Answer:
0.1 M
Explanation:
The overall balanced reaction equation for the process is;
IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)
Generally, we must note that;
1 mol of IO3^- require 6 moles of S2O3^2-
Thus;
n (iodate) = n(thiosulfate)/6
C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6
Concentration of iodate C(iodate)= 0.0100 M
Volume of iodate= V(iodate)= 26.34 ml
Concentration of thiosulphate= C(thiosulfate)= the unknown
Volume of thiosulphate=V(thiosulfate)= 15.51 ml
Hence;
C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)
0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M
The displacement of a bromine atom by an amine is a substituion reaction. Write out the mechanism of this reaction (2-->3) Why might you expect that the reaction you have performed, using t-BuNH2, to be much slower than the same reaction using methylamine
Answer:
An alkyl halide can undergo SN2 reaction with an amine
Explanation:
The displacement of a bromine atom by an an amine (step 2---> 3) in the reaction sequence is an example of an SN2 reaction in which the amine is the nucleophile.
The nitrogen atom of the amine which bears a lone pair of electrons functions as the nucleophile and attacks the electrophilic carbon atom of the alkyl halide displacing the bromide and creating a new Carbon-Nitrogen bond. An ammonium intermediate is immediately formed and the reaction is completed by the abstraction of a hydrogen by a base (such as excess amine present in the system).
This reaction is slower with t-BuNH2 because of steric hindrance and steric crowding in the transition state. SN2 reactions are faster with methylamine where the alkyl carbon is easily accessible.
The detailed mechanism of this reaction has been attached to this answer.
A certain radioactive nuclide has a half life of 1.00 hour(s). Calculate the rate constant for this nuclide. s-1 Calculate the decay rate for 1.000 mole of this nuclide. decays s-1
Answer:
k= 1.925×10^-4 s^-1
1.2 ×10^20 atoms/s
Explanation:
From the information provided;
t1/2=Half life= 1.00 hour or 3600 seconds
Then;
t1/2= 0.693/k
Where k= rate constant
k= 0.693/t1/2 = 0.693/3600
k= 1.925×10^-4 s^-1
Since 1 mole of the nuclide contains 6.02×10^23 atoms
Rate of decay= rate constant × number of atoms
Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms
Rate of decay= 1.2 ×10^20 atoms/s
The diagram shows two waves.
How do the frequencies of the waves compare?
Wave A has a lower frequency because it has a
smaller amplitude.
Wave A has a higher frequency because it has a
shorter wavelength.
The waves have the same frequency because they
have the same wavelength.
The waves have the same frequency because they
have the same amplitude.
Answer:
Wave A has a higher frequency because it has a shorter wavelength.
Explanation:
The frequency of a wave and the wave length are related by the following equation:
Velocity (v) = wave length (λ) x frequency (f)
v = λf
If we make frequency (f) the subject of the above equation, we will have:
f = v/λ
Let the velocity (v) be constant.
f = v/λ
f & 1/λ
From the equation above,
We can see that the frequency (f) is inversely proportional to the wavelength (λ).
This implies that a wave with a high frequency, will have a short wavelength and a wave with a short frequency will have a longer wavelength.
Now considering wave A and B in the diagram above,
Wave A will have a higher frequency because it has a shorter wavelength as explained above.
Answer:
it is the second option
Explanation:
Qualitatively estimate the relative melting points for each of the solids, and rank them in decreasing order.
Rank from highest to lowest melting point. To rank items as equivalent, overlap them.
sodium chloride
graphite
solid ammonia
Answer:
Graphite> sodium chloride> solid ammonia
Explanation:
Melting points of solids has a lot to do with the nature of intermolecular forces in the solid. A substance melts when the intermolecular forces holding the crystal lattice has been overcome such that that the crystal structure of the solid just collapses.
Graphite consists of covalently bonded layers of carbon atom which form a giant lattice. The melting point of graphite is very high because of the fact that the strong covalent bonds that hold the carbon atoms together in the layers require a lot of heat energy to break. Grapoghite melts at about 3600°C
Sodium chloride is an ionic compound that melts at about 801°C. The lattice is composed of alternate sodium and chloride ions.
Solid ammonia is held together by much weaker intermolecular interaction hence it has a melting point of about −77.73 °C.
Match each property of a liquid to what it indicates about the relative strength of the intermolecular forces in that liquid.
Strong intermolecular forces
Weak intermolecular forces
Answer:
Strong intermolecular forces: an increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point.
Weak intermolecular forces: a decrease in viscosity, a decrease in surface tension, an increase in vapor pressure and an increase in boiling point.
Explanation:
Intermolecular forces are forces of attraction or repulsion between neighboring molecules in a substance. These intermolecular forces inclde dispersion forces, dipole-dipole interactions, hydrogen bonding, and ion-dipole forces.
The strength of the intermolecular forces in a liquid usually affects the various properties of the liquid such as viscosity, surface tension, vapour pressure and boiling point.
Strong intermolecular forces in a liquid results in the following; an increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point of the liquid.
Weak intermolecular forces in a liquid results in the following; a decrease in viscosity, a decrease in surface tension, an increase in vapor pressure and an increase in boiling point of that liquid.
Strong intermolecular force is defined as the increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point while weak intermolecular forces define as the decrease in viscosity, a decrease in surface tension, an increase in vapor pressure, and an increase in boiling point.
Intermolecular forces are forces of attraction or repulsion between neighboring molecules in a substance. These intermolecular forces include as follows:-
Dispersion forcesDipole-dipole interactionsHydrogen bondingion-dipole forces.
Strong intermolecular forces in a liquid result in the following; an increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point of the liquid.
Weak intermolecular forces in a liquid result in the following; a decrease in viscosity, a decrease in surface tension, an increase in vapor pressure, and an increase in the boiling point of that liquid.
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Calculate the pH of the 1L buffer composed of 500 mL 0.60 M acetic acid plus 500 mL of 0.60 M sodium acetate, after 0.010 mol of NaOH is added (Ka HC2H3O2 = 1.75 x 10-5). Report your answer to the hundredths place.
Answer:
pH = 4.79
Explanation:
The pH of the acetic buffer can be determined using H-H equation:
pH = pKa + log [A⁻] / [HA]
Where pKa is -logKa = 4.76
pH = 4.76 + log [sodium Acetate] / [Acetic Acid]
Where [] can be taken as moles of each specie.
Thus, to find pH of the buffer we need to calculate molesof acetic acid and sodium acetate.
Initial moles:
Initial moles of acetic acid and sodium acetate are:
500mL = 0.500L ₓ (0.60moles / L) = 0.30 moles of both acetic acid and sodium acetate
Moles after reaction:
Now, 0.010 moles of NaOH are added to the buffer reacting with acetic acid, CH₃COOH, producing more acetate ion, as follows:
NaOH + CH₃COOH → CH₃COO⁻ + H₂O
That means after reaction moles of both species are:
Acetic acid: 0.30mol - 0.010mol (Moles that react) = 0.29 moles
Acetate: 0.30mol + 0.010mol (Moles produced) = 0.31 moles
Replacing in H-H equation:
pH = 4.76 + log [0.31] / [0.29]
pH = 4.79
If a radioactive isotope of thorium (atomic number 90, mass number 232) emits 6 alpha particles and 4 beta particles during the course of radioactive decay, what is the mass number of the stable daughter product?
Answer:
The mass number of the stable daughter product is 208
Explanation:
First thing's first, we have to write out the equation of the reaction. This is given as;
²³²₉₀Th → 6 ⁴₂α + 4 ⁰₋₁ β + X
In order to obtain the identity of X, we have to obtain it's mass numbers and atomic number.
There is conservation of matter so we expect the mass number to remain the same in both the reactant and products.
Mass Number
Reactant = 232
Product = (6* 4 = 24) + (4 * 0 = 0) + x = 24 + x
since reactant = product
232 = 24 + x
x = 232 - 24 = 208
Atomic Number
Reactant = 90
Product = (6* 2 = 12) + (4 * -1 = -4) + x = 8 + x
since reactant = product
90 = 8 + x
x = 90 - 8 = 82
Consider Zn + 2HCl → ZnCl2 + H2 (g). If 0.30 mol Zn is added to HCl, how many mol H2 are produced?
Answer:
0.3 mol
Explanation:
Assuming HCl is in excess and Zn is the limiting reagent,
from the balanced equation, we can see the mole ratio of Zn:H2 = 1:1,
which means, each mole of zinc reacted gives 1 mole of H2.
So, if 0.30 mol Zn is added, the no. of moles of H2 produced will also be 0.3 mol, since the ratio is 1:1.
14. Which group of diamagnetic transition metals exhibits trends in density and melting points that don't match the same trends seen in
other groups?
A. Group 3
B. Group 12
C. Group 7
D. Group 11
Answer:
Group 12
Explanation:
Group 12 transition metals are diamagnetic. They behave properties that distinguish them. They naturally have twelve electrons hence their outermost shell is fully filled.
Transition metals have high densities which increases down the group. However, the increase in density of transition elements of group 12 varies with temperature at a rate that is quite different from other transition elements. Hence the differences in the value of melting points and density changes by only a very small amount as you come down group 12 compared to other groups of transition elements.
A sample of an unknown gas effuses in 11.1 min. An equal volume of H2 in the same apparatus at the same temperature and pressure effuses in 2.42 min. What is the molar mass of the unknown gas
Answer:
Molar mass of the gas is 0.0961 g/mol
Explanation:
The effusion rate of an unknown gas = 11.1 min
rate of [tex]H_{2}[/tex] effusion = 2.42 min
molar mass of hydrogen = 1 x 2 = 2 g/m
molar mas of unknown gas = ?
From Graham's law of diffusion and effusion, the rate of effusion and diffusion is inversely proportional to the square root of its molar mass.
from
[tex]\frac{R_{g} }{R_{h} }[/tex] = [tex]\sqrt{\frac{M_{h} }{M_{g} } }[/tex]
where
[tex]R_{h}[/tex] = rate of effusion of hydrogen gas
[tex]R_{g}[/tex] = rate of effusion of unknown gas
[tex]M_{h}[/tex] = molar mass of H2 gas
[tex]M_{g}[/tex] = molar mass of unknown gas
substituting values, we have
[tex]\frac{11.1 }{2.42 }[/tex] = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]
4.587 = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]
[tex]\sqrt{M_{g} }[/tex] = [tex]\sqrt{2}[/tex]/4.587
[tex]\sqrt{M_{g} }[/tex] = 0.31
[tex]M_{g}[/tex] = [tex]0.31^{2}[/tex] = 0.0961 g/mol
The molar mass of the unknown gas will be "0.0961 g/mol".
Given:
Effusion rate of unknown gas,
[tex]R_g = 11.1 \ min[/tex]Effusion rate of [tex]H_2[/tex],
[tex]R_h = 2.42 \ min[/tex]Molar mass of hydrogen,
[tex]M_h = 1\times 2[/tex][tex]= 2 \ g/m[/tex]
According to the Graham's law, we get
→ [tex]\frac{R_g}{R_h} = \sqrt{\frac{M_h}{M_g} }[/tex]
By substituting the values, we get
→ [tex]\frac{11.1}{2.42} = \sqrt{\frac{2}{M_g} }[/tex]
→ [tex]4.587=\sqrt{\frac{2}{M_g} }[/tex]
→ [tex]\sqrt{M_g} = \sqrt{\frac{2}{4.587} }[/tex]
[tex]\sqrt{M_g} = 0.31[/tex]
[tex]M_g = 0.0961 \ g/mol[/tex]
Thus the above solution is right.
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The decomposition of H2O2 is first order in H2O2 and the rate constant for this reaction is 1.63 x 10-4 s-1. How long will it take for [H2O2] to fall from 0.95 M to 0.33 M?
Answer:
It will take 6486.92 minutes for [H2O2] to fall from 0.95 M to 0.33 M
Explanation:
The order of reaction is defined as the sum of the powers of the concentration terms in the equation. Order of a reaction is given by the number of atoms or molecule whose concentration change during the reaction and determine the rate of reaction.
In first order reaction;
[tex]In \dfrac{a}{a_o-x}= k_1 t[/tex]
where;
a = concentration at time t
[tex]a_o[/tex] = initial concentration
and k = constant.
[tex]In (\dfrac{0.33}{0.95})= -1.63 \times 10^{-4} \times t[/tex]
[tex]-1.05736933 = -1.63 \times 10^{-4} \times t[/tex]
[tex]t = \dfrac{-1.05736933}{ -1.63 \times 10^{-4} }[/tex]
t = 6486.92 minutes
Which of the following best describes hydrocarbons? a. Alkanes in which a hydrogen atom is replaced by a hydroxyl group b. Binary compounds of carbon and hydrogen c. Organic compounds containing water and carbon d. Covalently bonded carbon compounds which have intermolecular force attractions to hydrogen compounds e. Compounds which are formed by the reaction of a naturally occurring carbon-containing substance and water
Answer:
b. Binary compounds of carbon and hydrogen
Explanation:
Before proceeding, Hydrocarbons refers to organic chemical compounds composed exclusively of hydrogen and carbon atoms. This means the only elements present in an hydrocarbon are;
- Carbon
- Hydrogen
Looking through the options;
- Option A: This is wrong because the hydroxyl group contains oxygen and hydrocarbons contain only hydrogen and carbon.
- option B: This is correct. Binary compounds refers to compounds with just two elements.
- option C: This is wrong because water contains oxygen and hydrocarbons contain only hydrogen and carbon.
- option D: Carbon atoms can contain other elements so this option is wrong.
- option E: This also wrong because we had already gotten the correct option.
what are the similarities between amorphous solid and crystalline solid
Answer:
solid dont know
Explanation:
so sorry ask another
A buffered solution containing dissolved aniline, C6H5NH2, and aniline hydrochloride, C6H5NH3Cl, has a pH of 5.57 . A. Determine the concentration of C6H5NH+3 in the solution if the concentration of C6H5NH2 is 0.200 M. The pKb of aniline is 9.13. g
Answer:
[C₆H₅NH₃⁺] = 0.0399 M
Explanation:
This excersise can be easily solved by the Henderson Hasselbach equation
C₆H₅NH₃Cl → C₆H₅NH₃⁺ + Cl⁻
pOH = pKb + log (salt/base)
As we have value of pH, we need to determine the pOH
14 - pH = pOH
pOH = 8.43 (14 - 5.57)
Now we replace data:
pOH = pKb + log ( C₆H₅NH₃⁺/ C₆H₅NH₂ )
8.43 = 9.13 + log ( C₆H₅NH₃⁺ / 0.2 )
-0.7 = log ( C₆H₅NH₃⁺ / 0.2 )
10⁻⁰'⁷ = C₆H₅NH₃⁺ / 0.2
0.19952 = C₆H₅NH₃⁺ / 0.2
C₆H₅NH₃⁺ = 0.19952 . 0.2 = 0.0399 M
Determine the volumes of 0.10 M CH3COOH and 0.10 M CH3COONa required to prepare 10 mL of the following pH buffers: pH 4.7, pH 5.7. (Note: the pKa of CH3COOH
Answer:
pH 4.7: 5mL of 0.10 M CH3COOH and 5mL 0.10 M CH3COONa
pH 5.7: 0.91mL of 0.10 M CH3COOH and 9.09mL 0.10 M CH3COONa
Explanation:
pKa acetic acid, CH3COOH = 4.7
It is possible to determine pH of a buffer using H-H equation:
pH = pka + log [A⁻] / [HA]
For the acetic buffer,
pH = 4.7 + log [CH3COONa] / [CH3COOH]
As you want a pH 4.7 buffer:
4.7 = 4.7 + log [CH3COONa] / [CH3COOH]
1 = [CH3COONa] / [CH3COOH]
That means you need the same amount of both species of the buffer to make the pH 4.7 buffer. That is:
5mL of 0.10 M CH3COOH and 5mL 0.10 M CH3COONaFor pH 5.7:
5.7 = 4.7 + log [CH3COONa] / [CH3COOH]
1 = log [CH3COONa] / [CH3COOH]
10 = [CH3COONa] / [CH3COOH] (1)
That means you need 10 times [CH3COONa] over [CH3COOH]
And as you know:
10mL= [CH3COONa] + [CH3COOH] (2)
Replacing (1) in (2):
10 = 10mL + [CH3COOH] / [CH3COOH]
10[CH3COOH] = 10mL + [CH3COOH]
11[CH3COOH] = 10mL
[CH3COOH] = 0.91mL
And [CH3COONa] = 10mL - 0.91mL =
[CH3COONa] = 9.09mL
That is:
0.91mL of 0.10 M CH3COOH and 9.09mL 0.10 M CH3COONaThe volumes according to the pH are as follows:
(i) 5mL of 0.10 M CH₃COOH and 5mL 0.10 M CH₃COONa for pH 4.7
(ii) 0.91mL of 0.10 M CH₃COOH and 9.09mL 0.10 M CH₃COONa pH 5.7
Calculating the volume of chemicals needed:Given that pKa of acetic acid, CH₃COOH = 4.7
The pH of a buffer using the H-H equation is given by:
pH = pKa + log [A⁻] / [HA]
For the acetic buffer,
pH = 4.7 + log [CH₃COONa] / [ CH₃COOH]
4.7 = 4.7 + log [CH₃COONa] / [ CH₃COOH]
0 = log [CH₃COONa] / [ CH₃COOH]
takin antilog on both sides of the equation we get:
1 = [CH₃COONa] / [CH₃COOH]
It implies that the same amount of both species is needed to make the pH 4.7 buffer.
So,
5mL of 0.10 M CH₃COOH and 5mL 0.10 M CH₃COONa makes a buffer of pH 4.7
Similarly:
5.7 = 4.7 + log [CH₃COONa] / [CH₃COOH]
1 = log [CH₃COONa] / [CH₃COOH]
takin antilog on both sides of the equation we get:
10 = [CH₃COONa] / [CH₃COOH]
10[CH₃COOH] = [CH₃COONa]
It implies that we need 10 times [CH₃COONa] as much of [CH₃COOH]
We have to prepare 10 mL of buffer, so:
10mL= [CH₃COONa] + [CH₃COOH]
10mL = 11[CH₃COOH]
[CH₃COOH] = 0.91mL
So, [CH₃COONa] = 10mL - 0.91mL
[CH₃COONa] = 9.09mL
Therefore,
0.91mL of 0.10 M CH3COOH and 9.09mL 0.10 M CH3COONa is required to make a buffer of pH 5.7
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What are the solutions to the quadratic equation 2x2 + 10x - 48 = 0?
Answer:
x = 3 , x= -8Explanation:
[tex]2x^2+10x-48\\=2\left(x^2+5x-24\right)\\x^2+5x-24\\=\left(x^2-3x\right)+\left(8x-24\right)\\=x\left(x-3\right)+8\left(x-3\right)\\=\left(x-3\right)\left(x+8\right)\\=2\left(x-3\right)\left(x+8\right)\\2\left(x-3\right)\left(x+8\right)=0\\x-3=0\\x = 0+3\\x = 3\\x+8=0\\x+8-8=0-8\\x=-8\\x=3,\:x=-8[/tex]
1. Unas de las formas de producir nitrógeno gaseoso (N2) es mediante la oxidación de metilamina (CH3NH2), tal como se muestra en la siguiente reacción: CH3NH2 + O2 → CO2 + H2O + N2 Si reaccionan 0,5 mol de metil amina (CH3NH2) con 25,6 g de O2. Determine: a) Balancee la ecuación. (2 ptos) b) ¿Cuántos gramos de nitrógeno (N2) se pueden producir? (4 ptos) c) Si experimentalmente se obtuvieron 3,5 gramos de N2. Determine el porcentaje de rendimiento de la reacción. (4 ptos) Por favor es urgente!!!
Answer:
a) 4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂
b) m = 5,043 g
c) % = 69,4 %
Explanation:
a) La ecuación balanceada es la siguiente:
4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂
En el balanceo, se tiene en la relación estequiométrica que 4 moles de metilamina reacciona con 9 moles de oxígeno para producir 4 moles de dióxido de carbono, 10 moles de agua y 2 moles de nitrógeno.
b) Para determinar la masa de nitrógeno se debe calcular primero el reactivo limitante:
[tex]n_{O_{2}} = \frac{m}{M} = \frac{25,6 g}{31,99 g/mol} = 0,800 moles[/tex]
[tex]n_{CH_{3}NH_{2}} = \frac{4}{9}*0,800 moles = 0,356 moles[/tex]
De la ecuación anterior se tiene que la cantidad de moles de metilamina necesaria para reaccionar con 0,800 moles de oxígeno es 0,356 moles, y la cantidad de moles iniciales de metilamina es 0,5 moles, por lo tanto el reactivo limitante es el oxígeno.
Ahora, podemos calcular la masa de nitrógeno producida:
[tex]n_{N_{2}} = \frac{2}{9}*n_{O_{2}} = \frac{2}{9}*0,8 moles = 0,18 moles[/tex]
[tex]m_{N_{2}} = n_{N_{2}}*M = 0,18 moles*28,014 g/mol = 5,043 g[/tex]
Por lo tanto, se pueden producir 5,043 g de nitrógeno.
c) El redimiento de la reacción se puede calcular usando la siguiente fórmula:
[tex] \% = \frac{R_{r}}{R_{T}}*100 [/tex]
Donde:
[tex]R_{r}[/tex]: es el rendimiento real
[tex]R_{T}[/tex]: es el rendimiento teórico
[tex]\% = \frac{3,5}{5,043}*100 = 69,4[/tex]
Entonces, el procentaje de rendimiento de la reacción es 69,4%.
Espero que te sea de utilidad!
The amount of space an object takes up is called _____. gravity weight mass volume
A solution of malonic acid, H2C3H2O4, was standardized by titration with 0.0990 M NaOH solution. If 20.52 mL mL of the NaOH solution is required to neutralize completely 11.13 mL of the malonic acid solution, what is the molarity of the malonic acid solution
Answer:
0.0913 M
Explanation:
We'll begin by writing the balanced equation for the reaction.
This is given below:
H2C3H2O4 + 2NaOH —> C3H2Na2O4 + 2H2O
From the balanced equation above, we obtained the following:
The mole ratio of the acid (nA) = 1
The mole ratio of the base (nB) = 2
Data obtained from the question include:
Molarity of base, NaOH (Mb) = 0.0990 M
Volume of base, NaOH (Vb) = 20.52 mL
Volume of acid, H2C3H2O4 (Va) = 11.13 mL
Molarity of acid, H2C3H2O4 (Ma) =..?
The molarity of the acid, H2C3H2O4 can be obtained as follow:
MaVa/MbVb = nA/nB
Ma x 11.13 / 0.0990 x 20.52 = 1/2
Cross multiply
Ma x 11.13 x 2 = 0.0990 x 20.52 x 1
Divide both side by 11.13 x 2
Ma = (0.0990 x 20.52)/ (11.13 x 2)
Ma = 0.0913 M
Therefore, the molarity of malonic acid, H2C3H2O4 solution is 0.0913 M
Determine the volume occupied by 10 mol of helium at 27 ° C and 82 atm
please.
Answer:
3.00 L
Explanation:
Convert the pressure to Pascals.
P = 82 atm × (101325 Pa/atm)
P = 8,308,650 Pa
Convert temperature to Kelvins.
T = 27°C + 273
T = 300 K
Use ideal gas law:
PV = nRT
(8,308,650 Pa) V = (10 mol) (8.314 J/mol/K) (300 K)
V = 0.00300 m³
If desired, convert to liters.
V = (0.00300 m³) (1000 L/m³)
V = 3.00 L
Answer:
[tex]\large \boxed{\text{3.0 L}}[/tex]
Explanation:
[tex]\begin{array}{rcl}pV &=& nRT\\\text{82 atm} \times V & = & \text{10 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{300.15 K}\\82V & = & \text{246 L}\\V & = & \textbf{3.0 L} \\\end{array}\\\text{The volume of the balloon is $\large \boxed{\textbf{3.0 L}}$}[/tex]
Write the net ionic equation for any precipitation reaction that may be predicted when aqueous solutions of manganese(II) nitrate and sodium hydroxide are combined.
Answer:
Explanation:
Mn( NO₃ )₂ + 2Na OH = Mn( OH)₂ (s) ↓ + 2Na NO₃
Converting into ions
Mn⁺ + 2 NO₃⁻ + 2 Na⁺ + 2 OH⁻ = Mn( OH)₂ + 2 Na⁻ + 2 NO₃⁻
Cancelling out common terms
Mn⁺ + 2 OH⁻ = Mn( OH)₂
this is net ionic equation required.
A student mixes 43.8 mL of acetone (58.08 g/mol, 0.791 g/mL) with excess benzaldehyde and NaOH to produce 79.4 g of (1E,4E)-1,5-diphenylpenta-1,4-dien-3-one (234.29 g/mol). What is the percent yield of this student's experiment
Answer:
% yield of the student's experiment is
[tex]\frac{0.34}{0.60}[/tex] ˣ 100 = 56.67%
Explanation:
given
volume of acetone= 43.8 mL
molar weight of acetone = 58.08 g/mol
density of acetone = 0.791 g/mL
A student mixes 43.8 mL of acetone (58.08 g/mol, 0.791 g/mL)
43.8 mL = 43.8mL × 0.791g/mL
= 34.6458g ≈34.65g
1 mole of acetone = 58.08g
∴34.65g = 34.65g/58.08g
= 0.60mol
molecular weight of the product 1,5-diphenylpenta-1,4-dien-3-one = 234.29 g/mol
mole = mass/ molar weight
mole = 79.4g/ 234.29g/mol
mole(n) = 0.3389mol ≈ 0.34mol
1 mole of acetone will produce 1 mole of the product
∴0.60mol of acetone will produce 0.60mol of the product
but we get 0.34mol of the product
∴ % yield of the student's experiment is
[tex]\frac{0.34}{0.60}[/tex] ˣ 100 = 56.67%
g Increasing the number of unsaturations in a fatty acid ____________ the melting temperature of the fatty acid.
Answer:
Decreases
Explanation:
Fatty acid which have the double bond or triple bond are called unsaturated fatty acids. Because of the double or triple bond, unsaturated fatty acids are loosely packed and form some distance among molecules which lowers the melting point of unsaturated fatty acids.
So, if the unsaturation of fatty acid will increase, it leads to more branched and loosely packed molecules and decreases the melting temperature accordingly.
What element is primarily used in appliances to make electronic chips
A. Silicon (Si)
B. Nickel (Ni)
C. Copper (Cu)
D. Selenium (Se)
Answer:
Option A
Explanation:
Silicon (Obtained from Sand (SiO2)) is the element that is primarily used in appliances to make electronic chips.
Answer:
A. Silicon (Si)
Explanation:
Silicon (Si) is primarily used as a semiconductor material to make electronic chips.
It takes 242. kJ/mol to break a chlorine-chlorine single bond. Calculate the maximum wavelength of light for which a chlorine-chlorine single bond could be broken by absorbing a single photon. Round your answer to 3 significant digits. single by absorbing a significant digit.
Answer:
495nm
Explanation:
The energy of a photon could be obtained by using:
E = hc / λ
Where E is energy of a photon, h is Planck's constant (6.626x10⁻³⁴Js), c is speed of the light (3x10⁸ms⁻¹) and λ is wavelength.
The energy to break 1 mole of Cl-Cl bonds is 242kJ = 242000J. The energy yo break a single bond is:
242000J/mol ₓ (1mol / 6.022x10²³bonds) = 4.0186x10⁻¹⁹J/bond.
Replacing in the equation:
E = hc / λ
4.0186x10⁻¹⁹J = 3x10⁸ms⁻¹ₓ6.626x10⁻³⁴Js / λ
λ = 4.946x10⁻⁷m
Is maximum wavelength of light that could break a Cl-Cl bond.
Usually, wavelength is given in nm (1x10⁻⁹m / 1nm). The wavelength in nm is:
4.946x10⁻⁷m ₓ (1nm / 1x10⁻⁹m) =
495nmA vehicle travels 2345 meter in 35 second toward the evening sun in the West. What is its speed? A. 47 m/s West
Explanation:
Speed = 2345 ÷ 35 = 67m/s
a certain compound was found to contain 54.0 g of carbon and 10.5 grams of hydrogen. its relative molecular mass is 86.0. find the empirical and molecular formulas
Answer:
empirical formula = C3H7
molecular formula = C6H14
What mass of aluminum metal can be produced per hour in the electrolysis of a molten aluminum salt by a current of 21 A? Express your answer using two significant figures.
Answer
mass of aluminum metal= 7 .0497g of Al
Explanation:
current = 21 A
time = 1 hour = 60 X 60 = 3600 s
quantity of electricity passed = current X time = 21X 3600 = 75600 C
Following the electrolysis the below reaction will occur :
Al3+ + 3e- --------> Al
therefore, 3F i.e. 3 X 96500 C = 289500 C gives 1 mole of Al
so 1 C will produce 1/289500 moles of Al
so 108000 C will produce 1/289500 X 75600 = 0.2611 moles of Al
now 1 mole of aluminium weighs = 27 g/mole
so 0.2611 moles of Al = 0.2611 X 27 = 7 .0497 g
mass of aluminum metal= 7 .0497 g of Al
The mass of aluminum metal can be produced per hour in the electrolysis of a molten aluminum salt by a current of 21 A is 7.05 g
We'll begin by calculating the the quantity of electricity used. This can be obtained as follow:
Current (I) = 21 A
Time(t) = 1 h = 60 × 60 = 3600 s
Quantity of electricity (Q) =?Q = it
Q = 21 × 3600
Q = 75600 CFinally, we shall determine the mass of the aluminum metal produced. Al³⁺ + 3e —> AlRecall:
1 mole of Al = 27 g
1 electron (e) = 96500 C
Thus,
3 electrons = 3 × 96500 = 289500 C
From the balanced equation above,
289500 C of electricity produced 27 g of Al.
Therefore,
75600 C of electricity will produce = (75600 × 27) / 289500 = 7.05 g of Al
Thus, the mass of the aluminum metal obtained is 7.05 g
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What amounts of sodium benzoate would be required to prepare 2.5L of 0.35M benzoic buffer solution with a pH of 6.10? Ka of benzoic acid = 6.5 x 10-5 MW benzoic acid, HC7H5O2, is 122.01 MW sodium benzoate, NaC7H5O2, is 144.01
Answer:
Benzoic acid: 1.288g
Sodium benzoate: 124.48g
Explanation:
Benzoic acid, HC7H5O2 is in equilibrium with its conjugate base, C7H5O2⁻ producing a buffer. The pH of the buffer can be determined following H-H equation:
pH = pKa + log [C7H5O2⁻] / [HC7H5O2] (1)
Where pH is desire pH = 6.10 pKa is -log Ka = 4.187 and [] are molar concentrations of the buffer.
As you want to prepare 2.5L of a 0.35M of buffer, moles of buffer are:
2.5L ₓ (0.35mol / L) = 0.875moles of buffer.
And you can write:
0.875 moles = [C7H5O2⁻] + [HC7H5O2] (2)
Replacing (2) in (1)
pH = pKa + log [C7H5O2⁻] / [HC7H5O2]
6.10 = 4.187 + log [C7H5O2⁻] / [HC7H5O2]
1.913 = log [C7H5O2⁻] / [HC7H5O2]
81.846 = 0.875mol - [HC7H5O2] / [HC7H5O2]
81.846 [HC7H5O2] = 0.875mol - [HC7H5O2]
82.846 [HC7H5O2] = 0.875mol
[HC7H5O2] = 0.01056 molesAnd moles of the benzoate, [C7H5O2⁻]:
[C7H5O2⁻] = 0.875mol - 0.01056mol =
[C7H5O2⁻] = 0.8644molUsing molar mass of benzoic acid and sodium benzoate, amount of each compound you must add to prepare 2.5L of the buffer are:
Benzoic acid: 0.01056mol ₓ (122.01g/mol) = 1.288g
Sodium benzoate: 0.8644mol ₓ (144.01g/mol) = 124.482g
When the owners of some wells in Pallerla started using high-powered motors to
draw water from the wells, the owners of other wells noticed that their wells were
drying up. Discuss the possible solution to the problem solutions to the problem
Answer:
The possible solution is to balance the rate of water removal from the well to the rate of natural recharge of the well from its underground aquifer.
Explanation:
A well is an excavation in the earth, made with the aim of extracting water from the aquifers. The water from a well can be drawn up by the means of a pump, containers, such as buckets, or by hand. Aquifers can also be recharged through a well.
Well draw down occurs when water from the well is drained faster than it is naturally recharged from the aquifer. This can be as a result of over pumping, extended drought, among other factors. The use of the high-powered motor in this case, for pumping, might be the possible cause of the well drying up. The situation might have resulted from the pump drawing out water from the well at a rate tat exceeds the rate at which it is recharged naturally, causing the well water to start drying up. There's also a possibility that the well is pumped indiscriminately, possibly leading to wastage of water.
The solution to this problem is to give the well a time duration for it to recharge itself. Then, the rate of recharges should be calculated and determined by an hydrologist. When all these is done, a pump with a motor power that does not exceed the calculated recharge rate should be used in place of the high-powered motor. Also, water usage should be brought to the minimum level to prevent unnecessary pumping due to excessive, wasteful use of water.