soda may remain if the reaction is not completed)
6. Mass of the unknown mixture of baking soda = 1473 g
Mass loss from the mixture = 0.325 g
Required:
The percentage of NaHCO₃ in the mixture.
Solution:
The chemical in the mass loss from heating the NaHCO₃ = H₂CO₃
Molar mass of H₂CO₃ = 62.03 g/mol
[tex]\displaystyle \mathrm{Number \ of \ moles \ of \ H_2CO_3 \ produced} = \frac{0.325 \, g}{62.03 \, g/mol} \approx 5.2394 \times 10^{-3} \ moles[/tex]
The chemical reaction is presented as follows;
2NaHCO₃(s) [tex]\underrightarrow {\Delta \ Heated}[/tex] Na₂CO₃(s) + H₂CO₃(g)2 moles of NaHCO₃ produces 1 mole of H₂CO₃The number of moles of NaHCO₃ in the mixture is therefore;
2 × 5.2394 × 10⁻³ moles ≈ 1.04788 × 10⁻² molesMass of NaHCO₃ in the mixture is therefore
Mass of NaHCO₃ = 1.04788 × 10⁻² moles × 84.007 g/mol = 0.88029 g[tex]\displaystyle Percentage \ of \ NaHCO_3 \ in \ the \ mixture \ = \mathbf{ \frac{Mass \ of \ NaHCO_3}{Mass \ of \ mixture} \times 100}[/tex]
Which gives;
[tex]\displaystyle Percentage \ of \ NaHCO_3 \ in \ the \ mixture \ = \ \frac{0.88029 \, g}{1.473 \, g} \times 100 \approx \underline{ 59.76 \%}[/tex]Learn more here:
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A solution contains A13+ and Co2+. The addition of 0.3731 L of 1.735 M NaOH results in the complete precipitation of the
ions as Al(OH), and Co(OH)2. The total mass of the precipitate is 22.73 g. Find the masses of Al3+ and Co2+ in the solution.
Answer:
The correct answer is mass of Al3+ will be 3.23 grams and the mass of Co2+ will be 8.50 grams.
Explanation:
Based on the given information, 0.3731 L of 1.735 M of NaOH is added in a solution resulting in the precipitation of the ions as Al(OH)₃ and Co(OH)₂. Thus, the moles of NaOH will be molarity × V(L) = 1.735 × 0.3731 L = 0.647 moles.
The mass of the precipitate given is 22.73 grams.
Now let us assume that the mass of Al(OH)₃ will be x grams and the mass of Co(OH)₂ will be (22.73-x) grams
Therefore, the moles of Al(OH)₃ will be x grams/78 g/mol and as 3OH⁻ ions are needed so the moles will be 3x/78 mole.
And, the moles of Co(OH)₂ will be (22.73-x)grams/92.94 g/mol and as 2OH⁻ ions are needed so the moles will be 45.46-2x/92.94 moles.
Now the equation will become,
3x/78 + 45.46-2x/92.94 = 0.647 moles
0.03846 x + 0.489 - 0.02152 x = 0.647
0.01694 x + 0.489 = 0.647
0.01694 x = 0.158
x = 0.158/0.01694
x = 9.327 grams
Hence, the mass of Al(OH)₃ is 9.327 grams, and the mass of Al³⁺ will be,
= 9.327 gm/78 g/mol × 27 g/mol = 3.23 grams
Now the mass of Co(OH)₂ will be, (22.73 - 9.327) grams = 13.403 grams
the mass of Co²⁺ will be,
= 13.403 grams / 92.94 g/mol × 58.94 g/mol = 8.50 grams
Write the following isotope in nuclide notation: oxygen-14
Answer:
[tex]14\\8[/tex]O
Explanation:
The top number always represents the mass number.
The bottom number always represents the atomic number.
The element always goes after the numbers.
If charge is present, that comes after the element.
Using the provided table, determine the enthalpy for the reaction
2 NH3 (g) + 3 N20 (g) 4 N2 (g) + 3 H20 (1)
Answer:
ΔH°r = -1009.8 kJ
Explanation:
Let's consider the following balanced reaction.
2 NH₃(g) + 3 N₂O(g) ⇒ 4 N₂(g) + 3 H₂O(l)
We can calculate the standard enthalpy of the reaction (ΔH°r) using the following expression.
ΔH°r = [4 mol × ΔH°f(N₂(g)) + 3 mol × ΔH°f(H₂O(l))] - [2 mol × ΔH°f(NH₃(g)) + 3 mol × ΔH°f(N₂O(g))]
ΔH°r = [4 mol × 0 kJ/mol + 3 mol × (-285.8 kJ/mol)] - [2 mol × (-46.2 kJ/mol) + 3 mol × 81.6 kJ/mol]
ΔH°r = -1009.8 kJ
The NaOH solution is standardized (or its true concentration) is found by reacting it with KHSO4. One of the two products from when NaOH reacts with KHSO4 is H2O. The other product is is a salt consisting of what?
a. NaK (aq)
b. (aq)
c. NaS (aq)
d. None of the above
A solution is prepared by adding 6.24 g of benzene (C 6H 6, 78.11 g/mol) to 80.74 g of cyclohexane (C 6H 12, 84.16 g/mol). Calculate the mole fraction and molality of benzene in this solution.
Answer:
[tex]x_B=0.0769[/tex]
[tex]m=0.990m[/tex]
Explanation:
Hello,
In this case, we can compute the mole fraction of benzene by using the following formula:
[tex]x_B=\frac{n_B}{n_B+n_C}[/tex]
Whereas n accounts for the moles of each substance, thus, we compute them by using molar mass of benzene and cyclohexane:
[tex]n_B=6.24g*\frac{1mol}{78.11g}=0.0799mol\\ \\n_C=80.74g*\frac{1mol}{84.16g} =0.959mol[/tex]
Thus, we compute the mole fraction:
[tex]x_B=\frac{0.0799mol}{0.0799mol+0.959mol}\\ \\x_B=0.0769[/tex]
Next, for the molality, we define it as:
[tex]m=\frac{n_B}{m_C}[/tex]
Whereas we also use the moles of benzene but rather than the moles of cyclohexane, its mass in kilograms (0.08074 kg), thus, we obtain:
[tex]m=\frac{0.0799mol}{0.08074kg}=0.990mol/kg[/tex]
Or just 0.990 m in molal units (mol/kg).
Best regards.
Considering the definition of mole fraction and molality:
the mole fraction of benzene is 0.077.the molality of benzene is 0.9908 [tex]\frac{moles}{kg}[/tex].You know that:
Mass of benzene = 6.24 gramsMass of cyclohexane= 80.74 gramsMolar mass of benzene= 78.11 g/moleMolar mass of cyclohexane= 84.16 g/moleMole fractionThe molar fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution.
Being the molar mass and the mass in the solution of each compound, the number of moles of each compound can be calculated as:
Benzene: [tex]\frac{mass of benzene}{molar mass of benzene} =\frac{6.24 grams}{78.11 \frac{grams}{mole} } = 0.08 moles[/tex]Cyclohexane:[tex]\frac{mass of cyclohexane}{molar mass of ciclohexane} =\frac{80.74 grams}{84.16\frac{grams}{mole} } = 0.96 moles[/tex]So, the total moles of the solution can be calculated as:
Total moles = 0.08 moles + 0.96 moles = 1.04 moles
Finally, the mole fraction of benzene can be calculated as follow:
[tex]\frac{number moles of benzene}{total moles} =\frac{0.08 moles}{1.04 moles} = 0.077[/tex]
Finally, the mole fraction of benzene is 0.077.
MolalityMolality is the ratio of the number of moles of any dissolved solute to kilograms of solvent.
The Molality of a solution is determined by the expression:
[tex]Molality=\frac{number of moles of solute}{kilograms of solvent}[/tex]
In this case, you know:
number of moles of solute= 0.08 moles Mass of solvent = 80.74 g = 0.08074 kg (being 1000 g=1 kg)Replacing:
[tex]Molality benzene=\frac{0.08 moles}{0.08074 kg}[/tex]
Molality benzene= 0.9908 [tex]\frac{moles}{kg}[/tex]
Finally, the molality of benzene is 0.9908 [tex]\frac{moles}{kg}[/tex].
Learn more about:
mole fraction brainly.com/question/14434096?referrer=searchResults brainly.com/question/10095502?referrer=searchResults molality brainly.com/question/20366625?referrer=searchResults brainly.com/question/4580605?referrer=searchResults
The next few questions will walk you through solving the following problem: At a given temperature, a 5.0M solution of hydrazine (N2H4) as a pH of 11.34. Hydrazine is base.
A. What is the concentration of hydroxide ion at equilibrium?
B. What is the pK for hydrazine reacting with water at this temperature?
Answer:
A. [OH⁻] = 2.188x10⁻³M
B. pKb = 6.02
Explanation:
When hydrazine is in equilbrium with water, its reaction is:
N₂H₄(aq) + H₂O(l) ⇄ HN₂H₄⁺(aq) + OH⁻(aq)
Where Kb, is defined as the ratio between concentrations in equilibrium of the species, thus:
Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]
A. From pH, you can find [OH⁻], thus:
pH = -log [H⁺]
11.34 = -log [H⁺]
4.57x10⁻¹² = [H⁺]
As 1x10⁻¹⁴ = [OH⁻] [H⁺]
1x10⁻¹⁴ / 4.57x10⁻¹² = [OH⁻]
[OH⁻] = 2.188x10⁻³MB. Concentrations in equilibrium of the species are:
[N₂H₄] = 5.0M - X
[HN₂H₄⁺] = X
[OH⁻] = X
Where X is reaction coordinate
As [OH⁻] = 2.188x10⁻³M
X = 2.188x10⁻³M
Replacing:
[N₂H₄] = 5.0M - 2.188x10⁻³M = 4.9978M
[HN₂H₄⁺] = 2.188x10⁻³M
[OH⁻] = 2.188x10⁻³M
Replacing in Kb expression:
Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]
Kb = [2.188x10⁻³M] [2.188x10⁻³M] / [4.9978M]
Kb = 9.577x10⁻⁷
pKb is defined as -log Kb
pKb = -log 9.577x10⁻⁷
pKb = 6.02
A substance used as a cleaner and a fuel is 37.48% C, 49.93% O and 12.58% H by mass. A 0.2804-g sample of the substance occupies a volume of 250.0 mL when it is vaporized at 75o C and 1.00 atm of pressure.
R = 0.0821 L atm/ mol K
a) This compound can be made by combining gaseous carbon monoxide with hydrogen gas (with this compound as the only product). What is the maximum mass of this compound that can be prepared if 8.0 kg of hydrogen gas react with 59.0 kg of carbon monoxide gas?
b) If 59.6 kg of the product is actually produced, given the reaction described in (a), what is the percent yield?
c) This compound (the substance you identified in part a) is a potential replacement for gasoline. The products of the complete combustion of this fuel are the same as those for the complete combustion of a hydrocarbon (CO2 and H2O). Calculate the volume of CO2 produced at 27o C and 766 mmHg when 1.00 gallon of this fuel is completely combusted. The density of the fuel is 0.7914 g/mL. 1 gallon = 3.785 liters
d) A claim was made that this fuel is better for the environment because it produces less CO2 per gallon than gasoline, which can be represented by the formula C8H18 (octane). Is this claim true? Octane has a density of 0.6986 g/mL
Answer:
Explanation:
We shall find out the molecular formula of the substance .
Ration of number of atoms of C , O and H
= [tex]\frac{37.48}{12} :\frac{49.93}{16} :\frac{12.58}{1}[/tex]
= 3.12 : 3.12 : 12.58
= 1 : 1 : 4
volume of gas at NTP
= 250 x 273 / 350 mL .
= 195 mL .
Molecular weight of the substance = .2804 x 22400 / 195 g
= 32. approx
Let the molecular formula be
(COH₄)n
n x 32 = 32
n = 1
Molecular formula = COH₄
The compound appears to be CH₃OH
a )
CO + 2H₂ = CH₃OH
28g 4g 32g
59 8
For 8 kg hydrogen , CO required = 56 kg
CO is in excess . hydrogen is the limiting reagent .
mass of product formed
= 32 x 8 / 4
= 64 kg
b )
percentage yield = product actually formed / product to be formed theoretically x 100
= 59.6 x 100 / 64
= 93.12 %
c )
2CH₃OH + 3O₂ = 2CO₂ + 4H₂O .
64 g 2 x 22.4 L
Gram of gas in 1 gallon of fuel
= .7914 x 3785
= 2995.5 g
CO₂ produced at NTP by 2995.5 g CH₃OH
= 2 x 22.4 x 2995.5 / 64 L
= 2096.85 L
At 27° C and 766 mm Hg , this volume is equal to
2096.85 x 300 x 760 / 273 x 766
= 2286.18 L .
d )
C₈H₁₈ = 8CO₂
114g 8 x 22.4 L
gram of fuel per unit gallon
= .6986 x 3785
= 2644.2g
gram of CO₂ produced by 1 gallon of fuel at NTP
= 8 x 22.4 x 2644.2 / 114
= 4156.5 L
So it produces more CO₂ .
. Explain why, in the sample calculations, 0.1 g of the unknown produced a GREATER freezing point depression than~e same mass of naphthalene.
Answer
Naphthalene is a non electrolyte
If the unknown compound is an electrolyte it gives 2 or more ions in solution
( NaCl >> Na+ + Cl- => 2 ions
Ca(NO3)2 >> Ca2+ + 2 NO3- => 3 ions)
the f.p. lowering is directly proportional to the molal concentration of dissolved ions in the solution )
For naphthalene
delta T = 1.86 x m
for a salt that gives 2 ions
delta T = 1.86 x m x 2
hence the lowering in freezion point of unkown is greater then napthalene
Give the concentration of each type of ion in the following solutions:
a. 0.50 M CO(NO3)2
b. 1 M Fe(C1O4)3
Answer:
a.
[tex]M_{Co^{2+}}=0.5M\\ \\M_{NO_3^{-}}=1.0M[/tex]
b.
[tex]M_{Fe^{3+}}=1.0M\\ \\M_{ClO_4^{-}}=3.0M[/tex]
Explanation:
Hello,
a. In this case, the ions are cobalt (II) and nitrate, for which, one mole of cobalt (II) nitrate contains one mole of cobalt (II) and two moles of nitrate (see subscripts), therefore, concentrations turn out:
[tex]M_{Co^{2+}}=0.5\frac{molCo(NO_3)_2}{L}* \frac{1molCo}{1molCo(NO_3)_2}=0.5M\\ \\M_{NO_3^{-}}=0.5\frac{molCo(NO_3)_2}{L}* \frac{2molNO_3^{-}}{1molCo(NO_3)_2}=1.0M[/tex]
b. In this case, the ions are iron (III) and chlorate, for which one mole of iron (III) is contained in one mole of iron (III) chlorate and three moles of chlorate are in one mole of iron (III) chlorate (see subscripts), therefore, the concentrations turn out:
[tex]M_{Fe^{3+}}=1.0\frac{molFe(ClO_4)_3}{L}* \frac{1molFe^{3+}}{1molFe(ClO_4)_3}=1.0M\\ \\M_{ClO_4^{-}}=0.5\frac{molFe(ClO_4)_3}{L}* \frac{3molClO_4^{-}}{1molFe(ClO_4)_3}=3.0M[/tex]
Regards.
The solubility of O2 in water is approximately 0.00380 g L-1 of water when the temperature is 25.0°C and the partial pressure of gaseous oxygen is 760. torr. The oxygen gas above the water is replaced by air at the same temperature and pressure, in which the mole fraction of oxygen is 0.210. What will the solubility of oxygen in water be under these new conditions?
Answer:
The correct answer is 0.00080 gram per liter.
Explanation:
Based on the given information, the solubility of water is 0.00380 gram per liter, the temperature mentioned is 25 degree C, the partial pressure of oxygen gas is 760 torr, and the mole fraction of oxygen is 0.210. There is a need to determine the solubility of oxygen in water.
Based on Henry's law,
Solubility of oxygen gas = Henry's constant × partial pressure of oxygen gas
Henry's constant, K = solubility of oxygen gas / partial pressure of oxygen gas
= 0.00380 g/L × 1 mol/32 grams / 760 torr × 1 atm/760 torr
= 0.00012 mol/L/atm
= 0.00012 M/atm
Now the partial pressure of the oxygen gas = mole fraction of oxygen × atmospheric pressure
= 0.210 × 1 atm
= 0.210 atm
Now putting the values in Henry's law equation we get,
Solubility of oxygen gas = 0.00012 mol/L/atm × 0.210 at,
= 0.000025 mol/L × 32 gram/mol
= 0.00080 gram per liter
Determine the rate of a reaction that follows the rate law:
rate = k[A]”[B]", where:
k= 1.5
[A] = 1 M
[B] = 3 M
m = 2
n = 1
Answer:
k= 1.5
[A] = 1 M
[B] = 3 M
m = 2
n = 1
Explanation:
rate = k[A]”[B]"
The rate of the reaction is 4.5 mol L⁻¹s⁻¹.
What is meant by rate of a reaction ?Rate of a reaction is defined as the change in concentration of any one of the reactants or products of the reaction, in unit time.
Here,
The concentration of A, [A] = 1 M
The concentration of B, [B] = 3 M
The partial order with respect to A, m = 2
The partial order with respect to B, n = 1
The rate constant of the reaction, k = 1.5
The rate of the reaction,
r = k[A]^m [B}^n
r = 1.5 x 1² x 3
r = 4.5 mol L⁻¹s⁻¹
Hence,
The rate of the reaction is 4.5 mol L⁻¹s⁻¹.
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A volumetric flask contains 25.0 mL of a 14% m/V sugar solution. If 2.5 mL of this solution is added to 22.5 mL of distilled water, what is the % m/V of the new solution.
Answer:
The new solution is 1.4% m/V
Explanation:
The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.
We have 2.5 mL (V₁) of a concentrated solution and add it to 22.5 mL of distilled water. Assuming the volumes are additives, the volume of the new solution (V₂) is:
[tex]2.5 mL + 22.5 mL = 25.0 mL[/tex]
We want to prepare a dilute solution from a concentrated one, whose concentration is 14% m/V (C₁). We can calculate the concentration of the dilute solution (C₂) using the dilution rule.
[tex]C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{14\% m/V \times 2.5 mL}{25.0 mL} = 1.4 \% m/V[/tex]
The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.
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Question 23
1 pts
When solutions of AgNO3 and NaOH react, the balanced molecular equation is:
2 AgNO3(aq) + 2NaOH(aq) O--> Ag2O(s) + 2 NaNO3(aq) + H20(1)
How much Ag2O is produced when 0.200 g of AgNO3 and 0.200 g of NaOH react?
a. 0.127 g
c. 0.273 g
b. 0.136 g
d. 0.400 g
OB
OC
OA
OD
Answer:
Option B. 0.136 g
Explanation:
The balanced equation for the reaction is given below:
2AgNO3(aq) + 2NaOH(aq) —> Ag2O(s) + 2NaNO3(aq) + H2O(l)
Next, we shall determine the masses of AgNO3 and NaOH that reacted and the mass of Ag2O produced from the balanced equation. This is illustrated below:
Molar mass of AgNO3 = 108 + 14 + (16x3) = 170g/mol
Mass of AgNO3 from the balanced equation = 2 x 170 = 340g
Molar mass of NaOH = 23 + 16 + 1 = 40g/mol
Mass of NaOH from the balanced equation = 2 x 40 = 80g
Molar mass of Ag2O = (108x2) + 16 = 232g/mol
Mass of Ag2O from the balanced equation = 1 x 232 = 232g
Summary:
From the balanced equation above,
340g of AgNO3 reacted with 80g of NaOH to produce 232g of Ag2O.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
340g of AgNO3 reacted with 80g of NaOH.
Therefore, 0.2g of AgNO3 will react with = (0.2 x 80)/340 = 0.047g of NaOH.
From the calculations made above, only 0.047g out of 0.2g of NaOH given, reacted completely with 0.2g of AgNO3. Therefore, AgNO3 is the limiting reactant and NaOH is the excess reactant.
Now, we can calculate the mass of Ag2O produced from the reaction of 0.2g of AgNO3 and 0.2g of NaOH.
In this case, we shall use the limiting reactant because it will produce the maximum yield of Ag2O as all of it is used up in the reaction.
The limi reactant is AgNO3 and the mass of Ag2O produced can be obtained as follow:
From the balanced equation above,
340g of AgNO3 reacted to produce 232g of Ag2O.
Therefore, 0.2g of AgNO3 will react to produce = (0.2 x 232)/340 = 0.136g of Ag2O.
Therefore, 0.136g of Ag2O was produced from the reaction.
Consider the following reaction where Kc = 6.50×10-3 at 298 K: 2NOBr(g) 2NO(g) + Br2(g) A reaction mixture was found to contain 9.83×10-2 moles of NOBr(g), 5.44×10-2 moles of NO(g), and 4.13×10-2 moles of Br2(g), in a 1.00 liter container. Is the reaction at equilibrium? If not, what direction must it run in order to reach equilibrium? The reaction quotient, Qc
Answer:
This reaction isn't yet at an equilibrium. It must shift in the direction of the reactant (namely [tex]\rm NOBr\; (g)[/tex]) in order to reach an equilibrium.
For this mixture, the reaction quotient is [tex]Q_c = 0.0126[/tex].
Explanation:
A reversible reaction is at equilibrium if and only if its reaction quotient [tex]Q_c[/tex] is equal to the equilibrium constant [tex]K_c[/tex].
Start by calculating the equilibrium quotient [tex]Q_c[/tex] of this reaction. Given the reaction:
[tex]\rm 2\; NOBr\; (g) \rightleftharpoons 2\; NO\; (g) + Br_2\; (g)[/tex].
Let [tex][\mathrm{NOBr\; (g)}][/tex], [tex][\mathrm{NO\; (g)}][/tex], and [tex][\mathrm{Br_2\; (g)}][/tex] denote the concentration of the three species. The formula for the reaction quotient of this system will be:
[tex]\displaystyle Q_c = \frac{[\mathrm{NO\; (g)}]^2 \cdot [\mathrm{Br_2\; (g)}]}{[\mathrm{NOBr\; (g)}]^2}[/tex].
(Note, that in this formula, both [tex][\mathrm{NO\; (g)}][/tex] and [tex][\mathrm{NOBr\; (g)}][/tex] are raised to a power of two. That corresponds to the coefficients in the balanced reaction.)
Calculate the reaction quotient given the concentration of each species:
[tex]\displaystyle Q_c = \frac{[\mathrm{NO\; (g)}]^2 \cdot [\mathrm{Br_2\; (g)}]}{[\mathrm{NOBr\; (g)}]^2} \approx 1.26\times 10^{-2} = 0.0126[/tex].
(Note that the unit is ignored.)
Apparently, [tex]Q_c > K_c[/tex]. Since [tex]Q_c[/tex] and [tex]K_c[/tex] are not equal, this reaction is not at an equilibrium. If external factors like temperature stays the same,
Keep in mind that [tex]Q_c[/tex] denotes a quotient. To reduce the value of a quotient, one may:
reduce the value of the numerator, increase the value of the denominator, orboth.In [tex]Q_c[/tex], that means reducing the concentration of the products while increasing the concentration of the reactants. In other words, the system needs to shift in the direction of the reactants before it could reach an equilibrium.
How many moles of aqueous potassium ions and sulfate ions are formed when 63.7 g of K2SO4 dissolves in water
Answer:
WHEN 63.7 g OF K2SO4 IS DISSOLVED IN WATER, 0.73 MOLES OF POTASSIUM ION AND 0.366 MOLES OF SULFATE ION ARE FORMED.
Explanation:
Equation for the reaction:
K2SO4 + H20 ------->2 K+ + SO4^2-
When K2SO4 dissolves in water, potassium ion and sulfate ion are formed.
1 mole of K2SO4 produces 2 moles and 1 mole of SO4^2-
At STP, 1 mole of K2SO4 will be the molar mass of the substance
Molar mass of K2SO4 = ( 39 *2 + 32 + 16*4) g/mol
Molar mass = 174 g/mol
So therefore;
1 mole of K2SO4 contains 174 g and it produces 2 moles of potassium and 1 mole of sulfate ion
When 63.7 g is used; we have:
174 g = 2 moles of K+
63.7 g = ( 63.7 * 2 / 174) moles of K+
= 0.73 moles of K+
Forr sulfate ion, we have:
174 g = 1 mole ofSO4^2-
63.7 g = (63.7 * 1 / 174) moles of SO4^2-
= 0.366 moles of SO4^2-
In other words, when 63.7 g of K2SO4 is dissolved in water, 0.73 moles of potassium ion and 0.366 moles of sulfate ion are formed.
Why does the excess of base used in these eliminations favor the E2 over the E1 mechanism for elimination
Answer:
The base is involved in the rate determining step of an E2 reaction mechanism
Explanation:
Let us get back to the basics. Looking at an E1 reaction, the rate determining step is unimolecular, that is;
Rate = k [Carbocation] since the rate determining step is the formation of a carbonation.
For an E2 reaction however, the reaction is bimolecular hence for the rate determining step we can write;
Rate = k[alkyl halide] [base]
The implication of this is that an excess of either the alkyl halide or base will facilitate an E2 reaction.
Hence, when excess base is used, E2 reaction is favoured since the base is involved in its rate determining step. In an E1 reaction, the base is not involved in the rate determining step hence an excess of the base has no effect on an E1 reaction.
Therapeutic drugs generally need to have some hydrophobic and hydrophilic components to be able to effectively reach their target organs and tissues given there are aqueous and nonaqueous parts of the body. The degree to which a compound is hydrophobic and hydrophilic can be determined by measuring its relative solubility in water and octanol, C8H17OH, and water. To do this, a sample of the compound is added to a mixture of water and octanol and mixed well. Water and octanol are immiscible so after the mixture settles, the concentration of the compound in water and the concentration of the compound in octanol is measured. The ratio of the concentrations is called the partition ratio:
The question is incomplete as some part is missing:
concentration in octanol Partition Ratio = concentration in water
a) What are the intermolecular forces of attraction between octanol molecules? Explain.
b) Which of the intermolecular forces of attraction identified in (a) account for most of the interactions between octanol molecules? Explain. Use the immiscibility in water and the data included in figures 1 and 2 as evidence to support your answer.
c) Would a compound with a partition coefficient less than one be more hydrophobic or more hydrophilic than one with a partition coefficient greater than 10? Explain.
d) Would nonane (figure 2) be more soluble in water or octanol? Explain.
e) Draw another structure for a compound with the same chemical formula as nonane (CH20) that has a lower boiling point. Explain.
f) Are any of the C atoms in the structure you drew for CH20 sp?hybridized? Explain.
Octanol Boiling point = 195°C Figure 2 Nonane (CH20) Boiling point = 151°C
Answer:
1. The forces between octanol molecules would be attractive. These forces include Vanderwaal forces, H-bonds due to the presence of highly polar O-H group.
2. H-bonding ahould account for most of the attractive forces. The O-H bond should behave like and dipole, oxygen of one molecule attracts the hydrogen of the neighbouring molecule forming D-H...A links throughout (D stands for donor of H-Bond and A for acceptor for H-Bond).
3. Partition coefficient less than 1 will be more hydrophilic, generally drugs with low partition coefficients are regarded as hydrophilic. As parition coefficient of 10 mean more of the solute is dispersed in octanol as compared to water.
4. Nonane is non polar, so it would not dissolve in water. It follows the rule like dissolves like. Polar substances dissolve in polar solvents. 1-octanol is able to bind with water through hydrogen bonds thus its soluble in water but nonane doesn't. Nonane will forms a different layer from water.
5) no all carbons in 2-methyloctane are single bonded. Thus sp3 hybrid. A sp2 hybridised carbon would have a double bond C=C.
please help me I am begging you.. )))): PLEASE HELP ME ~~~~~~~~~~~~~~~~~~~~~~ A football player experiences acute pain in his knee. Which of the following methods can a doctor use to diagnose the reason for the pain? --_-_-____- A.) Use infrared radiation from warm objects to look inside the knees. B.) Use radio waves emitted by radioactive substances to look at bones. C.) Use radiations emitted by very hot objects to penetrate the skin and bones. D.) Use x‒ray radiation to see if there are any fractured bones.
Answer:
D. Use x-ray radiation to see if there are any fractured bones.
Explanation:
The football player may have fractured a bone while he was practicing or playing, so it is best for the doctor to check if the player broke his bone or fractured it.