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%
It was calculated that 4.3mL of 0.417 M HCl is required to titrate 11.9 mL of 0.151 M Mg(OH)2. Show evidence 2 HCl(aq) + Mg(OH)2(aq) → + MgCl2(aq) + 2 H2O(l)
Answer:
See explanation.
Explanation:
Hello,
In this case, for the described chemical reaction:
2 HCl(aq) + Mg(OH)2(aq) → MgCl2(aq) + 2 H2O(l)
We can notice there is a 2:1 molar ratio between the moles of hydrochloric acid and magnesium hydroxide, therefore, at the equivalence point:
[tex]n_{HCl}=2*n_{Mg(OH)_2}[/tex]
And in terms of volumes and concentrations we verify:
[tex]V_{HCl}M_{HCl}=2*V_{Mg(OH)_2}M_{Mg(OH)_2}[/tex]
So we use the given data to proof it:
[tex]4.3mL*0.417M=2*11.9mL*0.151M\\1.793=3.594[/tex]
Therefore, we can conclude the data is wrong by means of the 2:1 mole ratio that for sure was not taken into account. This is also supported by the fact that normalities are actually the same, but the nomality of magnesium hydroxide is the half of the hydrochloric acid normality since the acid is monoprotic and the base has two hydroxyl ions.
Best regards.
Why does a new period start on the periodic table, instead of the row continuing? A. A new period starts when a new energy shell starts. B. A new period starts when a new neutron cycle starts. C. None of these D. It is based on how many protons it has.
Answer:
B
Explanation:
All the elements in a period have valence electrons in the same shell. The number of valence electrons increases from left to right in the period. When the shell is full, a new row is started and the process repeats.
A new period starts when a new neutron cycle starts. Hence, option B is correct.
What is the period in the periodic table?A period in the periodic table is a row of chemical elements. All elements in a row have the same number of electron shells.
All the elements in a period have valence electrons in the same shell.
The number of valence electrons increases from left to right in the period.
When the shell is full, a new row is started and the process repeats.
Hence, option B is correct.
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When a hydrochloric acid solution is combined with a potassium hydroxide solution, an acid-base reaction occurs Write a balanced molecular equation for this reaction. Express your answer as a chemical equation. Identify all of the phases in your answer..
Answer:
HCl(aq) + KOH(aq) ⇒ KCl(aq) + H₂O(l)
Explanation:
Hydrochloric acid is an acid because it releases H⁺ in an aqueous solution.
Potassium hydroxide is a base because it releases OH⁻ in an aqueous solution.
When an acid reacts with a base they form a salt and water. This is a neutralization reaction. The neutralization reaction between hydrochloric acid and potassium hydroxide is:
HCl(aq) + KOH(aq) ⇒ KCl(aq) + H₂O(l)
4. What is the molar concentration and grams/Liter of a NaOH solution if 86 ml are titrated to an
endpoint by 375 ml of a solution of HCl that is .0175 M?
g/L: ________
Molarity: ____
Answer:
0.76 M
30 g/L
Explanation:
Step 1: Given data
Molarity of the acid (Ma): 0.175 MVolume of the acid (Va): 375 mLMolarity of the base (Mb): ?Volume of the base (Vb): 86 mLStep 2: Calculate the molarity of the base
We will use the following expression.
[tex]Ma \times Va = Mb \times Vb\\Mb = \frac{Ma \times Va}{Vb} = \frac{0.175M \times 375mL}{86mL} = 0.76 M[/tex]
Step 3: Calculate the concentration of the base in g/L
The molar mass of NaOH is 40.00 g/mol.
[tex]\frac{0.76mol}{L} \times \frac{40.00g}{mol} = 30 g/L[/tex]
Each energy sub level contains __________ number of electrons. For example, sub level D can hold up to _______ electrons. A. the same, 10 B. the same, 14 C. a different, 6 D. a different, 10
Answer:
the same and 14
Explanation:
ed tell chem
Answer: the same 10
Explanation:
Which of the following aqueous solutions are good buffer systems? . a. 0.12 M calcium hydroxide + 0.29 M calcium bromide . b. 0.25 M perchloric acid + 0.16 M sodium perchlorate . c. 0.34 M hydrocyanic acid + 0.27 M sodium cyanide .
Answer:
c. 0.34 M hydrocyanic acid + 0.27 M sodium cyanide .
Explanation:
A buffer is defined as the aqueous mixture of a weak acid with its conjugate base or vice versa. Based on the systems:
a. 0.12 M calcium hydroxide + 0.29 M calcium bromide. IS NOT A GOOD BUFFER SYSTEM because Ca(OH)₂ is a strong base.
b. 0.25 M perchloric acid + 0.16 M sodium perchlorate. IS NOT A GOOD BUFFER SYSTEM because perchloric acid is a strong acid
c. 0.34 M hydrocyanic acid + 0.27 M sodium cyanide. IS A GOOD BUFFER SYSTEM because HCN is a weak acid, and its conjugate base, CN⁻, is obtained in the dissolution of NaCN as Na⁺ and CN⁻ ions.
A 32.3-gram sample of gas is found to have a volume of 1.9 liters at 301 K and 1.21 atm. What is the molar mass of this gas? Show all of the work used to solve this problem.
Answer:
351.1g/mol
Explanation:
you can find the answer using The ideal gas equation
n= PV/RT
n=(1.21*1.9/0.082*301)mol
n=0.092 mol
molar mass=Mass/mole
m=32.3g/0.092mol
m=351.1g/mol
solution solution solution
Answer:
Oxygen present in food items makes then rancid due to the presence of oils and fats. If the food is flushed with nitrogen, it prevents it from being oxidised (the nitrogen acts as an antioxidant).
Hope it helps ! :)
Describe what happens when two substances at different temperatures cine into contact. Describe how the law of conservation of energy applies to this system
Answer:
The substance with the highest heat gives heat to the lowest temperature, equating both temperatures,
In this situation there is talk of giving up heat but not matter, it is here that the law of conservation of energy comes into play.
Explanation:
The law of conservation of energy talks about that energy is transformed and never lost between two substances or two bodies that interact with each other, these interactions can be heat exchanges, as in this example.
By what mechanism does cyclohexanol react when treated in sulfuric acid and what compound results?A) E 1; methoxycyclohexane B) E2: methoxycyclohexane C) SN 1; methoxycycloheXafle D) E2; cyclohexene E) E 1: cyclohexene
Answer:
E 1: cyclohexene
Explanation:
This reaction is an example of the dehydration of cyclic alcohols. The reaction proceeds in the following steps;
1) The first step of the process is the protonation of the cyclohexanol by the acid. This now yields H2O^+ attached to the cyclohexane ring.
2) the water molecule, which a good leaving group now leaves yielding a carbocation. This now leaves a cyclohexane carbocation which is highly reactive.
3) A water molecule now abstracts a proton from the carbon adjacent to the carbocation leading to the formation of cyclohexene and the regeneration of the acid catalyst. This is an E1 mechanism because it proceeds via a carbocation intermediate and not a concerted transition state, hence the answer.
A mixture with H2 and He exerts a total pressure of 0.48 atm. If there is 1.0 g of H2 and 1.0 g of He in the mixture, what is the partial pressure (in atmospheres) of hydrogen?
Answer:
Partial pressure of hydrogen H₂ = 0.32 atm
Explanation:
Given:
Total pressure = 0.48 atm
Find:
Partial pressure of hydrogen
Computation:
Number of mole of H₂ = 1 / 2 = 0.5 moles
Number of mole of He = 1 / 4 = 0.25 moles
Total moles = 0.5 + 0.25 = 0.75
Partial pressure of hydrogen H₂ = [moles / total moles] Total pressure
Partial pressure of hydrogen H₂ = [0.50 / 0.75]0.48 atm
Partial pressure of hydrogen H₂ = 0.32 atm
Answer: 0.32 atm
Explanation:
First convert the mass of H2 to moles using the molar mass.
(1.0 gram H2 ⋅ (1.0 mol H2 / 2.016 g H2)) ≈ 0.50 mol H2
Next, convert the mass of helium He to moles using the atomic mass.
(1.0 gram He ⋅ (1.0 mol He / 4.003 g He)) ≈ 0.25mol He
The total number of moles is about 0.75 moles . The partial pressure of a component of a gas mixture can be found by multiplying the mole fraction by the total pressure.
PH2 = XH2 × Ptotal
PH2 = (0.50 mol / 0.75 mol)(0.48 atm) = 0.32 atm
If Kb = 1.6 x 10−6, calculate the pH of solution of 0.0015 M morphine. Justify any approximations you make
Answer:
4.3
Explanation:
Data provided in the question as per the question is as follows
[tex]k_b = 1.6 \times 10^{-6}[/tex]
Now we use the log in both the sides
So,
[tex]Pk_b = -log\ k_b = -log (1.6 \times 10^{-6})[/tex]
[tex]Pk_b[/tex] = 5.795
And, C = 0.0015
log c = -2.823
Now pH is
[tex]= \frac{1}{2} (Pk_b - log\ c)[/tex]
[tex]= \frac{1}{2} (5.795 - (-2.823))[/tex]
= 4.3
Hence, the pH of the solution of 0.0015 M morphine is 4.3 and the same is to be considered by applying the above formulas and the calculations part
A mixture of krypton and nitrogen gases, at a total pressure of 711 mm Hg, contains 11.7 grams of krypton and 4.10 grams of nitrogen. What is the partial pressure of each gas in the mixture
Answer:
A. Partial pressure of krypton, Kr is 346.97 mmHg
B. Partial pressure of nitrogen, N2 is 364.03 mmHg.
Explanation:
Step 1:
Data obtained from the question. This include the following:
Total pressure (Pt) = 711 mmHg
Mass of Kr = 11.7 g
Mass of N2 = 4.10 g
Partial pressure of Kr =..?
Partial pressure of N2 =...?
Step 2:
Determination of the number of mole of krypton, Kr and nitrogen, N2. This is illustrated below:
Molar mass of Kr = 84g/mol
Mass of Kr = 11.7g
Mole of Kr =?
Mole = mass /Molar mass
Mole of Kr = 11.7/84 = 0.139 mole
Molar mass of N2 = 2x14 = 28g/mol
Mass of N2 = 4.10g
Mole of N2 =?
Mole = mass /Molar mass
Mole of N2 = 4.1/28 = 0.146 mole
Step 3:
Determination of the mole fraction for each gas. This is illustrated below:
Mole of Kr = 0.139 mole
Mole of N2 = 0.146 mole
Total mole = 0.139 + 0.146 = 0.285 mole
Mole fraction of Kr = mol of Kr/total mol
Mole fraction of Kr = 0.139/0.285
Mole fraction of Kr = 0.488
Mole fraction of N2 = mol of N2/total mol
Mole fraction of N2 = 0.146/0.285
Mole fraction of N2 = 0.512
A. Determination of the partial pressure of krypton, Kr.
This is illustrated below:
Total pressure (Pt) = 711 mmHg
Mole fraction of Kr = 0.488
Partial pressure of Kr =..?
Partial pressure = mole fraction x total pressure
Partial pressure of Kr = 0.488 x 711
Partial pressure of Kr = 346.97 mmHg
B. Determination of the partial pressure of nitrogen, N2
This is illustrated below:
Total pressure (Pt) = 711 mmHg
Mole fraction of N2 = 0.512
Partial pressure of N2 =?
Partial pressure = mole fraction x total pressure
Partial pressure of N2 = 0.512 x 711
Partial pressure of N2 = 364.03 mmHg
Average Molarity for HCl is .391
Average Molarity for NaOH is .0962
Volume for HCl is:
Trial 1 Your Answer: 14mL
Trial 2 Your Answer: 14mL
Trial 3 Your Answer: 14mL
Volume for NaOH is:
Trial 1: 34.26mL
Trial 2: 33.48mL
Trial 3: 33.84mL
Entry # mass tablet(g) mass antacid(g) Vol HCl(mL) Vol NaOH(mL)
#1: 1.515 0.9010 14.00 34.26
#2: 1.452 0.8370 14.00 33.48
#3: 1.443 0.8280 14.00 33.84
I need help finding the mmoles HCl/mg please.
Answer:
#1: 0.00144 mmolHCl/mg Sample
#2: 0.00155 mmolHCl/mg Sample
#3: 0.00153 mmolHCl/mg Sample
Explanation:
A antiacid (weak base) will react with the HCl thus:
Antiacid + HCl → Water + Salt.
In the titration of antiacid, the strong acid (HCl) is added in excess, and you're titrating with NaOH moles of HCl that doesn't react.
Moles that react are the difference between mmoles of HCl - mmoles NaOH added (mmoles are Molarity×mL added). Thus:
Trial 1: 0.391M×14.00mL - 0.0962M×34.26mL = 2.178 mmoles HCl
Trial 2: 0.391M×14.00mL - 0.0962M×33.48mL = 2.253 mmoles HCl
Trial 3: 0.391M×14.00mL - 0.0962M×33.84mL = 2.219 mmoles HCl
The mass of tablet in mg in the 3 experiments is 1515mg, 1452mg and 1443mg.
Thus, mmoles HCl /mg OF SAMPLE for each trial is:
#1: 2.178mmol / 1515mg
#2: 2.253mmol / 1452mg
#3: 2.219mmol / 1443mg
#1: 0.00144 mmolHCl/mg Sample#2: 0.00155 mmolHCl/mg Sample#3: 0.00153 mmolHCl/mg SampleA 25.0 mL sample of a solution of an unknown compound is titrated with a 0.115 M NaOH solution. A buffer region was found around a pH of 3.5. The unknown compound is
Answer:
The unknown compound is a weak acid.
Explanation:
Given that :
a 25 mL sample of a solution of an unknown compound is titrated with a 0.115 M NaOH solution.
A buffer region was found around a pH of 3.5. We know that a pH of 3.5 is a weak acid. So, it is likely to be an organic acid
Let assume the solution of the unknown sample to be CH₃COOH
Now :
25 mL of CH₃COOH reacted with 0.115 M of NaOH
The equation for the reaction will be :
CH₃COOH + NaOH -----> CH₃COONa + H₂O
at x mole of 0.115y M of
CH₃COOH NaOH is present
If NaOH was added in excess;
CH₃COOH + NaOH -----> CH₃COONa , NaOH will be lost then CH₃COOH and CH₃COONa will be present
Therefore;
At equilibrium : Only CH₃COONa will be present but if it is above equilibrium NaOH will be present because the pH will increase due to the presence of the strong base
A metal, M , of atomic mass 56 amu reacts with chlorine to form a salt that can be represented as MClx. A boiling point elevation experiment is performed to determine the subscript x , and therefore, the formula of the salt. A 22.9 g sample of the salt is dissolved in 100.0 g of water and the boiling point of the solution is found to be 375.93 K. Find the formula of the salt. Assume complete dissociation of the salt in solution g
Answer:
Formula for the salt: MCl₃
Explanation:
MClₓ → M⁺ + xCl⁻
We apply the colligative property of boiliing point elevation.
We convert the boiling T° to °C
375.93 K - 273K = 102.93°C
ΔT = Kb . m . i
where ΔT means the difference of temperature, Keb, the ebulloscopic constant for water, m the molality of solution (mol of solute/kg of solvent) and i, the Van't Hoff factor (numbers of ions dissolved)
ΔT = 102.93°C - 100°C = 2.93°C
Kb = 0.512 °C/m
We replace data: 2.93°C = 0.512 °C/m . m . i
i = x + 1 (according to the equation)
22.9 g / (56g/m + 35.45x) = moles of salt / 0.1kg = molality
We have calculated the moles of salt in order to determine the molar mass, cause we do not have the data. We replace
2.93°C = 0.512 °C/m . [22.9 g / (56g/m + 35.45x)] / 0.1kg . (x+1)
2.93°C / 0.512 m/°C = [22.9 g / (56g/m + 35.45x)] / 0.1kg . (x+1)
5.72 m = [22.9 g / (56g/m + 35.45x)]/ 0.1 (x+1)
5.72 . 0.1 / [22.9 g / (56g/m + 35.45x)] = x+1
0.572 / (22.9 g / (56g/m + 35.45x) = x+1
0.572 (56 + 35.45x) / 22.9 = x+1
0.572 (56 + 35.45x) = 22.9x + 22.9
32.03 + 20.27x = 22.9x + 22.9
9.13 = 2.62x
x = 3.48 ≅ 3
can somebody please help me asap !!!
Answer:
Option A. 1191.49 K
Explanation:
Data obtained from the question include:
The equation for the reaction is given below:
4HCl + O2 —> 2Cl2 + 2H2O
Enthalpy (H) = +280 KJ/mol = +280000 J/mol
Entropy (S) = +235 J/Kmol
Temperature (T) =..?
The temperature at which the reaction will be feasible can be obtained as follow:
Change in entropy (ΔS) = change in enthalphy (ΔH)/T
(ΔS) = (ΔH)/T
235 = 280000/T
Cross multiply
235 x T = 280000
Divide both side by 235
T = 280000/235
T = 1191.49 K
Therefore, the temperature at which the reaction will be feasible is 1191.49 K
What is the mass of 7.68 x 1024 molecules of phosphorus trichloride?
Answer:
THE MASS OF 7.68 *10^24 MOLECULES OF PHOSPHORUS TRICHLORIDE IS 1746.25 g.
Explanation:
Molar mass of PCl3 = ( 31 + 35.5 *3) = 137.5 g/mol
At 7.68 * 10^24 molecules, how many number of mole is present?
6.03 * 10^23 molecules = 1 mole
7.68*10^24 molecules = x mole
x mole = 7.68 *10^24 molecules/ 6.03 *10^23
x mole = 1.27 *10 moles
x mole = 12.7 moles
Using mole = mass / molar mass
mass = mole * molar mass
mass = 12.7 moles * 137.5 g/mol
mass = 1746.25 g
Hence, the mass of 7.68 *10^24 molecules is 1746.25 g
The graph below shows the half life values of parent isotopes
Based on the graph, it can be concluded that the isotope which is most likely to be found with 12.5% of its original amount in 42 billion years is
Answer:
Thorium-235
Explanation:
Half-life is defined as the time taken for a radioactive material to reduce to half of its original amount. If the original amount of a radioactive substance N is 100%, then;
1st half-life- 50% of N is left
2nd half life - 25% of N is left
3rd half-life- 12.5% of N is left
The half-life of Thorium-235 is 15 billion years, hence three half lives will take place in 45 billion years. Hence 12.5% of the original amount of Thorium-235 present will remain after about 42 billion years.
Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calculate the volume (in milliliters) of the 1.436 M stock NaOH solution needed to prepare 250.0 mL of 0.1342 M dilute NaOH solution.
Answer:
23.36mL of the stock solution are required.
Explanation:
A dilution consist in the addition of solvent to decreases the concentration of a stock solution (The solution more concentrated).
As you want to prepare 250.0mL = 0.2500L of a 0.1342M NaOH, moles of NaOH you require to make this concentration in this volume are:
0.2500L × (0.1342mol / L) = 0.03355 moles of NaOH you require in the diluted solution.
These moles comes from the 1.436M stock solution. The volume of the stock solution you need to add is:
0.03355moles NaOH × (1L / 1.436mol) = 0.02336L of the 1.436M solution =
23.36mL of the stock solution are required.A 2.0 g sample of hydrocarbon was burned in the calorimeter. The temperature rose from 29°c to 32°c and heat and combustion is 11. Kj/g. Thr heat capacity of the calorimeter is
Answer:
THE HEAT CAPACITY OF THE CALORIMETER IS 3666.67 J/C
Explanation:
Mass = 2 g
Temperature difference = 32 C - 29 C = 3 C
Heat of combustion = 11 kJ/g
Heat capacity of the calorimeter = unknown
It is important to note that the heat of combustion of the reaction is the heat absorbed by the calorimeter in raising the mixture by 3 C
So therefore,
Heat = heat capacity * temperature difference
Heat capacity = Heat / temperature difference
Heat capacoty = 11 000 J / 3 C
Heat capacity = 3666.67 J/ C
If the reaction consumes methane gas ( CH4 ) at a rate of 2.08 M/s, what is the rate of formation of H2 ? the balanced equation is CH4 + N2Cl4 = CCl4 + N2 + 2 H2
Answer:
4.16M/s
Explanation:
Based on the reaction:
CH₄ + N₂Cl₄ ⇄ CCl₄ + N₂ + 2H₂
1 mole of methane, CH₄, produce 2 moles of H₂.
That means whereas 1 mole of methane is consumed, 2 moles of H₂ are formed
Having this in mind, if you are consuming methane at a rate of 2.08M/s, the rate of formation of hydrogen must be twice this rate, because there are produced twice moles of H₂.
Thus, rate of formation of H₂ is:
2.08M/s ₓ 2 =
4.16M/s
The rate of formation of H2 is 4.16M/s
The calculation is as follows:Based on the reaction:
CH₄ + N₂Cl₄ ⇄ CCl₄ + N₂ + 2H₂
here
1 mole of methane, CH₄, produce 2 moles of H₂.
In the case when you are consuming methane at a rate of 2.08M/s, the rate of formation of hydrogen must be twice this rate, because there are produced twice moles of H₂.
Thus, rate of formation of H₂ is:
2.08M/s ( 2) = 4.16M/s
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According to valence bond theory, which orbitals overlap in the formation of the bond in HCl?
a) 1s on H and 3p on Cl
b) 1s on H and 4s on Cl
c) 1s on H and 2p on Cl
d) 2s on H and 3p on Cl
e) 2s on H and 2p on Cl
Determine the enthalpy change when 19.399999999999999 g of carbon is reacted with oxygen according to the reaction
Answer:
The enthalpy change is −636.9798 kJ
Explanation:
The equation of the reaction between Carbon and Oxygen can be represented as :
[tex]C_{(s)} + O_{2(g)} \to CO_{2_(g)} \ \ \ \ \ \ \Delta H = -394 \ \ kJ/mol[/tex]
molar mass of Carbon = 12 mole
when mass of 19.399999999999999 g of carbon reacted with oxygen;
the number of moles = mass/molar mass
the number of moles = 19.399999999999999/12
the number of moles = 1.6167 mol
Thus the enthalpy change when 1 mole of carbon react with oxygen = -394 kJ/mol
Therefore when 1.6167 moles react with oxygen; we have:
= 1.6167 mol × -394 kJ/mol
= −636.9798 kJ
Select the correct classification for the reaction.
4NH3(g) + 502(g) – 4NO(g) + 6H20(9)
ΔGrxn = 958 kJ
(answer on Edge)
To solve such this we must know the concept of redox reaction. Redox reaction is the one in which oxidation and reduction takes place simultaneously. The classification for the given reaction is redox reaction.
What is chemical reaction?
Chemical reaction is a process in which two or more than two molecules collide in right orientation and energy to form a new chemical compound. The mass of the overall reaction should be conserved. There are so many types of chemical reaction reaction like combination reaction, double displacement reaction.
The balanced equation is
4NH[tex]_3[/tex](g) + 50[tex]_2[/tex](g) – 4NO(g) + 6H[tex]_2[/tex]0(9)
In the given reaction, Nitrogen in NH[tex]_3[/tex] is getting reduced, Oxygen in 0[tex]_2[/tex] is getting oxidized.
Therefore, the classification for the given reaction is redox reaction.
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If the NaOH is added to 35.0 mL of 0.167 M Cu(NO3)2 and the precipitate isolated by filtration, what is the theoretical yield of the reaction?
Answer:
The correct answer is - 0.570 grams
Explanation:
The balanced chemical reaction is given by
Cu(NO3)2(aq) + 2NaOH(aq) --------> Cu(OH)2(s) + 2NaNO3(aq)
1.0 mole 2.0 mole 1.0 mole 2.0 mole
number of mol of Cu(OH)2,
n = Molarity * Volume
= [tex]35.0*0.167 = 5.845[/tex] millimoles
As clear in the equation, 1 mole of Cu(NO3)2 gives 1 mole of Cu(OH)2 , So, 5.845 millimoles of Cu(NO3)2 will produce 5.845 millimoles of Cu(OH)2
Mass of Cu(OH)2 = number of mol * molar mass
= [tex]97.5*5.845*10^-3[/tex]
= 0.570 grams
Thus, the correct answer is - 0.570 grams
Calculate the amount of heat required to convert 10.0 grams of ice at –20.°C to steam at 120.°C. (Sp. heat of H2O(s) = 2.09 J/g•°C, Sp. heat of H2O(l) = 4.18 J/g•°C, Sp. heat of H2O(gas) = 2.03 J/g•°C; heat of fusion of H2O(solid) = 333 J/g, heat of vaporization of H2O(liquid) = 2260 J/g).
Answer:
THE AMOUNT OF HEAT REQUIRED TO CONVERT ICE FROM -20 C TO STEAM AT 120 C IS 30 946 J OR 30.946 KJ OF HEAT.
Explanation:
Mass = 10 g
To convert 10 g of ice at -20°C to steam at 120°C, the heat involved is:
1. Heat involved in converting the ice from -20 °c to ice at 0 °C:
Heat = mass * specific heat of water solid * change in temperature
heat = 10g * 2.09 J/g°C * ( 0- (-20))
Heat = 10 * 2.09 * 20
heat = 418 J
2. Heat required to convert the ice from 0°C to water at 0°C:
Heat = mass * specific heat of fusion of water solid
Heat = 10 * 333
Heat = 3330 J
3. Heat required to convert water at 0 C to water at 100 C:
Heat = mass * specific heat of water * change in teperature
Heat = 10 * 4.18 * (100 -0)
Heat = 4180 J
4. Heat required to convert water at 100 C to steam at 100 C:
Heat = mass * specific heat of vaporization
Heat = 10 * 2260
Heat = 22600 J
5. Heat required to convert steam from 100 C to steam at 120 C:
Heat = mass * specific heat of water * change in temperature
Heat = 10 * 2.09 * (120 -100)
Heat = 10 * 2.09 * 20
Heat = 418 J
T
he heat required to convert 10 g of ice at -20 C to steam at 120 C is therefore the total of the individual heat of reactions
Total amount of heat = ( 418 J + 3330 J + 4180 J + 22600 J + 418 J)
Total heat = 30946 J
The substance used by homeowners and municipal workers to melt ice on sidewalks and roadways is usually calcium chloride rather than sodium chloride. Discuss two possible rea-sons for this preference.
Answer:
1. It dissolves much more ice faster than sodium chloride
2. Calcium chloride is more effective in melting ice at lower temperatures.
Explanation:
Salts are used to melt ice on roadways and sidewalks because they help to lower the freezing point of water.
Sodium chloride and calcium chloride are both salts used for this purpose but calcium chloride is usually preferred for the following two reasons:
1. It dissolves much more ice faster than sodium chloride: Calcium chloride dissolves much more ice faster than sodium chloride because when it dissociates, it produces three ions instead of the two produced when sodium chloride. Therefore, the heat of hydration of its ions is greater than that of sodium chloride.
2. Calcium chloride is more effective in melting ice at lower temperatures. It lowers the freezing point of water more than sodium chloride. Calcium chloride is able to lower the freezing point of water to about -52°C while sodium chloride only lowers it to about -6°C.
Al(NO3)3+H2SO4=HNO3+Al2(SO4)3
Hey there!:
2 Al(NO)₃+ 3 H₂SO₄ → 1 Al₂O₁₂S₃+ 6 HNO₃
Reagents : Al(NO₃)₃ and H₂SO₄
Products : Al₂O₁₂S₃ and HNO₃
Coefficients : 2 , 3 , 1 and 6
Hope this helps!
Which aqueous solution will have the highest boiling point temperature? A. 0.100 molal NiBr2(aq) B. 0.250 molal CH3OH(aq) C. 0.100 molal MgSO4(aq) D. 0.150 molal Na2SO4(aq) E. 0.150 molal NH4NO3(aq)
Answer: 0.150 m [tex]Na_2SO_4(aq)[/tex] will have highest boiling point.
Explanation:
Formula used for Elevation in boiling point :
[tex]\Delta T_b=i\times k_b\times m[/tex]
where
[tex]\Delta T_b=T_b-T^o_b[/tex]= elevation in boiling point
[tex[k_b[/tex] = boiling point constant
m = molality
i = Van't Hoff factor
A) 0.100 m [tex]NiBr_2[/tex]
i = 3 as [tex]NiBr_2\rightarrrow Ni^{2+}+2Br^-[/tex]
concentration will be [tex]3\times 0.100=0.300[/tex]
B) 0.250 m [tex]CH_3OH[/tex]
i = 1 as [tex]CH_3OH[/tex] is a non electrolyte
concentration will be [tex]1\times 0.250=0.250[/tex]
C) 0.100 molal [tex]MgSO_4(aq)[/tex]
i = 2 as [tex]MgSO_4\rightarrrow Mg^{2+}+SO_4^{2-}[/tex]
concentration will be [tex]2\times 0.100=0.200[/tex]
D. 0.150 molal [tex]Na_2SO_4(aq)[/tex]
i = 3 as [tex]Na_2SO_4\rightarrrow 2Na^{+}+SO_4^{2-}[/tex]
concentration will be [tex]3\times 0.150=0.450[/tex]
E. 0.150 molal [tex]NH_4NO_3(aq)[/tex]
i = 2 as [tex]NH_4NO_3\rightarrrow NH_4^{+}+NO_3^{-}[/tex]
concentration will be [tex]2\times 0.150=0.300[/tex]
The solution having the highest concentration of ions will have the highest boiling point and thus 0.150 m [tex]Na_2SO_4(aq)[/tex] will have highest boiling point.
The aqueous solution that would have the highest temperature at boiling point would be:
D). 0.150 molal Na2SO4(aq)
What is a boiling point?The boiling point is described as the temperature at which the solution starts boiling or the vapor pressure becomes equivalent to the provided external/outer pressure.
To determine the elevation in boiling point, we will use:
Δ[tex]T_{b}[/tex] [tex]= i[/tex] × [tex]k_{b}[/tex] × [tex]m[/tex]
with
[tex]T_{b}[/tex] [tex]= T_{b} - T^{0}_{b}[/tex]
[tex]k_b[/tex] [tex]=[/tex] constant of boiling point
Using this formula,
0.150 molal Na2SO4(aq)
Given,
[tex]i = 3[/tex]
[tex]Na2So4[/tex] will have
[tex]2Na^{+}[/tex] [tex]+[/tex] [tex]SO^{2-}_{4}[/tex]
So,
Concentration [tex]= 3[/tex] × [tex]0.15[/tex][tex]0[/tex]
[tex]= 0.45[/tex][tex]0[/tex]
∵ 0.150 molal [tex]Na2SO4[/tex]Na2SO4(aq) has the maximum concentration.
Thus, option D is the correct answer.
Learn more about "Aqueous solution" here:
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