Answer:
C) as ionic strength increases, the value of the activity coefficient increases.
Explanation:
The effective concentration of ions available for reactions is known as the activity of the ion.
The activity coefficient important in chemistry because it accounts for the deviation of a solution from ideal behaviour.
The activity of a chemical species is defines as the product of concentration and activity coefficient.
Following the Debye–Hückel limiting law; log γ = −0.509z2I1/2. The ionic strength of a solution tends to increase as the activity coefficient (γ) of the ion decreases.
How many moles of (CH3)3NH+ are in 6.0 g of (CH3)3NH+?
Answer:
0.1 mol
Explanation:
6/(15*3+15)
0.1 mol moles of (CH3)3NH+ are in 6.0 g of (CH3)3NH+
What is mole?
The mole, symbol mol, exists as the SI base unit of the amount of substance. The quantity amount of substance exists as a measure of how many elementary entities of a provided substance exist in an object or sample.A mole corresponds to the mass of a substance that includes 6.023 x 1023 particles of the substance. The mole exists the SI unit for the amount of a substance. Its symbol stands mol.
The compound trimethylamine, (CH3 )3N, exists as a weak base when dissolved in water.
A mole exist expressed as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole exists as a convenient unit to utilize because of the great number of atoms, molecules, or others in any substance.
To find the amount of the substance (CH3)3NH+ to calculate its molar mass:
M((CH3)3NH+) = (12+3)*3 + 14+1 = 60 g/mol
n((CH3)3NH+) = m/M
m((CH3)3NH+) = 6g
Thus,
n((CH3)3NH+) = 6g/60 g/mol = 0.1 mol
Hence,
n((CH3)3NH+) = 0.1 mol
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a) Define typical polyfunctional acid ?
b) Show the equations of dissociation mechanism of phosphoric acid as an example.
c) Write the equation for calculating the [H3O*].
a) A polyfunctional acid is an acid that has more than one functional group.
b) The equations of dissociation of phosphoric acid are:
H₃PO₄ + H₂O ⇄ H₂PO₄⁻ + H₃O⁺ H₂PO₄⁻ + H₂O ⇄ HPO₄²⁻ + H₃O⁺ HPO₄²⁻ + H₂O ⇄ PO₄³⁻ + H₃O⁺c) The equation for calculating the concentration of H₃O⁺ is [tex] [H_{3}O^{+}] = (\frac{K_{1}K_{2}K_{3}[H_{3}PO_{4}]}{[PO_{4}^{-3}]})^{1/3} [/tex]
a) A polyfunctional acid can be defined as an acid that has more than one functional group. Phosphoric acid (H₃PO₄) is an example of polyfunctional acid since it is composed of three hydroxyl groups joined to a phosphorus atom, which is also joined to an oxygen atom by a double bound. In that structure, the three hydrogen atoms of the hydroxyl groups give the acidic behavior to this compound.
b) Phosphoric acid has three equations of dissociation:
H₃PO₄ + H₂O ⇄ H₂PO₄⁻ + H₃O⁺ (1)H₂PO₄⁻ + H₂O ⇄ HPO₄²⁻ + H₃O⁺ (2)HPO₄²⁻ + H₂O ⇄ PO₄³⁻ + H₃O⁺ (3)The dissociation constants for the three above equations are:
[tex] K_{1} = \frac{[H_{2}PO_{4}^{-}][H_{3}O^{+}]}{[H_{3}PO_{4}]} [/tex] (4)
[tex] K_{2} = \frac{[HPO_{4}^{2-}][H_{3}O^{+}]}{[H_{2}PO_{4}^{-}]} [/tex] (5)
[tex] K_{3} = \frac{[PO_{4}^{3-}][H_{3}O^{+}]}{[HPO_{4}^{2-}]} [/tex] (6)
c) We can calculate the concentration of H₃O⁺ for each equilibrium with the equations (4), (5), and (6).
The general reaction of dissociation of phosphoric acid is given by the sum of equations (1), (2), and (3):
H₃PO₄ + 3H₂O ⇄ PO₄³⁻ + 3H₃O⁺ (7)
The concentration of H₃O⁺ for the total dissociation reaction (eq 7) can be found as follows:
[tex] K_{t} = \frac{[PO_{4}^{-3}][H_{3}O^{+}]^{3}}{[H_{3}PO_{4}]} [/tex] (8)
Where:
[tex] K_{t} = K_{1}*K_{2}*K_{3} [/tex]
Hence, by knowing the dissociation constants K₁, K₂ and K₃, and the concentrations of PO₄³⁻ and H₃PO₄, the [H₃O⁺] is:
[tex][H_{3}O^{+}] = (\frac{K_{1}K_{2}K_{3}[H_{3}PO_{4}]}{[PO_{4}^{-3}]})^{1/3}[/tex]
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9. Consider a magnesium atom with charge +2. How many overall electrons are on this particle?
Hint: Magnesium's atomic number is 12.
10
12
14
A solution is made by dissolving 5.84 grams of NaCl in enough distilled water to give a final volume of 1.00 L. What is the molarity of the solution
Group of answer choices
0.0250 M
0.400 M
0.100 M
1.00 M
Answer:
Explanation:
1. A solution is made by dissolving 5.84g of NaCl is enough distilled water to a give a final volume of 1.00L. What is the molarity of the solution? a. 0.100 M b. 1.00 M c. 0.0250 M d. 0.400 M 2. A 0.9% NaCl (w/w) solution in water is a. is made by mixing 0.9 moles of NaCl in a 100 moles of water b. made and has the same final volume as 0.9% solution in ethyl alcohol c. a solution that boils at or above 100°C d. All the above (don't choose this one) 3. In an exergonic process, the system a. gains energy b. loses energy c. either gains or loses energy d. no energy change at all
Answer:
[tex]\boxed {\boxed {\sf 0.100 \ M }}[/tex]
Explanation:
Molarity is a measure of concentration in moles per liter.
[tex]molarity = \frac{moles \ of \ solute}{liters \ of \ solution}}[/tex]
The solution has 5.84 grams of sodium chloride or NaCl and a volume of 1.00 liters.
1. Moles of SoluteWe are given the mass of solute in grams, so we must convert to moles. This requires the molar mass, or the mass of 1 mole of a substance. These values are found on the Periodic Table as the atomic masses, but the units are grams per mole, not atomic mass units.
We have the compound sodium chloride, so look up the molar masses of the individual elements: sodium and chlorine.
Na: 22.9897693 g/mol Cl: 35.45 g/molThe chemical formula (NaCl) contains no subscripts, so there is 1 mole of each element in 1 mole of the compound. Add the 2 molar masses to find the compound's molar mass.
NaCl: 22.9897693 + 35.45 = 58.4397693 g/molThere are 58.4397693 grams of sodium chloride in 1 mole. We will use dimensional analysis and create a ratio using this information.
[tex]\frac {58.4397693 \ g\ \ NaCl} {1 \ mol \ NaCl}[/tex]
We are converting 5.84 grams to moles, so we multiply by that value.
[tex]5.84 \ g \ NaCl *\frac {58.4397693 \ g\ NaCl} {1 \ mol \ NaCl}[/tex]
Flip the ratio. It remains equivalent and the units of grams of sodium chloride cancel.
[tex]5.84 \ g \ NaCl *\frac {1 \ mol \ NaCl}{58.4397693 \ g\ NaCl}[/tex]
[tex]5.84 *\frac {1 \ mol \ NaCl}{58.4397693 }[/tex]
[tex]0.09993194823 \ mol \ NaCl[/tex]
2. MolarityWe can use the number of moles we just calculated to find the molarity. Remember there is 1 liter of solution.
[tex]molarity= \frac{moles \ of \ solute}{liters \ of \ solution}[/tex]
[tex]molarity= \frac{ 0.09993194823 \ mol \ NaCl}{1 \ L}[/tex]
[tex]molarity= 0.09993194823 \ mol \ NaCl/L[/tex]
3. Units and Significant FiguresThe original measurements of mass and volume have 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 9 in the ten-thousandths place tells us to round the 9 to a 0, but then we must also the next 9 to a 0, and the 0 to a 1.
[tex]molarity \approx 0.100 \ mol \ NaCl/L[/tex]
1 mole per liter is 1 molar or M. We can convert the units.
[tex]molarity \approx 0.100 \ M \ NaCl[/tex]
The molarity of the solution is 0.100 M.
Calculate the numerical value of the equilibrium constant, Kc, for the reaction below if the equilibrium concentrations for CO, H2 , CH4 and H2O are 0.989 M, 0.993 M, 1.078 M and 0.878 M, respectively. (calculate your answer to three sig figs)
CO(g) + 3 H2(g) ⇌ CH4(g) + H2O(g)
Kc = [CH4]×[H2O] / [CO]×[H2]^3
Kc = 1.078×0.878 / (0.989×0.933^3)
Kc = 0.977
The numerical value of the equilibrium constant, Kc, for the given reaction is found to be 0.977.
What is Equilibrium constant?The Equilibrium constant may be defined as the numerical value that significantly indicated the correlation between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a definite temperature.
According to the question, the reaction is as follows:
[tex]CO +3H_2[/tex] ↔ [tex]CH_4+ H_2O[/tex].
The equilibrium concentrations are 0.989 M, 0.993 M, 1.078 M and 0.878 M, respectively.
Now, the equilibrium constant is calculated by the following formula:
Kc = [CH4]×[H2O] / [CO]×[tex][H_2]^3[/tex]= 1.078×0.878 / (0.989×0.93[tex]3^3[/tex]).
= 0.9464/(0.989 × 0.8121)
= 0.977.
Therefore, the numerical value of the equilibrium constant, Kc, for the given reaction is found to be 0.977.
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6. In a particular atom, an electron moves from n = 3 to the ground state (n = 1), emitting a photon with frequency 5.2 x 1015 Hz as it does so. What is the difference in energy between n = 3 and n = 1 in this atom? g
Answer: The question wants you to determine the energy that the incoming photon must have in order to allow the electron that absorbs it to jump from
n
i
=
2
to
n
f
=
6
.
A good starting point here will be to calculate the energy of the photon emitted when the electron falls from
n
i
=
6
to
n
f
=
2
by using the Rydberg equation.
1
λ
=
R
⋅
(
1
n
2
f
−
1
n
2
i
)
Here
λ
si the wavelength of the emittted photon
R
is the Rydberg constant, equal to
1.097
⋅
10
7
m
−
1
Plug in your values to find
1
λ
=
1.097
⋅
10
7
.
m
−
1
⋅
(
1
2
2
−
1
6
2
)
1
λ
=
2.4378
⋅
10
6
.
m
−
1
This means that you have
λ
=
4.10
⋅
10
−
7
.
m
So, you know that when an electron falls from
n
i
=
6
to
n
f
=
2
, a photon of wavelength
410 nm
is emitted. This implies that in order for the electron to jump from
n
i
=
2
to
n
f
=
6
, it must absorb a photon of the same wavelength.
To find the energy of this photon, you can use the Planck - Einstein relation, which looks like this
E
=
h
⋅
c
λ
Here
E
is the energy of the photon
h
is Planck's constant, equal to
6.626
⋅
10
−
34
.
J s
c
is the speed of light in a vacuum, usually given as
3
⋅
10
8
.
m s
−
1
As you can see, this equation shows you that the energy of the photon is inversely proportional to its wavelength, which, of course, implies that it is directly proportional to its frequency.
Plug in the wavelength of the photon in meters to find its energy
E
=
6.626
⋅
10
−
34
.
J
s
⋅
3
⋅
10
8
m
s
−
1
4.10
⋅
10
−
7
m
E
=
4.85
⋅
10
−
19
.
J
−−−−−−−−−−−−−−−−−
I'll leave the answer rounded to three sig figs.
So, you can say that in a hydrogen atom, an electron located on
n
i
=
2
that absorbs a photon of energy
4.85
⋅
10
−
19
J
can make the jump to
n
f
=
6
.
Explanation:
100 mL of 0.2 mol/L sodium carbonate solution and 200 mL of 0.1 mol/L calcium nitrate solution are mixed together. Calculate the mass of calcium carbonate that would precipitate and the concentration of the sodium nitrate solution that will be produced.
Answer:
Explanation:
Na2CO3+Ca(NO3)2=CaCO3+2NaNO3
nNa2CO3=0.02
nCa(NO3)2=0.02
mCaCO3=0.02*100=2 gram
nNaNo3=0.04
Cm=2/15
From the calculation, the mass of the product is 2 g.
What is a reaction?A chemical reaction occurs when two more substances are mixed together. In this case, the reaction is shown by; Ca(NO3)2 + Na2CO3 ----> CaCO3(s) + 2NaNO3.
Number of moles of Na2CO3 = 100/1000 L * 0.2 mol/L = 0.02 moles
Number of moles of Ca(NO3)2 = 200/1000 L * 0.1 mol/L = 0.02 moles
Since the reaction is equimolar, amount of the product = 0.02 moles * 100 g/mol = 2 g
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The half life of radium-226 is 1600 years. If you have 200 grams of radium today how many grams would be present in 8000 years?
Answer:
Half life is the time taken by a radio active isotope to reduce by half of its original amount. Radium-226 has a half life of 1602 years meaning that it would take 1602 years for a mass of radium to reduce by half.
Number of half lives in 9612 years = 9612/1602 = 6 half lives
New mass = Original mass x (1/2)n where n is the number of half lives.
Therefore, New mass= 500 x (1/2)∧6
= 500 x 0.015625
= 7.8125 g
Hence the mass of radium after 9612 years will be 7.8125 grams.
Explanation:
Answer:
[tex]\boxed {\boxed {\sf 6.25 \ grams}}[/tex]
Explanation:
We are asked to find the mass of a sample of radium-226 after half-life decay. We will use the following formula:
[tex]A= A_o *\frac{1}{2}^{\frac{t}{h}}[/tex]
In this formula, [tex]A_o[/tex] is the initial amount, t is the time, and h is the half-life.
For this problem, the initial amount is 200 grams of radium-226, the time is 8,000 years, and the half-life is 1,600 years.
[tex]\bullet \ A_o= 200 \ g \\\\bullet \ t= 8,000 \ \\\bullet \ h= 1,600[/tex]
Substitute the values into the formula.
[tex]A= 200 \ g * \frac{1}{2} ^{\frac{8.000}{1,600}[/tex]
Solve the fraction in the exponent.
[tex]A= 200 \ g * \frac{1}{2}^{5}[/tex]
Solve the exponent.
[tex]A= 200 \ g *0.03125[/tex]
[tex]A= 6.25 \ g[/tex]
In addition, we can solve this another way. First, we find the number of half-lives by dividing the total time by the half-life.
8,000/1,600= 5 half-livesEvery half-life, 1/2 of the mass decays. Divide the initial mass in half, then that result in half, and so on 5 times.
1. 200 g/2= 100 g2. 100 g / 2 = 50 g3. 50 g / 2 = 25 g 4. 25 g / 2 = 12.5 g5. 12.5 g / 6.25 gAfter 8,000 years, 6.25 grams of radium-226 remains.
Oxygen is composed of three isotopes: oxygen-16, oxygen-17 and oxygen-18 and has an average atomic mass of 15.9982 amu. Oxygen-17 has a mass of 16.988 amu and makes up 0.032% of oxygen. Oxygen-16 has a mass of 15.972 amu and oxygen-18 has a mass of 17.970 amu. What is the percent abundance of oxygen-18?
Answer:
The percent abundance of oxygen-18 is 1.9066%.
Explanation:
The average atomic mass of oxygen is given by:
[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]
Where:
m: is the atomic mass
%: is the percent abundance
Since the sum of the percent abundance of oxygen isotopes must be equal to 1, we have:
[tex] 1 = \%_{16} + \%_{17} + \%_{18} [/tex]
[tex] 1 = x + 3.2 \cdot 10^{-4} + \%_{18} [/tex]
[tex] \%_{18} = 1 - x - 3.2 \cdot 10^{-4} [/tex]
Hence, the percent abundance of O-18 is:
[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]
[tex]15.9982 = 15.972*x + 16.988*3.2 \cdot 10^{-4} + 17.970*(1 - 3.2 \cdot 10^{-4} - x)[/tex]
[tex] x = 0.980614 \times 100 = 98.0614 \% [/tex]
Hence, the percent abundance of oxygen-18 is:
[tex]\%_{18} = (1 - 3.2 \cdot 10^{-4} - 0.980614) \times 100 = 1.9066 \%[/tex]
Therefore, the percent abundance of oxygen-18 is 1.9066%.
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Is a 4p S 4s transition allowed in sodium? If so, what is its wavelength? If not, why not? b. Is a 3d S 4s transition allowed in sodium? If so, what is its wavelength? If not, why not? g
Answer:
a) 4p ⇒ 4s transition is Allowed
b) 3d ⇒ 4s transition not allowed
Explanation:
a) 4p ⇒ 4s transition
This transition is allowed because for a 4p state; l = 1 and for a 4s state I = 0
hence Δl = 1 - 0 = 1
Energy of 4p ( Ei ) = 3.75eV
Energy of 4s ( E2 ) = 3.19 eV
where : λ = 1240 eV nm / ( E₂ - E₁ )
= 2214 nm ≈ 2.214 μm
b) 3d ⇒ 4s transition
This transition is not allowed
a 3d state , l = 2 while for 4s state l = 0
hence Δl = 2 - 0 = 2
therefore the transition is not allowed
A hot pot of water is set on the counter to cool. After a few minutes it has lost 495 J of heat energy. How much heat energy has the surrounding air gained?
_____unit_____
Answer:
495 J
Explanation:
When the hot pot was set on the counter to cool, heat energy was lost from the pot. Note that according to the first law of thermodynamics, heat is neither created nor destroyed.
This implies that, the heat energy lost from the pot must be gained by the surrounding air. Therefore, if 495 J of energy is lost from the pot, then 495 J of energy is gained by the surrounding air.
convert 100kcals to kilojoules
Answer:
Explanation:
418.4kj is the correct answer
15.27
The following equilibria were attained at 823 K:
COO(s) + H2() Co(s) + H2O(g) K = 67
COO(s) + CO(8) = Co(s) + CO2(8) K = 490
Based on these equilibria, calculate the equilibrium con-
stant for
H2(g) + CO2(g) = CO(g) + H2O(g) at 823 K.
The equilibrium constant for the reaction is K = 0.137
We obtain the equilibrium constant considering the following equilibria and their constants:
COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
COO(s) + CO(g) → Co(s) + CO₂(g) K₂ = 490
We write the first reaction in the forward direction because we need H₂(g) in the reactants side:
(1) COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
Then, we write the second reaction in the reverse direction because we need CO₂(g) in the reactants side. Thus, the equilibrium constant for the reaction in the reverse direction is the reciprocal of the constant for the reaction in the forward direction (K₂):
(2) Co(s) + CO₂(g) → COO(s) + CO(g) K₂ = 1/490
From the addition of (1) and (2), we obtain:
COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
+
Co(s) + CO₂(g) → COO(s) + CO(g) K₂ = 1/490
-------------------------------------------------
H₂(g) + CO₂(g) → CO(g) + H₂O(g)
Notice that Co(s) and COO(s) are removed that appear in the same amount at both sides of the chemical equation.
Now, the equilibrium constant K for the reaction that is the sum of other two reactions is calculated as the product of the equilibrium constants, as follows:
K = K₁ x K₂ = 67 x 1/490 = 67/490 = 0.137
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Different vinegars can be 5-20% acetic acid solutions and have been used for medicinal purposes for thousands of years. If a person takes 2.0 tablespoons of vinegar a day and the Molarity of the vinegar is .84 M, then how many grams of acetic acid (HC2H3O2) will be consumed? 1 Tablespoon is 15 mL.
.013 g
.026 g
.76 g
1.5 g
Answer:
1.5g
Explanation:
Remember that Molarity = (#moles of solute)/(#liters of solution)
This problem informs us that the Molarity of the vinegar is 0.84 and that the solution is 15mL.
First let's get your SI units to the correct ones.
15mL (1L/1000mL) = 0.015L
Molarity = (#moles of solute)/(#liters of solution) ~
(Molarity)(#liters of solution) = #moles of solute
(0.84M)(.015L) = 0.0126moles of acetic acid per tablespoon
2 tablespoons a day = 0.0126moles*2 = 0.0252 moles of acetic acid.
Now that we have the # of moles of acetic acid we need to get our answer into grams. The molecular weight of HC2H3O2 is 60g/mole.
0.0252mole HC2H3O2 (60g HC2H3O2/1mole HC2H3O2) = 1.512g ~ 1.5g HC2H3O2.
How is magma formed?
Answer:
“Magma” is exclusively found and formed beneath the earth’s surface. Once magma is on or above the surface of the earth it is referred to as “lava.” Magma is typically formed by extreme temperature melting solid rock within the earth. Pressure and rock composition can also affect magma formation. High pressure can help magma be “squeezed” from partially molten rock. Likewise, as rocks are usually composed of different minerals with different melting points, magma formation from rocks is usually only partial and uneven.
Explanation:
Please help fast
All four referenced Greek thinkers: Democritus, Aristotle, Archimedes, and Anaxagoras, observed Nature and argued for his theory of
the composition of matter and natural laws. Only one of them tested his hypothesis and proposed a natural laws based on reproducible
observations, controlled experiments, and mathematical reasoning. All others used logic and thought experiments, as philosophers do,
to support their theories. Who is the experimental scientist in this group?
O Democritus
O Aristotle
O Archimedes
O Anaxagoras
Answer:
Anaxagoras was perhaps the first literate person to attempt to explain physical phenomena rationally, basing his ideas upon careful observations and simple experiments. This is fundamental to modern science and is the sine qua non of environmental study.
Determine the kinds of intermolecular forces that are present in each of the following. Part A Xe Check all that apply. dispersion forces dipole-dipole forces hydrogen bonding Request Answer Part B N2 Check all that apply. dispersion forces dipole-dipole forces hydrogen bonding Request Answer Part C CO Check all that apply. dispersion forces dipole-dipole forces hydrogen bonding Request Answer Part D HF Check all that apply. dispersion forces dipole-dipole forces hydrogen bonding
Answer:
Part A
dispersion forces
Part B
dispersion forces
Part C
dispersion forces
dipole-dipole forces
Part D
dispersion forces
dipole-dipole forces
hydrogen bonding
Explanation:
Dispersion forces occur in all molecules. They result from momentary shifts in the electron cloud of molecules which induces a dipole in another molecule. This induced dipole eventually spreads throughout the molecule.
For Xe which is a noble gas and N2 which is a diatomic molecule, dispersion forces is the only kind of intermolecular force present in the molecule.
CO is a polar molecule hence in addition to dispersion forces, dipole-dipole forces also exist in the molecule.
HF is a polar molecule hence it possesses dipole-dipole forces in addition to dispersion forces. In this molecule, hydrogen is bonded to a highly electronegative atom (fluorine). Hence, hydrogen bonding is a dominant intermolecular interaction in the molecule.
Please help me! I am a bit stuck on this.
3. Oxalic acid, C H20, is a toxic substance found in rhubarb leaves. When mixed with sufficient
quantities of a strong base, this weak diprotic acid loses two protons to form a polyatomic ion
called oxalate, C2022. Write a balanced equation that describes the reaction between oxalic acid
and sodium hydroxide.
Answer:
COOHCOOH + 2OH⁻ ⇄ C₂O₄²⁻ + 2H₂O
Explanation:
The reaction of oxalic acid with a strong base like sodium hydroxide is the following:
COOHCOOH + OH⁻ ⇄ COOHCOO⁻ + H₂O (1)
In this first reaction, the oxalic acid loses one proton. In a second reaction with NaOH, the ion COOHCOO⁻ loses its second proton to form ion oxalate as follows:
COOHCOO⁻ + OH⁻ ⇄ C₂O₄²⁻ + H₂O (2)
The general reaction between oxalic acid and NaOH is (eq 1 + eq 2):
COOHCOOH + 2OH⁻ ⇄ C₂O₄²⁻ + 2H₂O
I hope it helps you!
Dung dịch nào sau đây chỉ chứa các ion (bỏ qua sự điện li của nước, các chất điện li mạnh phân li hoàn toàn)?
A. HBr, Na2S, Mg(OH)2, Na2CO3.
B. H2SO4, NaOH, NaCl, HF.
C. HNO3, H2SO4, KOH, K2SiO3.
D. Ca(OH)2, KOH, CH3COOH, NaCl.
Answer:
Dung dịch nào sau đây chỉ chứa các ion (bỏ qua sự điện li của nước, các chất điện li mạnh phân li hoàn toàn)?
A. HBr, Na2S, Mg(OH)2, Na2CO3.
B. H2SO4, NaOH, NaCl, HF.
C. HNO3, H2SO4, KOH, K2SiO3.
D. Ca(OH)2, KOH, CH3COOH, NaCl.
A solution is prepared by dissolving 6.60 g of an nonelectrolyte in water to make 550 mL of solution. The osmotic pressure of the solution is 1.84 atm at 25 °C. The molecular weight of the nonelectrolyte is ________ g/mol.
Answer:
160 g/mol
Explanation:
Step 1: Calculate the molarity of the solution
We will use the following expression.
π = M × R × T
where,
π: osmotic pressure of a nonelectrolyteM: molarityR: ideal gas constantT: absolute temperature (25 °C = 298 K)M = π / R × T
M = 1.84 atm / (0.0821 atm.L/mol.K) × 298 K = 0.0752 mol/L
Step 2: Calculate the moles of solute in 550 mL (0.550 L)
0.550 L × 0.0752 mol/L = 0.0413 mol
Step 3: Calculate the molecular weight of the nonelectrolyte
0.0413 moles weigh 6.60 g.
6.60 g/0.0413 mol = 160 g/mol
How are all planets in the solar system similar?
A. They have a gas atmosphere.
B. They have a water atmosphere.
C. They have a gas-surface composition.
D. They have a rock surface composition.
THIS IS FOR SCIENCE!!!!!!
PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
They have a rock surface composition.
Explanation:
I hope this will help you
385 x 42.13 x 0.079 is (consider significant figures):
385 x 42.13 x 0.079 = 1281.38395
A solution is made by dissolving 0.565 g of potassium nitrate in enough water to make up 250. mL of solution. What is the molarity of this solution?
Please explain and show work.
[tex]\\ \large\sf\longmapsto KNO_3[/tex]
[tex]\\ \large\sf\longmapsto 39u+14u+3(16u)[/tex]
[tex]\\ \large\sf\longmapsto 53u+48u[/tex]
[tex]\\ \large\sf\longmapsto 101u[/tex]
[tex]\\ \large\sf\longmapsto 101g/mol[/tex]
Now
[tex]\boxed{\sf No\:of\:moles=\dfrac{Given\:mass}{Molar\:mass}}[/tex]
[tex]\\ \large\sf\longmapsto No\:of\:moles=\dfrac{0.565}{101}[/tex]
[tex]\\ \large\sf\longmapsto No\:of\:moles=0.005mol[/tex]
We know
[tex]\boxed{\sf Molarity=\dfrac{Moles\:of\:solute}{Vol\:of\:Solution\:in\:L}}[/tex]
[tex]\\ \large\sf\longmapsto Molarity=\dfrac{0.005}{\dfrac{250}{1000}L}[/tex]
[tex]\\ \large\sf\longmapsto Molarity=\dfrac{0.005}{0.250}[/tex]
[tex]\\ \large\sf\longmapsto Molarity=0.02M[/tex]
[tex] \: \: \: \: \: \: \: \: \: [/tex]
Calculate the number of iron atoms contained in 434.52 g of iron: Question 7 options: 2.620 x 10 26 Fe atoms 5.769 x 10 24 Fe atoms 4.685 x 10 24 Fe atoms 3.223 x 10 25 Fe atoms 1.169 x 10 23 Fe atoms
Answer:
[tex]\boxed {\boxed {\sf 4.685 \times 10^{24} \ Fe \ atoms}}[/tex]
Explanation:
We are asked to convert grams of iron to atoms of iron.
1. Convert Grams to MolesFirst, we convert grams to moles. We use the molar mass or the mass in 1 mole of a substance. This is found on the Periodic Table because the molar mass is equal to the atomic mass, but the units are grams per mole instead of atomic mass units. Look up iron's molar mass.
Fe: 55.84 g/molWe convert using dimensional analysis, so we must set up a ratio using the molar mass.
[tex]\frac {55.84 \ g \ Fe}{ 1 \ mol \ Fe}[/tex]
We are converting 434.52 grams to moles, so we multiply by this value.
[tex]434.52 \ g \ Fe *\frac {55.84 \ g \ Fe}{ 1 \ mol \ Fe}[/tex]
Flip the ratio so the units of grams of iron cancel.
[tex]434.52 \ g \ Fe *\frac { 1 \ mol \ Fe}{55.84 \ g \ Fe}[/tex]
[tex]434.52 *\frac { 1 \ mol \ Fe}{55.84}[/tex]
[tex]\frac { 434.52}{55.84} \ mol \ Fe[/tex]
[tex]7.781518625 \ mol \ Fe[/tex]
2. Convert Moles to AtomsNext, we convert moles to atoms. We use Avogadro's Number or 6.02 ×10²³. It is the number of particles (atoms, molecules, formula units, etc) in 1 mole of a substance. For this problem, the particles are atoms of iron. We set up another ratio using this number.
[tex]\frac {6.02 \times 10^{23} \ atoms \ Fe }{ 1 \ mol \ Fe}[/tex]
Multiply by the number of moles we calculated.
[tex]7.781518625 \ mol \ Fe *\frac {6.02 \times 10^{23} \ atoms \ Fe }{ 1 \ mol \ Fe}[/tex]
The units of moles of iron cancel.
[tex]7.781518625 *\frac {6.02 \times 10^{23} \ atoms \ Fe }{ 1 }[/tex]
[tex]4.68447421 \times 10^{24} \ atoms \ Fe[/tex]
The correct answer choice is Choice 3: 4.685 × 10²⁴ atoms of iron.
QUESTION 11
Identify the reaction type.
KOH + HNO3 -> H2O + KNO3
O combustion
O decomposition
O combination
O single displacement
O double displacement
Suppose that you add 24.3 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 3.14 oC compared to pure benzene. What is the molar mass (in g/mol) of the unknown compound
Solution :
We know that :
[tex]$\Delta T_f = k_f.m$[/tex] and [tex]$m=\frac{w_2}{m_2 \times w_1}$[/tex]
Then, [tex]$\Delta T_f = k_f.\frac{w_2}{m_2.w_1}$[/tex] ..................(1)
Where,
[tex]w_1[/tex] = amount of solvent (in kg)
[tex]w_2[/tex] = amount of solute (in kg)
[tex]m_2[/tex] = molar mass of solute (g/mole)
[tex]m[/tex] = molality of solution (mole/kg)
Given :
[tex]\Delta T_f[/tex] = [tex]3.14\ ^\circ C[/tex], [tex]k_f= 5.12\ ^\circ C/m[/tex]
[tex]=5.12 \ ^\circ C/mole/kg[/tex]
[tex]=5.12 \ ^\circ C \ kg/mole[/tex]
[tex]w_1[/tex] = 0.250 kg, [tex]w_2[/tex] = 24.3 g
Then putting this values in the equation is (1),
[tex]$3.14 = \frac{5.12 \times 24.3}{m_2 \times 0.250}$[/tex]
[tex]$m_2 = \frac{5.12 \times 24.3}{3.14 \times 0.250}$[/tex]
[tex]m_2= 158.49[/tex] g/mole
So, the molar mass of the unknown compound is 158.49 g/mole.
What Volume of silver metal will weigh exactly 2500.0g. The density of silver
Answer:
cm3 = 2500.0 g / 10.5 g/cm3 = 238 cm3
Arrange the following compounds in order of increasing reactivity (least reactive first.) to electrophilic aromatic substitution:.
Bromobenzene Nitrobenzene Benzene Phenol
a. Bromobenzene < Nitrobenzene < Benzene < Phenol
b. Nitrobenzene < Bromobenzene < Benzene < Phenol
c. Phenol < Benzene < Bromobenzene < Nitrobenzene
d. Nitrobenzene < Benzene < Bromobenzene < Phenol
Answer:
Nitrobenzene < Bromobenzene < Benzene < Phenol
Explanation:
Aromatic compounds undergo electrophilic aromatic substitution reaction in the presence of relevant electrophiles. Certain substituents tend to increase or decrease the tendency of an aromatic compound towards electrophilic aromatic substitution reaction.
Substituents that increase the electron density around the ring such as in phenol tends to make the ring more reactive towards electrophilic substitution. Halogens such as bromine has a -I inductive effect as well as a +M mesomeric effect.
However the -I(electron withdrawing effect) of the halogens supersedes the +M electron donation due to mesomeric effect.
Putting all these together, the order of increasing reactivity of the compounds towards electrophilic aromatic substitution is;
Nitrobenzene < Bromobenzene < Benzene < Phenol
Select the correct relationship among the concentrations of species present in a 1.0 M aqueous solution of the weak acid represented by HA. A. [H2O] > [HA] > [A-] > [H3O ] > [OH-] B. [H2O] > [A-] ~ [H3O ] > [HA] > [OH-] C. [HA] > [H2O] > [A-] > [H3O ] > [OH-] D. [H2O] > [HA] > [A-] ~ [H3O ] > [OH-] E. [HA] > [H2O] > [A-] ~ [H3O ] > [OH-]
Answer:
D
Explanation:
We have to bear in mind that the acid is a weak acid. A weak acid does not dissociate completely in solution. We will have more concentration of undissociated acid than A^- and H3O^+ and OH^- in the system at equilibrium.
Being a weak acid, there is maximum concentration of water molecules followed by that of undissiociated acid.
Hence, for this solution, the concentration of ions in solution follows the order;
[H2O] > [HA] > [A-] ~ [H3O ] > [OH-]
