Answer: A) 4C2+S8 -> 4CS2
Explanation:
The chemical equation
4C2+S8 -> 4CS2 is flawed because the equation is not balanced i.e the carbon atom at the reactant is not equal to that of the product and for a chemical reaction to be a balanced equation, the number of atoms of elements at the reactants must be equal to that of the product.
As you can see, there are 8 atoms of carbon(C) at the reactant while we have only 4 carbon atoms at the product therefore making the equation unbalanced.
The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate constant, ????,k, is 6.1×10−8 s−1.6.1×10−8 s−1. What is the value of the rate constant at 785.0 K?
Answer : The rate constant at 785.0 K is, [tex]1.45\times 10^{-2}s^{-1}[/tex]
Explanation :
According to the Arrhenius equation,
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
or,
[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]K_1[/tex] = rate constant at [tex]600.0K[/tex] = [tex]6.1\times 10^{-8}s^{-1}[/tex]
[tex]K_2[/tex] = rate constant at [tex]785.0K[/tex] = ?
[tex]Ea[/tex] = activation energy for the reaction = 262 kJ/mole = 262000 J/mole
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex] = initial temperature = [tex]600.0K[/tex]
[tex]T_2[/tex] = final temperature = [tex]785.0K[/tex]
Now put all the given values in this formula, we get:
[tex]\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}][/tex]
[tex]K_2=1.45\times 10^{-2}s^{-1}[/tex]
Therefore, the rate constant at 785.0 K is, [tex]1.45\times 10^{-2}s^{-1}[/tex]
The rate constant at the new temperature is 1.81 ×10^−2 s−1.
We have the reaction written as follows; C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g).
We have to use the formula;
ln(k2/k1) =Ea/R(1/T1 - 1/T2)
Now;
k2 = ?
k1 = 6.1×10−8 s−1
Ea = 262 ×10 ^3 J/mol
T1= 600.0 K
T2 = 785.0 K
R = 8.314 J/K. mol
ln (k2/6.1×10−8) = 262 ×10 ^3 /8.314 (1/600.0 - 1/785.0)
ln (k2/6.1×10−8) = 12.6
k2/6.1×10−8 = e^12.6
k2 = 6.1×10−8 (e^12.6)
k2 = 1.81 ×10^−2 s−1.
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For the reaction A (g) → 2 B (g), K = 14.7 at 298 K. What is the value of Q for this reaction at 298 K when ∆G = -20.5 kJ/mol?
The value of Q for the reaction A (g) → 2 B (g) at 298 K when ΔG = -20.5 kJ/mol is approximately 0.00069.
To find the value of the reaction quotient Q when ΔG is known, we use the relationship between Gibbs free energy change, the equilibrium constant K, and Q.
The equation is: ΔG = ΔG° + RT ln(Q)
Where:
ΔG = -20.5 kJ/mol (given)ΔG° = standard free energy changeR = 8.314 J/mol·K (0.008314 kJ/mol·K)T = 298 K (given)ln = natural logarithmFirst, we need to calculate ΔG°. This can be done using the equilibrium constant (K): ΔG° = -RT ln(K)
Given K = 14.7:
ΔG° = - (0.008314 kJ/mol·K) x (298 K) * ln(14.7)ΔG° = - (2.475 kJ/mol)Now we substitute ΔG and ΔG° back into the equation: ΔG = ΔG° + RT ln(Q)
-20.5 kJ/mol = -2.475 kJ/mol + (0.008314 kJ/mol·K) x (298 K) x ln(Q)
Solving for ln(Q):
-20.5 + 2.475 = (0.008314 x 298) x ln(Q)
-18.025 = 2.477 x ln(Q)
ln(Q) = -18.025 / 2.477
ln(Q) ≈ -7.28
Finally, taking the exponent to find Q: Q = [tex]e^(^-^7^.^2^8^)[/tex]
Q ≈ 0.00069
To find the reaction quotient (Q) when ΔG = -20.5 kJ/mol, we calculate ΔG° using the given value of K and the equation ΔG° = -RT ln K. Then, we use the relationship ΔG = ΔG° + RT ln Q to solve for Q, resulting in Q =0.0037.
To determine the reaction quotient (Q) for the reaction A(g) → 2 B(g) at 298 K when ΔG = -20.5 kJ/mol, we use the relationship between ΔG, Q, and K. The equation is ΔG = ΔG° + RT ln Q. We rearrange to find Q: ln Q = (ΔG - ΔG°) / RT, where ΔG° = -RT ln K
First, calculate ΔG°:
ΔG° = -RT ln K = - (8.314 J/(mol·K) * 298 K * ln 14.7)Therefore, Q = 0.0037
Propane gas, C3H8, is sometimes used as a fuel. In order to measure its energy output as a fuel a 1.860 g sample was combined with an excess of O2 and ignited in a bomb calorimeter. After the reaction, it was found that the temperature of the calorimeter had increased from 25.000C to 26.061C. The calorimeter contained 1.000 kg of water. The heat capacity of the calorimeter was 4.643 kJ/C. Determine the heat of reaction, in kJ/mol propane. The reaction was:
Answer:
The heat of the reaction is 105.308 kJ/mol.
Explanation:
Let the heat released during reaction be q.
Heat gained by water: Q
Mass of water ,m= 1kg = 1000 g
Heat capacity of water ,c= 4.184 J/g°C
Change in temperature = ΔT = 26.061°C - 25.000°C=1.061 °C
Q=mcΔT
Heat gained by bomb calorimeter =Q'
Heat capacity of bomb calorimeter ,C= 4.643 J/g°C
Change in temperature = ΔT'= ΔT= 26.061°C - 25.000°C=1.061 °C
Q'=CΔT'=CΔT
Total heat released during reaction is equal to total heat gained by water and bomb calorimeter.
q= -(Q+Q')
q = -mcΔT - CΔT=-ΔT(mc+C)
[tex]q=-1.061^oC(1000 g\times 4.184J/g^oC+4.643 J/^oC )=-4,444.15J=-4.444 kJ[/tex]
Moles of propane =[tex]\frac{1.860 g}{44 g/mol}=0.0422 mol[/tex]
0.0422 moles of propane on reaction with oxygen releases 4.444 kJ of heat.
The heat of the reaction will be:
[tex]\frac{4.444 kJ}{0.0422 mol}=105.308 kJ/mol[/tex]
The kilogram is the SI unit for mass.
True
False
True, the kilogram is used as a unit for mass in the the international system of units (SI) for scientific measurements.
Answer: True
Explanation:
Mass is defined as the amount of matter contained in the body.
Its units are kg, gram, milligram which are inter convertible.
S.I or M.K.S system has seven fundamental units which are used to find derived units
1) Mass - Kilogram
2) Length - meter
3) Time - Seconds
4) Electric Current - Ampere
5) Amount of substance - Moles
6) Intensity of light - Candela
7) Temperature - Kelvin
Thus the kilogram is the SI unit for mass.
You need a 30% alcohol solution. On hand, you have a 200 mL of a 10% alcohol mixture. You also have 55% alcohol mixture. How much of the 55% mixture will you need to add to obtain the desired solution?
Answer: 160 ml
Explanation:
The expression used will be :
[tex]C_1V_1+C_2V_2=C_3V_3[/tex]
where,
[tex]C_1[/tex] = concentration of Ist alcohol solution= 10%
[tex]C_2[/tex] = concentration of 2nd alcohol solution= 55%
[tex]V_1[/tex] = volume of Ist alcohol solution = 200 ml
[tex]V_2[/tex] = volume of 2nd alcohol solution= v ml
[tex]C_3[/tex] = concentration of resulting alcohol solution= 30%
[tex]V_2[/tex] = volume of resulting alcohol solution= (v+200) ml
Now put all the given values in the above law, we get the volume of added.
[tex](10\times 200)+(55\times v)=(30\times (v+200)ml)[/tex]
By solving the terms, we get :
[tex]v=160ml[/tex]
Therefore, the volume of 55% mixture needed to be added to obtain the desired solution is 160 ml.
It takes 155. Kj/mol to break a fluorine-fluorine single bond. Calculate the maximum wavelength of light for which a flouine-flouring single bond could be broken by absorbing a single photon
Answer:
The maximum wavelength of light will be 773.46 nm.
Explanation:
Energy required to break a fluorine-fluorine single bond = 155 kJ/mol
For 1 mole of fluorine gas = 155 kJ = 155000 J
For [tex]6.022\times 10^{23}[/tex] molecules = 155000 J
Then energy to break 1 molecule = [tex]\frac{155000 J}{6.022\times 10^{23}}=2.57\times 10^{-19} J[/tex]
Energy of the photon E =[tex]2.57\times 10^{-19} J[/tex]
Wavelength of the light =[tex]\lambda [/tex]
[tex]E=\frac{hc}{\lambda }[/tex]
[tex]2.57\times 10^{-19} J=\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{\lambda }[/tex]
[tex]\lambda =7.7346\times 10^{-7} m=773.46 nm[/tex]
The maximum wavelength of light will be 773.46 nm.
PLEASE ANSWER FAST
If iron pyrite, FeS2, is not removed from coal, oxygen from the air will combine with both the iron and the sulfur as coal burns. If a furnace burns an amount of coal containing 198.20 g of FeS2, how much SO2 (an air pollutant) is produced?
4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2
Select one:
a. 211.7
b. 52.92
c. 590.8
d. 582.1
First we need to calculate the number of moles of FeS[tex]_{2}[/tex]:
number of moles = mass (grams) / molecular mass (g/mol)
number of moles of FeS[tex]_{2}[/tex] = 198.2/120 = 1.65 moles
From the chemical reaction we deduce that:
if 4 moles of FeS[tex]_{2}[/tex] produces 8 moles of SO[tex]_{2}[/tex]
then 1.65 moles of FeS[tex]_{2}[/tex] produces X moles of SO[tex]_{2}[/tex]
X = (1.65×8)/4 = 3.3 moles of SO[tex]_{2}[/tex]
Now we can calculate the mass of SO[tex]_{2}[/tex]:
mass (grams) = number of moles × molecular mass (grams/mole)
mass of SO[tex]_{2}[/tex] = 3.3×64 = 211.2 g
A container of carbon dioxide (CO2) has an initial temperature of 170 K with a pressure of 50 kPa. When the container is heated the pressure is measured at 283 kPa. The volume is constant at 12 L throughout. What is the final temperature of the carbon dioxide (CO2)?
Answer: T2= 962.2 K
Explanation:
The ideal gases is often written like PV=nRT, where P is pressure, V is volume, n is moles, R is the universal constant of the gases and T is Temperature.
So, in this problem there is a container that is a closed system, therefore n is constant and volume too. The initial point is 1 and the final point is 2, so
V1=V2 ⇒
[tex]\frac{n1RT1}{P1} = \frac{n2RT2}{P2} \\\\\frac{T1}{P1} = \frac{T2}{P2} \\\\T2= \frac{T1P2}{P1} =\frac{170 K 283KPa}{50 KPa} =962.2 K[/tex]
Which of the following would increase the rate of a chemical reaction between hydrochloric acid (HCl) and solid zinc metal (Zn)?
A. Decreasing the concentration of HCl
B. Pulverizing the zinc metal into a fine powder
C. Performing the reaction at a lower temperature
D. Decreasing the amount of Zn
Answer:
B. Pulverizing the zinc metal into a fine powder
Answer: B. Pulverizing the zinc metal into a fine powder
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
[tex]Zn+2HCl\rightarrow ZnCl_2+H_2[/tex]
A. Decreasing the concentration of [tex]HCl[/tex] and [tex]Zn[/tex]: Thus rate of the reaction would decrease on decreasing the concentration of hydrochloric acid and zinc.
B. Pulverizing the zinc metal into a fine powder : If the solute particles is present in smaller size, more it can take part in the chemical reaction due to large surface area, hence increasing the rate of reaction.
C. Performing the reaction at a lower temperature. Decreasing the temperature means the energy of the particles is less and thus lesser reactants would cross the energy barrier and thus lesser will be the rate.
An aqueous solution containing 5.0 g of solute in 100 ml is extracted with three 25 ml portion of diethyl ether. what is the total amount of solute that will be extracted by the ether, k = 1.0 ?
First, in this case, we define the K constant as the solubility of the solute in water divided by the solubility of the solute in ether.
K = (X grams of solute / 75 mL of ethyl ether) / (5 g of solute / 100 mL of water)
K = (X / 75) / (5 / 100)
1 = (X / 75) / (5 / 100)
5 / 100 = X / 75
0,05 = X /75
X = 0,05 × 75 = 3.75 g of solute that will be extracted by the ether
Answer:
The correct answer is 2.44 grams.
Explanation:
The partition coefficient, K = concentration of solute in ether / concentration of solute in water
As partition coefficient is 1, therefore, the concentration of solute in both the solvents would be similar.
Thus, when 25 milliliters of ether is added to 100 ml of water Wether/25 = Wwater/100
However, Wwater + Wether = 5.00 g
Wwater = 5.00 g - Wether
So, Wether/25 = 5.00 - Wether/100
Wether = 1.00 g
Thus, Wwater = 5.00 - 1.00 = 4.00 grams
So, for the first time, the solute extracted by ether is 1.0 gram
Now add 25 milliliters of ether to 4.00 grams of solute of 100 ml water,
Wwater + Wether = 4.00 g
Wwater = 4.00 g - Wwater
So, Wether/25 = 4.00 - Wether/100
Wether = 0.8
Thus, Wwater = 4.0 - 0.8 = 3.2 grams
So, for the second time the amount of solute extracted by ether is 0.8 gram.
Now, add 25 ml of ether to 3.2 grams of solute of 100 ml water
Wwater + Wether = 3.20 g
Wwater = 3.20 - Wether
So, Wether/25 = 3.20 - Wether/100
Thus, Wwater = 3.20 - 0.64 = 2.56 grams
So, for the third time, the amount of solute extracted by ether is 0.64 grams.
Therefore, total weight of solute extracted by ether is 1.00 + 0.80 + 0.64 = 2.44 grams.
When a 35.07 g sample of phosphorus reacts with oxygen a 71.00 g sample of phosphorus oxide is formed. What is the percent composition of the compound? What is the empirical formula for this compound?
Answer:
Percentage composition of phosphorus is 43.66%.
Percentage composition of oxygen is 56.66%.
[tex]P_2O_5[/tex]is the empirical formula for this compound.
Explanation:
Moles of phosphorus =[tex]\frac{35.07 g}{31 g/mol}=1.1312 mol[/tex]
Moles of phosphorus oxide =[tex]\frac{71.00 g}{31x+16 y g/mol}=z[/tex]
1.1312 moles of moles of phosphorus gives z moles of phosphorus oxide.
The from 1 mol phosphorus :
[tex]\frac{z}{1.1312 }[/tex] moles of phosphorus oxide...(1)
[tex]2xP+yO_2\rightarrow 2P_xO_y[/tex]
According to reaction, 2x moles of phosphorus gives 2 mol of phosphorus oxide.
The from 1 mol phosphorus :
[tex]\frac{2}{2x}[/tex] moles of phosphorus oxide...(2)
(1)=(2)
[tex]\frac{z}{1.1312 }=\frac{2}{2x}=\frac{71.00 g}{31x+16 y g/mol}[/tex]
[tex]x=\frac{2}{5}y[/tex]
The molecular formula of the phosphorus oxide :[tex]P_xO_y[/tex]:
[tex]P_{\frac{2y}{5}}O_{y}=P_2O_5[/tex]
The empirical formula is simplest ratio of elements in a compound.
[tex]P_2O_5[/tex]is the empirical formula for this compound.
Percentage composition of phosphorus =
[tex]P\%=\frac{2\times 31 g/mol}{2\times 31 g/mol+5\times 16g/mol}\times 100[/tex]
[tex]P\%=43.66\%[/tex]
Percentage composition of oxygen=
[tex]O\%=\frac{5\times 16 g/mol}{2\times 31 g/mol+5\times 16g/mol}\times 100[/tex]
[tex]O\%=56.33\%[/tex]
The percent composition of the phosphorus oxygen compound is 49.4% phosphorus and 50.6% oxygen. The empirical formula, based on the simplest ratio of moles of each component, is P2O4.
Explanation:Firstly, to obtain the percent composition of the phosphorus oxide compound, we divide the mass of phosphorus by the total mass of the compound, then multiply the quotient by 100. Thus, using the given masses: (35.07 g P/71.00 g P-Oxide) * 100 = 49.4% Phosphorus.
By subtraction, we can find the percentage of Oxygen: 100% - 49.4 % = 50.6% Oxygen. This gives us the percent composition of the phosphorus oxygen compound.
Next, we need to identify the empirical formula from these percentages. We convert percentages to grams (which doesn't change our values because percentages are based on a total of 100), and then we convert from grams to moles using the atomic masses of phosphorus (P) and oxygen (O): 35.07 g P * (1 mol P/30.97 g) = 1.13 mol P and 35.93 g O * (1 mol O/16.00 g) = 2.24 mol O.
To find the empirical formula, we then divide each mole quantity by the smallest obtained mole value (1.13), to generate the simplest mole ratio: P1O1.98, which we round to P1O2 due to natural variation. Therefore, the empirical formula for this phosphorus oxygen compound is P1O2 or more commonly written as P2O4.
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Compounds A and B react to form compounds C and D according to the equation: aA + bB → cC + dD. Under which conditions will the rate law be given by the equation: rate = k[A]a[B]b? A. The reaction takes place in one step. B. The reaction is endothermic. C. The reaction is exothermic. D. The reaction involves more than one step.
Answer: A. The reaction takes place in one step.
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
Molecularity of the reaction is defined as the number of atoms, ions or molecules that must colloid with one another simultaneously so as to result into a chemical reaction.
Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.
Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.
[tex]aA=bB\rightarrow cC+dD[/tex]
[tex]Rate=k[A]^a[B]^b[/tex]
k= rate constant
a= order with respect to A
b = order with respect to B
Answer:
anwer is a
Explanation:
How do the interactions that are broken in water when it is boiled compare with those broken when water is electrolyzed? Boiling water breaks intermolecular attractions and electrolysis breaks covalent bonds. Boiling water breaks covalent bonds and electrolysis breaks intermolecular attractions. Boiling water and electrolysis of water break covalent bonds. Boiling water and electrolysis of water break intermolecular forces.
Answer:
Boiling water breaks intermolecular attractions and electrolysis breaks covalent bonds.
Explanation:
When water boils, hydrogen bonds are broken between adjacent water molecules. The hydrogen bond is an intermolecular bond between adjacent oxygen and hydrogen atoms of water molecules.
During electrolysis, water dissociates in the presence of electric current. Here, ions are formed in the process. Therefore, covalent bonds are broken here.
Boiling water breaks the intermolecular hydrogen bonds between water molecules, causing it to change from liquid to gas. Electrolysis on the other hand, breaks the covalent bonds within individual water molecules, splitting them into hydrogen and oxygen.
Explanation:There is a distinct difference between the type of interactions broken when water is boiled and when it's electrolyzed. Boiling water primarily breaks the intermolecular attractions, specifically hydrogen bonds that exist between water molecules, thereby allowing water molecules to escape into the air as steam or water vapor. On the other hand, electrolysis of water involves breaking of the covalent bonds within individual water molecules, thereby splitting water into its constituent hydrogen and oxygen atoms. This process requires more energy compared to boiling water due to the strength of covalent bonds when compared to hydrogen bonds.
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Carbon dioxide and an unknown gas start to effuse from a container and the carbon dioxide takes 4.69 times as long to escape as the unknown gas. What is the identity of the unknown gas? 7) A) Br2 B) H2 C) HCl D) CO E) NO2
Answer: The unknown gas is B) [tex]H_2[/tex] .
Explanation: The problem is based on Graham's law of effusion rates of gases.
Carbon dioxide takes 4.69 times as long to escape as the unknown gas. It means the unknown gas is lighter than carbon dioxide since the lighter gas takes less time to escape.
From Graham's law, "Rate of effusion of a gas is inversely proportional to the molar mass of the gas."
Molar mass of carbon dioxide is 44.0 gram per mol. Let's say the molar mass of the unknown gas is M.
[tex]\frac{rate_A}{rate_C_O_2}=\sqrt{\frac{44.0}{M}}[/tex]
[tex]4.69=\sqrt{\frac{44.0}{M}}[/tex]
Do the square to both sides:
[tex]21.9961={44.0}{M}}[/tex]
[tex]M=\frac{44.0}{21.9961}[/tex]
M = 2.00
2.00 gram per mol is the molar mass of hydrogen gas. So, the correct option is B) [tex]H_2[/tex] .
Final answer:
Using Graham's Law of Effusion, the unknown gas effusing 4.69 times faster than carbon dioxide is identified as hydrogen (H2), which has a much lower molar mass, making answer B) H2 the correct choice.
Explanation:
The question involves the application of Graham's Law of Effusion, which relates the rates of effusion of two gases to their molar masses. Given that carbon dioxide takes 4.69 times as long to escape as the unknown gas, we can use the inverse relationship between the rate of effusion and the square root of the molar mass (rate ≈ 1/sqrt(molar mass)) to identify the unknown gas.
The molar mass of carbon dioxide (CO₂) is approximately 44 g/mol.
By setting up a ratio using Graham's Law, (rate of unknown gas/rate of CO₂) = sqrt(molar mass of CO₂/molar mass of unknown gas), and substituting the given rates, we can solve for the molar mass of the unknown gas:
(1/4.69)² = (44 g/mol) / (molar mass of unknown gas)
(1/22.0361) = (44 g/mol) / (molar mass of unknown gas)
Molar mass of unknown gas = 44 g/mol * 22.0361 = 969.8684 g/mol
Saturated fatty acids are different to unsaturated fatty acids because they ________. (A) Have no C=C double bonds(B) Have an even number of carbon atoms(C) Exhibit free rotation about the carbon-carbon bonds in the hydrocarbon tail (D) Have short hydrophobic tails
The volume of 7.91 M HCl needed to make 196.1 mL of 2.13 M HCl is ____.
Select one:
a. 52.8
b. 728
c. 0.198
d. 3490
Answer:
a. 52.8
Explanation:
To find the number of moles of HCl we use the relation M₁V₁=M₂V₂
where M₁ is the initial molarity, M₂ the new molarity, V₁ the initial volume used, and V₂ the final volume obtained.
M₁=7.91 M
M₂=2.13 M
V₁=?
V₂=196.1 mL
Replacing these values in the relationship.
M₁V₁=M₂V₂
7.91 M× V₁=2.13 M×196.1 mL
V₁=(2.13 M×196.1 mL)/7.91 M
=52.8 mL
Which region in the IR spectrum could be used to distinguish between butanoic acid and 2-butanone?
a. 1680-1750 cm-1
b. 3200-3600 cm-1
c. 1600 cm1
d. 2500-3300 cm-1
Answer:2500-3300[tex]cm^{-1}[/tex]
Explanation:We can distinguish between 2-butanone and butanoic acid by analyzing the spectra .
The structure of the two compounds are slightly different as butanoic acid has an OH group whereas for 2-butanone there is no OH group present.
Please refer the attachments for structure of compounds.
Both the compounds will show carbonyl stretching frequency at around 1700-1750cm^{-1}[/tex] due to the presence of carbonyl group in both the compounds.
A broad intense peak at around 3000[tex]cm^{-1}[/tex] will be observed only for butanoic acid due to the presence of OH group and 2-butanone would not show any broad intense peak at around 3000[tex]cm^{-1}[/tex].
So by analyzing the IR spectra and identifying the intense broad peak of OH we can easily distinguish between butanoic acid and 2-butanone.
Due to hydrogen bonding in between two carboxylic acid molecules ,peak broadening for OH group in butanoic acid takes place.
Generally the ketones show IR stretching in around 1700-1730cm^{-1}[/tex] due to the presence of carbonyl group.
Generally carboxylic acids show IR stretching for carbonyl group at around 1700-1760cm^{-1}[/tex] and a broad intense peak for OH at around 2500-3000cm^{-1}[/tex].
The C=O stretching region in the IR spectrum can be used to distinguish between butanoic acid and 2-butanone.
Explanation:In the IR spectrum, the region that could be used to distinguish between butanoic acid and 2-butanone is the C=O stretching region. The C=O stretching vibration band of saturated aliphatic ketones, such as 2-butanone, appears at around 1715 cm-1. However, butanoic acid contains a carboxyl group (COOH), which exhibits a broad and intense peak in the 1680-1750 cm-1 range.
Sodium borohydride is stable in acidic aqueous solutions. True or False
Answer:
The answer is False.
Explanation:
When mixed with an acidic liquid etc. a violent reaction happens creating Hydrogen Gas.
The statement that sodium borohydride is stable in acidic aqueous solutions is false.
Sodium borohydride is a very useful reducing agent in chemistry. It is able to carry out many important organic transformations where reduction reactions are involved.
However, sodium borohydride is unstable in acid solutions. It decomposes in acidic and even neutral solutions to release hydrogen gas. Therefore, the statement that sodium borohydride is stable in acidic aqueous solutions is false.
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Identify the missing coefficient in the following equation:
3
2
1
0
Answer:
2
Explanation:
In balancing nuclear reactions the mass number and atomic numbers are usually conserved. This implies that from the given equation, the sum of the number of the subscript on the right hand side must be equal to that on the left hand side. This also applies to the superscript:
For the mass numbers(superscript):
235 + 1 = 1 + 139 + 95
236 = 235
This is not balanced
For the atomic number:
92 + 0 = 0 + 53 + 39
92 = 92
This is balanced.
We simply inspect to see how to balance the mass number.
By putting a coefficient of 2 behind the neutron atom, the equation becomes balanced.
The volume of 7.91 M HCl needed to make 196.1 mL of 2.13 M HCl is ____.
Select one:
a. 52.8
b. 728
c. 0.198
d. 3490
Answer:
a) 52.8
Explanation:
M1V1 = M2V2
(7.91 M)(x ml) = (2.13 M) (196.1 ml)
(7.91M) (xml) = 417.693 M.ml
x ml = 417.693/ 7.91
x = 52.8
Answer:
A. 52.8
Explanation:
We make use of the dilution formula to solve this
[tex]M_1\times V_1 =M_2\times V_2[/tex]
Where ([tex]M_1[/tex] and [tex]M_2[/tex] are the initial molarity and final molarity and [tex]V_1[/tex] and [tex]V_2[/tex] are the initial volume and final volume)
Plugging into the values in the formula
[tex]M_1\times V_1 =M_2\times V_2[/tex]
[tex]7.91M \times V_1=2.13M \times 196.1mL[/tex]
[tex]V_1=\frac {(2.13M \times 196.1mL)}{7.91M}[/tex]
= 52.8 mL is the Answer
Molality is a unit of concentration that measures the moles of solute per
liter of solution.
liter of solvent.
kilogram of solution.
kilogram of solvent.
Answer : The correct option is, kilogram of solvent.
Explanation :
Molality : It is defined as the number of moles of solute present in one kilogram of solvent.
Or, we can say that the number of moles of solute per kilogram of solvent.
The unit of molality is mole per kilogram or mole/Kg.
The formula will be :
[tex]\text{Molality}=\frac{\text{Moles of solute}}{\text{Mass of solvent in Kg}}[/tex]
Hence, the molality is a unit of concentration that measures the moles of solute per kilogram of solvent.
Part complete When a 235 92U nucleus is bombarded by neutrons (10n) it undergoes a fission reaction, resulting in the formation of two new nuclei and neutrons. The following equation is an example of one such fission process: 235 92U+10n→AZBa+9436Kr+310n
Enter the isotope symbol for the barium (Ba) nucleus in this reaction.In another process in which 235 92U undergoes neutron bombardment, the following reaction occurs:235 92U+10n --> AZSr+143 54Xe+310nEnter the isotope symbol for the strontium (Sr) nucleus in this reaction.
The isotope symbol for the barium (Ba) nucleus in the given reaction is 13756Ba. The isotope symbol for the strontium (Sr) nucleus in the second reaction is 8838Sr.
Explanation:In the given nuclear fission reaction, the isotope symbol for the barium (Ba) nucleus is 13756Ba. This can be determined by balancing the atomic numbers and mass numbers on both sides of the reaction.
In the second reaction, the isotope symbol for the strontium (Sr) nucleus is 8838Sr. Again, this can be obtained by balancing the atomic numbers and mass numbers.
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Which formula equation shows a reversible reaction?
a. 2Na+F2–> 2NaF
b. CaCO3–> CaO+CO2
c. NH4CI(s)—–NH3(
g.+HCI(
g. 2H2O2(aq)–pt—> 2H2O(I)+O2(g)
Answer:
NH₄Cl(s) ⇄ NH₃ (g) + HCl(g)Explanation:
A reversible reaction is indicated by using a double arrow ⇄
The upper arrow (→), from left to right, indicates the direct or forward reaction, which goes from left to right.
In the direct reaction, the reactants are the substances shown on the left side of the equation, and the products are the substances shown on the right side.
The lower arrow (←), from right to left, indicates the reverse reaction, which goes from right to left.
In the reverse reaction, the reactants are the substances shown of the right side and the products are the substances shown of the left side.
Summarizing:
Forward reaction: NH₄Cl(s) → NH₃ (g) + HCl(g)Reverse reaction NH₄Cl(s) ← NH₃ (g) + HCl(g)Reversible reaction NH₄Cl(s) ⇄ NH₃ (g) + HCl(g)In the moment that the rates of both forward and reverse reactions are equal it is said that the equilibrium has been reached.
Answer:
just did edge test:
c. NH4CI(s)<—–>NH3( g)+HCI(g)
Select the phrase that best describes scientific findings communicated through popular media.
A) generally free of technical jargon
B) peer reviewed
C) commonly used by most scientists
D) extremly reliable
Answer:
The correct answer is option A) generally free of technical jargon
Explanation:
Hello!
Let's solve this!
When we read popular magazines or newspapers, we see scientific findings. They generally do not use scientific or technical vocabulary so that more people can understand it. They are also not the most reliable sources, since to look for reliable information, we have to look for scientific journals.
After the analysis we conclude that the correct answer is option A) generally free of technical jargon
You have 750 grams of water at 80° Celsius. Which of the following would lower the temperature of the water by 10° Celsius? (1 point)
adding 750 grams of water at 50° Celsius
adding 325 grams of water at 60° Celsius
adding 750 grams of water at 60° Celsius
adding 1000 grams of water at 55° Celsius
Answer:
adding 750 grams of water at 60° Celsius .
Explanation:
We can calculate the amount of heat lost from 750 grams of water at 80°C to be lowered by 10°C using the relation:Q = m.c.ΔT,
Where, Q is the amount of heat lost by water (Q = ??? J).
m is the mass of water (m = 750.0 g).
c is the specific heat capacity of the water (c = 4.18 J/g.°C).
ΔT is the temperature difference (ΔT = final T - initial T = - 10.0°C, the temperature of water is lowered by 10.0°C).
∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(- 10.0°C) = - 31350.0 J = -31.350 kJ.
Now, we can calculate the Q that is gained by the different added amounts of water:
adding 750 grams of water at 50° Celsius :ΔT = 70.0°C - 50.0°C = 20.0°C,
∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(20.0°C) = 62700.0 J = 62.70 kJ.
adding 325 grams of water at 60° Celsius :ΔT = 70.0°C - 60.0°C = 10.0°C,
∴ Q = m.c.ΔT = (325.0 g)(4.18 J/g.°C)(10.0°C) = 13585.0 J = 13.585 kJ.
adding 750 grams of water at 60° Celsius :ΔT = 70.0°C - 60.0°C = 10.0°C,
∴ Q = m.c.ΔT = (750.0 g)(4.18 J/g.°C)(10.0°C) = 31350.0 J = 31.350 kJ.
adding 1000 grams of water at 55° Celsius:ΔT = 70.0°C - 55.0°C = 15.0°C,
∴ Q = m.c.ΔT = (1000.0 g)(4.18 J/g.°C)(15.0°C) = 62700.0 J = 62.70 kJ.
So, the right choice is:adding 750 grams of water at 60° Celsius
How many moles of al2o3 can be produced from the reaction of 10.0 g of al and 19.0 of o2?
Answer:
0.185moles of Al₂O₃
Explanation:
Mass of Al = 10g
Mass of O₂ = 19g
Equation of the reaction: 4Al + 3O₂ → 2Al₂O₃
This is the balanced reaction equation.
Solution
From the given parameters, the reactant that would determine the extent of the reaction is Aluminium. It is called the limiting reagent. Oxygen is in excess and it is in an unlimited supply.
Working from the known mass to the unknown, we simply solve for the number of moles of Al using the mass given.
Then from the equation, we can relate the number of moles of Al to that of Al₂O₃ produced:
Number of moles of Al = [tex]\frac{mass}{molar mass}[/tex]
= [tex]\frac{10}{27}[/tex]
= 0.37mol
From the equation:
4 moles of Al produced 2 moles of Al₂O₃
0.37 mole will yield: [tex]\frac{2 x 0.37}{4}[/tex] = 0.185moles of Al₂O₃
0.74 moles of aluminium oxide can be produced from the reaction of 10.0 g of aluminium and 19.0 of oxygen.
What is the equation of the reaction?The equation of the reaction between aluminium and oxygen gas to produce aluminium oxide is given below as follows:
[tex]4Al + 3O₂ → 2Al₂O₃[/tex]
From the equation of reaction 4 moles of aluminium reacts with 3 moles of oxygen to produce 2 moles of aluminium oxide.
The moles of reactants is calculated using:
moles = mass/molar massmolar mass of aluminium = 27.0 g
molar mass of oxygen = 32.0 g
moles of aluminium = 10/27 = 0.37 moles
moles of oxygen = 19/32 = 0.59
Aluminium is the limiting reactant
Thus, moles of aluminium oxide produced = 0.37 × 2 = 0.74 moles
Therefore, 0.74 moles of aluminium oxide can be produced from the reaction of 10.0 g of aluminium and 19.0 of oxygen.
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We know that for a given reaction when the temperature increases from 100 k to 200 k the rate constant doubles. What is the activation energy in kj/mol
Answer:
The activation energy of the reaction is 1.152 kJ/mol.
Explanation:
Activation energy is the minimum amount which is absorbed by the reactant molecules to undergo chemical reaction.
Initial temperature of reaction = [tex]T_1=100 K[/tex]
Final temperature of reaction = [tex]T_2=200 K[/tex]
Initial rate of the reaction at 100 k = [tex]K_1=k[/tex]
Final rate of the reaction at 200 k = [tex]K_2=2k[/tex]
Activation energy is calculated from the formula:
[tex]\log\frac{K_2}{K_1}=\frac{E_a}{2.303\times R}(\frac{T_2-T_1}{T_1T_2})[/tex]
R = Universal gas constant = 8.314 J/ K mol
[tex]\log\frac{2k}{k}=\frac{E_a}{2.303\times 8.314 J/K mol}(\frac{200 K-100K}{200 K\times 100K})[/tex]
[tex]E_a=1,152.772 J/mol=1.152 kJ/mol[/tex]
The activation energy for this chemical reaction can be calculated using the Arrhenius equation and the provided information about the doubling of the rate constant as the temperature increases from 100 K to 200 K. An alternative two-point form of the Arrhenius equation can also be used for this calculation.
Explanation:The activation energy of a reaction can be calculated by utilizing the Arrhenius equation, which relates the effect of temperature on the rate constant, k of a reaction: k = Ae−Ea/RT. In this equation, R is the ideal gas constant (8.314 J/mol/K), T is the temperature in Kelvin, Ea is the activation energy in joules per mole, A is a constant known as the frequency factor, and e is a mathematical constant.
Given the data that when the temperature increases from 100 K to 200 K the rate constant (k) doubles, we can use an alternative approach to calculate the activation energy. This involves rearranging the Arrhenius equation in a two-point form and inserting the provided temperatures and rate constants. Substitution and subsequent calculation will yield the activation energy value in kJ/mol.
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Consider the following reaction, equilibrium concentrations, and equilibrium constant at a particular temperature. Determine the equilibrium concentration of H2O(g). C2H4(g) + H2O(g) ⇌ C2H5OH(g) Kc = 9.0 × 103 [C2H4]eq = 0.015 M [C2H5OH]eq = 1.69 M
The equilibrium concentration of H2O(g) in the given reaction is approximately 12.91 × 10^-6 M.
Explanation:To determine the equilibrium concentration of H2O(g) in the given reaction, we need to use the equilibrium constant expression and the equilibrium concentrations of the other species. The equilibrium constant expression for the reaction C2H4(g) + H2O(g) ⇌ C2H5OH(g) is Kc = [C2H5OH]/([C2H4][H2O]). Given that [C2H4]eq = 0.015 M and [C2H5OH]eq = 1.69 M, we can substitute these values into the expression to solve for the equilibrium concentration of H2O(g).
Kc = [C2H5OH]/([C2H4][H2O])
9.0 × 10^3 = 1.69/((0.015)([H2O]))
[H2O] = 1.69/((0.015)(9.0 × 10^3))
[H2O] ≈ 12.91 × 10^-6 M
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The equilibrium concentration of H2O(g) is approximately 0.0125 M.
The equilibrium concentration of H2O(g) given the reaction C2H4(g) + H2O(g) ⇌ C2H5OH(g), the equilibrium constant (Kc), and the equilibrium concentrations of C2H4 and C2H5OH at a particular temperature. We start by writing the expression for the equilibrium constant:
Kc = [C2H5OH]/[C2H4][H2O]
Then, we plug in the known values:
9.0 × 10^3 = 1.69/[0.015][H2O]
Now, we solve for [H2O]:
[H2O] = 1.69/(9.0 × 10^3 × 0.015)
[H2O] ≈ 1.69/(135) ≈ 0.0125 M
Match the chemical reactions in column b with their properties listed I. Column a. 1 the reaction has at least two reactants and one product 2the reaction has one reactant and at least two products 3 oxygen is one of the reactants and a large amount of energy is released 4 one or more atoms replaces a part of a compound 5 the products are always salt and water A. Neutralization B. Combustion C. Decomposition D synthesis E. Replacement
Answer:
Explanation:
1. The reaction has at least two reactants and one product
Synthesis
Synthesis is a sort of combination reaction in which a single product is formed from two or more reactants:
A + B → C
2. The reaction has one reactant and at least two products
Decomposition
In decomposition reaction, a compound breaks down into individual elements or compounds.
A → B + C
3. Oxygen is one of the reactants and a large amount of energy is released
Combustion
In combustion reaction, oxygen combines with a fuel source to produce energy.
4. One or more atoms replaces a part of a compound
Replacement
In this type of reaction, one substance replaces another we say it is a single replacement reaction. Some other types involves the actual exchange of partners.
5. The products are always salt and water
Neutralization
Neutralization reaction is a reaction in which acids reacts with bases to produce salts and water.
Final answer:
Chemical reactions are matched with their properties: Synthesis is a reaction with multiple reactants and one product, Decomposition has one reactant and multiple products, Combustion involves oxygen and energy release, Replacement consists of atoms substituting part of a compound, and Neutralization results in salt and water.
Explanation:
When matching the chemical reactions in Column B with their properties listed in Column A, the following connections can be made:
Synthesis or combination reactions involve multiple reactants combining to make a single product.Decomposition reactions occur when a single substance reacts to produce several products.Combustion reactions always include oxygen as a reactant and result in the release of a large amount of energy, typically producing carbon dioxide and water.Replacement reactions involve one or more atoms taking the place of part of a compound.Neutralization reactions are specific types of reactions where an acid and a base react to produce salt and water.Therefore, the correct matches are:
1 - D (Synthesis)2 - C (Decomposition)3 - B (Combustion)4 - E (Replacement)5 - A (Neutralization)The starting molecule for the krebs cycle is
Answer:
acetyl CoA
Explanation:
The starting molecule for the krebs cycle is acetyl CoA.