Answer: Gas–liquid–solid triple point
The single combination of pressure and temperature at which liquid water, solid ice, and water vapor can coexist in a stable equilibrium occurs at approximately 273.1575 K (0.0075 °C; 32.0135 °F) and a partial vapor pressure of 611.657 pascals (6.11657 mbar; 0.00603659 atm).
Explanation:
It represents the equilibrium between the liquid and gas phases. The point on this curve where the vapor pressure is 1 atm is the normal boiling point of the substance. The vapor-pressure curve ends at the critical point (B), which is at the critical temperature and critical pressure of the substance.
Which of the following is an example of a mechanical wave?
O A. A light ray
B. A seismic wave
C. A radio wave
D. An X-ray
Answer:
A seismic wave
Explanation:
It requires a medium for its propagation.
A chemist titrates of a butanoic acid solution with solution at . Calculate the pH at equivalence. The of butanoic acid is__________ .Round your answer to decimal places.
Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of solution added.
Answer:
pH = 8.75
Explanation:
100.0mL of a 0.8108M of a butanoic acid (HC₃H₇CO₂, pKa 4.82) solution is titrated with 0.0520M KOH.
The reaction is:
HC₃H₇CO₂ + KOH → H₂O + KC₃H₇CO
Moles of butanoic acid are:
0.1000L × (0.8108mol / L) = 0.08108 moles of butanoic
For a complete reaction, volume of KOH must be added is the volume in which 0.08108 moles of KOH are added, that is:
0.08108 mol × (L / 0.0520mol) = 1.56L of KOH.
Total volume in equilibrium is 1.56L + 0.10L = 1.66L
That means concentration of butanoic acid is:
0.08108 mol / 1.66L = 0.04884M HC₃H₇CO₂
At equivalence point, there is just C₃H₇CO⁻ in solution
Kb of butanoic acid is:
C₃H₇CO⁻ + H₂O ⇄ HC₃H₇CO₂ + OH⁻
Kb = Kw / Ka
Ka = 10^-pKa
Ka = 1.51x10⁻⁵
Kb = 1x10⁻¹⁴ / 1.51x10⁻⁵ = 6.61x10⁻¹⁰
The equilibrium of Kb is:
Kb = 6.61x10⁻¹⁰ = [HC₃H₇CO₂] [OH⁻] / [C₃H₇CO⁻]
As at equivalence point there is just C₃H₇CO⁻, the equilibrium concentrations are:
[C₃H₇CO⁻] = 0.04884M - X
[HC₃H₇CO₂] = X
[OH⁻] = X
Replacing in Kb:
6.61x10⁻¹⁰ = X² / [0.04884M - X]
0 = X² + 6.61x10⁻¹⁰X - 3.23x10⁻¹¹
Solving for X:
X = -5.68x10⁻⁶ → False solution. There is no negative concentrations
X = 5.683x10⁻⁶ → Right solution.
As [OH⁻] = X, [OH⁻] = 5.683x10⁻⁶.
pOH = - log [OH⁻]
pOH = 5.245
pH = 14 - pOH
pH = 8.75Which of the following is evidence for a physical change? A) burning B) fizzing C) evaporating D) rusting
Answer:c
Explanation: rusting, burning and fuzzing are all examples of chemical reactions/changes.
Gravity on Earth is 9.8 m/s2, and gravity on the Moon is 1.6 m/s2.
so if the mass of an object on earth is 40 kilograms what is the mass on the moon.
and how much does it way
Answer:
Mass is the same but it weights 64 Newtons
Explanation:
First of Mass is the same in any sort of gravity. Now let's calculate weight
W = MG
where W = Weight
M = Mass
G = Gravity
W = (40kg)(1.6)
W = 64
Sorry for the spelling mistakes, hope this helps
Answer:6.61kilo
Explanation: fdfv
A laser is used in eye surgery to weld a detached retina back into place. The wavelength of the laser beam is 503 nm, while the power is 1.4 W. During surgery, the laser beam is turned on for 0.070 s. During this time, how many photons are emitted by the laser?
Answer:
Number of proton emmitted by laser=[tex]2.48*10^17proton[/tex]
Explanation:
Energy is the ability to cause change; power is directly proportional to energy and its the rate energy is utilized.
Power=energy/time.
First we need to calculate the total energy used which is equal to the total power utilized.
E(total)= P( total) = 1.4W × 0.070 s =[tex]0.098J[/tex]
CHECK THE ATTACHMENT FOR THE REMAINING DETAILED CALCULATION
Which of these tasks would a geologist be most likely to perform?
A. Determining the species of a recently collected specimen
O B. Hypothesizing how pieces of ancient pottery were used
O C. Creating a new kind of material using polymers
O D. Determining the best method to extract underground natural gas
SUBMIT
Answer:
Explanation:
O B. Hypothesizing how pieces of ancient pottery were used
The functional groups in an organic compound can frequently be deduced from its infrared absorption spectrum. A compound containing C, H, and O exhibits intense absorption at 1720 cm-1. No additional information is available. List possible classes for which there is positive evidence.
Relative absorption intensity: (s)=strong, (m)=medium, (w)=weak.
What functional class(es) does the compound belong to?
List only classes for which evidence is given here. Attach no significance to evidence not cited explicitly.
Do not over-interpret exact absorption band positions. None of your inferences should depend on small differences like 10 to 20 cm-1.
a. alkane (List only if no other functional class applies.)
b. alkene h. amine
c. terminal alkyne i. aldehyde or ketone
d. internal alkyne j. carboxylic acid
e. arene k. ester
f. alcohol l. nitrile
g. ether
Answer:
The class of this compound is aldehyde or ketone (i).
Explanation:
Absorption peak at 1720 cm-1 shows the presence of a carbonyl group, possibly an aldehyde or ketone with C=O bond.
Further information on molecular formula would be required for structural elucidation.
At high temperatures one mole of hydrogen gas reacts with one mole of bromine gas to form hydrogen bromide. At a given temperature the equilibrium constant is 57.6. If at the same temperature, a mixture of 4.67 × 10^-3M bromine gas, 2.14 × 10^−3 hydrogen gas, and 2.40 × 10^−2M hydrogen bromide gas is made, then:
a. the system is at equilibrium.
b. the system is far from equilibrium and will shift to form more hydrogen gas.
c. the system is far from equilibrium and will shift to form more hydrogen bromide gas.
d. nothing can be deduced since we do not know whether the reaction is endothermic or exothermic.
e. nothing can be deduced since we do not know whether the equilibrium constant is Kc or Kp.
Answer:
a. the system is at equilibrium.
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]H_2+Br_2\rightleftharpoons 2HBr[/tex]
Thus, the law of mass action is given by:
[tex]Keq=\frac{[HBr]^2}{[H_2][Br_2]} =57.6[/tex]
Nonetheless, for the given point of 4.67 × 10^-3M bromine gas, 2.14 × 10^−3 hydrogen gas, and 2.40 × 10^−2M hydrogen bromide gas we should compute the reaction quotient in order to know whether the direction of the reaction is to left or to right, thus:
[tex]Q=\frac{[HBr]^2}{[H_2][Br_2]} =\frac{(2.40x10^{-2})^2}{(4.67x10^{-3})(2.14x10^{-3})} \\\\Q=57.6[/tex]
Therefore, since Keq=Q, we say that the system is at equilibrium, for that reason, the answer is a.
Best regards.
Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a copper(II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of . Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to significant digits.
Answer:
Concentration of Copper (II) Sulfate in the original sample in mol/L = 0.0035 M
Concentration of Copper (II) Sulfate in the original sample in g/L = 0.56 g/L
Explanation:
Complete Question
Fe(s) + CuSO₄(aq) → Cu(s) + FeSO₄(aq)
Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant figures.
Solution
Noting that the precipitate is Copper as it is the only solid by-product of this reaction.
89 mg of Copper is produced from this reaction.
We convert this into number of moles for further stoichiometric calculations
Mass of Copper = 89 mg = 0.089 g
Molar mass of Copper = 63.546 amu
Number of moles of Copper produced from the reaction = (0.089/63.546) = 0.0014005602 = 0.001401 mole
From the stoichiometric balance of the reaction,
1 mole of Copper is produced from 1 mole of Copper (II) Sulfate
0.001401 mole of Copper will be produced similarly from 0.001401 mole of Copper (II) Sulfate.
Number of moles of Copper (II) Sulfate in the original sample = 0.001401 mole
Concentration of Copper (II) Sulfate in the original sample in mol/L = (Number of moles) ÷ (Volume in L)
Number of moles = 0.001401 mole
Volume in L = (400/1000) = 0.4 L
Concentration of Copper (II) Sulfate in the original sample in mol/L = (0.001401/0.4) = 0.0035025 mol/L = 0.0035 mol/L to 2 s.f.
Concentration in g/L = (Concentration in mol/L) × (Molar Mass)
Concentration in mol/L = 0.0035025 M
Molar mass of Copper (II) Sulfate = 159.609 g/mol
Concentration of Copper (II) Sulfate in the original sample in g/L = 0.0035025 × 159.609 = 0.559 g/L = 0.56 g/L to 2 s.f
Hope this Helps!!!!
The concentration of the original copper solution is 0.035 M.
The equation of the reaction is;
Fe(s) + CuSO4(aq) -------> FeSO4(aq) + Cu(s)
Number of moles of copper obtained = 89 × 10^-3g/63.5 = 0.0014 moles
Since the reaction is 1:1, the number of moles of copper sulfate that reacted is c.
From the question, we are told that the volume of solution is 400.mL or 0.04L.
Hence, the concentration of the solution is; number of moles /volume
= 0.0014 moles/0.04L = 0.035 M
Learn more: https://brainly.com/question/9352088
Missing parts;
Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant figures.
For the iodine trichloride molecule: a. Determine the number of valence electrons for each atom in the molecule b. Draw the Lewis Dot structure c. Describe why the molecule is drawn this way (i.e. any extra rules/steps needed?) d. Show the polarity of each bond and for the molecule by drawing in the dipole +à
Answer:
Explanation:
a. Determine the number of valence electrons for each atom in the molecule
In this case we both atoms are halogens. Therefore we will have 7 electrons for each atom.
b. Draw the Lewis Dot structure
In this case, the formula is [tex]ICl_3[/tex], so the central atom would be "I" and the "Cl" atoms would be placed around "I". See figure 1
c. Describe why the molecule is drawn this way (i.e. any extra rules/steps needed?)
In this specific case, the "I" atom don't follow the octet rule. We will have an expanded octet for iodine (more than 8 electrons).
d. Show the polarity of each bond and for the molecule by drawing in the dipole +d
The negative dipole would be placed in the atom with higher electronegativity, in this case "Cl". The positive dipole would be placed in the atom with low electronegativity, in this case "I".
I hope it helps!
One of the reagents below gives predominantly 1,2 addition (direct addition) while the other gives predominantly 1,4 addition (conjugate addition). a) Which major organic product is the result of 1,2 addition? ---Select--- b) Draw the skeletal structure of major organic product A
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
The correct option is reagent B
b
The skeletal structure of major organic product A is shown on the third uploaded image
Explanation:
The mechanism of the reaction for A and B are shown on the second the second reaction and looking at this we can see that the reagent that predominately gives 1,2 addition is reagent B
Using the following balanced chemical equation 8 H2 + S8à 8 H2S. Determine the mass of the product (molar mass = 34.08g/mol) if you start with 1.35 g of hydrogen and 6.86 g of S8 (Molar mass = 256.5 g/mole).
Answer: 7.29 g of [tex]H_2S[/tex] will be produced from the given masses of both reactants.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]{\text{Moles of} H_2}=\frac{1.35g}{2.01g/mol}=0.672moles[/tex]
[tex]\text{Moles of} S_8=\frac{6.86g}{256.5g/mol}=0.0267moles[/tex]
[tex]8H_2+S_8\rightarrow 8H_2S[/tex]
According to stoichiometry :
1 mole of [tex]S_8[/tex] require = 8 moles of [tex]H_2[/tex]
Thus 0.0267 moles of [tex]S_8[/tex] will require=[tex]\frac{8}{1}\times 0.0267=0.214moles[/tex] of [tex]H_2[/tex]
Thus [tex]S_8[/tex] is the limiting reagent as it limits the formation of product and [tex]H_2[/tex] is the excess reagent.
As 1 mole of [tex]S_8[/tex] give = 8 moles of [tex]H_2S[/tex]
Thus 0.0267 moles of [tex]S_8[/tex] give =[tex]\frac{8}{1}\times 0.0267=0.214moles[/tex] of [tex]H_2S[/tex]
Mass of [tex]H_2S=moles\times {\text {Molar mass}}=0.214moles\times 34.08g/mol=7.29g[/tex]
Thus 7.29 g of [tex]H_2S[/tex] will be produced from the given masses of both reactants.
A scientist mixed 25.00 mL of 2.00 M KOH with 25.00 mL of 2.00 M HBr. The temperature of the mixed solution rose from 22.7 oC to 31.9 oC. Calculate the enthalpy change for the reaction in kJ/mol HBr, assuming that the calorimeter loses negligible heat, that the volumes are additive, and that the solution density is 1.00 g/mL, and that its specific heat is 4.184 J/g.oC.
Answer:
38.493 KJ/mol
Explanation:
Equation of reaction; HBr + KOH ---> KBr + H2O
Heat evolved = mass * specific heat capacity * temperature rise
Mass of solution = density * volume
Mass = 1.00 g/ml*50 ml = 50g
Temperature rise = 31.9 - 22.7 = 9.2 °C
Heat evolved = 50 * 4.184 * 9.2 = 1924.64 J
From the equation of reaction, 1 mole of HBr reacts with 1mole of KOH to produce 1 mole of H20
Number of moles of HBr involved in the reaction = molar concentration * volume (L)
Molar concentration = 2.0 M, volume = 25 ml = 0.025 L
Number of moles = 2.0 M * 0.025 L= 0.05 moles
Therefore, 0.05 moles of HBr reacts with 0.05 moles of KOH to produce 0.05 moles of H20
Enthalpy change per mole of HBr = 1924.64 J/0.05 moles = 38492.8 J/mol = 38.493 KJ/mol
Consider the following specific heats of metals. Metal Specific Heat Aluminum 0.897 J/(g°C) Magnesium 1.02 J/(g°C) Lithium 3.58 J/(g°C) Silver 0.237 J/(g°C) Gold 0.129 J/(g°C) If the same amount of heat is added to 25.0 g of each of the metals, which are all at the same initial temperature, which metal will have the lowest temperature?
Answer:
Lithium
Explanation:
The specific geat capacity of a substance is the energy required to raise 1 unit of that substance by one degree.
Heat energy (Q) = mc∇t
Q = heat energy
M = mass of the substance
c = specific heat capacity
∇t = change in temperature of the substance
Generally, increase in the specific heat capacity will lead to a lower final temperature likewise decrease in the specific heat capacity will lead to increase the final temperature of the substance.
From the data above, we can take just two specific heat capacity and test this theory.
Assuming we have a
Mass = 25g
Heat energy applied (Q) = 1 J
Initial temperature (T1) = 10°C
Final temperature (T2) = ?
Q = mc∇t
Q = mc (T2 - T1)
For Lithium, specific heat capacity = 3.58J/g°C
1 = 25 × 3.58 (T2 - 10)
Solve for T2
1 = 89.5 (T2 - 10)
1 = 89.5T2 - 895
89.5T2 = 896
T2 = 896 / 89.5
T2 = 10.011°C
For Magnesium (Mg) specific heat capacity = 1.02J/g°C
Q = mc∇t
1 = 25 × 1.02 × (T2 - 10)
1 = 25.5 (T2 - 10)
1 = 25.5T2 - 255
Solve for T2
25.5T2 = 256
T2 = 10.039°C
Notice the trend that decrease in the specific heat capacity leads to increase in the final temperature.
Try and continue for the elements and see how it works.
A stock solution will be prepared by mixing the following chemicals together:
3.0 mL of 0.00200 M KSCN
10.0 mL of 0.200 M Fe(NO3)3
17.0 mL of 0.5 M HNO3
Determine the molar concentration of Fe(NO3)3 in the stock solution.
Answer:
0.067M Fe(NO3)3
Explanation:
A stock solution is a concentrated solution that is diluted to prepare the solutions that you will use.
The volume of the stock solution is 3.0mL + 10.0mL + 17.0mL= 30.0mL.
The ratio between volume of the aliquot (10.0mL) and total volume (30.0mL) is called dilution factor, that is: 30.0mL / 10.0mL = 3
That means the Fe(NO3)3 is diluted 3 times. That means the molar concentration of the stock solution is:
0.200M / 3 =
0.067M Fe(NO3)3Rubidium is ______ potassium in the periodic table. lodine is ______ bromine in the periodic table. Therefore, the rubidium ion is __________ than the potassium ion, and the iodine ion is___________ than the bromide ion. The _______ the distance between the rubidium ion and the iodide ion is the potassium ion and the bromide ion. Therefore, the energy associated with the interaction between rubidium and iodide is________ atomic radius means that than that between , and the lattice energy of potassium bromide is ________ more exothermidc.
Answer:
The given blanks can be filled with below, below, larger, larger, larger, larger and smaller.
Explanation:
In the periodic table, rubidium comes below the potassium, and iodine comes below bromine. Therefore, it can be said that the ion of rubidium is larger in comparison to potassium ion, and similarly the ion of iodine is larger in comparison to the ion of bromine.
When the atomic radius is larger it signifies that the distance in between the ion of iodine and the ion of rubidium is larger in comparison to that between the ion of potassium and the ion of bromine. Thus, smaller energy is associated with the interaction between iodine and rubidium, and potassium bromide's lattice energy is more exothermic.
g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete the mass balance expressions for the sodium and arsenate species in the final solution.
Answer:
The concentration of the sodium and arsenate ions at the end of the reaction in the final solution
[Na⁺] = 0.05512 M
[HAsO₄²⁻] = 0.00185 M
[AsO₄³⁻] = 0.01714 M
Explanation:
Complete Question
A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.
Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O
From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)
Concentration in mol/L = (Number of moles) ÷ (Volume in L)
Number of moles = (Concentration in mol/L) × (Number of moles)
For Na₂HAsO₄
Concentration in mol/L = 0.03798 M
Volume in L = (500/1000) = 0.50 L
Number of moles = 0.03798 × 0.5 = 0.01899 mole
For NaOH
Concentration in mol/L = 0.03428 M
Volume in L = (500/1000) = 0.50 L
Number of moles = 0.03428 × 0.5 = 0.01714 mole
Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.
Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O
0.01899 0.01714 0 0 (At time t=0)
(0.01899 - 0.1714) | 0 → 0.01714 0.01714 (end)
0.00185 | 0 → 0.01714 0.01714 (end)
Hence, at the end of the reaction, the following compounds have the following number of moles
Na₂HAsO₄ = 0.00185 mole
This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction
NaOH = 0 mole
Na₃AsO₄ = 0.01714 moles
This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction
H₂O = 0.01714 moles
So, at the end of the reaction
Na⁺ has 0.0037 + 0.05142 = 0.05512 mole
(HAsO₄)²⁻ has 0.00185 mole
(AsO₄)³⁻ has 0.01714 mole.
And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L
Hence, the concentration of the sodium and arsenate ions at the end of the reaction is
[Na⁺] = 0.05512 M
[HAsO₄²⁻] = 0.00185 M
[AsO₄³⁻] = 0.01714 M
Hope this Helps!!!
Answer:
[tex]\rm [Na^{+}]= \text{0.055 12 mol/L}[/tex]
[tex]\rm [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}] = \text{0.018 99 mol/L}[/tex]
Explanation:
The overall equation for the reaction is
Na₂HAsO₄ + NaOH ⟶ Na₃AsO₄ + H₂O
1. Mass balance for Na
All the Na⁺ comes from the Na₂HAsO₄ and the NaOH.
The mass balance equation for Na is
[tex]\rm c_{Na^{+}} = 2[Na^{+}]_{Na_{2}HAsO_{4}} + [Na^{+}]_{NaOH}[/tex]
At the moment of mixing and before the reaction started, the total volume had doubled, so the concentrations of each component were halved.
[Na₂HAsO₄] = ½ × 0.037 98 =0.018 99 mol·L⁻¹
[NaOH] = ½ × 0.034 28 = 0.017 14 mol·L⁻¹
[tex]\rm c_{Na^{+}} = 2\times 0.01899 + 0.01714 = 0.03798 + 0.01714\\c_{Na^{+}}= \textbf{0.055 12 mol/L}[/tex]
2. Mass balance for arsenate species
All the arsenate species come from the Na₂HAsO₄.
The reactions involved are
HAsO₄²⁻+ OH⁻ ⇌ AsO₄³⁻ + H₂O
HAsO₄²⁻ + H₂O ⇌ H₂AsO₄⁻ + OH⁻
H₂AsO₄⁻ + H₂O ⇌ H₃AsO₄ + OH⁻
The mass balance equation for arsenate species is
[tex]\rm c_{\text{arsenate}} = [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}][/tex]
At the moment of mixing, the concentration of Na₂HAsO₄ had halved.
[Na₂HAsO₄] = ½ × 0.039 78 = 0.018 99 mol·L⁻¹
[tex]\rm [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}] = \textbf{0.018 99 mol/L}[/tex]
please help!!!! Chem question
Answer : The net ionic equation will be,
[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]
Explanation :
In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
The given balanced ionic equation will be,
[tex]Ba(OH)_2(aq)+H_2SO_4(aq)\rightarrow 2H_2O(aq)+BaSO_4(s)[/tex]
The ionic equation in separated aqueous solution will be,
[tex]Ba^{2+}(aq)+2OH^-(aq)+2H^{+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2H^+(aq)+2OH^{-}(aq)[/tex]
In this equation, [tex]H^+\text{ and }OH^-[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]
what is the sign of Mercury
Answer:
The answer is Hg.
Explanation:
Symbol for Mercury is Hg.
g Identify the type of reaction [(combination (c), decomposition (d), combustion (co), single replacement (sr), double replacement (dr) , or neutralization (n)] and write a balanced equation. If there is no reaction, write NR. Aqueous ammonium phosphate reacts with aqueous calcium sulfide.
Answer: [tex]2(NH_4)_3PO_4(aq)+3CaS(aq)\rightarrow Ca_3(PO_4)_2(aq)+3(NH_4)_2S(s)[/tex] is a double displacement reaction (dr).
Explanation:
A double displacement reaction (dr) is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.
[tex]2(NH_4)_3PO_4(aq)+3CaS(aq)\rightarrow Ca_3(PO_4)_2(aq)+3(NH_4)_2S(s)[/tex]
Combination reaction (C) is defined as the reaction where substances combine in their elemental state to form a single compound.
Single displacement reaction (sr) is defined as the reaction where more reactive element displaces a less reactive element from its chemical reaction.
Decomposition reaction (d) is defined as the reaction where a single substance breaks down into two or more simpler substances.
Combustion (Co) is a type of chemical reaction in which hydrocarbons burn in the presence of oxygen to form carbon dioxide and water along with heat.
A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?
Answer:
0.136g
Explanation:
A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?
[tex]Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)[/tex]
Initial mole of Co(NO3)2 [tex]=\frac{mass}{molar mass}[/tex]
[tex]=\frac{5.00}{182.94} \\\\=0.02733mol[/tex]
Mole of Co(NO3)2 in final solution
[tex]=\frac{4.00}{100}\times 0.02733\\\\=0.04\times 0.02733\\\\= 0.001093mol[/tex]
Mole of NO3- in final solution = 2 x Mole of Co(NO3)2
[tex]=2\times 0.001093\\\\=0.002186mol[/tex]
Mass of NO3- in final solution is mole x Molar mass of NO3
[tex]=0.002186\times62.01\\\\=0.136g[/tex]
The final solution contains 0.24 g of nitrate ion.
Number of moles of Co(NO3)2 = 5.00 g/183 g/mol = 0.027 moles
Number of moles = concentration × volume
concentration = Number of moles /volume
Volume of solution = 100 mL or 0.1 L
concentration = 0.027 moles/0.1 L = 0.27 M
Using the dilution formula;
C1V1 = C2V2
C1 = 0.27 M
V1 = 4.00 mL
C2 = ?
V2 = 275. mL
C2 = C1V1/V2
C2 = 0.27 × 4.00/ 275
C2 = 0.0039 M
Number of moles of NO3- ion in Co(NO3)2 = 0.0039 M × 62 g/mol = 0.24 g
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When an unsymmetrical alkene such as propene is treated with N-bromosuccinimide in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. Is this Markovnikov or non-Markovnikov orientation
When an unsymmetrical alkene such as propene is treated with N-bromosuccinimide in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. This reaction describes a non-Markovnikov orientation.
In the reaction between an unsymmetrical alkene (such as propene) and N-bromosuccinimide (NBS) in the presence of aqueous dimethyl sulfoxide (DMSO), the major product is formed with the bromine atom bonded to the less highly substituted carbon atom of the alkene.
In Markovnikov's addition, the major product is formed by adding the electrophile (in this case, the bromine atom) to the carbon atom with more hydrogen atoms bonded to it. However, the given reaction exhibits non-Markovnikov selectivity, as the bromine atom adds to the less substituted carbon atom.
This non-Markovnikov selectivity can be attributed to the presence of DMSO, which acts as a polar solvent and helps generate a bromine radical (Br•). The radical intermediate can then undergo reaction with the alkene, leading to the observed regioselectivity where the bromine atom adds to the less substituted carbon. This process is known as a radical addition reaction.
Hence, the reaction demonstrates a non-Markovnikov orientation due to the addition of the bromine atom to the less highly substituted carbon atom of the propene molecule.
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The compound ClF contains Group of answer choices polar covalent bonds with partial negative charges on the Cl atoms. ionic bonds. nonpolar covalent bonds. polar covalent bonds with partial negative charges on the F atoms.
Answer:
polar covalent bonds with partial negative charges on the F atoms.
Explanation:
A covalent bond could be polar or nonpolar depending on the relative electro negativity difference between the two bonding atoms. In this case, the bonding atoms are chlorine and fluorine.
In the Pauling's scale, fluorine has an electro negativity value of 3.98 while chlorine has an electro negativity value of 3.16. The difference in electro negativity between the two atoms is about 0.82. This magnitude of electro negativity difference between the two bonding atoms correspond to the existence of a polar covalent bond in the molecule.
The direction of the dipole depends on the relative electro negativity values of the two bonding atoms. Since fluorine is more electronegative than chlorine, the fluorine atom will be partially negative and the chlorine atom will be partially positive accordingly.
The compound ClF (chlorine monofluoride) contains polar covalent bonds with partial negative charges on the F atoms. Therefore, option D is correct.
In ClF, chlorine (Cl) is more electronegative than fluorine. As a result, the shared electrons in the Cl-F bond are pulled closer to the chlorine atom, creating a partial negative charge on the fluorine atoms and a partial positive charge on the chlorine atom.
This polarity in the Cl-F bond gives the molecule an overall polarity, making it a polar molecule. Thank you for pointing out the error, and I apologize for any confusion caused.
Thus, option D is correct.
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One proposed mechanism of the reaction of HBr with O2 is given here. HBr + O2 → HOOBr (slow) HOOBr + HBr → 2HOBr (fast) HOBr + HBr → H2O + Br2 (fast) What is the equation for the overall reaction?
Answer:
4 HBr + O2 → + 2H2O + 2Br2
Explanation:
Based on the following reaction mechanism:
HBr + O2 → HOOBr (slow)
HOOBr + HBr → 2HOBr (fast)
HOBr + HBr → H2O + Br2 (fast)
The equation for the overall reaction is the sum of the three reactions in which intermediaries of reaction (HOBr and HOOBr are canceled). That is 1 + 2 + 2*(3):
HBr + O2 + HOOBr + HBr + 2HOBr + 2HBr → HOOBr + 2HOBr + 2H2O + 2Br2
4 HBr + O2 → + 2H2O + 2Br2Beeing this reaction the equation of the overall reaction.
what is a mitochondrion
Explanation:
Mitochondria (sing. mitochondria) are organelles, or parts of the eukaryote cell. They are in the cytoplasm, not the nucleus. They make the most cell supply of adenosine triphosphate (ATP), a molecule that cells use as an energy source. ... This means that mitochondria are known as '' the powerhouse of the cell'' or ''cell strength".
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When vinylcyclohexane is treated with in dichloromethane, the major product is (2-bromo ethylidene)cyclohexane . Account for the formation of this product by drawing the structure of the most stable radical intermediate. Include all valence lone pairs in your answer. Include all valence radical electrons in your answer.
Answer:
Explanation:
Vinylcyclohexane is an example of a cyclic hydrocarbon where the vinyl group (-CH=CH₂ ) attaches itself to an end of a cyclohexane in ring form thereby giving rise to a vinylcyclohexane. The vinyl group are ethylene with a reduction in one hydrogen atom given them the name vinyl.
SOo, when vinylcyclohexane is treated with NBS ( i.e N-Bromosuccinimide a chemical reagent used in organic reactions) ; the bromine in the NBS reacts with the cyclohexane thereby giving rise to a allyl radical first. The allyl radical is resonance stabilized radical with an unpaired electron on the allylic carbon . As a result of stabilization ; a more stable substituted cycloalkene is formed as an intermediate .
This stable substituted cycloalkene intermediate then finally react with a bromine ion to give a major product known as ; (2-bromo ethylidene)cyclohexane.
The diagram emphasizing more on the above explanation can be seen in the attached image below
How many moles of CO2 are produced when 84 0 mol O2 completely react?
Answer:
Explanation:boom
Volume of water is 35 cm3 and mass of water is 60 gram, what is the density of the water.
Answer:
p = 1.714 g/cm3
Explanation:
Density Equation
p=mV
Where:
p = density
m = mass = 60g
V = volume = 35cm3
p = 60g x 35cm3
p = 1.714 g/cm3
p=1.714g/cm³
Explanation:
v=35cm³
m=60g
P=mass/volume (density formula)
=60/35
=1.714g/cm³
how many molecules (not moles) of NH3 are produced from 5.25x10^-4 g of H2?
due in a few, please help. will mark as brainliest
Answer:
not 100% but i think its 1.57x10^20
Explanation:
5.25x10^-4g / 2.016g
2.60x10^-4 x 6.022x10^23= 1.56x10^20 molecules
What is Keq for the reaction 2HCl(9) = H2(g) + Cl2(g)?
Answer:
Keq= [(Cl2) (H2)] / (HCl)^2
Explanation:
Equilibrium Constant, Keq, is written as products/reactants.
So it's going to be Keq= [(Cl2) (H2)] / (HCl)^2
