Answer:
A neutralization reaction between an acid and a base will produce a salt and water. In a neutralization reaction, the acid will produce H+ ions that react to neutralize the OH- ions produced by the base, forming neutral water. ... A weak acid + a strong base in an acid/base neutralization reaction will form a basic salt
Explanation:
When I was a boy, Uncle Wilbur let me watch as he analyzed the iron content of runoff from his banana ranch. A 25.0-mL sample was acidified with nitric acid and treated with excess KSCN to form a red complex. (KSCN itself is colorless.) The solution was then diluted to 100.0 mL and put in a variable-path length cell. For comparison, a 10.0-mL reference sample of 6.74 times 10^-4 M Fe^3+ was treated with HNO_3 and KSCN and diluted to 50.0 mL, The reference was placed in a cell with a 1.00-cm light path. The runoff sample exhibited the same absorbance as the reference when the path length of the runoff cell was 2.41 cm. What was the concentration of iron in Uncle Wilbur's runoff?
Answer:
C = 2.24x10⁻⁴ M
Explanation:
The concentration of iron in Uncle Wilbur's runoff can be calculated using Beer-Lambert law:
[tex] A = \epsilon*C*l [/tex] (1)
Where:
A: is the absorbance of the compound
ε: is the molar absorptivity of the compound
C: is the concentration of the compound
l: is the optical path length
Since the runoff sample exhibited the same absorbance as the reference sample, we can find the concentration using equation (1):
[tex] \epsilon*C_{1}*l_{1} = \epsilon*C_{2}*l_{2} [/tex] (2)
Where:
Subscripst 1 and 2 refer to Uncle Wilbur's runoff and to reference sample, respectively.
l₁ = 2.41 cm
l₂ = 1.00 cm
We can find C₂ as follows:
[tex] C_{2} = \frac{C_{2i}*V_{i}}{V_{f}} [/tex] (3)
Where:
[tex]C_{2i}[/tex]: is the initial concentration of the reference sample = 6.74x10⁻⁴ M
[tex]V_{i}[/tex]: is the initial volume = 10.0 mL
[tex]V_{f}[/tex]: is the final volume = 50.0 mL
[tex] C_{2} = \frac{6.74 \cdot 10^{-4} M*10.0 mL}{50.0 mL} = 1.35 \cdot 10^{-4} M [/tex]
Now, we can find C₁ using equation (2):
[tex] C_{1} = \frac{C_{2}*l_{2}}{l_{1}} = \frac{1.35 \cdot 10^{-4} M*1.00 cm}{2.41 cm} = 5.60 \cdot 10^{-5} M [/tex]
Finally, since the runoff solution was diluted to 100.0 mL, the initial concentration can be calculated using equation (3) for [tex]C_{1i}[/tex]:
[tex]C_{1i} = \frac{C_{1}*V_{f}}{V_{i}} = \frac{5.60 \cdot 10^{-5} M*100.0 mL}{25.0 mL} = 2.24 \cdot 10^{-4} M[/tex]
Therefore, the concentration of iron in Uncle Wilbur's runoff is 2.24x10⁻⁴ M.
I hope it helps you!
PLEAS HELP I HAVE LIMITED TIME!!
Which compound is the limiting reagent?
Select one:
O2
O H20
H2
Cannot be determined.
Answer:
H2 is the limiting reactant.
Explanation:
From the diagram above:
H2 => White ball
O2 => Red ball
Before the reaction
H2 => White ball => 10
O2 => Red ball => 7
After the reaction
H2O => White and red ball => 10
O2 => 2
From the simple illustration above, we can see that all the H2 were used up in the reaction but there are left over of O2.
This simply means that H2 is the limiting reactant as all of it is used up in the reaction while O2 is the excess reactant as there are leftover.
Which type of semiconductor is created by doping with atoms that contain more valence electrons than the semiconductor material ?
Answer:
N-Type Semiconductor
Explanation:
Draw the reaction mechanism the reaction between 2-methylpent-2-ene and chlorine. State the type of reaction. Label the electrophile, intermediate and final product.
Answer:
Explanation:
CH₃C(CH₃)=CH-CH₂-CH₃ ( 2 - methylpent-2-ene)
Cl₂ ⇒ Cl⁺ + Cl⁻
Cl⁺( electrophile)
CH₃C(CH₃)=CH-CH₂-CH₃ + Cl⁺ ( electrophile) ⇒ CH₃C⁺(CH₃)-C(Cl)H-CH₂-CH₃ ( intermediate )
CH₃C⁺(CH₃)-C(Cl)H-CH₂-CH₃ + Cl⁻ ⇒ CH₃CCl(CH₃)-C(Cl)H-CH₂-CH₃ ( final product )
It is electrophilic addition reaction.
If 200 mL of 3M CaCO3 is diluted to 250 mL, what is the new molarity?
Answer: The new molarity is 2.4 M
Explanation:
According to the dilution law,
[tex]C_1V_1=C_2V_2[/tex]
where,
[tex]C_1[/tex] = concentration of pure solution = 3 M
[tex]V_1[/tex] = volume of pure solution = 200 ml
[tex]C_2[/tex] = concentration of diluted solution= ?
[tex]V_2[/tex] = volume of diluted solution= 250 ml
Putting in the values:
[tex]3M\times 200ml=M_2\times 250ml[/tex]
[tex]M_2=2.4M[/tex]
Thus the new molarity is 2.4 M
A 38.8 gram piece of metal absorbs 181J as it temperature increases from 25.0 degree celsius to 36.0 degree celsius. What is the specific heat of the metal?
__________________________J/g degree C
Answer: 0.424 J/g°C
Explanation:
For this problem, we would have to manipulate the equaiton for heat, q=mCT. Specific heat is the C in the equation. Since we are looking for specific heat, we manipulate the equation so that it says C=.
[tex]C=\frac{q}{m(deltaT)}[/tex]
*I didn't know how to type in delta so I just wrote the word delta, but pretend you see a Δ.
Now that we have our equation, we can plug in our values and solve.
[tex]C=\frac{181J}{(38.8g)(36-25°C)}[/tex]
*Please ignore the capital A in the equation. It pops up every time I type in the ° sign.
[tex]C=0.424J/g°C[/tex]
Question 5 Tungsten is a solid phase of tungsten still unknown to science. The only difference between it and ordinary tungsten is that Tungsten forms a crystal with an fcc unit cell and a lattice constant . Calculate the density of Tungsten .
Complete Question
The complete question is shown on the first uploaded image
Answer:
The density is [tex]\rho = 21.1 \ g/cm^3[/tex]
Explanation:
From the question we are told that
The lattice constant is [tex]a = 0.387 nm = 0.387 *10^{-9} \ m[/tex]
Generally the volume of the unit cell is [tex]V = a^3[/tex]
=> [tex]V = [0.387 *10^{-9}]^2[/tex]
[tex]V = 5.796 *10^{-29} \ m^3[/tex]
Converting to [tex]cm^3[/tex] We have [tex]5.796 *10^{-29} * 1000000 = 5.796 *10^{-23} cm^3[/tex]
The molar mass of Tungsten is constant with a value [tex]Z = 184 g/mol[/tex]
One mole of Tungsten contains [tex]6.022*10^{23}[/tex] unit cells
Where [tex]6.022*10^{23}[/tex] is a constant for the number of atom in one mole of a substance(Tungsten) which is known as Avogadro's constant
Now for FCC distance the number of atom per unit cell is n = 4
Mass of Tungsten (M) = [tex]= \frac{Z * n }{1 \ mole \ of \ Tungsten}[/tex]
=> Mass of Tungsten (M) = [tex]= \frac{184 * 4 }{6.023*10^{23}}[/tex]
=> Mass of Tungsten (M) = [tex]= 1.222*10^{-21} \ g[/tex]
Now
The density of Tungsten is
[tex]\rho = \frac{M}{V}[/tex]
substituting values
[tex]\rho = \frac{1.222*10^{-21}}{5.796*10^{-23}}[/tex]
[tex]\rho = 21.1 \ g/cm^3[/tex]
Classify each titration curve as representing a strong acid titrated with a strong base, a strong base titrated with a strong acid, a weak acid titrated with a strong base, a weak base titrated with a strong acid, or a polyprotic acid titrated with a strong base. A. Strong acid/ strong base.B. Strong base/ strong acid.C. Weak acid/ strong base.D. Weak base/ strong acid.E. Polyprotic acid/ strong base.
Answer:
Acid base titration curves shows the pH at equivalence point
Explanation:
Since the images were not shown, I will proceed to give a general description of the following acid-base titration curves:
In a strong acid-strong base titration, the acid and base will react to form a neutral solution. At the equivalence point of the reaction, hydronium (H+) and hydroxide (OH-) ions will react to form water, leading to a pH of 7.
The titration curve reflects the strengths of the corresponding acid and base. If one reagent is a weak acid or base and the other is a strong acid or base, the titration curve is irregular, and the pH shifts less with small additions of titrant near the equivalence point.
Polyprotic acids are able to donate more than one proton per acid molecule, in contrast to monoprotic acids that only donate one proton per molecule. In the titration curve of a polyptotic acid and a strong base, The curve starts at a higher pH than a titration curve of a strong base. There is always a steep climb in pH before the first midpoint. Gradually, the pH increases until it passes the midpoint; Right before the equivalence point there is a very sharp increase in pH.
Answer:
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Explanation:
Classify each amino acid according to whether its side chain is predominantly protonated or deprotonated at a pH of 7.40. The pKa values of the Asp, His, and Lys side chains are 3.65, 6.00, and 10.53, respectively.
Answer and Explanation:
7.40 becomes lower or shorter than the pKa of functional groups or chains of all the 3 amino acids respectively Lys, His, as well as Asp, although since pH provided throughout the statement. However, all of the covalent chains or bonds of the three compounds or acids (amino) will serve a purpose in the pro-toned pattern.
The pH during which everyone's binding sites will mostly keep track of customer for someone's comparison is:
Lys:
[tex]pH > 10.53[/tex]
His:
[tex]pH > 6.00[/tex]
Asp:
[tex]pH > 3.65[/tex]
How many moles of NaOH will be required to produce 375.4 g of any Na2S04
Answer:
5.28 moles of NaOH.
Explanation:
Step 1:
Determination of the number of mole in
375.4 g of any Na2S04.
Molar mass of Na2SO4 = (2x23) + 32 + (16x4) = 142g
Mass of Na2SO4 = 375.4g
Number of mole of Na2SO4 =..?
Mole = Mass /Molar Mass
Number of mole of Na2SO4 = 375.4/142 = 2.64 moles
Step 2:
The balanced equation for the reaction.
H2SO4 + 2NaOH —> Na2SO4 + 2H2O
Step 3:
Determination of the number of mole of NaOH needed for the reaction.
From the balanced equation above,
2 moles of NaOH reacted to produce 1 mole of Na2SO4.
Therefore, Xmol of NaOH will react to produce 2.64 moles of Na2SO4 i.e
Xmol of NaOH = 2 x 2.64
Xmol of NaOH = 5.28 moles
Therefore, 5.28 moles of NaOH is needed for the reaction.
Draw a mechanism for the reaction of methylamine with 2-methylpropanoic acid. Draw any necessary curved arrows. Show the products of the reaction. Include any nonzero formal charges and all lone pairs of electrons. Indicate which side of the reaction is favored at equilibrium.
Answer:
See figure 1
Explanation:
On this case we have a base (methylamine) and an acid (2-methyl propanoic acid). When we have an acid and a base an acid-base reaction will take place, on this specific case we will produce an ammonium carboxylate salt.
Now the question is: ¿These compounds can react by a nucleophile acyl substitution reaction? in other words ¿These compounds can produce an amide?
Due to the nature of the compounds (base and acid), the nucleophile (methylamine) doesn't have the ability to attack the carbon of the carbonyl group due to his basicity. The methylamine will react with the acid-producing a positive charge on the nitrogen and with this charge, the methylamine loses all his nucleophilicity.
I hope it helps!
Which is the best method for collection of co2
Answer:
The best way is to burn fuel with oxygen instead of airA 1.34 mole sample of LiCl dissolves in water, The volume of the final
solution is 0.86L. Find the Molarity of the Solution.
Answer:
[tex]1.56 \,\,mol/L[/tex]
Explanation:
Molarity of the solution measures number of moles of a solute per litre of solution.
Molarity = volume of solution in litres/number of moles of solute dissolved in solution
Volume of solution in litres = 0.86 L
Also, 1.34 mole sample of LiCl dissolves in water
So,
Molarity of the Solution = [tex]\frac{1.34}{0.86}=1.56 \,\,mol/L[/tex]
Steam reforming of methane (CH4 ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a tank with of methane gas and of water vapor, and when the mixture has come to equilibrium measures the amount of carbon monoxide gas to be 18 mol . Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.
The question is incomplete, here is the complete question:
Steam reforming of methane [tex](CH_4)[/tex] produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125 L tank with 20 mol of methane gas and 10 mol of water vapor at 38 degrees celsius. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of hydrogen gas to be 18 mol . Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.
Answer: The equilibrium constant for the reaction is [tex]3.99\times 10^{-2}[/tex]
Explanation:
We are given:
Initial moles of methane gas = 20 moles
Initial moles of water vapor = 10 moles
Equilibrium moles of carbon monoxide = 18 moles
Volume of the tank = 125 L
The chemical equation for the reaction of methane and water vapor follows:
[tex]CH_4(g)+H_2O(g)\rightleftharpoons CO(g)+3H_2(g)[/tex]
Initial: 20 10
At eqllm: 20-x 10-x x 3x
Evaluating the value of 'x':
[tex]\Rightarrow 3x=18\\x=6[/tex]
So, equilibrium moles of methane gas = (20 - x) = [20 - 6] = 14 mol
Equilibrium moles of water vapor = (10 - x) = [10 - 6] = 4 mol
Equilibrium moles of carbon monoxide gas = x = 6 mol
The expression of equilibrium constant for the above reaction follows:
[tex]K_{eq}=\frac{[H_2]^3[CO]}{[CH_4][H_2O]}[/tex]
We are given:
[tex][H_2]=\frac{18}{125}=0.144M[/tex]
[tex][CO]=\frac{6}{125}=0.048M[/tex]
[tex][CH_4]=\frac{14}{125}=0.112M[/tex]
[tex][H_2O]=\frac{4}{125}=0.032M[/tex]
Putting values in above expression, we get:
[tex]K_{eq}=\frac{(0.144)^3\times 0.048}{0.112\times 0.032}\\\\K_{eq}=3.99\times 10^{-2}[/tex]
Hence, the equilibrium constant for the reaction is [tex]3.99\times 10^{-2}[/tex]
Tenormin, a member of the group of drugs known as beta-blockers, is used to treat high blood pressure and improve survival after a heart attack. It works by slowing down the heart to reduce its workload. Which atom in Tenormin is the most basic?
Answer:
Oxygen
Explanation:
In the compound tenormin, there are two highly electronegative atoms capable of accepting electrons; oxygen and nitrogen. Oxygen is more electronegative than nitrogen.
However, the oxygen atom in tenormin is bonded to carbon in a carbonyl bond. Recall that the carbonyl bond is polar and the direction of the dipole is towards the oxygen atom. Looking at the structure of tenormin, it is clear that the electron density of the bond tends towards the oxygen atom of the carbonyl group. Electron density is withdrawn from the adjacent nitrogen atom of the amine group via mesomeric and inductive mechanism towards the more electronegative oxygen atom.
On the other side of the structure, there are two oxygen atoms. These oxygen atoms are more electronegative than nitrogen thus they are more basic.
Hence the oxygen atom is the most basic atom in the compound tenormin.
A metal M react with sulphur ro form MS. If 3.6g of M reacts with 0.09mol of sulphur to form MS . what is the name of M.
Answer:
Calcium
Explanation:
We must first put down the reaction equation as this will serve as a guide in solving the question.
M(s) + S(s) -----> MS(s)
According to the reaction equation; 1 mole of metal reacts with 1 mole of sulphur. Hence 0.09 moles of sulphur reacts with 0.09 moles of metal.
Now recall that;
Number of moles (n) = mass(m)/ molar mass(M)
Since
mass of metal reacted= 3.6g
Number of moles of metal= 0.09 moles
Then;
Making molar mass of metal the subject of the formula;
Molar mass of metal = mass of metal / number of moles of metal
Molar mass of metal = 3.6g /0.09 moles
Molar mass of metal= 40 gmol-1
The metal having a molar mass of 40gmol-1 is calcium, therefore the metal is calcium.
Nitric oxide, NO, is made from the oxidation of NH3, and the reaction is represented by the equation: 4NH3 + 5O2 → 4NO + 6H2O What mass of O2 would be required to react completely with 6.87 g of NH3?
Answer:
16.16g of O2.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
4NH3 + 5O2 → 4NO + 6H2O
Next, we shall determine the mass of NH3 and O2 that reacted from the balanced equation. This is illustrated below:
Molar mass of NH3 = 14 + (3x1) = 17g/mol
Mass of NH3 from the balanced equation = 4 x 17 = 68g
Molar Mass of O2 = 16x2 = 32g/mol
Mass of O2 from the balanced equation = 5 x 32 = 160g.
From the balanced equation above,
68g of NH3 reacted with 160g of O2.
Now, we can calculate the mass of O2 that will be required to react completely with 6.87 g of NH3. This is illustrated below:
From the balanced equation above,
68g of NH3 reacted with 160g of O2
Therefore, 6.87g of NH3 will react with = (6.87 x 160)/68 = 16.16g of O2.
Therefore, 16.16g of O2 is needed for the reaction.
Mass of oxygen required for the reaction is 16.16 g.
The equation of the reaction is;
4NH3 + 5O2 → 4NO + 6H2O
Since Number of moles = Mass/molar mass
Molar mass of ammonia = 17 g mol
Number of moles of NH3 reacted = 6.87 g/17 g mol = 0.404 moles
From the reaction equation;
4 moles of ammonia reacts with 5 moles of oxygen
0.404 moles reacts with 0.404 moles × 5 moles/4 moles = 0.505 moles
Mass of oxygen = 0.505 moles × 32 g/mol = 16.16 g
Learn more: https://brainly.com/question/165414
1) How many kJ are absorbed when 45.2 g of water at 31.3 oC is heated to 76.9 oC? 2) Calculate the total heat released in kcal when 72.1 g water at 25.2 oC is cooled to 0 oC and freezes. 3) How many kilojoules are required to heat 55,500 mg of gold with specific heat = 0.129 J/g oC is heated from 24.6 oC to 123.4 oC? 4) Calculate the heat needed in kcal to change 45.6 g of water at 100 oC to change into steam.
Answer:
1. Q = 8.66 KJ
2. Q = 7.58 Kcal
3. Q = 0.71 KJ
4. Q = 24.31 Kcal
Explanation:
1. The quantity of heat absorbed can be determined by:
Q = mcΔθ
where: Q is the quantity of heat absorbed or released, m is the mass of the substance, c is the specific heat capacity of water = 4.2 j/g[tex]^{0}C[/tex] and Δθ is the change in temperature.
= 45.2 × 4.2 × (76.9 - 31.3)
= 8656.704
∴ Q = 8.66 KJ
The quantity of heat absorbed is 8.66 KJ.
2. Q = mcΔθ + mL
Where L is the latent heat of fusion of ice = 334 J.
= m(cΔθ + L)
= 72.1(4.2 × 25.2 + 334)
Q = 31712.464 J
= 7579.466 calories
The total heat released is 7.58 Kcal.
3. Q = mcΔθ
= 55.5 × 0.129 × (123.4 - 24.6)
= 707.3586
The quantity of heat required to increase the temperature of gold is 0.71 KJ.
4. Q = mL
Where: L is the specific latent heat of vaporization = 533 calories.
Q = 45.6 × 533
= 24304.8
The quantity of heat required to change water to steam is 24.31 Kcal.
Which describes the correct procedure when converting a number from scientific notation to standard notation? If the power of 10 is positive, move the decimal point to the left. If the power of 10 is positive, move the decimal point to the right. If the number being converted is greater than 10, move the decimal point to the left. If the number being converted is greater than 10, move the decimal point to the right.
Answer:
option 2Explanation:
If the power of ten is positive, you move the decimal place to the right when converting a number from scientific to standard notation
For example:
5.6894 * 10^9
10^9 is 1 with 9 zeros
5.6894 * 1000000000
5689400000.
it moved right
If the power of 10 is positive, move the decimal point to the left.
If the power of 10 is positive, move the decimal point to the right.
If the number being converted is greater than 10, move the decimal point to the left.
If the number being converted is greater than 10, move the decimal point to the right.
This is option 2
The correct answer is if the power of 10 is positive, move the decimal point to the right.
What is scientific notation?A frequently-used floating-point system in which integers are expressed as products of a number between 1 and 10 multiplied by a power of 10.
Learn more about scientific notation here:-
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Amylose is a "polysaccharide" that plants use to store energy. It is made of repeating subunits of C6H10O5. If a particular amylose molecule has 2537 of these subunits, what is its molecular formula? What is its molar mass? What is the empirical formula of amylose?
Answer:
See exaplanation
Explanation:
For this question, we several sub-questions, lets start with the first one:
What is its molecular formula?
For this question we have to multiply the subunit by the number of subunits in the polysaccharide, so:
[tex](C_6H_1_0O_5)*2537=C_1_5_2_2_2H_2_5_3_7_0O_1_2_6_8_5[/tex]
What is its molar mass?
For this question, we have to find the molar mass of 1 subunit and then multiply by the number of subunits:
Atomic masses: C: 12 g/mol H: 1 g/mol O: 16 g/mol
Now we can multiply, the atomic masses by the number of atoms, so:
[tex](12*6)+(10*1)+(16*5)=~162~g/mol[/tex]
If we take into account the number of subunits:
[tex](162~g/mol)*2537=410994~g/mol[/tex]
What is the empirical formula of amylose?
In this case, we have to remember that the empirical formula is the smallest number of atoms. In other words, we have to simplify the formula. Therefore, the smallest formula is the subunit formula: [tex]C_6H_1_0O_5[/tex].
I hope it helps
A gas at 350 K and 12 atm has a molar volume 12 per cent larger than that calculated from the perfect gas law. Calculate (i) the compression factor under these conditions and (ii) the molar volume of gas. Which are dominating in the sample, the attractive or repulsive forces
Answer:
(i)The compression factor is 1.12
(ii) The molar volume of the gas is 2.68 L/mol
Since the compression factor is greater than 1, the attractive forces are dominating.
Explanation: Please see the attachments below
Calculate ΔHrxnΔHrxn for the following reaction: CH4(g)+2O2(g)→CO2(g)+2H2O(l)CH4(g)+2O2(g)→CO2(g)+2H2O(l) Use the following reactions and given ΔHΔH values. CH4(g)+O2(g)→CH2O(g)+H2O(g)CH4(g)+O2(g)→CH2O(g)+H2O(g), ΔH=−ΔH=−284 kJkJ CH2O(g)+O2(g)→CO2(g)+H2O(g)CH2O(g)+O2(g)→CO2(g)+H2O(g), ΔH=−ΔH=−527 kJkJ H2O(l)→H2O(g)H2O(l)→H2O(g), ΔH=ΔH= 44.0 kJ
Answer: the enthalpy of reaction is, -155 kJ
Explanation:-
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The final reaction is:
[tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)[/tex] [tex]\Delta H_{rxn}=?[/tex]
The intermediate balanced chemical reaction will be,
(1) [tex]CH_4(g)+O_2(g)\rightarrow CH_2O(g)+H_2O(g)[/tex] [tex]\Delta H_1=-284kJ[/tex]
(2) [tex]CH_2O(g)+O_2(g)\rightarrow CO_2(g)+H_2O(g)[/tex] [tex]\Delta H_2=-527kJ[/tex]
(3) [tex]H_2O(l)\rightarrow H_2O(g)[/tex] [tex]\Delta H_3=44.0kJ[/tex]
Now multiplying (3) by 2 and adding all the equations, we get :
[tex]\Delta H_{rxn}=\Delta H_1+\Delta H_2+2\times \Delta H_3[/tex]
[tex]\Delta H_{rxn}=(-284)+(-527)+2\times (44)[/tex]
[tex]\Delta H_{rxn}=-155kJ[/tex]
Therefore, the enthalpy of reaction is, -155 kJ
A solution consists of 0.27 M MgCl2 and 0.7 M CuCl2. Calculate the concentration of hydroxide ion needed to separate the metal ions. The Ksp of Mg(OH)2 is 6.3x10-10, and the Ksp of Cu(OH)2 is 2.2x10-20. Multiply your answer by 106 and enter that number to 2 decimal places.
Answer:
The concentration of hydroxide ion needed to precipitate Cu²⁺ whereas Mg²⁺ still in solution is 1.77x10⁻¹⁰
Explanation:
Based on the different Ksp of the hydroxides of the metal ions it is possible to precipitate 1 metal as hydroxide whereas the other still in solution.
Ksp of both hydroxides is:
Mg(OH)₂ ⇄ Mg²⁺ + 2OH⁻
Ksp = [Mg²⁺] [OH⁻]² = 6.3x10⁻¹⁰
Cu(OH)₂ ⇄ Cu²⁺ + 2OH⁻
Ksp = [Cu²⁺] [OH⁻]² = 2.2x10⁻²⁰
Replacing with the concentration of the metals:
[0.27] [OH⁻]² = 6.3x10⁻¹⁰
[OH⁻] = 4.83x10⁻⁵M
That means the precipitation of Mg²⁺ begins when [OH⁻] is 4.83x10⁻⁵M
[0.7] [OH⁻]² = 2.2x10⁻²⁰
[OH⁻] = 1.77x10⁻¹⁰
As the Cu²⁺ needs a low concentration of OH⁻ to begin precipitation,
the concentration of hydroxide ion needed to precipitate Cu²⁺ whereas Mg²⁺ still in solution is 1.77x10⁻¹⁰Suppose the reaction between nitric oxide and bromine proceeds by the following mechanism: step elementary reaction rate constant (g) (g) (g) (g) (g) (g) Suppose also ≫. That is, the first step is much faster than the second.
Write the balanced chemical equation for the overall chemical reaction:
Write the experimentally-observable rate law for the overall chemical reaction.
Note: your answer should not contain the concentrations of any intermediates.
Answer:
Koverall [NO]^2 [Br2]
Balanced chemical reaction equation;
2NO + Br2 ⇄2NOBr
Explanation:
Consider the first step in the reaction;
NO(g) + Br2(g) ⇄ NOBr2(g) fast
The second step is the slower rate determining step
NOBr2(g) + NO(g) ⇄ 2NOBr(g)
Given that k1= [NOBr2]/[NO] [Br2]
k2= [NOBr2] [NO]
The concentration of the intermediate is now;
[NOBr2]= k1[NO][Br2]
It then follows that overall rate of reaction is
Rate= k1k2[NO]^2 [Br2]
Since k1k2=Koverall
Rate= Koverall [NO]^2 [Br2]
Answer:
NOBr_2(g)+NO(g)----> 2NOBr(g)
Explanation:
Answer retrieved from ALEKS
Aqueous solutions of barium chloride and silver nitrate are mixed to form solid silver chloride and aqueous barium nitrate. The complete ionic equation contains which of the following species (when balanced in standard form)?
A. NO (aq)
B. 2Ba (aq)
C. 2Ag (aq)
D. CI(aq)
Answer:
Option C
Explanation:
Consider the ionic equation of this chemical equation. We are given barium chloride and silver nitrate as the reactants, and silver chloride and barium nitrate as the products. We can thus conclude that the ionic equation ( not balanced yet ) should be as follows -
Ba( 2 + ) + Cl ( - ) + Ag ( + ) + NO3 ( - ) ------> AgCl + Ba( 2 + ) + NO3( - )
As you can see these compounds are present in aqueous solutions, and are thus dissociated.
______________________________________________________
Now let us take a look at the number of elements on the reactant and product sides, and balance this chemical equation out -
Ba( 2 + ) + 2Cl ( - ) + 2Ag ( + ) + 2NO3 ( - ) ------> 2AgCl + Ba( 2 + ) + 2NO3( - )
Solution = Option C!
2 MnO4−(aq) + 10 Cl−(aq) + 16 H+(aq) → 5 Cl2(g) + 2 Mn2+(aq) + 8 H2O(l) Oxidizing agent: Reducing agent:
Answer:
MnO4 is the oxidizing agent.
Cl is the reducing agent.
Explanation:
To know which is oxidizing agent and which is reducing agent, let us calculate the change in oxidation number of each element in the equation below:
2MnO4−(aq) + 10Cl−(aq) + 16H+(aq) → 5Cl2(g) + 2Mn2+(aq) + 8H2O(l)
Please note:
1. The oxidation state of oxygen is always –2 except in peroxide where it is –1
2. The oxidation state of Hydrogen is always +1 except in hydrides where it is –1
3. The oxidation state of chlorine is always –1
For Mn:
At left hand side
MnO4 = –1
Mn + 4O = –1
O = –2
Mn + (4 x –2) = –1
Mn – 8 = –1
Collect like terms
Mn = –1 + 8
Mn = +7
At the right hand side, Mn is +2.
The oxidation state of Mn changes from +7 to +2.
For Cl:
At the left hand side
Cl = –1
At the right hand side
Cl = 0
The oxidation state of Cl changes from – 1 to 0.
The oxidation state of O and H are unchanged.
Since the oxidation state of Mn changes from +7 to +2 i.e reduced, MnO4 is the oxidizing agent.
Since the oxidation state of Cl changes from – 1 to 0 i.e increase, Cl is the reducing agent.
A 1.8 g mass of fructose is added to 0.100 kg of water and it is
found that the freezing point has decreased by 0.186 °C. Given
that the Kf value of water is 1.86 °C kg/mol, what is the molar
mass of fructose (van't Hoff factor, i = 1)?
g/mol
Round your answer to the nearest whole number. Do not include units in your answer
The molar mass of fructose will be "180 g/mol". To understand the calculation check below.
Molar mass and MolalityAccording to the question,
Fructose mass = 1.8 g
Water's mass = 0.100 kg
Molal freezing point depression constant, [tex]K_f[/tex] = 1.86°C/m
Freezing point change, Δ[tex]T_f[/tex] = 0.186°C
Freezing point constant, [tex]K_f[/tex] = 1.86°C/m
We know the relation,
→ Δ[tex]T_f[/tex] = i × [tex]K_f[/tex] × m
or,
= i × [tex]K_f[/tex] × [tex]\frac{Fructose \ mass}{Fructose \ molar \ mass\times Solvent's \ mass}[/tex]
By substituting the values, we get
0.186 = 1 × 1.86 × [tex]\frac{1.8}{Fructose \ molar \ mass\times 0.100}[/tex]
By applying cross-multiplication,
Molar mass = 180 g/mol
Thus the above approach is right.
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A 1.8 g mass of fructose is added to 0.100 kg of water and it is found that the freezing point has decreased by 0.186 °C. The molar mass of fructose is 180 g/mol.
Given:
ΔTf = 0.186 °C
Kf = 1.86 °C kg/mol
m = mass of fructose/mass of water
mass of fructose = 1.8 g
mass of water = 0.100 kg = 100 g
Use the equation:
ΔTf = Kf mi
Where:
ΔTf is the change in freezing point (in °C)
Kf is the cryoscopic constant of water (in °C kg/mol)
m is the molality of the solute (in mol/kg)
i is the van't Hoff factor
m = (mass of fructose) / (mass of water)
m = 1.8 g / 100 g
m = 0.018 mol/kg
Rearrange the equation to solve for the molar mass (M):
ΔTf = Kfmi
Substitute the values in the above equation:
0.186 = 1.86 × 0.018 × 1 / M
By applying cross multiplication,
Molar mass (M) = 180 g/mol
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Consider four beakers labeled A, B, C, and D, each containing an aqueous solution and a solid piece of metal. Identity the beakers in which a chemical reaction will occur and those in which no reaction will occur. Drag each item to the correct bin.
1. Mn(s) Ca(NO3)2(aq)
2. KOH(aq) Fe(s)
3. Pt(NO3)2(aq) Cu(s)
4. Cr(s) H2SO4(aq)
A. Reaction
B. Non-reaction
Consider the following data for five hypothetical element: Q, W, X, Y, and Z. Rank the elements from most reactive to least reactive.
Combination Observation of reaction
Q + W+ Reaction occurs
X + Z+ No reaction
W + Z+ Reaction occurs
Q+ + Y Reaction occurs
Place the element symbol from most to least reactive.
1. Q
2. W
3. X
4. Y
5. Z
A. Most reactive
B. Least reactive
Use the following reactions to arrange the elements A, B, C, and D in order of their rodox reactivity from most reactive to least reactive. Rank from most reactive to least reactive.
1. A + D+ righ tarrow A+ + D
2. C+ + D righ tarrow C + D+
3. B+ + D righ tarrow B + D+
4. B + C+ righ tarrow B+ + C
A. Most reactive
B. Least reactive
Answer:
A Reaction
3. Pt(NO₃)₂(aq) + Cu(s)
4. Cr(s) + H₂SO₄(aq)
B Non Reaction
1. Mn(s) + Ca(NO₃)₂(aq)
2. KOH(aq) + Fe(s)
Y > Q > W > Z > X
Explanation:
The first question is whether a reaction will occur base on the chemical equation below.
1. Mn(s) + Ca(NO₃)₂(aq)
2. KOH(aq) + Fe(s)
3. Pt(NO₃)₂(aq) + Cu(s)
4. Cr(s) + H₂SO₄(aq)
Firstly, some element are more reactive than others , base on this criteria element can be arranged base on it reactivity .
1. Mn(s) + Ca(NO₃)₂(aq)
This reaction will not occur because Mn cannot displace Ca in it compound. Usually, more reactive element displaces less reactive element.
2. KOH(aq) + Fe(s)
The reaction will not occur since Iron is less reactive and lower in the reactivity series than potassium . So iron won't be able to displace potassium.
3. Pt(NO₃)₂(aq) + Cu(s)
Copper is more reactive than platinum so it will displace platinum easily . The reaction will definitely occur.
4. Cr(s) + H₂SO₄(aq)
Chromium is higher up in the reactivity series than hydrogen so, it will definitely displace hydrogen in it compound . The reaction will occur in this case.
Base on the reaction
Q + W+ Reaction occurs
Since the reaction occurred element Q is more reactive as it displace element w from it compound.
X + Z+ No reaction
No reaction occurred because element x is less reactive than z therefore, it cannot displace z from it compound.
W + Z+ Reaction occurs
Element w is more reactive than z as it displaces z form it compound.
Q+ + Y Reaction occurs
Element Y is more reactive than element Q as it displaces Q from it compound.
Therefore, the order of reactivity from the most reactive to the least reactive will be Y > Q > W > Z > X
A. The beakers in which a chemical reaction will occur:
3. Pt(NO₃)₂(aq) + Cu(s)
4. Cr(s) + H₂SO₄(aq)
B. The beakers in which there is no reaction:
1. Mn(s) + Ca(NO₃)₂(aq)
2. KOH(aq) + Fe(s)
C. The elements from most reactive to least reactive is:
Y > Q > W > Z > X
Solving for each part:A.
1. Mn(s) + Ca(NO₃)₂(aq)
2. KOH(aq) + Fe(s)
3. Pt(NO₃)₂(aq) + Cu(s)
4. Cr(s) + H₂SO₄(aq)
Elements can be arranged on the basis of reactivity:
1. Mn(s) + Ca(NO₃)₂(aq)
This reaction will not occur because Mn cannot displace Ca in it compound. Usually, more reactive element displaces less reactive element.
2. KOH(aq) + Fe(s)
The reaction will not occur since Iron is less reactive and lower in the reactivity series than potassium . So iron won't be able to displace potassium.
3. Pt(NO₃)₂(aq) + Cu(s)
Copper is more reactive than platinum so it will displace platinum easily . The reaction will definitely occur.
4. Cr(s) + H₂SO₄(aq)
Chromium is higher up in the reactivity series than hydrogen so, it will definitely displace hydrogen in it compound . The reaction will occur in this case.
According to reactions:
Q + W→ Reaction occurs
Since the reaction occurred element Q is more reactive as it displace element w from it compound.
X + Z →No reaction
No reaction occurred because element X is less reactive than Z therefore, it cannot displace z from it compound.
W + Z→ Reaction occurs
Element W is more reactive than Z as it displaces Z form it compound.
Q + Y →Reaction occurs
Element Y is more reactive than element Q as it displaces Q from it compound.
Thus, the order of reactivity from the most reactive to the least reactive will be: Y > Q > W > Z > X
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A liquid solvent is added to a flask containing an insoluble solid. The total volume of the solid and liquid together is 93.0 mL. The liquid solvent has a mass of 33.7 g and a density of 0.865 g/mL. Determine the mass of the solid given its density is 3.50 g/mL
Answer:
Mass of solid = 189.141 gram
Explanation:
Given:
Total volume = 93 ml
Mass of liquid = 33.7 gram
Density of liquid = 0.865 g/ml
Density of solid = 3.50 g/ml
Find:
Mass of solid = ?
Computation:
Volume of liquid = Mass of liquid / Density of liquid
Volume of liquid = 33.7 / 0.865
Volume of liquid = 38.9595 ml
Volume of solid = Total volume - Volume of liquid
Volume of solid = 93 - 38.9595
Volume of solid = 54.0405 ml
Mass of solid = Volume of solid × Density of solid
Mass of solid = 54.0405 ml × 3.50 g/ml
Mass of solid = 189.141 gram
The standard free energy change in physiological conditions (G') for the reaction catalyzed by malate dehydrogenase in the citric acid cycle malate + NAD+ oxaloacetate + NADH + H+ is ~+29 kJmol-1 Calculate the actual G' at 37C if Keq' is 1.02 × 10-5
Answer: The actual value of [tex]\Delta G[/tex] is -618 J/mol
Explanation:
Relation of ree energy change and equilibrium constant
[tex]\Delta G=\Delta G^0+2.303\times RT\times \log K_{eq}[/tex]
where,
[tex]\Delta G[/tex] = Free energy change
[tex]\Delta G^o[/tex] = standard free energy change = +29 kJ/mol =
R = universal gas constant
T = temperature = [tex]37^0C=(37+273)K=310K[/tex]
[tex]K_{eq}[/tex] = equilibrium constant = [tex]1.02\times 10^{-5}[/tex]
[tex]\Delta G=+29000J/mol+2.303\times 8.314J/Kmol\times 310K\times \log (1.02\times 10^{-5})[/tex]
[tex]\Delta G=29000J/mol-29619J/mol=-618J/mol[/tex]
The actual value of [tex]\Delta G[/tex] is -618 J/mol
