Answer:
pH = 12.61
Explanation:
First of all, we determine, the milimoles of base:
0.120 M = mmoles / 300 mL
mmoles = 300 mL . 0120 M = 36 mmoles
Now, we determine the milimoles of acid:
0.200 M = mmoles / 100 mL
mmoles = 100 mL . 0.200M = 20 mmoles
This is the neutralization:
HCOOH + OH⁻ ⇄ HCOO⁻ + H₂O
20 mmol 36 mmol 20 mmol
16 mmol
We have an excess of OH⁻, the ones from the NaOH and the ones that formed the salt NaHCOO, because this salt has this hydrolisis:
NaHCOO → Na⁺ + HCOO⁻
HCOO⁻ + H₂O ⇄ HCOOH + OH⁻ Kb → Kw / Ka = 5.55×10⁻¹¹
These contribution of OH⁻ to the solution is insignificant because the Kb is very small
So: [OH⁻] = 16 mmol / 400 mL → 0.04 M
- log [OH⁻] = pOH → 1.39
pH = 14 - pOH → 12.61
Indicate the types of forces that are involved between the solute and solvent when forming a homogeneous solution between CH3CH2CH2CH2CH3 and CH3CH2CH2CH2CH2CH3.
a. dispersion forces
b. dipole-dipole forces
c. hydrogen bonding
d. ion-dipole forces
2. Indicate the types of forces that are involved between the solute and solvent when forming a homogeneous solution between LiNO3 and H2O.
a. dispersion forces
b. dipole-dipole forces
c. hydrogen bonding
d. ion-dipole forces
Answer:
1) dispersion forces
2) ion-dipole forces
Explanation:
For the formation of a solution of the substances, CH3CH2CH2CH2CH3 and CH3CH2CH2CH2CH2CH3, dispersion forces are involved. Recall that the both are organic substances hence they are both molecular in nature. They are not only molecular, they are nonpolar molecules. This implies that only dispersion forces are the possible intermolecular forces.
For the formation of a solution of LiNO3 and H2O, the dipoles of water interacts strongly with the ions Li^+ and NO3^- thereby separating the ions in solution. The positive dipole of water interacts with NO3^- while the negative dipole of water interacts with Li^+.
Therefore, the interaction between nonpolar molecules is dispersion forces while between polar molecules is ion-dipole forces.
Dispersion forcesIon-dipole forcesDispersion Forces:
They are the weakest form of intermolecular force. The attraction between molecules is temporary.
it is due to temporary assymetric distribution of electrons in the molecules. It is found in the non-polar molecules.For example- Interaction between [tex]\bold { CH_3CH_2CH_2CH_2CH_3 }[/tex] and [tex]\bold { CH_3CH_2CH_2CH_2CH_2CH_3 }[/tex]
Ion-dipole forces:
The attraction between molecules is due to the interaction between opposite ions of molecules.
The distribution of electrons is asymmeric. This interaction is seen between polar molecules.For example- Interaction between [tex]\bold { LiNO_3 }[/tex] and [tex]\bold { H_2O }[/tex]
Therefore, the interaction between nonpolar molecules is dispersion forces while between polar molecules is ion- dipole forces.
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You add 50.0 g of ice initially at ‒20.0 °C to 1.00 x 102 mL warm water at 67.0 °C. When all the ice melts, the water temperature is found to be somewhere above 0 °C. Calculate the final temperature of the water.
Answer:
[tex]T_2=17.8\°C[/tex]
Explanation:
Hello,
In this case, we can solve this problem by noticing that the heat lost by the warm water is gained by the ice in order to melt it:
[tex]Q_{water}=-Q_{ice}[/tex]
In such a way, the cooling of water corresponds to specific heat and the melting of ice to sensible heat and specific heat also that could be represented as follows:
[tex]m_{water}Cp_{water}(T_2-T_{water})=-m_{ice}\Delta H_{melting,ice}-m_{ice}\Cp_{ice}(T_2-T_{ice})[/tex]
Thus, specific heat of water is 4.18 J/g°C, heat of melting is 334 J/g and specific heat of ice is 2.04 J/g°C, thus, we can compute the final temperature as shown below:
[tex]m_{water}Cp_{water}(T_2-T_{water})+m_{ice}Cp_{ice}(T_2-T_{ice})=-m_{ice}\Delta H_{melting,ice}\\\\T_2=\frac{-m_{ice}\Delta H_{melting,ice}+m_{water}Cp_{water}T_{water}+m_{ice}Cp_{ice}T_{ice}}{m_{water}Cp_{water}+m_{ice}Cp_{ice}} \\\\T_2=\frac{-50.0*334+100*4.18*67+50.0*2.04*-20.0}{100*4.18+50.0*2.04} \\\\T_2=17.8\°C[/tex]
Best regards.
Explain why the following chemical equation represents a Lewis acid-base reaction. H+ + NH3 —>NH4+
Answer:
Due to an electron-pair acceptor and donor.
Explanations:
Lewis acid can be defined as an electron-pair acceptor. An example is Hydrogen ion(H+). This is because it is a proton and it distributes positive charge which means that it accepts electrons(negative charge).
Lewis base can be defined as an electron-pair donor. This is because it donates electrons to be accepted by the proton. An example is ammonia(NH3).
Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide.4HCl(aq)+MnO2(s)⟶MnCl2(aq)+2H2O(l)+Cl2(g)A sample of 36.9 g MnO2 is added to a solution containing 48.9 g HCl.What is the limiting reactant?MnO2HClWhat is the theoretical yield of Cl2?theoretical yield:g Cl2If the yield of the reaction is 74.7%, what is the actual yield of chlorine?actual yield:g Cl2
Answer:
HCl is limiting reactant
Theoretical yield: 23.8g Cl₂
Actual yield: 17.6g C₂
Explanation:
Based on the reaction:
4HCl(aq)+MnO2(s)⟶MnCl2(aq)+2H2O(l)+Cl2(g)
4 moles of HCl reacts per mole of MnO₂ to produce 1 mole of MnCl₂ and Cl₂ and 2 moles of water.
To find the limiting reactant you must know the moles of each reactant and knowing that 4 moles of HCl reacts per mole of MnO₂ you can sikve this problem, thus:
Moles HCl (Molar mass: 36.46g/mol): 48.9g ₓ (1mol / 36.46g/mol) =
1.341 moles HCl
Moles MnO₂ (Molar mass: 86.937g/mol): 36.9g ₓ (1mol / 86.937g) =
0.424 moles MnO₂
For a complete reaction of 0.424 moles of MnO₂ you require:
0.424moles MnO₂ ₓ (4 moles HCl / 1 mole MnO₂) = 1.696 moles of HCl.
As you have just 1.341 moles of HCl. HCl is limiting reactant.
Theoretical yield means, in the reaction, that 4 moles of HCl will produce 1 mole of Cl₂. As moles of HCl are 1.341:
1.341 moles HCl ₓ (1 mole Cl₂ / 4 moles HCl) = 0.33525 moles Cl₂
In grams (Molar mass Cl₂: 70.9g/mol):
Theoretical yield: 0.33525 moles Cl₂ ₓ (70.9g / mol) = 23.8g Cl₂
As yield of reaction is 74.7%, the real mass of Cl₂ you obtain (Actual yield) is:
23.8g Cl₂ ₓ 74% = 17.6g C₂
which conditions make snow
Answer:
It depends on the weather.
Explanation:
Like rain and hail, snow comes from the water vapor that forms the clouds, but obviously its formation is different: snow forms when the temperature is low in the atmosphere. In these conditions the water vapor drops freeze and when they collide form tiny crystals that join together to form snowflakes, which fall to Earth when they are heavy enough.
calculate the potential energy per ion for an infinite one dimensional crystal which will alternate sodium and chloride ions
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]
Nitrosyl bromide, NOBr, is formed from NO and Br2.
2 NO(g) + Br2(g) 2 NOBr(g)
Experiments show the reaction is second order in NO and first order in Br2.
(a) What is the rate equation for the reaction?
rate = k[NO]2[Br2]
rate = k[NO][Br2]
rate = k[NO]2[Br2]2
rate = k[NO]2[Br]2
(b) How does the reaction rate change if the concentration of Br2 is changed from 0.0022 mol/L to 0.0066 mol/L?
the rate will double
the rate will remain unchanged
the rate will triple
(c) What is the change in the reaction rate if the concentration of NO is changed from 0.0024 mol/L to 0.0048 mol/L?
the rate will increase by a factor of 4
the rate will remain unchanged
the rate will also decrease by a factor of 1/2
Answer:
(a) What is the rate equation for the reaction?
rate = k[NO]2[Br2]
Explanation:
(b) How does the reaction rate change if the concentration of Br2 is changed from 0.0022 mol/L to 0.0066 mol/L?
the rate will doubleNitrosyl bromide, NOBr, is formed from NO and Br2.
(a) Rate = [tex]k[NO]2[Br2][/tex]
(b) The rate will triple Nitrosyl bromide, NOBr, is formed from NO and Br2.
(c) The rate will triple increase by a factor of 4
Calculation of Nitrosyl bromide
(a) The reaction is second order in NO ---> [tex]rate , r = k[NO]^2[/tex]
the reaction is first order in Br2 --> rate ,[tex]r = k[Br2]^1[/tex]
Therefore, from the above two total rate equation is , rate , [tex]r = k [NO]^2[/tex][tex][Br2]^1[/tex]
--> [tex]r = k[NO]^2 [Br2][/tex]
(b)The rate law is rate , [tex]r = k[NO]^2 [Br2][/tex]
------> rate , r is directly proportional to [Br2]
----->[tex]r / r’ = [Br2] / [Br2]’[/tex]
r = initial rate
r’ = final rate
[Br2] = initial conc = 0.0022 mol/L
[Br2]’ = final conc = 0.0066 mol/L
Plug the values we get[tex]r/r’ = 0.0022 / 0.0066[/tex]
r / r’ = 0.33
----> r’ = r / 0.33 = 3r
Therefore, the rate will triple
-------------------------------------------------------------------------------
(c) The rate law is rate ,[tex]r = k[NO]^2 [Br2][/tex]
------> rate , r is directly proportional to [tex][NO]^2[/tex]
-----> r / r’ =[tex][NO]^2 / [NO]’^2[/tex]
r = initial rate[tex]r / r’ = [NO]^2 / [NO]’^2[/tex]
r’ = final rate
[NO] = initial conc = 0.0024 mol/L
[NO]’ = final conc = 0.0048 mol/L
Plug the values we get [tex]r/r’ = 0.0024 / 0.0048[/tex]
[tex]r / r’ = (0.0024^2) / (0.0048)^2[/tex]
= [tex](0.5)^2[/tex]
=[tex]0.25[/tex]
----> r’ = [tex]r / 0.25 = 4r[/tex]
Therefore, the rate will triple increase by a factor of 4
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What molar ratio of sodium acetate to acetic acid should be used to prepare a buffer with pH = 4.5? Ka acetic acid = 1.8 x 10-5.
Answer: molar ratio: 0.58
Explanation:
This is a buffer problem. We would use the equation [tex]pH=pKa+log(\frac{base}{acid} )[/tex]. Since we were given Ka and pH, we can find the ratio.
pKa=-log(Ka)
pKa=-log(1.8×10⁻⁵)
pKa=4.74
Now that we have pKa, we can plug in our values and find the molar ratio. Instead of writing base/acid, let's use x.
4.5=4.74+log(x)
-0.24=log(x)
10⁻⁰⁻²⁴=x
x=0.58
Consider the 1H NMR spectrum for the following compound:
CH3CH2CH3
Predict the first-order splitting patterns of the indicated proton. This question uses specific splitting patterns instead of the often used generic term "multiplet."
a. doublet of quartets
b. triplet of triplets
c. septet
d. quartet of quartets
e. quintet
Answer:
See explanation
Explanation:
In this case, we have to check the neighbors of each carbon in the molecule. In propane, we have two types of carbons (see figure 1) (blue and red ones). The red carbons are equivalent. (Both have the same neighbors). Now, we can analyze each carbon:
Blue carbon
In the blue carbon, we have 6 hydrogens neighbors (three on each methyl). If we follow the n+1 rule, we will have:
6+1= 7
For the blue carbon, we will have a Septet.
Red carbons
In the red carbon, we have only 2 neighbors (the carbon in the middle only have 2 hydrogens). If we follow the n+1 rule, we will have:
2+1=3
For the red carbon, we will have a triplet.
See figure 2
I hope it helps!
The melting point of sodium chloride is 801°C. The melting point of chlorine is -101°C. Explain, in terms of structure and bonding, the difference between the melting points of these two substances.
Answer:
It's harder to melt sodium chloride because of its density and stability.
Explanation:
The bond between Na+ and Cl- is an ionic bond, meaning its an ionic compound. Ionic compounds have more inter molecular forces(forces involving multiple molecules). Chlorine has much less of these forces than NaCl and isn't as dense.
Answer:
It's harder to melt sodium chloride because of its density
Explanation:
The following thermochemical equation is for the reaction of Fe 3 O 4 (s) with hydrogen (g) to form iron and water vapor Fe 3 O 4 (s)+4H 2 (g) 3Fe(s)+4H 2 O(g) When 64.6 grams of Fe 3 O 4 (s) react with excess hydrogenſe) how much heat is absorbed?
Answer:
41.3kJ of heat is absorbed
Explanation:
Based in the reaction:
Fe₃O₄(s) + 4H₂(g) → 3Fe(s) + 4H₂O(g) ΔH = 151kJ
1 mole of Fe3O4 reacts with 4 moles of H₂, 151kJ are absorbed.
63.4g of Fe₃O₄ (Molar mass: 231.533g/mol) are:
63.4g Fe₃O₄ × (1mol / 231.533g) = 0.274moles of Fe₃O₄
These are the moles of Fe₃O₄ that react. As 1 mole of Fe₃O₄ in reaction absorb 151kJ, 0.274moles absorb:
0.274moles of Fe₃O₄ × (151kJ / 1 mole Fe₃O₄) =
41.3kJ of heat is absorbed
How fast are the atoms moving if the temperature of a gas is cold?
A. very, very slowly
B. they are stagnant
C. very, very quickly
Answer:
i think option a is correct answer because when there is low temperature then the kinetic enegry will be very less and the atoms moves very slowly.
Answer:
A. very, very slowly
Explanation:
A is the answer because atoms will move faster in hot gas than in cold gas.
4Ga + 3S2 ⇒ 2Ga2S3
How many moles of Sulfur are needed to react with 100.0 grams of Gallium?
Answer:
1.076 mol (corrected to 2 d.p.)
Explanation:
Take the atomic mass of Ga be 69.7.
since no. of moles = mass/ molar mass,
no. of moles of Ga used = 100.0 / 69.7
= 1.43472023 mol
From the balanced equation, the mole ratio of Ga:S2 = 4:3, which means every 4 moles of Ga can react completely with 3 moles of S2.
So, let the no. of moles of Sulphur required be y.
[tex]\frac{4}{3} =\frac{ 1.43472023}{y}[/tex]
4 y = 1.43472023 x 3
y = 1.076 mol (corrected to 2 d.p.)
A gas originally at 27 °C and 1.00 atm pressure in a 3.3 L flask is cooled at constant pressure until the temperature is 11 °C. The new volume of the gas is ________ L.
Answer:
THE NEW VOLUME OF THE GAS IS 3.124 L
Explanation:
Initial Temperature = 27°C = 27 + 273 K = 300 K
Pressure = constant
Initial Volume = 3.3 L
Final Temperature = 11 °C = 11+ 273 K = 284 K
Final volume = unknown
Using Charles' law, at constant pressure, the volume of a given gas is directly proportional to the temperature.
Mathematically,
V1 / T1 = V2 / T2
Re-arranging the formula by making V2 the subject of the formula
V2 = V1 T2 / T1
V2 = 3.3 * 284 / 300
V2 = 3.124 L
The new volume of the gas is therefore 3.124 L at 11°C
The new volume is 3.1 L
The parameters given in the question are
T1= 27°c
convert to kelvin
= 273 + 27
= 300k
T2= 11°c
convert to kelvin
= 273+11
= 284k
Volume 1= 3.3 L
V1/T1= V2/T2
3.3/300= V2/284
Cross multiply
284×3.3= 300×V2
937.2= 300V2
V2= 937.2/300
= 3.1
Hence the new volume is 3.1 L
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Calculate the pH of a solution that has [H3O+]=4.3 x 10^-5 M
Answer:
pH = 4.4
Explanation:
pH = -log[H₃O⁺]
= -log(4.3 × 10⁻⁵)
pH = 4.37 = 4.4
I'm actually learning this in my chemistry class right now lol. Hope this helps though. :)
explain why you would expect that the Hneutzn for HBr reacting with KOH and the Hneutzn for HNO3 reacting with NaOH to be identical
Answer:
The net ionic equations for both the reactions are same so they have identical ΔH neutralization
H⁺ + OH⁻ ------ H₂O
Explanation:
The heat of neutralization for HBr reacting with KOH is similar to the heat of neutralization for HNO3 reacting with NaOH because they are both strong acid - strong base reactions.
A strong acid is an acid that ionizes completely in solution while a strong base is a base that ionizes completely in solution. A strong acid or base achieves almost 100% ionization in solution. A weak acid or base only ionizes to a very small extent in water.
In both cases of HBr reacting with KOH and HNO3 reacting with NaOH, we have a strong acid - strong base reaction hence the heat of neutralization of both reaction will be identical. Secondly, the both reactions have the same net ionic equation; H^+(aq) + OH^-(aq) -----> H2O(l)
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When the temperature in a room increases from 25°C to 33°C, changes from a solid to a liquid.
Answer:
true
Explanation:
Because ice melts if the temperature increasese
Answer:
Gallium
Explanation:
edg 2020
A student followed the procedure of this experiment to determine the percent NaCl in a commercial bleaching solution that was found in the basement of an abandoned house. The student diluted 50.00 mL of the commercial bleaching solution to 250 mL in a volumetric flask, and titrated a 20-ml aliquot of the diluted bleaching solution. The titration required 35.46 mL of 0.1052M Na S,O, solution. A faded price label on the gallon bottle read $0.79. The density of the bleaching solution was 1.10 g ml.
(1) Calculate the mass of commercial bleaching solution titrated. 1.1 450=559
(2) Determine the percent NaOCl in the commercial bleaching solution.
(3) Calculate the mass of one gallon of the commercial bleaching solution.
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what elements are designated by each symbol or group of symbols? a) CU and Cu, b) Os and OS, c) Ni and NI, d) BIN, BiN, and BIn
Answer:
a) CU: corresponds to carbon and uranium. Cu: corresponds to copper.
b) Os: corresponds to osmium. OS: Corresponds to oxygen and sulfur.
c) Ni: corresponds to nickel. NI: corresponds to nickel and iodine.
d) BIN: corresponds to boron, iodine and nitrogen. BiN: corresponds to bismuth and nitrogen. BIn: corresponds to boron and indium.
Explanation:
Hello.
Here, for establishing the elements in each numeral, we must remember that elements could have one capital letter only or one capital letter and one lower case only, say, nitrogen (N) or selenium (Se). In such a way, we have:
a) CU: corresponds to carbon and uranium. Cu: corresponds to copper.
b) Os: corresponds to osmium. OS: Corresponds to oxygen and sulfur.
c) Ni: corresponds to nickel. NI: corresponds to nickel and iodine.
d) BIN: corresponds to boron, iodine and nitrogen. BiN: corresponds to bismuth and nitrogen. BIn: corresponds to boron and indium.
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The equilibrium constant, K, for the following reaction is 2.44×10-2 at 518 K: PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 15.3 L container at 518 K contains 0.300 M PCl5, 8.55×10-2 M PCl3 and 8.55×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 8.64 L?
Answer:
[PCl₅] = 0.5646M
[PCl₃] = 0.1174M
[Cl₂] = 0.1174M
Explanation:
In the reaction:
PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)
K equilibrium is defined as:
K = 2.44x10⁻² = [PCl₃] [Cl₂] / [PCl₅]
The initial moles of each compound when volume is 15.3L are:
PCl₅ = 0.300mol/L×15.3L = 4.59mol
Cl₂ = 8.55x10⁻²mol/L×15.3L = 1.308mol
PCl₃ = 8.55x10⁻²mol/L×15.3L = 1.308mol
At 8.64L, the new concentrations are:
[PCl₅] = 4.59mol / 8.64L = 0.531M
[PCl₃] = 1.308mol / 8.64L = 0.151M
[Cl₂] = 1.308mol / 8.64L = 0.151M
At these conditions, reaction quotient, Q, is:
Q = [0.151M] [0.151M] / [0.531M]
Q = 4.29x10⁻²
As Q > K, the reaction will shift to the left producing more reactant, that means equilibrium concentrations are:
[PCl₅] = 0.531M + X
[PCl₃] = 0.151M - X
[Cl₂] = 0.151M - X
Where X is reaction coordinate.
Replacing in K expression:
2.44x10⁻² = [0.151M - X] [0.151M - X] / [0.531M + X]
1.296x10⁻² + 2.44x10⁻²X = 0.0228 - 0.302X + X²
0 = 9.84x10⁻³ - 0.3264X + X²
Solving for X:
X = 0.293 → False solution. Produce negative concentrations
X = 0.0336M → Right solution.
Replacing:
[PCl₅] = 0.531M + 0.0336
[PCl₃] = 0.151M - 0.0336
[Cl₂] = 0.151M - 0.0336
[PCl₅] = 0.5646M[PCl₃] = 0.1174M[Cl₂] = 0.1174MA solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would precipitate out of solution first?
A. Na3PO4.
B. Ag3PO4.
C. Ca3(PO4)2
When the second cation just starts to precipitate, what percentage of the first cation remains in solution?
Answer:
C. [tex]Ca_3(PO_4)_2[/tex] will precipitate out first
the percentage of [tex]Ca^{2+}[/tex]remaining = 12.86%
Explanation:
Given that:
A solution contains:
[tex][Ca^{2+}] = 0.0440 \ M[/tex]
[tex][Ag^+] = 0.0940 \ M[/tex]
From the list of options , Let find the dissociation of [tex]Ag_3PO_4[/tex]
[tex]Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}[/tex]
where;
Solubility product constant Ksp of [tex]Ag_3PO_4[/tex] is [tex]8.89 \times 10^{-17}[/tex]
Thus;
[tex]Ksp = [Ag^+]^3[PO_4^{3-}][/tex]
replacing the known values in order to determine the unknown ; we have :
[tex]8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}][/tex]
[tex]\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}][/tex]
[tex][PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}[/tex]
[tex][PO_4^{3-}] =1.07 \times 10^{-13}[/tex]
The dissociation of [tex]Ca_3(PO_4)_2[/tex]
The solubility product constant of [tex]Ca_3(PO_4)_2[/tex] is [tex]2.07 \times 10^{-32}[/tex]
The dissociation of [tex]Ca_3(PO_4)_2[/tex] is :
[tex]Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}[/tex]
Thus;
[tex]Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2[/tex]
[tex]2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2[/tex]
[tex]\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2[/tex]
[tex][PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}[/tex]
[tex][PO_4^{3-}]^2 = 2.43 \times 10^{-29}[/tex]
[tex][PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}[/tex]
[tex][PO_4^{3-}] =4.93 \times 10^{-15}[/tex]
Thus; the phosphate anion needed for precipitation is smaller i.e [tex]4.93 \times 10^{-15}[/tex] in [tex]Ca_3(PO_4)_2[/tex] than in [tex]Ag_3PO_4[/tex] [tex]1.07 \times 10^{-13}[/tex]
Therefore:
[tex]Ca_3(PO_4)_2[/tex] will precipitate out first
To determine the concentration of [tex][Ca^+][/tex] when the second cation starts to precipitate ; we have :
[tex]Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2[/tex]
[tex]2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2[/tex]
[tex][Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}[/tex]
[tex][Ca^{2+}]^3 =1.808 \times 10^{-7}[/tex]
[tex][Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}[/tex]
[tex][Ca^{2+}] =0.00566[/tex]
This implies that when the second cation starts to precipitate ; the concentration of [tex][Ca^{2+}][/tex] in the solution is 0.00566
Therefore;
the percentage of [tex]Ca^{2+}[/tex] remaining = concentration remaining/initial concentration × 100%
the percentage of [tex]Ca^{2+}[/tex] remaining = 0.00566/0.0440 × 100%
the percentage of [tex]Ca^{2+}[/tex] remaining = 0.1286 × 100%
the percentage of [tex]Ca^{2+}[/tex]remaining = 12.86%
In a typical situation, emissions from which category of activities comprise the largest amount of CO2 release into the atmosphere?
Answer:
Emission of CO2 from burning of fossil fuels (coal, natural gases and oil) is the activity that releases the largest amount of CO2 into the atmosphere.
Explanation:
The emission of CO2 is a major world crisis which is one of the single most contributor to global warming and rapid climate change. Fossil fuels produced from the remains of living organisms millions of years ago, under intense heat and pressure, has been the major fuel source for human industrial and domestic activities. Industries rely heavily on fossil fuels to power their machines and for industrial heating, movement and transportation now heavily depends on fossil fuel energy, and even basic heating and cooking process now depends on fossil fuel. Some of this activities rely on fossil fuel directly, or indirectly; by depending on power generated from power houses that generates electricity from the combustion of fossil fuels in huge generators. The after effect of all these energy demand is that more CO2 than ever is constantly released into the atmosphere.
Two moles of copper (II) sulfate,CuSO4,contains blank moles of O
Answer:
8 mol
Explanation:
Step 1: Given data
Moles of copper (II) sulfate: 2 mol
Chemical formula of copper (II) sulfate: CuSO₄
Step 2: Establish the molar ratio of copper (II) sulfate to oxygen
According to the chemical formula, the molar ratio of copper (II) sulfate to oxygen is 1:4.
Step 3: Calculate the moles of O in 2 mol of CuSO₄
[tex]2molCuSO_4 \times \frac{4molO}{1molCuSO_4} = 8molO[/tex]
What is the core charge of helium and why?
Answer:
Formula for effective nuclear charge is as follows. So, for He atom value of S = 0.30 because the electrons are present in 1s orbital. Therefore, calculate the effective nuclear charge for helium as follows. Thus, we can conclude that the effective nuclear charge for helium is 1.7
Explanation:
The effective nuclear charge experienced by a 1s electron in helium is +1.70.
The rate at which two methyl radicals couple to form ethane is significantly faster than the rate at which two tert-butyl radicals couple. Offer two explanations for this observation.
Answer:
1. stability factor
2. steric hindrance factor
Explanation:
stability of ethane is lesser to that of two tert-butyl, so ethane will be more reactive and faster.
ethane is less hindered and more reactive, while two tert-butyl is more hindered and less reactive
1. How many hydrogen atoms are in an alkane that has 7 carbon atoms?
2. An alkane contains 18 H atoms. Name the compound.
Answer:
1- there are 16 hydrogen atoms. ( heptane C7H16)
2- The compound is octane (C8H18)
explanation
to determine any atom in an alkane, this formula is used Cn H 2n +2, where n is the number of carbon atoms in the molecule.
to find the compound in Q2 we'll apply the numbers:
C nH₂n+2
Select the correct classification for the reaction
2Li(s) + Br2() - 2LIBr(s)
Double displacement, decomposition, single displacement, combustion, combination
Answer:
This is a combination reaction because Li and Br are combining.
Study this image.
The Continental crust and Lithosphere on the right are moving left toward the Continental crust and Lithosphere on the left that are moving right and pushing down below the Asthenosphere. A layer of Ancient oceanic crust is beneath the Lithosphere on the right.
Which feature forms at this plate boundary?
mountains
rift valleys
volcanoes
island chai
Answer:
Mountains:)
Explanation:
It's the only feature in this list that has an oceanic crust below the Lithosphere
I hope this helps you! Mark as brainliest, please! :)
Mountains are formed at this plate boundary, the bulk of earthquakes are brought on by small-scale movement along plate boundaries, hence option A is correct.
What is plate boundary?A plate boundary is defined as a three-dimensional surface or region where the velocity of one lithospheric plate in relation to the one next to it significantly changes.
Plate boundaries near the edge of the lithospheric plates might be conservative, divergent, or convergent. Wide zones of distortion are frequently observed at plate borders due to the connection between two plates.
Tectonic plate barrier where the plates are moving in opposite directions, either horizontally or vertically, rather than towards or away from one another, additionally known as a conservative plate boundary.
Therefore, option A is correct.
Learn more about plate boundary, here:
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Which statement is true: Mg -> Mg2+ + 2e-
(1) Mg gains 2 electrons (2) Mg2+ loses 2 electrons (3) mg loses 1 electron (4) mg loses 2 electrons
Answer:
(2) Mg²⁺ loses 2 electrons.
Explanation:
Let's consider the following half-reaction.
Mg ⇒ Mg²⁺ + 2 e⁻
Magnesium is a metal that loses 2 electrons to fulfill the octet rule. When magnesium loses electrons, the oxidation number increases from 0 to +2. This kind of half-reaction is known as oxidation, and has to be accompanied by a reduction of another species.
3. A student carries out the clay-catalyzed dehydration of cyclohexanol starting with 10 moles of cyclohexanol and obtains 500 mL of pure cyclohexene product. Calculate the student’s percent yield for this reaction. Show your work.
Answer:
[tex]49.45~%[/tex]
Explanation:
In this case, we have to start with the chemical reaction:
[tex]C_6H_1_2O~->~C_6H_1_0~+~H_2O[/tex]
So, if we start with 10 mol of cyclohexanol ([tex]C_6H_1_2O[/tex]) we will obtain 10 mol of cyclohexanol ([tex]C_6H_1_0[/tex]). So, we can calculate the grams of cyclohexanol if we calculate the molar mass:
[tex](6*12)+(10*1)=82~g/mol[/tex]
With this value we can calculate the grams:
[tex]10~mol~C_6H_1_0\frac{82~g~C_6H_1_0}{1~mol~C_6H_1_0}=820~g~C_6H_1_0[/tex]
Now, we have as a product 500 mL of [tex]C_6H_1_0[/tex]. If we use the density value (0.811 g/mL). We can calculate the grams of product:
[tex]500~mL\frac{0.811~g}{1~mL}=405.5~g[/tex]
Finally, with these values we can calculate the yield:
[tex]%~=~\frac{405.5}{820}x100~=~49.45%[/tex]%= (405.5/820)*100 = 49.45 %
See figure 1
I hope it helps!
Based on the data given, the percentage yield of the student's work is 49.45 %.
What is the equation of the reaction?The equation of the clay-catalyzed dehydration of cyclohexanol is given below:
C₆H₁₂O ----> C₆H₁₀ + H₂OFrom the equation of the reaction, 1 mole of cyclohexanol yields 1 mole of cyclohexene.
Therefore 10 moles of cyclohexanol should yield 10 moles of cyclohexene.
What is the moles of cyclohexene obtained?First we determine the mass of cyclohexene obtained.
Mass = density * volume
volume of cyclohexene = 500 mL
density = 0.811 g/mL
mass of cyclohexene = 500 * 0.811
mass of cyclohexene = 405.5 g
Number of moles of cyclohexene = mass/molar mass
molar mass of cyclohexene = 82 g
moles of cyclohexene = 405.5 g/82
moles of cyclohexene = 4.945 moles
What is the percentage yield?Percentage yield = actual yield /expected yield * 100%Percentage yield = 4.945/10 * 100%
Percentage yield = 49.45%
Therefore, the percentage yield of the student's work is 49.45 %.
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