Answer:
One carbon atom and the nitrogen atom have nonzero formal charges.
Explanation:
The compound Acetonitrile has sixteen valence electrons as is easily San from its structure. It contains a carbon nitrogen triple bond with a lone pair of electrons on nitrogen. All atoms satisfy the octet rule and there is no hyper valent atom in the molecule.
The formal charge an carbon and nitrogen is calculated as follows;
No. of valence electron on atom - [non bonded electrons + no. of bonds]
Therefore, for carbon and nitrogen, we have;
formal charge on carbon = 4 - (0 + 4) = 0
formal charge on nitrogen = 5 - (2 + 3) = 0
Hence carbon and nitrogen both possess zero formal charges.
please help guys the question is
give reasons
a. we have to separate the mixture
b. All impure substances are not harmful.
c. A mixture of iron fillings and sand can be separated by using a magnet
d. A sentences "shake before well use" is written on the bottle of the medicine.
Answer:
(a )people separate mixtures in order to ger a specific substance that they need.
In the Lewis structure of AB4 where B is more electronegative than A. Both are main group elements where A has 8 valence electrons and each B has 7 valence electrons.
Required:
a. What is the total number of valence electrons?
b. How many lone pairs are in the molecule?
Answer:
1. 36
2. Two
Explanation:
The Lewis structure shows the valence electrons present in a compound. Usually the valence electrons are shown as dot structures around the symbol of the elements involved in the compound.
For a compound AB4 where B is more electronegative than A and A has 8 electrons in its valence shell, there will be thirty six valence electrons on the outermost shell of the molecule.
There are six electron pair domains present in the molecule, four bond pairs and two lone pairs. The molecule is in a square planar geometry.
Answer: a- 36 valence electrons
b- 14 lone pairs
Explanation:
Valence is equal to A + 4B = 8 + 4(7)
With 4 bonds between A and the 4 B, that is 36 valence minus 8 electrons in those pairs = 28. 28 is 14 lone pairs.
Predict the reactants of this chemical reaction. That is, fill in the left side of the chemical equation. Be sure the equation you submit is balanced.
_______ → Ba(ClO)2 + H2O(l)
Answer:
2HClO(aq) + Ba(OH)₂(aq) → Ba(ClO)₂(aq) + 2H₂O(l)
Explanation:
The reaction corresponds to a neutralization reaction between an acid and a base, as follows:
2HClO(aq) + Ba(OH)₂(aq) → Ba(ClO)₂(aq) + 2H₂O(l)
From the equation above we have that the acid HClO reacts with the base Ba(OH)₂ to obtain a salt Ba(ClO)₂ and water.
In the balanced reaction, we have that 2 moles of HClO react with 1 mol of Ba(OH)₂ to produce 1 mol of Ba(ClO)₂ and 2 moles of water.
I hope it helps you!
How many grams of H2O will be formed when 32.0 g H2 is mixed with 73.0 g of O2 and allowed to react to form water
hope this helps u
pls mark as brainliest .-.
Which of the following contains a nonpolar covalent bond?
O A. Co
B. NaCl
O C. 02
O D. HE
Answer:
The answer is o2
Explanation:
I took the test
g Which ONE of the following pairs of organic compounds are NOT pairs of isomers? A) butanol ( CH3-CH2-CH2-CH2-OH ) and diethyl ether ( CH3–CH2–O–CH2–CH3 ) B) isopentane ( (CH3)2-CH-CH2-CH3 ) and neopentane ( (CH3)4C ) C) ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 ) D) acrylic acid ( CH2=CH-COOH ) and propanedial ( OHC–CH2–CHO ) E) trimethylamine ( (CH3)3N ) and propylamine ( CH3-CH2-CH2-NH2 )
Answer:
ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 )
Explanation:
Isomers are compounds that have the same molecular formula but different structural formulas. Hence any pair of compounds that can be represented by exactly the same molecular formula are isomers of each other.
If we look at the pair of compounds; ethanolamine ( H2N-CH2-CH2-OH ) and acetamide ( CH3-CO-NH2 ), one compound has molecular formula, C2H7ON while the other has a molecular formula, C2H5ON, hence they are not isomers of each other.
3,3-dibromo-4-methylhex-1-yne
Explanation:
see the attachment. hope it will help you...Which of the following is a salt that will form from the combination of a strong base with a weak acid?
Select the correct answer below:
A. NaHCO3
B. H2O
C. CH3CO2H
D. NH4Cl
Answer:
A. NaHCO₃
Explanation:
NaHCO₃ ⇒ NaOH + H₂CO₃
NaOH is a strong base and H₂CO₃ is a weak acid. Therefore, NaHCO₃ is a salt of a strong base-weak acid reaction. The salt is basic because carbonic acid (H₂CO₃) is a weak acid so it remains undissociated. So, there is a presence of additional OH⁻ ions that makes the solution basic.
Hope that helps.
Which of the following combinations will result in a reaction that is spontaneous at all temperatures?
Negative enthalpy change and negative entropy change
Negative enthalpy change and positive entropy change
Positive enthalpy change and negative entropy change
Positive enthalpy change and positive entropy change
PLS EXPLAIN WHAT EACH MEANS AND THE VARIABLES AND THE EXPLANATION BEHIND IT
Answer:
[tex]\huge\boxed{Option \ 2}[/tex]
Explanation:
A reaction is spontaneous at all temperatures by the following combinations:
=> A negative enthalpy change ( [tex]\triangle H < 0[/tex] )
=> A positive entropy change ( [tex]\triangle S > 0[/tex] )
See the attached file for more better understanding!
from Gibbs Equation, [tex] \Delta G = \Delta H - T\Delta S [/tex]
reaction is spontaneous if $\Delta G$ is negative.
so, first option is not valid at high temperature, ($-h+ts$)
second, is always a spontaneous reaction, ($-h-ts$)
third, is never spontaneous ($+h+ts$)
4th is similar to second, spontaneous at higher temperatures ($+h-ts$)
A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an elevation where the pressure is 0.73 atm, then the new volume is 1.8 L. What is the temperature (in °C) of the air at this elevation?
Answer:
The temperature of the air at this given elevation will be 53.32425°C
Explanation:
We can calculate the final temperature through the combined gas law. Therefore we will need to know 1 ) The initial volume, 2 ) The initial temperature, 3 ) Initial Pressure, 4 ) Final Volume, 5 ) Final Pressure.
Initial Volume = 1.2 L ; Initial Temperature = 25°C = 298.15 K ; Initial pressure = 1.0 atm ; Final Volume = 1.8 L ; Final pressure = 0.73 atm
We have all the information we need. Now let us substitute into the following formula, and solve for the final temperature ( T[tex]_2[/tex] ),
P[tex]_1[/tex]V[tex]_1[/tex] / T[tex]_1[/tex] = P[tex]_2[/tex]V[tex]_2[/tex] / T[tex]_2[/tex],
T[tex]_2[/tex] = P[tex]_2[/tex]V[tex]_2[/tex]T[tex]_1[/tex] / P[tex]_1[/tex]V[tex]_1[/tex],
T[tex]_2[/tex] = 0.73 atm [tex]*[/tex] 1.8 L [tex]*[/tex] 298.15 K / 1 atm [tex]*[/tex] 1.2 L = ( 0.73 [tex]*[/tex] 1.8 [tex]*[/tex] 298.15 / 1 [tex]*[/tex] 1.2 ) K = 326.47425 K,
T[tex]_2[/tex] = 326.47425 K = 53.32425 C
"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"
Answer:
245.66g of NH₄Cl is the mass we need to add to obtain the desire pH
Explanation:
The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:
pH = pKa + log [A⁻] / [HA]
Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical
First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.
pKa NH₃/NH₄⁺pKb = - log Kb
pKb = -log 1.8x10⁻⁵ = 4.74
pKa = 14 - pKb
pKa = 14 - 4.74
pKa = 9.26
Moles NH₃2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃
H-H equation:pH = pKa + log [NH₃] / [NH₄Cl]
8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]
-1.06 = log [0.400 moles] / [NH₄Cl]
0.0087 = [0.400 moles] / [NH₄Cl]
[NH₄Cl] = 0.400 moles / 0.0087
[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):
4.59 moles NH₄Cl ₓ (53.491g / mol) =
245.66g of NH₄Cl is the mass we need to add to obtain the desire pH
9
What might happen if acidic chemicals were emitted into
the air by factories? Choose the best answer.
A
The acid would destroy metallic elements in the air
B
The acid would be neutralized by bases within clouds
C
Acid rain might destroy ecosystems and farmland
D
Violent chemical reactions would take place within the
atmosphere
co search
O
BI
Consider these metal ion/metal standard reduction potentials Cd2+(aq)|Cd(s) Zn2+(aq)|Zn(s) Ni2+(aq)|Ni(s) Cu2+(aq)|Cu(s) Ag+(aq)|Ag(s) -0.40 V -0.76 V ‑0.25 V +0.34 V +0.80 V Based on the data above, which species is the best reducing agent?
Answer:
The best reducing agent is Zn(s)
Explanation:
A reducing agent must to be able to reduce another compound, by oxidizing itself. Consequently, the oxidation potential must be high. The oxidation potential of a compound is the reduction potential of the same compound with the opposite charge. Given the reduction potentials, the best reducing agent will be the compound with the most negative reduction potential. Among the following reduction potentials:
Cd₂⁺(aq)|Cd(s) ⇒ -0.40 V
Zn²⁺(aq)|Zn(s) ⇒ -0.76 V
Ni²⁺(aq)|Ni(s) ⇒‑0.25 V
Cu²⁺(aq)|Cu(s) ⇒ +0.34 V
Ag⁺(aq)|Ag(s) ⇒ +0.80 V
The most negative is Zn²⁺(aq)|Zn(s) ⇒ -0.76 V
From this, the most reducing agent is Zn. Zn(s) is oxidized to Zn²⁺ ions with the highest oxidation potential (0.76 V).
What is the ph of 0.36M HNO3 ?
Answer:
0.44
Explanation:
We know that the pH of any acid solution is given by the negative logarithm of its hydrogen ion concentration. Hence, if I can obtain the hydrogen ion concentration of any acid, I can obtain its pH.
For the acid, HNO3, [H^+] = [NO3^-]= 0.36 M
pH= -log [H^+]
pH= - log[0.36]
pH= 0.44
Heterocyclic aromatic compounds undergo electrophilic aromatic substitution in a similar fashion to that undergone by benzene with the formation of a resonance-stabilized intermediate. Draw all of the resonance contributors expected when the above compound undergoes bromination
Answer:
See explanation
Explanation:
When we talk about electrophilic substitution, we are talking about a substitution reaction in which the attacking agent is an electrophile. The electrophile attacks an electron rich area of a compound during the reaction.
The five membered furan ring is aromatic just as benzene. This aromatic structure is maintained during electrophilic substitution reaction. The attack of the electrophile generates a resonance stabilized intermediate whose canonical structures have been shown in the image attached.
Half-cells were made from a nickel rod dipping in nickel sulfate solution and a copper rod dipping in copper sulfate solution. The cells were combined to construct a voltaic electrochemical cell. Sketch the cell and label anode and cathode with charges, electrode material and electrolyte solutions, half-reactions and overall reaction, give direction of electron flow and movement of ions.
Answer:
Check the Attachment.
Half-reactions:
Anode: (OXIDATION) Ni --> Ni2+ + 2e-
Cathode: (REDUCTION) Cu2+ +2e- --> Cu
Overall reaction: Ni + Cu2+ --> Ni2+ + Cu
Explanation:
Overall, reaction is basically Anode + Cathode, where electrons on both sides cancel out (if not, you need to multiply the equation in a way you can cancel them out).
Hope this helps.
Write the equations that represent the first and second ionization steps for sulfuric acid (H2SO4) in water.
Answer:
[tex]H_2SO_4(aq)\rightarrow H^+(aq)+HSO_4^-\\\\HSO_4^-(aq)\rightarrow H^+(aq)+SO_4^{2-}rightarrow[/tex]
Explanation:
Hello,
In this case, given that the sulfuric acid is a diprotic acid (two hydrogen ions) we can identify two ionization reactions, the first one, showing up the dissociation of the first hydrogen to yield hydrogen sulfate ions and the second one, showing up the dissociation of the hydrogen sulfate ions to hydrogen ions and sulfate ions by separated as shown below:
[tex]H_2SO_4(aq)\rightarrow H^+(aq)+HSO_4^-\\\\HSO_4^-(aq)\rightarrow H^+(aq)+SO_4^{2-}[/tex]
They are have one-sensed arrow, since sulfuric acid is a strong acid.
Regards.
The equations that represent the first and second ionization steps for sulfuric acid in water are H₂SO₄→HSO₄+H⁺ & HSO₄⁻→SO₄⁻+H⁺ respectively.
What is ionization reaction?Ionization reactions are those reactions in which atom or molecule will convert into ion by bearing a positive or negative charge on itself.
In water in the following way ionization of sulphuric acid takes place:
In the first ionization step one hydrogen atom (H⁺) will loose from the sulphuric acid molecule as:H₂SO₄ → HSO₄⁻ + H⁺
In the second ionization step another hydrogen atom will also loose and we get the sulphate ion (SO₄⁻) and one proton (H⁺) as:HSO₄⁻ → SO₄⁻ + H⁺
Hence, two steps are shown above.
To know more about ionization reaction, visit the below link:
https://brainly.com/question/1445179
Calculate the molarity of a solution containing 29g of glucose (C 6 H 12 O 6 ) dissolved in 24.0g of water. Assume the density of water is 1.00g/mL.
Answer:
whats the ph ofpoh=9.78
Explanation:
When 91.96g of Na reacts with 32.o g of O2 how many grams of NaO2 are produced
Answer:
123.96 g Na₂O
Explanation:
4 Na + O₂ ⇒ 2 Na₂O
You first need to find the limiting reagent. Convert the reactants to moles and see which produces the least amount of product using the mole ratios in the chemical equation.
(91.96 g Na)/(22.99 g/mol Na) = 4 mol Na
(4 mol Na) × (2 mol Na₂O/4 mol Na) = 2 mol Na₂O
(32.0 g O₂)/(32.0 g/mol) = 1 mol O₂
(1 mol O₂) × (2 mol Na₂O/1 mol O₂) = 2 mol Na₂O
Since they both produce the same amount of product, you don't need to pick a limiting reagent. Now, convert moles of Na₂O to grams.
(2 mol Na₂O) × (61.98 g/mol Na₂O) = 123.96 g Na₂O
When we react a weak acid with a strong base of equal amounts and concentration, the component of the reaction that will have the greatest effect on the pH of the solution is:______.
a. the acid.
b. the base.
c. the conjugate acid.
d. the conjugate base.
Answer:
d. the conjugate base.
Explanation:
The general reaction of a weak acid, HA, with a strong base YOH, is:
HA + YOH → A⁻ + H₂O + Y⁻
Where A⁻ is the conjugate base of the weak acid and Y⁻ usually is a strong electrolyte.
That means after he complete reaction you don't have weak acid nor strong base, just conjugate base that will be in equilibrium with water, thus (Strong electrolyte doesn't change pH:
A⁻ + H₂O ⇄ HA + OH⁻
As the equilibrium is producing OH⁻, the pH of the solution is being affected for the conjugate base
Right option:
d. the conjugate base.Assume that you are provided with the following materials:
• Strips of metallic zinc, metallic copper, metallic iron
• 1M aqueous solutions of ZnSO4, CuSO4, FeSO4, and aqueous iodine(I2)
• Other required materials to create Voltaic cells such as beakers, porous containers, graphite rods, a voltmeter, and a few wires with alligator clips.
In this modified version of the lab, after thoroughly studying the lab hand out and watching the videos,identify 4 different combinations of Voltaic cells that are possible to be created with the above materials.For each cell created, include the following details.
A) Which electrode was the anode,and which was the Cathode?
B) The anode and cathode half reactions.
C) Balanced equation for each cell you propose to construct.
D) Calculated Eocelle Short hand notation (line notation) for each cell (be sure to include the inactive electrode if needed).
Answer:
See explanation
Explanation:
First voltaic cell;
Zn(s)|Zn^2+(aq)||Cu^2+(aq)|Cu(s)
Anode;
Zinc
Cathode;
Copper
Oxidation half equation;
Zn(s)------> Zn^2+(aq) + 2e
Reduction half equation;
Cu^2+(aq) +2e -----> Cu(s)
Overall; Zn(s) + Cu^2+(aq) -----> Zn^2+(aq) + Cu(s)
E°cell = 0.34 -(-0.76) =1.1 V
Second voltaic cell;
Zn(s)|Zn^2+(aq)||Fe^2+(aq)|Fe(s)
Anode;
Zinc
Cathode;
Iron
Oxidation half equation;
Zn(s)------> Zn^2+(aq) + 2e
Reduction half equation;
Fe^2+(aq) +2e -----> Fe(s)
Overall; Zn(s) + Fe^2+(aq) -----> Zn^2+(aq) + Fe(s)
E°cell = (-0.44) -(-0.76) = 0.32 V
Third voltaic cell;
Fe(s)|Fe^2+(aq)||Cu^2+(aq)|Cu(s)
Anode;
Iron
Cathode;
Copper
Oxidation half equation;
Fe(s)------> Fe^2+(aq) + 2e
Reduction half equation;
Cu^2+(aq) +2e -----> Cu(s)
Overall; Fe(s) + Cu^2+(aq) -----> Fe^2+(aq) + Cu(s)
E°cell = 0.34 -(-0.44) = 0.78 V
Fourth voltaic cell
Cu(s)|Cu^2+(aq)||I2(aq)|C(s)|I^-(aq)
Anode;
Copper
Cathode;
Graphite rod
Oxidation half equation;
Cu(s)------> Cu^2+(aq) + 2e
Reduction half equation;
I2(aq) +2e -----> 2I^-(aq)
Overall; Cu(s) + I2(aq) -----> Cu^2+(aq) + 2I^-(aq)
E°cell = 0.54 -0.34 = 0.20 V
A hypothetical metal crystallizes with the face-centered cubic unit cell. The radius of the metal atom is 198 picometers and its molar mass is 195.08 g/mol. Calculate the density of the metal in g/cm3.
Answer:
7.38 g/cm³ is the density of the metal
Explanation:
In a Face-centered cubic unit cell you have 4 atoms. Also, the edge length is √8×r (r is radius of the atom).
To solve this problem, we need first to calculate the volume of the unit cell and then, with molar mass calculate the mass of 4 atoms. As density is the ratio between mass and volume we can obtain this value.
Volume of the unit cellVolume = a³
a = √8×r
(r = 198x10⁻¹²m)
a = 5.6x10⁻¹⁰ m
Volume = 1.756x10⁻²⁸ m³
1m = 100cm → 1m³ = (100cm)³:
1.756x10⁻²⁸ m³× ((100cm)³ / 1m³) =
1.756x10⁻²² cm³ → Volume of the unit cell in cm³Mass of the unit cell:There are 4 atoms of gold:
4 atoms × (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles of gold
As 1 mole weighs 195.08g:
6.64x10⁻²⁴ moles of gold × (195.08g / mol) =
1.296x10⁻²¹g is the mass of the unit cellDensity of the metal:1.296x10⁻²¹g / 1.756x10⁻²² cm³ =
7.38 g/cm³ is the density of the metalThe density of the metal is 7.40 g/cm³
In cubic crystal system, face-centered cubic FFC is the name given to sort of atom arrangement observed in which structure is made up of atoms organized in a cube with a portion of an atom in each corner and six extra atoms in the center of each cube face.
It is expressed by using the formula:
[tex]\mathbf{\rho = \dfrac{Z \times M}{N_A\times a^}}[/tex]
where;
[tex]\rho[/tex] = density of the metalZ = atoms coordination no = 4 (for FCC)Molar mass (M) = 195.8 g/molAvogadro's constant (NA) = 6.022 × 10²³ /mola = edge lengthFor face-centered cubic FFC;
The edge length [tex]\mathbf{a =2 \sqrt{2}\times r }[/tex]
[tex]\mathbf{a =2 \sqrt{2}\times 198 \ pm }[/tex]
[tex]\mathbf{a =560.0285 \ pm }[/tex]
a = 5.60 × 10⁻⁸ cm
Replacing it into the previous equation, we have:
[tex]\mathbf{\rho = \dfrac{4 \times 195.8}{6.022 \times 10^{23} \times( 5.60 \times 10^{-8} )^3}}[/tex]
[tex]\mathbf{\rho = 7.40\ g/cm^3 }[/tex]
Learn more about face-centered cubic arrangement here:
https://brainly.com/question/14786352?referrer=searchResults
A student sets up the following equation to convert a measurement. The (?) Stands for a number the student is going to calculate. Fill in the missing part of this equation. (0.030 cm^3) x ? =m^3
Answer:
\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 3.0 \times 10^{-7} \text{ m}^{3}
Explanation:
0.030 cm³ × ? = x m³
You want to convert cubic centimetres to cubic metres, so you multiply the cubic centimetres by a conversion factor.
For example, you know that centi means "× 10⁻²", so
1 cm = 10⁻² m
If we divide each side by 1 cm, we get 1 = (10⁻² m/1 cm).
If we divide each side by 10⁻² m, we get (1 cm/10⁻² m) = 1.
So, we can use either (10⁻² m/1 cm) or (1 cm/10⁻² m) as a conversion factor, because each fraction equals one.
We choose the former because it has the desired units on top.
The "cm" is cubed, so we must cube the conversion factor.
The calculation becomes
[tex]\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 0.30 \times 10^{-6}\text{ m}^{3} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}\\\\\textbf{0.30 cm}^{\mathbf{3}} \times \left (\dfrac{\mathbf{10^{-2}}\textbf{ m}}{\textbf{1 cm}}\right )^{\mathbf{3}} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}[/tex]
A chemist prepares a solution of sodium nitrate by measuring out of sodium nitrate into a volumetric flask and filling the flask to the mark with water.
Calculate the concentration in mol/L of the chemist's sodium nitrate solution. Round your answer to 3 significant digits.
Answer:
5.74M or 5.74 mol/L (to 3 sign. fig.)
Explanation:
The molar mass of NaNO3 is 85g/mol, which means that:
1 mole of NaNO3 - 85g
? moles - 122.0g
= 122/85 = 1.44 moles
Concentration in mol/L = no. of moles (moles) ÷ volume (L)
[tex]\frac{1.44}{0.250}[/tex] = 5.74M or 5.74 mol/L (to 3 sign. fig.)
I hope the steps are clear and easy to follow.
Question 14 (5 points)
What's the acid ionization constant for an acid with a pH of 2.11 and an equilibrium
concentration of 0.30 M?
O A) 4.87x10-8
B) 1.99x10-6
C) 3.32x10-4
OD) 2.01x10-4
Answer:
D) 2.01 x 10⁻⁴ .
Explanation:
pH = 2.11
[ H⁺ ] = [tex]10^{-2.11}[/tex]
Let the acid be HA
It will ionise as follows .
HA ⇄ H⁺ + A⁻
in equilibrium .30 [tex]10^{-2.11}[/tex] [tex]10^{-2.11}[/tex]
Acid ionisation constant Ka = [tex]\frac{(10^{-2.11})^2}{0.3}[/tex]
= 2 x 10⁻⁴
Answer:
D) 2.01 x 10⁻⁴ is correct!
Explanation:
I got it in class!
Hope this Helps!! :))
is a polyprotic acid. Write balanced chemical equations for the sequence of reactions that carbonic acid can undergo when it's dissolved in water.
Answer:
H₂CO₃ H₂O ⇄ HCO₃⁻ + H₃O⁺ Ka1
HCO₃⁻ + H₂O ⇄ CO₃⁻² + H₃O⁺ Ka2
CO₃⁻² + H₂O ⇄ HCO₃⁻ + OH⁻ Kb1
HCO₃⁻ + H₂O ⇄ H₂CO₃ + OH⁻ Kb2
Explanation:
Formula for carbonic acid is: H₂CO₃
It is a dyprotic acid, because it can release two protons. We can also mention that is a weak one. The equilibrums are:
H₂CO₃ H₂O ⇄ HCO₃⁻ + H₃O⁺ Ka1
HCO₃⁻ + H₂O ⇄ CO₃⁻² + H₃O⁺ Ka2
When the conjugate strong bases, carbonate and bicarbonate take a proton from water, the reactions are:
CO₃⁻² + H₂O ⇄ HCO₃⁻ + OH⁻ Kb1
HCO₃⁻ + H₂O ⇄ H₂CO₃ + OH⁻ Kb2
Notice, that bicarbonate anion can release or take a proton to/from water. This is called amphoteric,
a boy capable of swimming 2.1m/a in still water is swimming in a river with a 1.8 m/a current. At what angle must he swim in order to end up directly opposite his starting point?
Answer:
The boy must swim at an angle of 59°northwest to get to a position directly opposite his starting point.
Explanation:
To get to a point directly opposite his starting point, the boy must travel at an angle x, in a direction northwest of his starting point. The speed of the boy and the speed of the river current forms a right-angled triangle with an an opposite side of 1.8 m/a and a hypotenuse of 2.1 m/a having an angle x.
Sin x = opp/ hyp
Sin x = 1.8/2.1
x = sin⁻¹ (1.8/2.10
x = 58.99
x = 59°
Therefore, the boy must swim at an angle of 59° in the northwesterly direction to get to a position directly opposite his starting point.
Calculate the amount of heat that must be absorbed by 10.0 g of ice at –20°C to convert it to liquid water at 60.0°C. Given: specific heat (ice) = 2.1 J/g·°C; specific heat (water) = 4.18 J/g·°C; ΔH fus = 6.0 kJ/mol.
Answer:
The amount of heat to absorb is 6,261 J
Explanation:
Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.
The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.
So:
Heat required to raise the temperature of ice from -20 °C to 0 °CBeing the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).
In this case, m= 10 g, specific heat of the ice= 2.1 [tex]\frac{J}{g*C}[/tex] and ΔT=0 C - (-20 C)= 20 C
Replacing: Q= 10 g*2.1 [tex]\frac{J}{g*C}[/tex] *20 C and solving: Q=420 J
Heat required to convert 0 °C ice to 0 °C waterThe heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:
Q= m* ΔHfusion
In this case, being 1 mol of water= 18 grams: Q= 10 g*[tex]6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}[/tex]= 3.333 kJ= 3,333 J (being kJ=1,000 J)
Heat required to raise the temperature of water from 0 °C to 60 °CIn this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 [tex]\frac{J}{g*C}[/tex] and ΔT=60 C - (0 C)= 60 C
Replacing: Q= 10 g*4.18 [tex]\frac{J}{g*C}[/tex] *60 C and solving: Q=2,508 J
Finally, Qtotal= 420 J + 3,333 J + 2,508 J
Qtotal= 6,261 J
The amount of heat to absorb is 6,261 J
The amount of heat to absorb is 6,261 J.
Calculation for heat:Heat required to raise the temperature of ice from -20 °C to 0 °C.
The formula for specific heat is used to calculate the amount of heat
Q = c * m * ΔT
Where,
Q =heat exchanged by a body,
m= mass of the body
c= specific heat
ΔT= change in temperature
Given:
m= 10 g,
specific heat of the ice= 2.1
ΔT=0 C - (-20 C)= 20 C
On substituting the values:
Q= 10 g*2.1 *20 C
Q=420 J
Heat required to convert 0 °C ice to 0 °C water.
The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:
Q= m* ΔHfusion
Heat required to raise the temperature of water from 0 °C to 60 °C
m= 10 g,
Specific heat of the water= 4.18
ΔT=60 C - (0 C)= 60 C
On substituting:
Q= 10 g*4.18 *60 C
Q=2,508 J
Thus, Qtotal= 420 J + 3,333 J + 2,508 J
Qtotal= 6,261 J
The amount of heat to absorb is 6,261 J
Find more information about Specific heat here:
brainly.com/question/13439286
An electrolysis cell has two electrodes. Which statement is correct? A. Reduction takes place at the anode, which is positively charged. B. Reduction takes place at the cathode, which is positively charged. C. Reduction takes place at the dynode, which is uncharged. D. Reduction takes place at the cathode, which is negatively charged. E. Reduction takes place at the anode, which is negatively charged.
Answer:
D. Reduction takes place at the cathode, which is negatively charged.
Explanation:
In an electrolytic cell there are two electrodes; the cathode and the anode. The anode is the positive electrode while the cathode is the negative electrode. Oxidation occurs at the anode while reduction occurs at the cathode.
At the anode, species give up electrons and become positively charged ions while at the cathode species accept electrons and become reduced.
Im really confused and select all that apply questions scare me.
Answer:
The 3rd one
Explanation:
