Answer:
pH of the buffer is 3.75
Explanation:
It is possible to find pH of a buffer using Hendersson-Hasselbalch equation:
pH = pKa + log [A⁻] / [HA]
Where [A⁻] is molar concentration of the conjugate base and [HA] concentration of the weak acid
In the lactic buffer, pKa = 3.86. Lactic acid is the weak acid and its conjugate base is tha lactate salt. H-H equation for this buffer is:
pH = 3.86 + log [Lactate] / [Lactic acid]
Replacing with the concentrations of the problem:
pH = 3.86 + log [0.10M] / [0.13M]
pH = 3.75
pH of the buffer is 3.75
A 135 g sample of H20 at 85°C is cooled. The water loses a total of 15 kJ of energy in the cooling
process. What is the final temperature of the water? The specific heat of water is 4.184 J/g.°C.
A. 112°C
B. 58°C
C. 70°C
D. 84°C
E. 27°C
Answer:
B. 58°C
Explanation:
Hello,
In this case, the relationship among heat, mass, specific heat and temperature for water is mathematically by:
[tex]Q=mCp\Delta T=mCp(T_2-T_1)[/tex]
In such a way, solving for the final temperature [tex]T_2[/tex] we obtain:
[tex]T_2=T_1+\frac{Q}{mCp}[/tex]
Therefore, we final temperature is computed as follows, considering that the involved heat is negative as it is lost for water:
[tex]T_2=85\°C+\frac{-15kJ*\frac{1000J}{1kJ} }{135g*4.184\frac{J}{g\°C} }\\\\T_2=58\°C[/tex]
Thereby, answer is B. 58°C .
Regards.
"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
11.
What is the value of AH in k cal/mol for the following reaction?
CH3CH2CH3 + Br2
CH3 -CH-CH3 + HBr
Br
(1)
-12
(2)
-13
(3)
-15
(
4)
-16
Answer:
(1) -12 Kcal/mol
Explanation:
Our answer options for this question are:
(1) -12 Kcal/mol
(2) -13 Kcal/mol
(3) -15 Kcal/mol
(4) -16 Kcal/mol
With this in mind, we can start with the chemical reaction (Figure 1). In this reaction, two bonds are broken, a C-H and a Br-Br. Additionally, a C-Br and a H-Br are formed.
If we want to calculate the enthalpy value, we can use the equation:
ΔH=ΔHbonds broken-ΔHbonds formed
If we use the energy values reported, its possible to calculate the energy for each set of bonds:
ΔHbonds broken
C-H = 94.5 Kcal/mol
Br-Br = 51.5 Kcal/mol
Therefore:
105 Kcal/mol + 53.5 Kcal/mol = 146 Kcal/mol
ΔHbonds formed
C-Br = 70.5 Kcal/mol
H-Br = 87.5 Kcal/mol
Therefore:
70.5 Kcal/mol + 87.5 Kcal/mol = 158 Kcal/mol
ΔH of reaction
ΔH=ΔHbonds broken-ΔHbonds formed=(146-158) Kcal/mol = -12 Kcal/mol
I hope it 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,
What is the initial temperature (°C) of a system that has the pressure decreased by 10 times while the volume increased by 5 times with a final temperature of -123°C?
Answer:
27°C or 300K
Explanation
We were told that the pressureof the system decreased by 10 times implies that P2= P1/10
Where P2=final pressure
P1= initial pressure
Wew were also told that the volume of the system increased by 5 times this implies that V2= 5×V1
Where T2= final temperature =-123C= 273+(-123C)=150K
T1= initial temperature
But from gas law
PV=nRT
As n and R are constant
P1V1/T1 = P2V2/T2
T1= P1V1T2/P2V2
T1=2×T2
T1=2×150
T1=300K
=300-273
=27°C
the initial temperature (°C) of a system is 27°C
How did Earth come to have an oxygen atmosphere?A.Precambrian rocks released oxygen into the atmosphere.B.Volcanoes released oxygen into the atmosphere.C.Early organisms created oxygen from other gases in the atmosphere.D.Oxygen was the primary gas originally in Earth's atmosphere.
Which of the following buffer systems would be the best choice to create a buffer with pH 9.10?
a) HF/KF (pKa = 3.14)
HNO2/KNO2 (pKa = 3.39)
NH3/NH4Cl (pKa = 9.25)
HClO/KClO (pKa = 7.46)
b) for the best buffer system, calculate the ratio of the molarities of the buffer components required to make the buffer
c) for the best buffer system, calculate the ratio of the masses of the buffer components required to make 1.00 L of the buffer
Answer:
a) NH3/NH4Cl (pKa = 9.25)
b) [tex]\frac{[Base]}{[Acid]} =0.708[/tex]
c)
Explanation:
Hello,
a) In this case, for a buffering capacity, if we want to select the best buffer, we should ensure that the buffer's pKa approaches the desired pH, therefore, since the buffer NH3/NH4Cl has a pKa of 9.25 that is very close to the desired pH of 9.10, we can pick it as the best choice.
b) In this case, we use the Henderson-Hasselbach equation in order to compute the molar ratio:
[tex]pH=pKa+log(\frac{[Base]}{[Acid]} )\\\\log(\frac{[Base]}{[Acid]} )=9.10-9.25=-0.15\\\\\frac{[Base]}{[Acid]} =10^{-0.15}\\\\\frac{[Base]}{[Acid]} =0.708[/tex]
c) Finally, for the ratio of masses, we use the molar mass of both ammonia as the base (17 g/mol) and ammonium chloride as the acid (53.45 g/mol) to compute it, assuming 1.00 L as the volume of the solution:
[tex]\frac{m_{Base}}{m_{Acid}} =0.708\frac{molBase}{molAcid}*\frac{17gBase}{1molBase} *\frac{1molAcid}{53.45gAcid}\\ \\\frac{m_{Base}}{m_{Acid}} =0.225[/tex]
Regards.
The best choice to create a buffer with pH 9.10 is NH₃/NH₄Cl (pKa=9.25), ratio of molarities and masses for NH₃/NH₄Cl are 0.708 & 0.225 respectively.
How do we calculate the pH of buffer solution?pH of buffer solution will be calculated by using the Henderson Hasselbalch equation as:
pH = pKa + log([base]/[acid])
From the above reaction it is clear that valu of pH is directly proportional to the value of pKa. So, the pKa value for NH₃/NH₄Cl is comparatively high which will close to the 9.10 pH.Ratio of the molarities for the NH₃/NH₄Cl buffer solution will be calculated by using the Henderson Hasselbalch equation as:log([NH₃]/[NH₄Cl]) = 9.10 - 9.25 =
log([NH₃]/[NH₄Cl]) = -0.15
[NH₃]/[NH₄Cl] = [tex]10^{-0.15}[/tex] = 0.708
Ratio of masses for the NH₃/NH₄Cl buffer solution will be calculated by using the below equation as:M = n/V, where
M = molarity
V = volume = 1L
n = moles = W(mass) / M(molar mass)
Mass(NH₃)/Mass(NH₄Cl) = 0.708 {(mol of NH₃×17g of NH₃NH₃) /
(mol of NH₄Cl×53.45g of NH₄Cl)
Mass(NH₃)/Mass(NH₄Cl) = 0.225
Hence required values are calculated above.
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The half-life of Zn-71 is 2.4 minutes. If one had 100.0 g at the beginning, how many grams would be left after 7.2 minutes has elapsed? Report your answer to 1 decimal place.
Answer:
12.5g
Explanation:
Half life = 2.4 Minutes.
The half life of a compound is the time it takes to decay to half of it's original concentration or mass.
Time lapsed= 7.2 minutes. This is equivalent to 3 half lives ( 3 * 2.4)
Initial mass = 100g
First half life;
100g --> 50g
Second half life;
50g --> 25g
Third half life;
25g --> 12.5g
The amount of Zn-71 that remains after 7.2 mins has elapsed is 12.5 g
We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:
Half-life (t½) = 2.4 mins
Time (t) = 7.2 mins
Number of half-lives (n) =?[tex]n = \frac{t}{t_{1/2}} \\\\n = \frac{7.2}{2.4} \\\\[/tex]
n = 3Thus, 3 half-lives has elapsed.
Finally, we shall the amount remaining. This can be obtained as follow:Original amount (N₀) = 100 g
Number of half-lives (n) = 3
Amount remaining (N) =?[tex]N = \frac{N_{0}}{2^{n}} \\\\N = \frac{100}{2^{3 }}\\\\N = \frac{100}{8}\\\\[/tex]
N = 12.5 gThus, the amount of Zn-71 that remains after 7.2 mins is 12.5 g
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when the temperature of an ideal gas is increased from 27C to 927C then kinetic energy increases by
Answer:
The rms speed of its molecules becomes. (T) has become four times. Therefore, v_(rms) will become two times,...
A sample of a hydrocarbon is found to contain 7.99g carbon and 2.01g hydrogen. What is the empirical formula for this compound
Answer:
The empirical formulae for the compound is CH3.
Determine whether each phrase describes carboxylic acids or esters.a. Do not form hydrogen bonds amongst themselves and have higher vapor pressureb. Form hydrogen bonds amongst themselves and have lower vapor pressurec. Notable for their pleasant fragrancesd. Their reactions with base are kn. own as saponificationse. Usually have a sour odorf. Their reactions with base are known as neutralizations
Explanation:
a. Esters
b. Carboxylic acids
c. Esters (ethyl hexanoate smells like pineapple)
d. Carboxylic acids (produces a carboxylic salt)
For the given phrases the following description is better.
a. Esters
b. Carboxylic acids
c. Esters (ethyl hexanoate smells like pineapple)
d. Carboxylic acids (produces a carboxylic salt)
Esters and carboxylic acids:An ester is a synthetic compound got from a corrosive in which somewhere around one - OH hydroxyl bunch is supplanted by an - O-alkyl (alkoxy) bunch, as in the replacement response of a carboxylic acid and a liquor.
Carboxylic acid is any of a class of natural mixtures in which a carbon (C) particle is clung to an oxygen (O) molecule by a twofold bond and to a hydroxyl bunch (―OH) by a solitary bond.
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Explain why, when the guanidino group of arginine is protonated, the double-bonded nitrogen is the nitrogen that accepts the proton. There is a scheme of a reversible reaction, where one equivalent of the reactant reacts with two equivalents of H plus. The reactant is H2NCNHCH2CH2CH2CHCO minus, with an NH group, with a lone pair at the N atom, double-bonded to the first (from left to right) carbon, an NH2 group attached to the fifth carbon, an O atom double-bonded to the sixth carbon and a lone pair of electrons at the first and the second N atoms of the chain. The product has the same structure as the reactant, except that not an NH group with a lone pair, but an NH2 plus group is double-bonded to the first carbon. In addition, an NH3 plus group is attached to the fifth carbon instead of the NH2 group.
Answer:
Due to the resonance structures
Explanation:
In the question:
"Explain why, when the guanidino group of arginine is protonated, the double-bonded nitrogen is the nitrogen that accepts the proton. There is a scheme of a reversible reaction, where one equivalent of the reactant reacts with two equivalents of H plus"
We have to take into account the structure of the amino acid arginine. In which, we have the amino and the carboxylic groups in the right and the guanidine group in the left.
In this group, we have a central carbon with three nitrogen atoms around and a double bond with the nitrogen on the top. This nitrogen on the top will accept the proton because the structure produced will have a positive charge on this nitrogen. Then, the double bond with the carbon can be delocalized into the nitrogen producing a positive charge in the carbon.
In this structure (the carbocation), we can have several resonance structures. In the blue option, we can produce a double bond with the nitrogen on the right. In the purple option, we can produce a double bond with the nitrogen on the left.
In conclusion, if the nitrogen in the top on the guanidine group accepts an hydrogen atom and we will have several resonance structures that can stabilize the molecule. Due to this, the nitrogen in the top its the best option to accept hydrogens.
See figure 1
I hope it helps!
Solid sodium iodide is slowly added to a solution that is 0.0050 M Pb 2+ and 0.0050 M Ag +. [K sp (PbI 2) = 1.4 × 10 –8; K sp (AgI) = 8.3 × 10 –17] Calculate the Ag + concentration when PbI 2 just begins to precipitate.
Answer:
[Ag⁺] = 5.0x10⁻¹⁴M
Explanation:
The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:
Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸
Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷
The PbI₂ just begin to precipitate when the product [Pb²⁺] [I⁻]² = 1.4x10⁻⁸
As the initial [Pb²⁺] = 0.0050M:
[Pb²⁺] [I⁻]² = 1.4x10⁻⁸
[0.0050] [I⁻]² = 1.4x10⁻⁸
[I⁻]² = 1.4x10⁻⁸ / 0.0050
[I⁻]² = 2.8x10⁻⁶
[I⁻] = 1.67x10⁻³So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:
[Ag⁺] [I⁻] = 8.3x10⁻¹⁷
[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷
[Ag⁺] = 5.0x10⁻¹⁴Mo prepare vegetables for finishing by grilling, sautéing, pan frying, deep frying, or stewing, you should _______ them to cook them to partial doneness
Answer:
To prepare vegetables for finishing by grilling, sautéing, pan frying, deep frying, or stewing, you should parboil them to cook them to partial doneness.
Calculate the [H+] and pH of a 0.0010 M acetic acid solution. The Ka of acetic acid is 1.76×10−5. Use the method of successive approximations in your calculations.
Answer:
[tex][H^+]=0.000123M[/tex]
[tex]pH=3.91[/tex]
Explanation:
Hello,
In this case, dissociation reaction for acetic acid is:
[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]
For which the equilibrium expression is:
[tex]Ka=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]
Which in terms of the reaction extent [tex]x[/tex] could be written as:
[tex]1.74x10^{-5}=\frac{x*x}{[CH_3COOH]_0-x}=\frac{x*x}{0.0010M-x}[/tex]
Thus, solving by using a solver or quadratic equation we obtain:
[tex]x_1=0.000123M\\\\x_2=-0.000141M[/tex]
And clearly the result is 0.000123M, which also equals the concentration of hydronium ion in solution:
[tex][H^+]=0.000123M[/tex]
Now, the pH is computed as follows:
[tex]pH=-log([H^+])=-log(0.000123)\\\\pH=3.91[/tex]
Best regards.
g Use the References to access important values if needed for this question. A researcher took 2.592 g of a certain compound containing only carbon and hydrogen and burned it completely in pure oxygen. All the carbon was changed to 7.851 g of CO2, and all the hydrogen was changed to 4.018 g of H2O . What is the empirical formula of the original compound
Answer:
Empirical formula is: C₂H₅
Explanation:
The chemical equation of burning of a compound that conatins only Carbon and Hydrogen is:
CₓHₙ + O₂ → XCO₂ + n/2H₂O
That means the moles of CO₂ produced are the moles of Carbon in the compound and moles of hydrogen are twice moles of water. Empirical formula is the simplest ratio between moles of each element in the compound. Thus, finding molse of C and moles of H we can find empirical formula:
Moles C and H:
Moles C = Moles CO₂:
7.851g CO₂ ₓ (1mol / 44g) = 0.1784 moles CO₂ = Moles C
Moles H = 2 Moles H₂O
4.018g H₂O ₓ (1mol / 18.01g) = 0.2231 * 2 = 0.4417 moles H
Ratio C:H
The ratio between moles of hydrogen and moles of Carbon are:
0.4417 moles H / 0.1784 moles C = 2.5
That means there are 2.5 moles of H per mole of Carbon. As empirical formula must be given only in whole numbers,
Empirical formula is: C₂H₅In the following net ionic equation, identify each reactant as either a Bronsted-Lowry acid or a Bronsted-Lowry base.
HF-(aq) + H2O(l) rightarrow F-(aq) + H3O(aq)
B-L_____ B-L_____
The formula of the reactant that acts as a proton donor is_____.
The formula of the reactant that acts as a proton acceptor is_______.
Answer:
Bronsted lowry base = Proton acceptor = H2O
Bronsted lowry acid = Proton donor = HF-
Explanation:
The equation is given as;
HF-(aq) + H2O(l) --> F-(aq) + H3O(aq)
A bronsted lowry base is any specie that can accept hydrogen ion (proton) from another molecule.
Basically a bronsted lowry base is a proton acceptor while a bronsted lowry acid is a proton donor.
In the reaction above, upon comparing both the reactants and products;
Bronsted lowry base = Proton acceptor = H2O
Bronsted lowry acid = Proton donor = HF-
A solution is known to contain only one type of cation. Addition of Cl1- ion to the solution had no apparent effect, but addition of (SO4)2- ion resulted in a precipitate. Which cation is present
Answer:
We can have: Calcium, strontium, or barium
Explanation:
In this case, we have to remember the solubility rules for sulfate [tex]SO_4~^-^2[/tex] and the chloride [tex]Cl^-[/tex]:
Sulfate
All sulfate salts are SOLUBLE-EXCEPT those also containing: Calcium, silver, mercury (I), strontium, barium or lead.([tex]Ca^+^2~,Ag^+~,Hg_2^+^2~,Sr^+2~,Ba^+^2~,Pb^+^2[/tex]), which are NOT soluble.
Chloride
All chloride salts as SOLUBLE-EXCEPT those also containing: lead, silver, or mercury (I). ([tex]Pb^+^2~,Ag^+~,Hg_2~^+^2[/tex]), which are NOT soluble.
If we the salt formed a precipitated with the sulfate anion, we will have as possibilities "Calcium, silver, mercury (I), strontium, barium or lead". If We dont have any precipitated with the Chloride anion we can discard "Silver, mercury (I), lead" and our possibilities are:
"Calcium, strontium, or barium".
I hope it helps!
What is the compound formed from the combination of the base and a hydrogen ion
Answer:
Water
Explanation:
When a base react to and hydrogen ion, we can produce water.
According to these equation
H⁺ + OH⁻ ⇄ H₂O Kw: 1×10¹⁴
Remember that OH⁻ is determined by a strong base.
This reaction is called neutralization. You can also produce water with a weak base, because OH⁻ are released. For example, let's mention ammonia which is a weak base, it takes protons from water (H⁺)
NH₃ + H₂O ⇄ OH⁻ + NH₄⁺ Kb
When the ammonium ion (acid), reacts to a base, you produce water.
NH₄⁺ + NaOH → NH₃ + H₂O + Na⁺
When equation for neutralization of HBr by Ca(OH)2 is correctly balanced, how many molecules of water will be formed
Answer:
When equation for neutralization of HBr by Ca(OH)₂ is correctly balanced, 1.2046*10²⁴ molecules of water will be formed
Explanation:
A neutralization reaction is one in which an acid (or acidic oxide) reacts with a base (or basic oxide). In the reaction a salt is formed and in most cases water is formed. A Salt is an ionic compound formed by the union of ions and cations through ionic bonds.
In the reactions of a strong acid (those substances that completely dissociate) with a strong base (they dissociate completely, giving up all their OH-), the complete neutralization of the species is carried out:
2 HBr (aq) + Ca(OH)₂ (s) → CaBr₂ (aq) + 2 H₂O (l)
The reaction is already balanced, complying with the law of conservation of matter. This law states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reactants must be equal to the number of atoms present in the products.
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), 2 moles of water H₂O are formed.
On the other hand, Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023 * 10²³ particles per mole. Avogadro's number applies to any substance.
Then you can apply the following rule of three: if 1 mole of H₂O contains 6.023*10²³ molecules, 2 moles of H₂O, how many molecules does it contain?
[tex]amount of molecules=\frac{2moles*6.023*10^{23}molecules }{1 mole}[/tex]
amount of molecules= 1.2046*10²⁴ molecules
When equation for neutralization of HBr by Ca(OH)₂ is correctly balanced, 1.2046*10²⁴ molecules of water will be formed
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.
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It would require ? Liters of water to dissolve 36 grams of the substance.
The correct answer is 3 liters
Explanation:
If a substance has a solubility of [tex]12 \frac{grams}{liter}[/tex], this means in 1 liter, the grams that can be dissolved are 12 grams. Now, considering Justin and Ellie need to dissolve 36 grams to calculate the number of liters just divide the total of grams into 12 as each liter dissolves only 12 grams. The process is shown below:
36 grams (the amount that will be dissolved) ÷ 12 (grames dissolved per liter) = 3 liters (liters to dissolved 36 grams)
Answer:
It would be 3 liters
Explanation:
What does Avogadro's number represent? A. How much 1 atom of a substance weighs. B. The number of molecules in 1 mole of a substance. C. The number of molecules in 1 atomic number of a substance. D. How much 1 mole of a substance weighs.
Answer:
B, the number of molecules in 1 mole of a substance
How many milliliters of 7.10 M hydrobromic acid solution should be used to prepare 5.50 L of 0.400 M HBr
Answer:
310 mL
Explanation:
Step 1: Given data
Initial concentration (C₁): 7.10 MInitial volume (V₁): ?Final concentration (C₂): 0.400 MFinal volume (V₂): 5.50 LStep 2: Calculate the initial volume
We have a concentrated HBr solution and we want to prepare a diluted one. We can do so using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 0.400 M × 5.50 L / 7.10 M
V₁ = 0.310 L = 310 mL
Which of the following is the balanced reaction, given the rate relationships below.
a. rate = − 13 Δ[CH4] Δt = − 12 Δ[H2O] Δt = − Δ[CO2] Δt = 14 Δ[CH3OH] Δt
b. rate = − 12 Δ[N2O5] Δt = 12 Δ[N2] Δt = 15 Δ[O2] Δt
c. rate = − 12 Δ[H2] Δt = − 12 Δ[CO2] Δt = − Δ[O2] Δt = 12 Δ[H2CO3] Δt
Answer:
a. [tex]3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH[/tex]
b. [tex]2N_2O_5\rightarrow 2N_2 + 5O_2[/tex]
c. [tex]2H_2+2CO_2+O_2\rightarrow 2H_2CO_3[/tex]
Explanation:
Hello,
In this case, since those rate relationships have the stoichiometric coefficient at the denominators of the fractions preceding each ratio and the negative terms account for reactants and positive for products, we have:
a. [tex]3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH[/tex]
b. [tex]2N_2O_5\rightarrow 2N_2 + 5O_2[/tex]
c. [tex]2H_2+2CO_2+O_2\rightarrow 2H_2CO_3[/tex]
Best regards.
Plz help!!!! Solve this by using factor labeling
Answer:
the answer is 2,000 nickels.
Explanation:
we multiplied 100 by 100, because there are 100 cents in a dollar, and we divided 10,000 by 5, because there are 5 cents in a nickel.
Convert 59800 kilograms to pounds
Answer:
131836.43 pounds
Explanation:
one kilogram is 2.20462 pounds. multiply 2.20462 by 59800
Answer: 131836.43
Formula: Multiply the mass value by 2.205
59800×2.205=131836.43
The new hybrid car can get 51.5 km/gal. It has a top speed of 40000.00 cm/min and is 4m long. How fast can the car go in m/hr?
Answer:
The anawer of this question is 0.024 m/h
Explanation:
Other explanations of the question are additional.
What is the molarity of 4 g of NaCl dissolved in 100mL of water?
Answer:
[tex]M=0.684M[/tex]
Explanation:
Hello,
In this case, considering that the solution is formed by NaCl as the solute and water as the solvent, we can compute the molarity as shown below:
[tex]M=\frac{mol_{solute}}{V_{solution}}[/tex]
Whereas the volume of the solution must be in liters. In such a way, since the addition of sodium chloride does not significantly changes the volume of the solution we can say it remains in 100 mL (0.100 L) and the moles of sodium chloride are computed by using its molar mass (58.45 g/mol):
[tex]mol_{solute}=4g*\frac{1mol}{58.45g} =0.0684mol[/tex]
Therefore, the molarity is:
[tex]M=\frac{0.0684mol}{0.100L} \\\\M=0.684\frac{mol}{L}=0.684M[/tex]
Regards.
An actacide tablet containing Mg(OH)2 (MM = 58.3g / (mol)) is titrated with a 0.100 M solution of HNO3. The end point is determined by using an indicator. Based on 20.00mL HNO3 being used to reach the endpoint, what was the mass of the Mg * (OH) in the antacid tablet? * 0.0583 g 0.583 5.83 g 58.3 g
Answer:
0.0583g
Explanation:
The equation of the reaction is;
2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)
From the question, number of moles of HNO3 reacted= concentration × volume
Concentration of HNO3= 0.100 M
Volume of HNO3 = 20.00mL
Number of moles of HNO3= 0.100 × 20/1000
Number of moles of HNO3 = 2×10^-3 moles
From the reaction equation;
2 moles of HNO3 reacts with 1 mole of Mg(OH)2
2×10^-3 moles reacts with 2×10^-3 moles ×1/2 = 1 ×10^-3 moles of Mg(OH)2
But
n= m/M
Where;
n= number of moles of Mg(OH)2
m= mass of Mg(OH)2
M= molar mass of Mg(OH)2
m= n×M
m= 1×10^-3 moles × 58.3 gmol-1
m = 0.0583g
