Answer:
pH = - log [concentration]
pH = - log (0.0043M)
pH = 2.37
For the following reaction, draw the major organic product and select the correct IUPAC name for the organic reactant. If there is more than one major product, both may be drawn.
When drawing hydrogen atoms on a carbon atom, either include all hydrogen atoms or none on that carbon atom, or your structure may be incorrect.
Select the correct IUPAC name for the organic reactant:
a) 2-methylbutene
b) 2-methyl-1-butene
c) 3-methyl-3-butened) 3-methylbutene
Answer:
The correct IUPAC name for the organic reactant is :
d) 3-methylbutene
Explanation:
Firstly the missing diagram is attached in the diagram below.
The objective of this question is to draw the major organic product and select the correct IUPAC name for the organic reactant. If there is more than one major product, both may be drawn.
From the image attached below; we would see the reaction that occurs between the alkene and the HBr (hydrobromic acid). What really occur in the reaction is that; in the presence of HBr with an alkene compound a secondary 2° carbocation is usually formed. This secondary 2° carbocation formed is usually unstable, so what we called an hydride shift occurs (Markovnikov's product) here to form a stable tertiary 3° carbocation.
The correct IUPAC name for the organic reactant is : 3-methylbutene
An electrochemical cell is constructed with a zinc metal anode in contact with a 0.052 M solution of zinc nitrate and a silver cathode in contact with a 0.0042 M solution of silver(I) nitrate. What is the value of Q to use in the Nernst equation for this cell
Answer:
Q = 12.38
Explanation:
The Nernst equation is given as; Ecell = E°cell - (2.303RT/nF) log Q ;where Q is the reaction quotient.
The reaction quotient, Q in a reaction, is the product of the concentrations of the products divided by the product of the concentrations of the reactants.
In an electrochemical cell, Q is the ratio of the concentration of the electrolyte at the anode to that of the electrolyte at the cathode.
Q = [anode]/[cathode]
therefore , Q = 0.052/0.0042 = 12.38
During chemical reaction 7.55gKI and 9.06g were allowed to react. How many grams of excess reagent are left over after the reaction is complete. Reaction: Pb(NO3)2(s) + 2KCI(s) > 2KNO3(s) + PbI(s)
Answer: 7.45 g of [tex]Pb(NO_3)_2[/tex] excess reagent are left over after the reaction is complete.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
a) [tex]{\text{Number of moles of} KI}=\frac{7.55g}{166g/mol}=0.045moles[/tex]
b) [tex]{\text{Number of moles of} Pb(NO_3)_2}=\frac{9.06g}{331.2g/mol}=0.027moles[/tex]
The balanced chemical reaction is :
[tex]Pb(NO_3)_2(s)+2KI(s)\rightarrow 2KNO_3(s)+PbI(s)[/tex]
According to stoichiometry :
2 moles of [tex]KI[/tex] require = 1 mole of [tex]Pb(NO_3)_2[/tex]
Thus 0.045 moles of [tex]KI[/tex] will require=[tex]\frac{1}{2}\times 0.045=0.0225moles[/tex] of [tex]Pb(NO_3)_2[/tex]
Thus [tex]KI[/tex] is the limiting reagent as it limits the formation of product and [tex]Pb(NO_3)_2[/tex] is the excess reagent as (0.045-0.0225) = 0.0225 moles are left
Mass of [tex]Pb(NO_3)_2=moles\times {\text {Molar mass}}=0.0225moles\times 331.2g/mol=7.45g[/tex]
Thus 7.45 g of [tex]Pb(NO_3)_2[/tex] of excess reagent are left over after the reaction is complete.
WILL GIVE BRAINLIEST!!!!! will give brainliest!!!!! ---------Write the molecular equation, complete ionic equation and net ionic equation of Barium nitrate reacting with potassium carbonate.
Answer:
Molecular:
Ba(NO3)2 + K2CO3 -> BaCO3 + 2KNO3
Complete ionic:
Ba2+ + 2NO3- + 2K+ + CO3 2- -> BaCO3 + 2K+ + 2NO3-
Net ionic:
Ba2+ + CO3 2- - > BaCO3
Explanation:
Molecular consists of all species reacting.
Complete ionic consists of all ionic species (ex. K+, NO3-) separated, as well as any compounds that didn't dissociate into ions (BaCO3 doesn't dissolve).
Net ionic doesn't include spectator ions (in this case, nitrate and potassium) and only species that aren't present on both sides of the arrow (barium and carbonate become a solid precipitate, so the ions aren't present as products, making them appear in the net ionic equation).
uses of sodium chloride in daily life
Answer:
sodium chloride can be used as salt
extraction sodium metal by electrolysis
a common chemical in laboratory experiments
Answer:
sodium chloride can be used as preservatives,
in preserving foods.
CAN SOMEONE HELP ME!!
Solutions of each of the hypothetical acids in the following table are prepared with an initial concentration of 0.100 M. Which of the four solutions will have the lowest pH and be most acidic? Explain please.
Acid pKa
HA 4.00
HB 7.00
HC 10.00
HD 11.00
a. HA
b. HB
c. HC
d. HD
e. All will have the same pH because the concentrations are the same.
Answer: HA has lowest pH and it is the most acidic as compared to the rest of given acids.
Explanation:
We know that relation between [tex]pK_a[/tex] and [tex]K_a[/tex] is as follows.
[tex]pK_a = -log K_a[/tex]
This means that more is the value of [tex]K_a[/tex], smaller will be the [tex]pK_a[/tex]. Also, more is the value of [tex]K_a[/tex] smaller will be the pH of a solution.
As, larger is the value of [tex]K_a[/tex] more negative will be the [tex]pK_a[/tex] value. Hence, stronger will be the acid.
In the given options, HA has the smallest [tex]pK_a[/tex] value.
Therefore, we can conclude that HA has lowest pH and it is the most acidic as compared to the rest of given acids.
What is cell culturing?
a technique that uses specific antibodies to visualize features of cells
a technique that visualizes how specific genes are used within a cell
a technique in which cells are purposefully grown under specific conditions
an imaging technology used to study features smaller than the human eye can see
Answer:
a technique in which cells are purposefully grown under specific conditions
Explanation:
Answer:
its c
Explanation:
correct edge2020
calculate the moles of 25.2 g Na2S2O8
Answer:
To calculate the moles we must first find the molar mass M
M (Na2S2O8) = (23*2) + (32*2) + (16*8)
= 46 + 64 + 168
= 278g/mol
Molar mass = mass/moles
moles =mass / molar mass
= 25.2/278
= 0.0906mol
Hope this helps.
How many moles of solute are contained in the following solution: 15.25 mL of a 2.10 M CaCl₂
Answer:
0.032moles
Explanation:
2.10moles in 1000ml what about 15.25ml
(15.25×2.10)÷1000
0.032moles
Identify the Lewis acids and Lewis bases in the following reactions:
1. H+ + OH- <-> H2O Lewis acid: Lewis base:
2. Cl- + BCl3 <-> BCl4- Lewis acid: Lewis base:
3. K+ + 6H2O <-> K(H2O)6+ Lewis acid: Lewis base:
Answer: 1. [tex]H^++OH^-\rightarrow H_2O[/tex] Lewis acid : [tex]H^+[/tex], Lewis base : [tex]OH^-[/tex]
2. [tex]Cl^-+BCl_3\rightarrow BCl_4^-[/tex] Lewis acid : [tex]BCl_3[/tex], Lewis base : [tex]Cl^-[/tex]
3. [tex]K^++6H_2O\rightarrow K(H_2O)_6[/tex] Lewis acid : [tex]K^+[/tex], Lewis base : [tex]H_2O[/tex]
Explanation:
According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.
1. [tex]H^++OH^-\rightarrow H_2O[/tex]
As [tex]H^+[/tex] gained electrons to complete its octet. Thus it acts as lewis acid.[tex]OH^-[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]H^+[/tex].
2. [tex]Cl^-+BCl_3\rightarrow BCl_4^-[/tex]
As [tex]BCl_3[/tex] is short of two electrons to complete its octet. Thus it acts as lewis acid. [tex]Cl^-[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]BCl_3[/tex].
3. [tex]K^++6H_2O\rightarrow K(H_2O)_6[/tex]
As [tex]K^+[/tex] is short of electrons to complete its octet. Thus it acts as lewis acid. [tex]H_2O[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]K^+[/tex].
Which of the following bases is the WEAKEST? The base is followed by its Kb value. Group of answer choices HOCH2CH2NH2, 3.2 × 10-5 (CH3CH2)3N, 5.2 × 10-4 NH3, 1.76 × 10-5 C5H5N, 1.7 × 10-9 Since these are all weak bases, they have the same strength.
Answer:
C₅H₅N being the weakest base
Explanation:
A weak base (B) is defined as a chemical compound that, in reaction with water, produce a small quantity of BH⁺
The general reaction is:
B + H₂O ⇄ BH⁺ + OH⁻ Where Kb is defined as:
Kb = [BH⁺] [OH⁻] / [B]
That means the smallest Kb is the weakest base because is producing the smallest quantity of BH⁺.
In the problem, the smallest Kb is C₅H₅N being the weakest base.
Which of the following is not an example of a mechanical wave?
A. Fans doing "The Wave" at a sporting event.
B. Sound waves coming out of the radio.
C. Water waves at hie beach.
D. Sunshine.
Answer:
Option D
Explanation:
A mechanical wave is a wave of energy that can travel long distances and could go through characteristics of matter such as solids, liquids, and gases. Mechanical waves can also travel through vacuums. A good example of a mechanical wave would be sound, sound is a wave spread through a object and can go through different types of matter. Which is why your answer is option D "sunshine." Light cannot go through a vacuum while sounds, and water can.
Hope this helps.
The mechanical wave example does not include the sunshine
What is mechanical waves ?It is the wave of energy that can travel long distances and considered the characteristics of matter like solids, liquids, and gases. It can also travel via vacuums. The Light cannot go via a vacuum while sounds, and water can go.
Learn more about sound here:https://brainly.com/question/16750970
How many moles of h2 can be formed if a 3.25g sample of Mg reacts with excess HCl
Answer:
0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl
Explanation:
The balanced reaction is:
Mg + 2 HCl → MgCl₂ + H₂
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles react:
Mg: 1 moleHCl: 2 molesMgCl₂: 1 moleH₂: 1 moleBeing:
Mg: 24. 31 g/moleH: 1 g/moleCl: 35.45 g/molethe molar mass of the compounds participating in the reaction is:
Mg: 24.31 g/moleHCl: 1 g/mole + 35.45 g/mole= 36.45 g/moleMgCl₂: 24.31 g/mole + 2*35.45 g/mole= 95.21 g/moleH₂: 2*1 g/mole= 2 g/moleThen, by stoichiometry of the reaction, the following quantities of mass participate in the reaction:
Mg: 1 mole* 24.31 g/mole= 24.31 gHCl: 2 moles* 36.45 g/mole= 72.9 gMgCl₂: 1 mole* 95.21 g/mole= 95.21 gH₂: 1 mole* 2 g/mole= 2 gThen you can apply the following rule of three: if by stoichiometry 24.31 grams of Mg form 1 mole of H₂, 3.25 grams of Mg how many moles of H₂ will they form?
[tex]moles of H_{2} =\frac{3.25 grams of Mg*1 mole of H_{2} }{24.31 grams of Mg}[/tex]
moles of H₂= 0.134
0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl
0.134 moles of H₂ are formed by the reaction of 3.25 g of Mg with excess HCl.
Let's consider the balanced equation between Mg and HCl.
Mg + 2 HCl ⇒ MgCl₂ + H₂
The molar mass of Mg is 24.3 g/mol. The moles corresponding to 3.25 g of Mg are:
[tex]3.25 g \times \frac{1mol}{24.3g} = 0.134 mol[/tex]
The molar ratio of Mg to H₂ is 1:1. The moles of H₂ formed by 0.134 moles of Mg are:
[tex]0.134 mol Mg \times \frac{1molH_2}{1molMg} = 0.134molH_2[/tex]
0.134 moles of H₂ are formed by the reaction of 3.25 g of Mg with excess HCl.
Learn more: https://brainly.com/question/9743981
When 200g of AgNO3 solution mixes with 150 g of NaI solution, 2.93 g of AgI precipitates, and the temperature of the solution rises by 1.34oC. Assume 350 g of solution and a specific heat capacity of 4.184 J/g•oC. Calculate H for the following: Ag+(aq) + I- (aq) → AgI(s)
Answer:
[tex]\Delta H=1962.3J[/tex]
Explanation:
Hello,
In this case, we can compute the change in the solution enthalpy by using the following formula:
[tex]\Delta H=mC\Delta T[/tex]
Whereas the mass of the solution is 350 g, the specific heat capacity is 4.184 J/g °C and the change in the temperature is 1.34 °C, therefore, we obtain:
[tex]\Delta H=350g*4.184\frac{J}{g\°C} *1.34\°C\\\\\Delta H=1962.3J[/tex]
It is important to notice that the mass is just 350 g that is the reacting amount and by means of the law of the conservation of mass, the total mass will remain constant, for that reason we compute the change in the enthalpy as shown above, which is positive due to the temperature raise.
Best regards.
differentiate between sol,aerosol and solid soluti
Answer:
Sol is a colloidal suspension with solid particles in a liquid. Foam is formed when many gas particles are trapped in a liquid or solid. Aerosol contains small particles of liquid or solid dispersed in a gas. While solid solution contain solid as solute in either solid, liquid or gas.
Consider each pair of compounds listed below and determine whether a fractional distillation would be necessary to separate them or if a simple distillation would be sufficient.
a. Ethyl acetate and hexane
b. Diethyl Ether and 1-butanol
c. Bromobenzene and 1,2-dibromobenzene
Empirical formula for compound of 2.17 mol N and 4.35 mol O
Answer:
Explanation:
ratio of moles of N and O in molecule =
N / O = 2.17 / 4.35
1/2
empirical formula = NO₂
Which of the following aqueous solutions are good buffer systems? . 0.24 M hydrochloric acid + 0.23 M sodium chloride 0.28 M ammonia + 0.35 M ammonium nitrate 0.16 M barium hydroxide + 0.28 M barium bromide 0.15 M nitrous acid + 0.14 M potassium nitrite 0.35 M calcium nitrate + 0.21 M calcium iodide
Answer: 0.28 M ammonia + 0.35 M ammonium nitrate and 0.15 M nitrous acid + 0.14 M potassium nitrite
Explanation:
Buffer solution is the solution which resists the change in the magnitude of the pH when small additions of either acid or base is added.
Acidic Buffer solutions consist of weak acid and its conjugate base usually mixed in relatively equal and large quantities.
Basic Buffer solutions consist of weak base and its conjugate acid usually mixed in relatively equal and large quantities.
Thus 0.28 M ammonia + 0.35 M ammonium nitrate ( weak base + conjugate acid) and 0.15 M nitrous acid + 0.14 M potassium nitrite (weak acid + conjugate base) are good buffer systems
The aqueous solutions that are good buffer systems are:
0.28 M ammonia + 0.35 M ammonium nitrate. 0.15 M nitrous acid + 0.14 M potassium nitrite.We want to determine which of the given solutions would make a good buffer.
What is a buffer?A buffer is a solution used to resist abrupt changes in pH when an acid or a base is added.
What kinds of buffers exist?Acidic buffer: formed by a weak acid and its conjugate base.Basic buffer: formed by a weak base and its conjugate acid.Which of the following aqueous solutions are good buffer systems?
0.24 M hydrochloric acid + 0.23 M sodium chloride. No, since HCl is a strong acid.0.28 M ammonia + 0.35 M ammonium nitrate. Yes, it would be a good basic buffer.0.16 M barium hydroxide + 0.28 M barium bromide. No, since Ba(OH)₂ is a strong base. 0.15 M nitrous acid + 0.14 M potassium nitrite. Yes, it would be a good acidic buffer.0.35 M calcium nitrate + 0.21 M calcium iodide. No, since no acids nor bases are present.The aqueous solutions that are good buffer systems are:
0.28 M ammonia + 0.35 M ammonium nitrate. 0.15 M nitrous acid + 0.14 M potassium nitrite.Learn more about buffers here: brainly.com/question/24188850
a gas obeys the equation of state p(v-b)=RT.for the gas b=0.0391L/mol.calculate the fugacity coefficient for the gas at 1000°c and 1000atm
Answer:
The fugacity coefficient is [tex][\frac{f}{p} ] = 1.45[/tex]
Explanation:
From the question we are told that
The gas obeys the equation [tex]p(v-b) = RT[/tex]
The value of b is [tex]b = b = 0.0391 \ L /mol[/tex]
The pressure is [tex]p = 1000 \ atm[/tex]
The temperature is [tex]T= 1000^oC = 1273 K[/tex]
generally
[tex]RT ln[\frac{f}{p} ] = \int\limits^{p}_{o} [ {v_{r} -v_{i}} ]\, dp[/tex]
Where [tex]\frac{f}{p}[/tex] is the fugacity coefficient
[tex]v_r[/tex] is the real volume which is mathematically evaluated from above equation as
[tex]v_r = \frac{RT}{p} + b[/tex]
[tex]v_r = \frac{RT}{p} + 0.0391[/tex]
and [tex]v_{i}[/tex] is the ideal volume which is evaluated from the ideal gas equation (pv = nRT , at n= 1) as
[tex]v_{i} = \frac{RT}{p}[/tex]
So
[tex]RT ln[\frac{f}{p} ] = \int\limits^{1000}_{o} [[ \frac{RT}{p} + 0.0391] - [\frac{RT}{p} ]} ]\, dp[/tex]
=> [tex]RT ln[\frac{f}{p} ] = \int\limits^{1000}_{o} [0.391 ]\, dp[/tex]
=> [tex]RT ln[\frac{f}{p} ] = [0.391p]\left | 1000} \atop {0}} \right.[/tex]
=> [tex]RT ln[\frac{f}{p} ] = 38.1[/tex]
So
[tex]ln[\frac{f}{p} ] = \frac{39.1}{RT}[/tex]
Where R is the gas constant with value [tex]R = 0.082057\ L \cdot atm \cdot mol^{-1}\cdot K^{-1}[/tex]
[tex][\frac{f}{p} ] = \frac{39.1}{ 2.303 *0.082057 * 1273}[/tex]
[tex][\frac{f}{p} ] = 1.45[/tex]
Consider the following reaction where Kc = 1.29×10-2 at 600 K: COCl2 (g) CO (g) + Cl2 (g) A reaction mixture was found to contain 0.104 moles of COCl2 (g), 4.66×10-2 moles of CO (g), and 3.76×10-2 moles of Cl2 (g), in a 1.00 liter container. Indicate True (T) or False (F) for each of the following:
1. In order to reach equilibrium COCl2(g) must be consumed.
A. True B. False
2. In order to reach equilibrium Kc must increase.
A. True B. False
3. In order to reach equilibrium CO must be consumed.
A. True B. False
4. Qc is greater than Kc.
A. True B. False
5. The reaction is at equilibrium. No further reaction will occur.
A. True B. False
Answer:
1. In order to reach equilibrium COCl₂(g) must be consumed.
B. False
2. In order to reach equilibrium Kc must increase.
B. False .
3. In order to reach equilibrium CO must be consumed.
A. True.
4. Qc is greater than Kc.
A. True
5. The reaction is at equilibrium. No further reaction will occur.
B. False.
Explanation:
Based on the reaction:
COCl₂(g) → CO (g) + Cl₂(g)
And Kc is defined as:
Kc = 1.29x10⁻² = [CO] [Cl₂] / [COCl₂]
Molar concentrations of each species are:
[COCl₂] = 0.104 moles of COCl₂ / 1L = 0.104M
[CO] = 4.66×10⁻² moles of CO / 1L = 4.66×10⁻²M
[Cl₂] = 3.76×10⁻² moles of Cl₂ / 1L = 3.76×10⁻²M
Replacing in Kc formula:
4.66×10⁻²M × 3.76×10⁻²M / 0.104M = 1.68x10⁻²
As the concentrations are not in equilibrium, 1.68x10⁻² is defined as the reaction quotient, Qc.
As Qc > Kc, the reaction will shift to the left producing more COCl₂ and consuming CO and Cl₂. Thus
1. In order to reach equilibrium COCl₂(g) must be consumed.
B. False
2. In order to reach equilibrium Kc must increase.
B. False . Kc is a constant that never change.
3. In order to reach equilibrium CO must be consumed.
A. True.
4. Qc is greater than Kc.
A. True
5. The reaction is at equilibrium. No further reaction will occur.
B. False. The reaction is in equilibrium when Qc = Kc
Calculate the percentage of the void space out of the total volume occupied by 1 mole of water molecules. The density of water is assumed to be 1.0 g/mL that is 1.0 g/cm3. The molar mass of water is 18.0 g/mol. The atomic radius of hydrogen is 37 pm and of oxygen is 73 pm. The formula for the volume of a sphere is 4/3(r3
Answer:
The percentage of the void space out of the total volume occupied is 93.11%
Explanation:
A mole of water contains 2 atoms of hydrogen and 1 atom of oxygen.
To calculate the volume of a mole of water, we calculate 2 times the volume of the hydrogen atom and 1 times the volume of the oxygen atom
Let's calculate this one after the other.
For the hydrogen, formula for the volume will be
[tex]V_{hydrogen[/tex] = 2 × 4/3 × π × [tex]r_{H}^{3}[/tex]
where [tex]r_{H}^{3}[/tex] = 37 pm which is read as 37 picometer (1 picometer = 10^-12 m) = 37 × [tex]10^{-12}[/tex] meters
Volume of the hydrogen = 8/3 × (37 × 10^-12)^3 = 4.05 * 10^-31
we multiply this by the avogadro's number = 6.02 * 10^23
= 4.05 * 10^-31 * 6.02 * 10^23 = 2.6 * 10^-8 m^3
We do same for thr oxygen, but this time we do not multiply the volume of the oxygen by 2 as we have only one atom of oxygen
Volume of oxygen = 4/3 * π * (73 * 10^-12) ^3 * avogadro's number = 9.81 * 10^-7 m^3
adding both volumes together, we have 1.24 * 10^-6 m^3 or simply 1.24 ml ( 0.01 m = 1 ml)
Dividing the molar mass of one mole of water by its density, we can get the volume of 1 mole of water
= (18g/mol)/(1 g/ml) = 18 ml/mol
Now we proceed to calculate the volume of void = Total volume - volume of molecule = 18 - 1.24 = 16.76 ml
Now, the percentage of void = volume of void/total volume * 100%
= 16.76/18 * 100% = 93.11%
Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 4.3 g of hexane is mixed with 7.14 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
We can produce 6.20 grams of CO2
Explanation:
Step 1: Data given
Mass of hexane = 4.3 grams
Molar mass of hexane = 86.18 g/mol
Mass of oxygen = 7.14 grams
Molar mass of oxygen = 32.0 g/mol
Step 2: The balanced equation
2C6H14 + 19O2 → 12CO2 + 14H2O
Step 3: Calculate moles
Moles = mass / molar mass
Moles hexane = 4.3 grams / 86.18 g/mol
Moles hexane = 0.0499 moles
Moles oxygen = 7.14 grams / 32.0 g/mol
Moles oxygen = 0.2231 moles
Step 4: Calculate the limiting reactant
For 2 moles hexane we need 19 moles O2 to produce 12 moles CO2 and 14 moles H2O
Oxygen is the limiting reactant. It will completely be consumed ( 0.2231 moles). Hexane is in excess. There will react 2/19 * 0.2231 = 0.02348 moles
There will be porduced 12/19 * 0.2231 = 0.1409 moles CO2
Step 5: Calculate mass CO2
Mass CO2 = moles CO2 * molar mass CO2
Mass CO2 = 0.1409 moles * 44.01 g/mol
Mass CO2 = 6.20 grams
We can produce 6.20 grams of CO2
Le Chatelier's Principle. For the reaction below, if the equilibrium concentrations were NH3 = 2 x 10-4, H3O+ = 2 x 10-4M and NH4+ = 18.0M, what is the equilibrium constant for the reaction and what would happen if you were to add some acid to this reaction? NH3 + H3O+ --> NH4+ + H2O
Answer:
Explanation:
NH3 + H3O+ --> NH4+ + H2O
equllibrium constant =K = [ H2O] [NH4+] / [NH3] [H3O+ ]
=
by inserting thier respecive values can you calcaulte, by the way coniseder [ H2O] =1 ,
A quantity of 2.00 × 102 mL of 0.662 M HCl is mixed with 2.00 × 102 mL of 0.331 M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the HCl and Ba(OH)2 solutions is the same at 22.00°C. For the process below, the heat of neutralization is −56.2 kJ/mol. What is the final temperature of the mixed solutions? H+(aq) + OH−(aq) → H2O(l)
Answer:
Final temperature of the solution = 26.43°C
Explanation:
Concentration of HCl = 0.662 M, Volume = 200 mL= 0.200 L
Concentration of Ba(OH)₂ = 0.331 M, Volume = 200 mL = 0.200 L
Initial temperature of solution = 22.00°C
Specific Heat capacity of water = 4.184 J/g°C
Heat of neutralization = -56.3 KJ/mol of H₂O produced.
The full calculations is found in the attachment below
Balance the following chemical equation:
NH4NO3
N20+
H2O
Answer:
NH4NO3 = N2O + 2(H2O)
Explanation:
there are 2 N, 4 H, 3 O
Answer:
NH4NO3=N2O+2H2O
Explanation:
N-2,O-3,H-4
In TLC chromatography of plant pigments, why do different pigments travel up the plate at different rates
Nitrogen is a group 15 element. What does being in this group imply about the structure of the nitrogen atom?
O A. Nitrogen has 15 valence electrons.
OB.
Nitrogen has 15 neutrons.
OC. Nitrogen has 5 valence electrons.
D.
Nitrogen has 5 neutrons.
Answer:
D. Nitrogen has 5 valence electrons.
Explanation:
Nitrogen is an element in group 5A of the periodic table. Elements in group 5A all contain just 5 valence electrons. (Electrons in the outer shell).
**Elements are organized into these groups in a periodic table based on the number of valence electrons which determines their charge. (Does not apply to transition metals)
2 Points
What is the voltage of an electrolytic cell with copper and magnesium
electrodes?
A. -2.71 v
B. 2.71 v
C. 2.03 V
D. -2.03 V
The voltage of an electrolytic cell with copper and magnesium electrodes is 2.71 v. Therefore, option B is correct.
What is an electrolytic cell ?The electrolytic cell is a type of cell that performs a redox reaction while using electrical energy. When electrical energy is applied, a redox reaction occurs in molten NaCl. It is therefore an electrolytic cell.
An electrolyte, two electrodes, and an electrolytic cell make up an electrolytic cell (a cathode and an anode). The electrolyte is typically a mixture of ions that have been dissolved in water or another solvent. Electrolytes can also be molten salts, like sodium chloride.
The standard reduction potential, E° of the metals are as below:
Mg²⁺ + 2e⁻ ⇌ Mg; E° = -2.372
Cu²⁺ + 2e⁻ ⇌ Cu; E° = +0.337
Therefore, magnesium has the the lower E°, it will serve as the anode in the electrolytic cell while copper will serve as the cathode.
At the anode; Mg ⇌ Mg²⁺ + 2e⁻, E° = -2.372
At the cathode; Cu²⁺ + 2e⁻ ⇌ Cu, E° = +0.337
EMF of the cell = E° cathode - E° anode
= 0.337 - (-2.372) = 2.71 V
Therefore, EMF of the cell is 2.71 V
Thus, option B is correct.
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• Briefly discuss the cause of errors in the measurements
Experiment predicted observation A student has two unopened cans containing carbonated water. Can A has been stored in the garage () and can B has been stored in the fridge (). The student opens one can at the time, both cans make a fizz.
A) The fizz will be the same for both cans
B) There is not enough information to predict which can will make the louder fizz
C) Can A will make a louder and stronger fizz than can B.
D) Can B will make a louder and stronger fizz than can A.
Answer:
Can A will make a louder and stronger fizz than can B.
Explanation:
Temperature has a direct effect on gas solubility. We know that carbonated water contains carbon dioxide dissolved in water. The extent of dissolution or solubility of this gas is dependent on the temperature of the system.
As the temperature of the system rises, the solubility of gas in solution decreases. It follows that can A, having been stored in a garage is definitely at a higher temperature than can B stored in the refrigerator.
Since solubility of gases decreases with increasing temperature, the carbon dioxide in can A will be less soluble than in can B. This will cause can A to make a louder and stronger fizz when opened than can B.
