Answer:
34.5 mL
Explanation:
Given, we ned to prepared buffer solution with pH = 4.00 , volume = 100.0 mL .
[C6H5COOH] = 0.100 M , [C6H5COO-] = 0.120 M , pKa = 4.20
First we need to calculate the ratio of the conjugate base and acid
We know, Henderson Hasselbalch equation
pH = pKa + log [C6H5COO-] /[C6H5COOH]
4.00 = 4.20 + log [C6H5COO-] /[C6H5COOH]
log [C6H5COO-] /[C6H5COOH] = 4.00-4.20
= - 0.20
Antilog from both side
[C6H5COO-] /[C6H5COOH] = 0.631
Now we are given the molarity of each acid and its conjugate base and we need to calculate for volume
Sum of volume = 100 mL = 0.100 L
volume of benzoic acid = x
volume of benzoate = 0.10 -x ,
so Volume of acid + volume of conjugate base = 0.100 L
[C6H5COO-] /[C6H5COOH] = 0.631
[C6H5COO-] = 0.631 * [C6H5COOH]
0.120 (0.1-x) = 0.631 *0.100x
So, x = 0.065
So, volume of benzoic acid = x = 0.0655 L
= 65.5 mL
So, volume of sodium benzoate = 0.1 -x
= 0.1-0.0655
= 0.0345 L
= 34.5 mL
So we need to mix 65.5 mL of 0.100 M benzoic acid and 34.5 mL of 0.120 M sodium benzoate form 100.0 mL of a pH=4.00 buffer solution.
Answer:
Volume benzoic acid= 77.74 mL
Volume sodium benzoate: = 22.26 mL
Explanation:
Step 1: Data given
Volume = 100.0 mL
pH = 4.00
Molarity benzoic acid = 0.100 M
pKa benzoic acid = 4.20
Molarity sodium benzoate = 0.220 M
Step 2
pH = pKa + log([base]/[acid])
4 = 4.2 + log([base]/[acid])
-0.2 = log([base]/[acid])
10^-0.2 = [base]/[acid]
0.63 = [base]/[acid]
Step 3
start with 100 mL 0.1 M benzoic acid
0.100 L* 0.1M = 0.01 moles.
Since base/acid = 0.63, add 0.63*0.01 moles base = 0.0063 moles base.
And 0.0063 moles of a 0.22 M salt solution are x mL:
0.22 moles = 1000 mL
1 mole = 1000/0.22 mL
0.0063 moles = 1000*0.0063/0.22 mL = 28.63 mL .
So a mixture of 100 mL 0.1 M bencoic acid + 28.63 mL of 0.22 benzoate has pH =4.00
Step 4: For a total volume of 100 mL we'll have:
Volume benzoic acid: 100*100/(128.63) = 77.74 mL
Volume sodium benzoate: 28.63*100/(128.63) = 22.26 mL
If there are 160 Calories in 28 g in 1 serving of Flamin' Hot Cheetos, how many gram(s) of Cheetos must you burn to raise the temperature of 20.0 mL of water 1oC?
Answer:
3.5 g
Explanation:
The density of water is 1 g/mL. The mass (m) corresponding to 20.0 mL is 20.0 g.
We can calculate the heat (Q) required to raise the temperature of 20.0 mL of water 1 °C (ΔT).
Q = c × m × ΔT = 1 cal/g.°C × 20.0 g × 1 °C = 20 cal
where,
c is the specific heat capacity of water
There are 160 calories in 28 g of Cheetos. The mass that releases 20 cal is:
20 cal × (28 g/160 cal) = 3.5 g
Final answer:
To raise the temperature of 20.0 mL of water by 1°C, 0.35 grams of Flamin' Hot Cheetos must be burned, utilizing the concept of specific heat capacity and energy content of the Cheetos.
Explanation:
The question asks how many grams of Flamin' Hot Cheetos one must burn to raise the temperature of 20.0 mL (which translates to 20.0 grams, assuming the density of water is 1 g/mL) of water by 1°C. To solve this, we first acknowledge that the specific heat capacity of water is about 1 calorie per gram per Celsius degree (1 cal/g°C). Therefore, to raise 20.0 grams of water by 1°C, it requires 20.0 calories. Given the conversion factor that 1 Calorie (kcal, with a capital 'C', equivalent to food calories) = 1000 calories (small 'c'), we deduce that 20.0 calories = 0.02 Kcal.
From the question's premise, 28 grams of Flamin' Hot Cheetos contain 160 Calories (160 Kcal). To find how many grams are needed to produce 0.02 Kcal, we set up a proportion: 160 Kcal per 28 grams = 0.02 Kcal per x grams. Solving for x, we find that x = (0.02 Kcal * 28 grams) / 160 Kcal = 0.35 grams of Flamin' Hot Cheetos are needed to raise the temperature of 20.0 mL of water by 1°C.
The neutralization of H 3 PO 4 with KOH is exothermic. H 3 PO 4 ( aq ) + 3 KOH ( aq ) ⟶ 3 H 2 O ( l ) + K 3 PO 4 ( aq ) + 173.2 kJ If 55.0 mL of 0.227 M H 3 PO 4 is mixed with 55.0 mL of 0.680 M KOH initially at 22.62 °C, predict the final temperature of the solution, assuming its density is 1.13 g/mL and its specific heat is 3.78 J/(g·°C). Assume that the total volume is the sum of the individual volumes.
Explanation:
It is known that,
No. of moles = Molarity × Volume
So, we will calculate the moles of [tex]H_{3}PO_{4}[/tex] as follows.
No. of moles = [tex]0.227 \times 0.055 L[/tex]
= 0.0125 mol
Now, the moles of KOH are as follows.
No. of moles = [tex]0.680 \times 0.055 L[/tex]
= 0.0374 mol
And, [tex]3 \times \text{moles of} H_{3}PO_{4}[/tex] = [tex]3 \times 0.0125[/tex]
= 0.0375 mol
Now, the balanced reaction equation is as follows.
[tex]H_{3}PO_{4}(aq) + 3KOH(aq) \rightarrow 3H_{2}O(l) + K_{3}PO_{4}(aq) + 173.2 kJ[/tex]
This means 1 mole of [tex]H_{3}PO_{4}[/tex] produces 173.2 kJ of heat. And, the amount of heat produced by 0.0125 moles of [tex]H_{3}PO_{4}[/tex] is as follows.
M = [tex]\frac{0.0125 mol \times 173.2 kJ}{1}[/tex]
= 2.165 kJ
Total volume of the solution = (55.0 + 55.0) ml
= 110 ml
Density of the solution = 1.13 g/ml
Mass of the solution = Volume × Density
= [tex]110 ml \times 1.13 g/ml[/tex]
= 124.3 g
Specific heat = 3.78 [tex]J/g^{o}C[/tex]
Now, we will calculate the final temperature as follows.
q = [tex]mC \times \Delta T[/tex]
2165 J = [tex]124.3 \times 3.78 \times (T - 22.62)^{o}C[/tex]
2165 - 469.854 = [tex]T - 22.62^{o}C[/tex]
17.417 = [tex]T - 22.62^{o}C[/tex]
T = [tex]40.04^{o}C[/tex]
Thus, we can conclude that final temperature of the solution is [tex]40.04^{o}C[/tex].
The final temperature of the solution after neutralization of H3PO4 with KOH is determined to be 27.24°C based on the enthalpy change, specific heat capacity of the solution, and the mass of the final mixture.
Explanation:To predict the final temperature of the solution after the neutralization reaction of H3PO4 with KOH, we must first determine the amount of heat released during the reaction using the enthalpy change given and then apply this to the specific heat capacity equation of the resulting solution. Given that the reaction is exothermic and releases 173.2 kJ per mole of H3PO4 reacted, we first need to figure out how many moles of H3PO4 are reacting. Then we calculate the heat (q) released using the molarity and volume of H3PO4.
Moles of H3PO4 = 0.227 M × 0.055 L = 0.012485 mol
Heat released (q) = 0.012485 mol × 173.2 kJ/mol = 2.16322 kJ
Now convert q to Joules (1 kJ = 1000 J) and calculate the resulting temperature change using the mass of the solution (density × volume), specific heat capacity, and heat equation (q = mcΔT). The total volume of the solution is 110.0 mL, yielding a total mass of 124.3 g (1.13 g/mL × 110 mL).
Converting q to Joules: q = 2.16322 kJ × 1000 J/kJ = 2163.22 J
Using the equation ΔT = q / (mc), where m is mass and c is specific heat capacity,
ΔT = 2163.22 J / (124.3 g × 3.78 J/g°C) = 4.62°C
Therefore, the final temperature of the solution is 22.62°C + 4.62°C = 27.24°C.
6. How many moles of Cu(NO3)2 are in 25 ml of a 0.35 M solution of Cu(NO3)2? a) How many moles of NO3- are present in 25 ml of the above solution? b) What volume of this solution should be used to get 0.050 moles of Cu(NO3)2?
Explanation:
Below is an attachment containing the solution.
Answer:
0.00875 moles Cu(NO3)2
a) 0.0175 moles NO3-
b) We should use 0.14L
Explanation:
Step 1: Data given
Volume = 25 mL = 0.025 L
Molarity of a Cu(NO3)2 solution = 0.35 M
Step 2: Calculate moles Cu(NO3)2
Moles Cu(NO3)2 = molarity * volume
Moles Cu(NO3)2 = 0.35 M * 0.025 L
Moles Cu(NO3)2 = 0.00875 moles
Step 3: Calculate moles NO3-
Cu(NO3)2 → Cu^2+ + 2NO3-
In 1 mol Cu(NO3)2 we have 2 moles NO3-
For 0.00875 moles Cu(NO3)2 we'll have 2*0.00875 = 0.0175 moles NO3-
Step 4: What volume of this solution should be used to get 0.050 moles of Cu(NO3)2?
Volume = moles / molarity
Volume = 0.050 moles / 0.35 M
Volume = 0.14L
uppose you perform a titration of an unknown weak acid solution. You start with 4.00 mL of the weak acid and find that it takes 12.8 mL of 0.0500 M NaOH to reach the equivalence point. What is the concentration of the unknown weak acid solution?
Explanation:
It is known that at the equivalence point,
Equivalent of acid = equivalent of base ......... (A)
and, Equivalent = [tex]\text{Molarity} \times n_{f} \times V[/tex]
Therefore, equivalent of acid is calculated as follows.
Equivalent of acid = [tex]M \times 1 \times 4[/tex] .......... (1)
Equivalent of base = [tex]0.05 \times 1 \times 12.8[/tex] ............ (2)
Hence, using equation (A) we will calculate the concentration as follows.
[tex]M \times 4 = 0.05 \times 12.8[/tex]
M = 0.16 M
Thus, we can conclude that concentration of the unknown weak acid solution 0.16 M.
Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), H2S(aq)⇌HS−(aq)+H+(aq), K1K1K_1 = 9.39×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), HS−(aq)⇌S2−(aq)+H+(aq), K2K2K_2 = 1.45×10−19, what is the equilibrium constant KfinalKfinalK_final for the following reaction? S2−(aq)+2H+(aq)⇌H2S(aq)
Answer: The value of [tex]K_{final}[/tex] for the net reaction is [tex]7.34\times 10^{25}[/tex]
Explanation:
The given chemical equations follows:
Equation 1: [tex]H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^+(aq.);K_1[/tex]
Equation 2: [tex]HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^+(aq.);K_2[/tex]
The net equation follows:
[tex]S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_{final}[/tex]
As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.
The value of equilibrium constant for net reaction is:
[tex]K_{final}=\frac{1}{K_1}\times \frac{1}{K_2}[/tex]
We are given:
[tex]K_1=9.39\times 10^{-8}[/tex]
[tex]K_2=1.45\times 10^{-19}[/tex]
Putting values in above equation, we get:
[tex]K_{final}=\frac{1}{(9.39\times 10^{-8})}\times \frac{1}{(1.45\times 10^{-19})}=7.34\times 10^{25}[/tex]
Hence, the value of [tex]K_{final}[/tex] for the net reaction is [tex]7.34\times 10^{25}[/tex]
The equilibrium constant for the reaction S2−(aq) + 2H+(aq) ⇌ H2S(aq) is 1.361×10−26.
Explanation:To find the equilibrium constant for the reaction S2−(aq) + 2H+(aq) ⇌ H2S(aq), we can use the equilibrium constant expressions for the given reactions and multiply them together. The equilibrium constant for the first reaction (K1) is 9.39×10−8, and for the second reaction (K2) it is 1.45×10−19. So, the equilibrium constant for the overall reaction (Kfinal) is the product of K1 and K2: Kfinal = K1 × K2 = 9.39×10−8 × 1.45×10−19 = 1.361×10−26.
Two nitro (NO_2) groups are chemically bonded to a patch of surface. They can't move to another location on the surface, but they can rotate (see sketch at right). It turns out that the amount of rotational kinetic energy each NO_2 group can have is required to be a multiple of epsilon, where epsilon = 1.0 times^-24 J. In other words, each NO_2 group could have epsilon of rotational kinetic energy, or 2 epsilon, or 3 epsilon, and so forth - but it cannot have just any old amount of rotational kinetic energy. Suppose the total rotational kinetic energy in this system is initially known to be 39 epsilon. Then, some heat is added to the system, and the total rotational kinetic energy rises to 59 epsilon. Calculate the change in entropy. Round your answer to 3 significant digits, and be sure it has the correct unit symbol.
Answer:
Explanation:
The detailed and step by step analysis is as shown in the atached file with appropriate substitution.
Consider the reaction C2H4(g) + H2O(g)CH3CH2OH(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.90 moles of C2H4(g) react at standard conditions. S°surroundings = J/K
Answer: The value of [tex]\Delta S^o[/tex] for the surrounding when given amount of ethene gas is reacted is 238.80 J/K
Explanation:
Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.
The equation used to calculate entropy change is of a reaction is:
[tex]\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}][/tex]
For the given chemical reaction:
[tex]C_2H_4(g)+H_2O(g)\rightarrow CH_3CH_2OH(g)[/tex]
The equation for the entropy change of the above reaction is:
[tex]\Delta S^o_{rxn}=[(1\times \Delta S^o_{(CH_3CH_2OH(g))})]-[(1\times \Delta S^o_{(C_2H_4(g))})+(1\times \Delta S^o_{(H_2O(g))})][/tex]
We are given:
[tex]\Delta S^o_{(CH_3CH_2OH(g))}=282.7J/K.mol\\\Delta S^o_{(C_2H_4(g))}=219.56J/K.mol\\\Delta S^o_{(H_2O(g))}=188.82J/K.mol[/tex]
Putting values in above equation, we get:
[tex]\Delta S^o_{rxn}=[(1\times (282.7))]-[(1\times (219.56))+(1\times (188.82))]\\\\\Delta S^o_{rxn}=-125.68J/K[/tex]
Entropy change of the surrounding = - (Entropy change of the system) = -(-125.68) J/K = 125.68 J/K
We are given:
Moles of ethene gas reacted = 1.90 moles
By Stoichiometry of the reaction:
When 1 mole of ethene gas is reacted, the entropy change of the surrounding will be 125.68 J/K
So, when 1.90 moles of ethene gas is reacted, the entropy change of the surrounding will be = [tex]\frac{125.68}{1}\times 1.90=238.80J/K[/tex]
Hence, the value of [tex]\Delta S^o[/tex] for the surrounding when given amount of ethene gas is reacted is 238.80 J/K
A turn-of-the-century chemist isolated an aromatic compound of molecular formula C6H4Br2. He carefully nitrated this compound and purified three isomers of formula C6H3Br2NO2. Propose structures for the original compound and the three nitrated derivatives.
Answer: the structures are shown in the image attached.
Explanation:
Before the invention of mordern spectroscopy, Korner's absolute method was commonly applied in determining whether a disubstituted benzene derivative was the ortho, meta, or para isomer. Korner's method depends on the addition of a third group, nitro groups are usually added. The number of isomers formed is then determined. The isomers are usually obtained only in small amounts and analysed to know the structure of the original compound.
Final answer:
The original compound is p-bromotoluene or 4-bromotoluene. The three nitrated derivatives are m-dinitrobenzene or 1,3-dinitrobenzene.
Explanation:
The turn-of-the-century chemist isolated a compound with the molecular formula C6H4Br2. After nitrating this compound, three isomers of formula C6H3Br2NO2 were obtained.
The original compound is called p-bromotoluene or 4-bromotoluene. It contains a methyl (CH3) group and the bromine atom is attached to the fourth carbon atom.
The three nitrated derivatives are called m-dinitrobenzene or 1,3-dinitrobenzene. These compounds have two nitro (NO2) groups attached to the benzene ring, at the first and third positions.
A student's calibration curve for Blue #1 at 635 nm yields a straight line described by the equation y = 0.194 x + 0.025. If the measured absorbance for a solution of unknown concentration of Blue #1 is 0.293, what is the concentration of the Blue #1 solution? Provide your response to three digits after the decimal.
Answer:
1.38 M is the concentration of blue solution.
Explanation:
The equation of calibration curve:
[tex]A=\epsilon \times l\times c[/tex]
where,
A = absorbance of solution
C = concentration of solution
l = path length
[tex]\epsilon[/tex] = molar absorptivity coefficient
The graph between A and c will straight line with slope equal to prduct of [tex]\epsilon and l[/tex].
y = 0.194 x + 0.025.
A = 0.194 c+ 0.025
Given the absorbance of solution :
A = 0.293
So, concentration at this value of absorbance will be= c
[tex]0.293=0.194 c+0.025[/tex]
[tex]c=\frac{0.293-0.025}{0.194}=1.381\approx 1.38[/tex]
1.38 M is the concentration of blue solution.
The concentration of the Blue #1 solution can be calculated by rearranging the equation of the calibration curve to solve for 'x' when 'y' is the measured absorbance, 0.293. The calculated concentration to three decimal places is approximately 1.381.
Explanation:The equation of the straight line formed by your calibration curve for Blue #1 at 635 nm is y = 0.194x + 0.025. In this equation, 'y' represents the measured absorbance and 'x' represents the concentration. If the measured absorbance, or 'y', is 0.293, you can calculate the concentration, or 'x', by rearranging the equation to solve for 'x'.
To isolate 'x', you would subtract 0.025 from each side of your equation.
0.293 - 0.025 = 0.194x, which simplifies to 0.268 = 0.194x. Finally, divide each side of the equation by 0.194 to solve for 'x':
0.268 / 0.194 = x
When you divide 0.268 by 0.194, you get approximately 1.381. Therefore the concentration of the Blue #1 solution to three decimal places is approximately 1.381.
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Oxaloacetic acid (2-ketosuccinic acid) is a very important intermediate in metabolism. The compound is involved in the citric acid cycle for energy production within the cell. However, the compound is unstable and slowly decomposes spontaneously. Draw the decomposition products.
Answer:
Carbon dioxide and acetoacetic acid
Explanation:
Oxaloacetic acid is a β-keto acid. It decarboxylates readily via a six-membered cyclic transition state.
The initial products are carbon dioxide and the enol of a ketone.
The enol is unstable and rapidly tautomerizes to the more stable keto form — acetoacetic acid.
Pyruvic acid and carbon dioxide were formed when oxyaloacetic acid decomposes, and the further discussion can be defined as follows:
Oxaloacetic acid: As oxyaloacetic acid first, pyruvic acid and carbon dioxide, are produced.This [tex]\bold{\beta}[/tex]-keto acid is oxaloacetic acid contains. Its six-membered cyclic transition state allows this one to decarboxylate easily.Carbon dioxide as well as the enol of a ketone are the first by-products.This enol is unstable and tautomerizes to the more stable keto form of acetoacetic acid very quickly.Learn more about the Oxaloacetic acid here:
brainly.com/question/21990765
The ________ binding of oxygen to hemoglobin is facilitated by changes in protein conformation upon oxygen binding to one subunit that affect the binding of oxygen to another subunit.
Answer:
Hi, you haven't provided the options to the question, so I will just give the answer in my own words and you can check with the options.
Answer is: The COOPERATIVE binding oxygen to hemoglobin is facilitated by changes in protein conformation upon oxygen binding to one subunit that affect the binding of oxygen to another subunit.
Explanation:
Hemoglobin, which is a complex protein contained within our red blood cells is an alternative source of transportation of oxygen through our body. With hemoglobin, we can carry 20 ml of oxygen in that same 100 ml of blood.
The way by which hemoglobin binds oxygen is referred to as cooperative binding. The binding of oxygen to hemoglobin makes it easier for more oxygen to bind.
For example: during the cooperative binding of oxygen to hemoglobin, using four oxygen molecules. The binding of one oxygen molecule causes a change in conformation allowing the 2nd, 3rd, and 4th molecules to bind more efficiently to the hemoglobin.
Therefore, the answer to the question is COOPERATIVE.
Hemoglobin exhibits cooperative binding of oxygen, characterized by easier binding of the second and third oxygen molecules, and visualized through an S-shaped oxygen dissociation curve.
Explanation:Cooperative Binding of Oxygen to Hemoglobin
The cooperative binding of oxygen to hemoglobin is facilitated by changes in protein conformation upon oxygen binding to one subunit that affect the binding of oxygen to another subunit. When oxygen binds to the heme group in one of the subunits within the hemoglobin molecule, this causes a conformational change facilitating the binding of the subsequent oxygen molecules. Initially, it is easier to bind the second and third oxygen molecules to hemoglobin than the first due to these structural changes. However, the binding of the fourth oxygen molecule is more difficult. This process can be visualized in the sigmoidal or S-shaped oxygen dissociation curve, which shows hemoglobin saturation with oxygen as the partial pressure of oxygen increases.
Furthermore, factors such as pH, temperature, and the presence of 2,3-bisphosphoglycerate also influence the ease of oxygen binding and dissociation from hemoglobin. For instance, fetal hemoglobin demonstrates a higher affinity for oxygen compared to adult hemoglobin. The dynamic nature of hemoglobin's affinity for oxygen is crucial for its role in transporting and releasing oxygen throughout the body, a process modulated by the partial pressure of oxygen in various environments such as the lungs and peripheral tissues.
428. mg of an unknown protein are dissolved in enough solvent to make of solution. The osmotic pressure of this solution is measured to be at . Calculate the molar mass of the protein. Round your answer to significant digits.
The question is incomplete, here is the complete question:
428. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution is measured to be 0.0766 atm at 25.0°C. Calculate the molar mass of the protein. Round your answer to 3 significant digits
Answer: The molar mass of protein is [tex]27.3\times 10^3g/mol[/tex]
Explanation:
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
[tex]\pi=iMRT[/tex]
Or,
[tex]\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT[/tex]
where,
[tex]\pi[/tex] = osmotic pressure of the solution = 0.0766 atm
i = Van't hoff factor = 1 (for non-electrolytes)
Mass of solute (protein) = 428 mg = 0.428 g (Conversion factor: 1 g = 1000 mg)
Volume of solution = 5.00 mL
R = Gas constant = [tex]0.0821\text{ L.atm }mol^{-1}K^{-1}[/tex]
T = temperature of the solution = [tex]25^oC=[273+25]=298K[/tex]
Putting values in above equation, we get:
[tex]0.0766=1\times \frac{0.428\times 1000}{\text{Molar mass of protein}\times 5}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\\text{Molar mass of protein}=\frac{1\times 0.428\times 1000\times 0.0821\times 298}{0.0766\times 5}=27340.4g/mol=27.3\times 10^3g/mol[/tex]
Hence, the molar mass of protein is [tex]27.3\times 10^3g/mol[/tex]
The enthalpy of reaction changes somewhat with temperature. Suppose we wish to calculate ΔH for a reaction at a temperature T that is different from 298 K. To do this, we can replace the direct reaction at T with a three-step process. In the first step, the temperature of the reactants is changed from T to 298 K. ΔH for this step can be calculated from the molar heat capacities of the reactants, which are assumed to be independent of temperature. In the second step, the reaction is conducted at 298 K with an enthalpy change ΔH°. In the third step, the temperature of the products is changed from 298 K to T. The sum of these three enthalpy changes is ΔH for the reaction at temperature T.
An important process contributing to air pollution is the following chemical reaction:
SO2(g) + ½O2(g) → SO3(g)
For SO2(g), the heat capacity cp is 39.9, for O2(g) it is 29.4, and for SO3(g) it is 50.7 J K⁻¹ mol⁻¹.
Calculate ΔH for the preceding reaction at 500 K, using the enthalpies of formation at 298.15 K from Appendix D.
Answer:
-99.8 kJ
Explanation:
We are given the methodology to answer this question, which is basically Kirchhoff law . We just need to find the heats of formation for the reactants and products and perform the calculations.
The standard heat of reaction is
ΔrHº = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants
where ν are the stoichiometric coefficients in the balanced equation, and ΔfHº are the heats of formation at their standard states.
Compound ΔfHº (kJmol⁻¹)
SO₂ -296.8
O₂ 0
SO₃ -395.8
The balanced chemical equation is
SO₂(g) + ½O₂(g) → SO₃(g)
Thus
Δr, 298K Hº( kJmol⁻¹ ) = 1 x (-395.8) - 1 x (-296.8) = -99.0 kJmol⁻¹
Now the heat capacity of reaction will be be given in a similar fashion:
Cp rxn = ∑ ν x Cp of products - ∑ ν x Cp of reactants
where ν is as above the stoichiometric coefficient in the balanced chemical equation.
Cprxn ( JK⁻¹mol⁻¹) = 50.7 - ( 39.9 + 1/2 x 29.4 ) = - 3.90
= -3.90 JK⁻¹mol⁻¹
Finally Δr,500 K Hº = Δr, 298K Hº + CprxnΔT
Δr,500 K Hº = - 99 x 10³ J + (-3.90) JK⁻¹ ( 500 - 298 ) K = -99,787.8
= -99,787.8 J x 1 kJ/1000 J = -99.8 kJ
Notice thie difference is relatively small that is why in some problems it is o.k to assume the change in enthalpy is constant over a temperature range, especially if it is a small range of temperatures.
If your unknow solid has an impurity that is insoluble in cyclohexane, will this impurity result in the molar mass of the unknown solid being recorded as high, low, or unaffected? Explain.
If your unknown solid has an impurity that is insoluble in cyclohexane the molar mass of the unknown solid being recorded as high.
What are impurities?Impurities are chemical substances inside a certain amount of liquid, gas, or solid, but are different from the chemical composition of the material itself.
When the unknown solid (solute) contains an impurity that is insoluble in cyclohexane, the impurity will not dissolve in the cyclohexane (solvent) completely.Also, there will be an increase in molecular weight obtained because the mass of the solute would now include the actual mass of solute that is changing and the excess mass of the impurity from the unknown solid this will increase the molar mass.
Therefore, If your unknown solid has an impurity that is insoluble in cyclohexane the molar mass of the unknown solid is recorded as high.
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If the impurity is insoluble in cyclohexane, it will not affect the molar mass of the unknown solid.
Explanation:If your unknown solid has an impurity that is insoluble in cyclohexane, it will not affect the molar mass of the unknown solid. This is because the impurity is insoluble and therefore does not dissolve in the cyclohexane solvent. The molar mass is determined by the mass of the solute and the number of moles present, and since the impurity does not dissolve, it does not contribute to the mass or molar mass of the unknown solid.
mpurities can significantly affect the structure and properties of a solid. Even a small amount of impurity can disrupt the regular lattice of crystal structures, altering physical properties like melting point and electrical conductivity, which is a crucial consideration in industries such as electronics that require high-purity materials.
List the following aqueous solutions in order of decreasing boiling point. 0.120 mm glucose, 0.050 m LiBrm LiBr, 0.050 m Zn(NO3)2m Zn(NO3)2 . Rank items in order of decreasing boiling point.
Answer:
0.050 m LiBrm LiBr < 0.120 mm glucose <0.050 m Zn(NO3)2m Zn(NO3)2
Explanation:
The above aqueous solutions show that LiBr low boiling point followed by glucose and Zinc.
using the equation of boiling point elevation
ΔTb = i×Kb×M
Making temperature constant at a value of 290K you can simple do a calculation to conform the boiling point.
looking the van hoff values for Znc = 3 , LiBr = 2 and glucose = 1
(glucose)ΔTb = i×Kb×M = 1×290k×0.120 = 34.8
(Zinc) = 3×290×0.050 = 43.5
(LiBr) = 2×290×0.050 =29
∴ in conclusion LiBr has a the lowest decreasing boiling point
Based on the concentration of the solutions, the rank order of the solutions in terms of decreasing boiling point is 0.050 m LiBr < 0.120 mm glucose < 0.050 m Zn(NO3)2.
What is the boiling point of a liquid solution?The boiling point of a liquid solution is the temperature at which the liquid begins to boil and change to vapor.
The concentration of each solution can be used to determine its boiling point.
Usng the equation of boiling point elevation:
ΔTb = i × Kb × Mwhere:
i is the Van’t Hoff factorKb is the ebullioscopic constantm is the molality of the soluteThe van hoff values for Zn(NO3)2 = 3 , LiBr = 2 and glucose = 1
Siince water is the solvent in all solutions,
Kb of water = 0.512 and boiling point of water = 100°C
Calculating the boiling point elevation:
For Glucose
ΔTb = 1 × 0.512 × 0.120 = 0.061°C
For Zn(NO3)2
ΔTb = 3 × 0.512 × 0.050 = 0.0768°C
For LiBr
ΔTb = 2 × 0.512 × 0.050 = 0.0512°C
LiBr elevates the boiling point of water the least.
Therefore, based on the concentration of the solutions, the rank order of the solutions in terms of decreasing boiling point is 0.050 m LiBr < 0.120 mm glucose < 0.050 m Zn(NO3)2.
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A tank contains 70 kg of salt and 1000 L of water. Water containing 0.4kgL of salt enters the tank at the rate 16 Lmin. The solution is mixed and drains from the tank at the rate 4 Lmin. A(t) is the amount of salt in the tank at time t measured in kilograms.
Answer:
[tex]A(t) = 1600 - e^{-(\frac{t}{250} - 7.33)}[/tex]
Explanation:
A(t) is the amount of salt in the tank at time t, measured in kilograms.
A(0) = 70
[tex]\frac{dA}{dt} =[/tex]rate in - rate out
=0.4* 16 - (A/1000)*4
= [tex]\frac{1600 - A}{250}[/tex] kg/min
[tex]\int\limits {\frac{1}{1600-A} } \, dA = \int\ {\frac{1}{250} } \, dt\\\\ -ln(1600-A) = \frac{t}{250} + C\\\\[/tex]
A(0) = 70
-ln (1600 - 70) = 0/250 + C
-7.33 = C
[tex]-ln(1600-A) = \frac{t}{250} - 7.33\\\\[/tex]
[tex]1600 - A = e^{-(\frac{t}{250} - 7.33)} \\\\A = 1600 - e^{-(\frac{t}{250} - 7.33)}[/tex][tex]1600-A = e^{-(\frac{t}{250} - 7.33)} \\\\ A = 1600 - e^{-(\frac{t}{250} - 7.33)}[/tex]
When the concentrations of CH 3 Br and NaOH are both 0.100 M, the rate of the reaction is 0.0030 M/s. What is the rate of the reaction if the concentration of CH 3 Br is doubled?
Answer : The rate of the reaction if the concentration of [tex]CH_3Br[/tex] is doubled is, 0.006 M/s
Explanation :
Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
The balanced equations will be:
[tex]CH_3Br+NaOH\rightarrow CH_3OH+NaBr[/tex]
In this reaction, [tex]CH_3Br[/tex] and [tex]NaOH[/tex] are the reactants.
The rate law expression for the reaction is:
[tex]\text{Rate}=k[CH_3Br][NaOH][/tex]
As we are given that:
[tex][CH_3Br][/tex] = concentration of [tex]CH_3Br[/tex] = 0.100 M
[tex][NaOH][/tex] = concentration of [tex]NaOH[/tex] = 0.100 M
Rate = 0.0030 M/s
Now put all the given values in the above expression, we get:
[tex]0.0030M/s=k\times (0.100M)\times (0.100M)[/tex]
[tex]k=0.3M^{-1}s^{-1}[/tex]
Now we have to calculate the rate of the reaction if the concentration of [tex]CH_3Br[/tex] is doubled.
[tex]\text{Rate}=k[CH_3Br][NaOH][/tex]
[tex]\text{Rate}=(0.3M^{-1}s^{-1})\times (2\times 0.100M)\times (0.100M)[/tex]
[tex]\text{Rate}=0.006M/s[/tex]
Thus, the rate of the reaction if the concentration of [tex]CH_3Br[/tex] is doubled is, 0.006 M/s
In vacuum, the speed of light is c= 2.998 x 108m/s. However, the speed generally decreases when light travels through media other than vacuum. Then, the speed is approximately c divided by a quantity called the medium's refractive index (7). In which of the following, would the frequency of 210nm-light be the lowest? (c = 2v) (1) H2, n = 1.0001 (2) CCl4, n = 1.461 (3) H20, n = 1.3330 (4) silicone oil, n = 1.520 (5) diamond, n = 2.419
Answer:
Light of wavelength 200 nm will have lowest frequency in while traveling through diamond.
Explanation:
Speed of the light in vacuum = c
Relation between speed of light , wavelength (λ) and frequency (ν);
[tex]\nu =\frac{c}{\lambda }[/tex]
Speed of light in a medium = c' = c/n
[tex]\nu =\frac{c'}{\lambda }=\frac{c}{n\times \lambda }[/tex]...[1]
Where : n = refractive index of a medium
So, the medium with greater value of refractive index lower the speed of light to greater extent.
From [1] , we can see that frequency of light is inversely proportional to the refractive index of the medium :
[tex]\nu \propto \frac{1}{n} [/tex]
This means that higher the value of refractive index lower will be the value of frequency of light in that medium or vice-versa.
According to question, light of wavelength 200 nm will have lowest frequency in while traveling through diamond because refractive index of diamond out of the given mediums is greatest.
Increasing order of refractive indices:
[tex]H_2<H_2O<CCL_4<\text{Silicon oil}< Diamond[/tex]
Decreasing order of frequency of light 210 nm in these medium :
[tex]H_2>H_2O>CCL_4>\text{Silicon oil}> Diamond[/tex]
Pure carbon dioxide (PCO2 = 1 atm) decomposes at high temperature. For the reaction system 2 〖CO〗_2 (g) ⇌2 CO(g)+ O_2 (g) Is this reaction endothermic? Calculate the value of Kp at each temperature.
Answer:
The reaction decomposes more as T increases, therefore it is ENDOTHERMIC, meaning it requires energy to form CO and O₂.
Kp for each specie...
Kp = CO^2 O2 / (CO2)^2
for
T = 1500
Assume 1 atm for CO2, after % dissociation
P-CO2 left = 1*(1-0.048/100)= 0.99952
P-CO formed = 1-0.99952 = 0.00048
P-O2 = (1-0.99952)/2 = 0.00024
so..
Kp = CO^2 O2 / (CO2)^2
Kp = (0.00048^2)(0.00024) / (0.99952^2) = 5.53*10^-11
T = 2500
Assume 1 atm for CO2, after % dissociation
P-CO2 left = 1*(1-17.6/100)= 0.824
P-CO formed = 1-0.824= 0.176
P-O2 = (1-0.176)/2 = 0.088
so..
Kp = CO^2 O2 / (CO2)^2
Kp = (0.176^2)(0.088) / (0.824^2) =0.0040
T = 3000
Assume 1 atm for CO2, after % dissociation
P-CO2 left = 1*(1-54.8/100)= 0.452
P-CO formed = 1-0.452= 0.548
P-O2 = (1-0.452)/2 = 0.274
so..
Kp = CO^2 O2 / (CO2)^2
Kp = (0.548^2)(0.274) / (0.452^2) =0.4027
Explanation:
old blooded animals modulate the fatty acid composition of their membranes as a function of temperature in order to A) ensure consistent membrane fluidity. B) maximize available fatty acids for metabolic use. C) adjust the membrane thickness and increase thermal insulation. D) compensate for decreasing cholesterol solubility. E) all of the above
Answer:
In order to ensure consistent membrane fluidity
Explanation:
Cold Blooded AnimalsCold-blooded animals rely on the temperature of the surrounding environment to maintain its internal temperature, their blood is not cold. Their body temperature fluctuates, based on the external temperature of the environment. If it is 30 °F outside, their body temperature will eventually normalize to 30 °F, as well. If it eventually rises to 120 °F, their body temperature will follow the same pattern to 120 °F.
Membrane fluidityThe cell membrane of cold-blooded animals contains cholesterol, which acts as a shield for the membrane. Membrane fluidity is enhanced by temperature, as the temperature increases membrane fluidity increases and it decreases when the temperature goes down.
Most cold-blooded animals adjust their feeding habit to contain more unsaturated fats from plants. This helps them to maintain their motor coordination and membrane fluidity during the long winter
Cold-blooded animals modulate the fatty acid composition of their membrane to stabilize their membrane fluidity
Old-blooded animals modulate the fatty acid composition of their membranes as a function of temperature to ensure consistent membrane fluidity, and cholesterol plays a role in maintaining appropriate fluidity across a range of temperatures.
Explanation:Old-blooded animals, also known as poikilothermic organisms, modulate the fatty acid composition of their membranes in response to temperature changes. This modulation helps them maintain consistent membrane fluidity, which is crucial for proper cell function. These animals increase the unsaturated fatty acid content of their cell membranes in colder temperatures and increase the saturated fatty acid content in higher temperatures. Additionally, cholesterol, which lies alongside phospholipids in the membrane, acts as a buffer to dampen the effects of temperature on the membrane, extending the range of temperature in which the membrane is appropriately fluid and functional. Cholesterol also serves other functions, such as organizing clusters of transmembrane proteins into lipid rafts.
6.8. Liquid methyl ethyl ketone (MEK) is introduced into a vessel containing air. The system temperature is increased to 55°C, and the vessel contents reach equilibrium with some MEK remaining in the liquid state. The equilibrium pressure is 1200 mm Hg. (a) Use the Gibbs phase rule to determine how many degrees of freedom exist for the system at equilibrium. State the meaning of your result in your own words. (b) Mixtures of MEK vapor and air that contain between 1.8 mole% MEK and 11.5 mole% MEK can ignite and burn explosively if exposed to a flame or spark. Determine whether or not the given vessel constitutes an explosion hazard.
Answer:
A) Degree of Freedom (DF) = 2 and this means that 2 variables are required to determine the state of the system
B) The vessel doesn't constitute an explosion hazard.
Explanation:
A) Degree of Freedom(DF)=2 + C - π
Where;
π = 2 (liquid and gas phase equilibrium)
C = 2 (air and MEK)
Thus, DF = 2 + 2 - 2 = 2
DF of 2 means that 2 variables are required to determine the state of the system
B) Using Antoine Equation;
Log(Pmek) = 6.97421 - (1209.6/(55 + 216)) = 2.51
Log(Pmek) = 2.51
Thus,Pmek = 10^(2.51) = 323.59mm of Hg or approximately 324mm of Hg.
So, fraction of mek in vapour phase = 324/1200 = 0.27 or 27%
Now this fraction is more than 11.5% in the question.
Therefore, the vessel contains no explosion hazard.
The number of degrees of freedom for the MEK and air system in the vessel at equilibrium is 2, meaning two variables can be varied independently without changing the number of phases. Without knowing the concentration of MEK in the vapor phase, we cannot determine whether the mixture is explosively hazardous.
Explanation:(a) The Gibbs phase rule is given by F = C - P + 2, where F is the number of degrees of freedom, C is the number of components and P is the number of phases. Here, we have two components (MEK and air), and two phases (liquid and vapor) existing in the vessel. So, the number of degrees of freedom F = 2 - 2 + 2 = 2. This means that two independent variables (such as temperature and pressure) can be varied independently without changing the number of phases.
(b) The mixture of MEK vapor and air in the vessel is potentially explosive if it contains between 1.8 mole% MEK and 11.5 mole% MEK. However, without knowing the exact concentration of MEK in the vapor phase, it is impossible to determine whether the mixture is explosively hazardous or not.
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Muscle physiologists study the accumulation of lactic acid [CH3CH(OH)COOH] during exercise. Food chemists study its occurrence in sour milk, beer, wine, and fruit. Industrial microbiologists study its formation by various bacterial species from carbohydrates. A biochemist prepares a lactic acid-lactate buffer by mixing 225 mL of 0.85 M lactic acid (Ka = 1.38 × 10−4) with 435 mL of 0.68 M sodium lactate. What is the buffer pH?
Answer:
4.1
Explanation:
First, we will calculate the moles of each species.
Lactic acid: 0.225 L × 0.85 mol/L = 0.19 molLactate: 0.435 L × 0.68 mol/L = 0.30 molThe volume of the mixture is 0.225 L + 0.435 L = 0.660 L
The concentration of the species in the buffer are:
Lactic acid: 0.19 mol/0.660 L = 0.29 MLactate: 0.30 mol/0.660 L =0.45 MWe can find the pH of the buffer using the Henderson-Hasselbach equation.
pH = pKa + log [lactate] / [lactic acid]
pH = -log 1.38 × 10⁻⁴ + log 0.45 M / 0.29 M
pH = 4.1
In acidic aqueous solution, the purple complex ion Co(NH3)5Br2+ undergoes a slow reaction in which the bromide ion is replaced by a water molecule, yielding the pinkish-orange complex ion : Co(NH3)5(H2O)3+
Co(NH3)5Br2+Purple(aq)+H2O(l)? Co(NH3)5(H2O)3+Pinkish?orange(aq)+Br?(aq...
The reaction is first order in Co(NH3)5Br2+, the rate constant at 25 ?C is 6.3�10?6 s?1, and the initial concentration of Co(NH3)5Br2+ is 0.100 M.
A; What is its molarity after a reaction time of 19.0h ?
B; How many hours are required for 69% of the Co(NH3)5Br2+ to react?
A. The molarity after a reaction time of 19.00 hours is 0.0988M.
B. 10.12 hours are required for 69% of the [tex][Co(NH_3)_5]Br^{2+}[/tex].
Given data:
Rate constant (k) is [tex]6.3 * 10^{-6} s^{-1}[/tex].
Initial concentration of [tex][Co(NH_3)_5]Br^{2+}[/tex] is 0.100 M.
The rate law for a first-order reaction is R = k * [A]
Where,
R is the rate of the reaction.
k is the rate constant.
[A] is the concentration of the reactant.
A.
To find the molarity after a reaction time of 19.0 hours,
[tex]ln([A]^t / [A]^o) = -kt\\{[A]^t} = [A]^o * e^{-kt}\\{[A]^t} = 0.100 M * e^{-6.3 * 10^{-6} * (19.0 hours * 3600 s/hour)}\\{[A]^t} = 0.0988 M[/tex]
B.
[tex]ln([A]^t / [A]^0) = -kt\\ln(0.69) = -kt[/tex]
Solve for t:
[tex]t = -ln(0.69) / k[/tex]
Substitute the value of k:
[tex]t = -ln(0.69) / (6.3 * 10^{-6} s^{-1})[/tex]
[tex]t = 36441.49 seconds[/tex]
On converting seconds to hours:
t ≈ 10.12 hours
Therefore,
A. The molarity after a reaction time of 19.00 hours is 0.0988M.
B. The time required to react 69% of reactant is 10.12hours.
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The molarity of Co(NH3)5Br2+ after 19 hours is 0.073 M. Approximately 35.75 hours are required for 69% of the Co(NH3)5Br2+ to react.
Explanation:This question pertains to the concept of chemical kinetics, specifically the first order reaction. The molarity after a certain period can be calculated using the formula for first order kinetics: [A] = [A0]e^-kt, where [A0] is the initial concentration, k is the rate constant, t is time, and [A] is the concentration at time t.
A; After 19.0 hours (or 68400 seconds, converted from hours to seconds), the molarity of Co(NH3)5Br2+ can be calculated as follows: [Co(NH3)5Br2+] = (0.100 M)e^-(6.3×10^-6 s^-1 × 68400 s) = 0.073 M.
B; To find the time when 69% of the Co(NH3)5Br2+ has reacted, you need to calculate the time when 31% of the initial molarity remains (as 100% - 69% = 31%). Hence, using the same equation, you can rearrange to find t = -(1/k)ln([A]/[A0]) = -(1/(6.3×10^-6 s^-1)) ln(0.031/0.100) = Approx. 128,680 seconds, or about 35.75 hours.
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Consider the acid dissociation behavior of carbonic acid, H 2 CO 3 . A p H gradient from 0 to 14 is given. Below a p H equal to p K a 1 which is 6.351, the predominant form is H 2 C O 3. Above a p H equal to p K a 2 which is 10.329, the predominant form is C O 3 2 minus. Between the two p K a values, the predominant form is H C O 3 minus. What is the predominant species present at pH 6.73 ?
Complete Question
The complete question is shown on the first uploaded image
Answer:
The predominate specie is [tex]HCO_3^-[/tex]
Explanation:
From the question we can see that [tex]pk_{a1}(6.351) < pH(6.73) pk_{a2}(10.329)[/tex]
Hence looking at the question we can see that the predominant specie is [tex]HCO_3^-[/tex]
Final answer:
At a pH of 6.73, which is between the pKa₁ (6.351) and pKa₂ (10.329) of carbonic acid, the predominant form of carbonic acid is HCO³⁻ (bicarbonate ion).
Explanation:
The question pertains to the acid dissociation of carbonic acid (H₂CO₃) and which form is predominant at a certain pH. Carbonic acid is a diprotic weak acid, which means it can donate two protons (H⁺), and has two dissociation steps each with different pKa values.
The pKa₁ of carbonic acid is 6.351, and below this pH, the predominant form of carbonic acid is H₂CO₃. The pKa₂ is 10.329, and above this pH, the carbonate ion (CO₃²⁻) is predominant. Between these pKa values, the bicarbonate ion (HCO³⁻) is the predominant species. Therefore, at a pH of 6.73, which lies between the two pKa values, the predominant form is HCO³⁻ (bicarbonate ion).
The concentration of Mg2+ in a sample of coastal seawater is 1.09 g/kg seawater. What is the molarity of Mg2+ in this seawater with a density of 1.02 g/mL?
Answer: 0.0458 mol/L
Explanation:
in the attachment
The molarity of seawater for Mg²⁺ 0.044844 M
Convert the concentration of Mg²⁺ from g/kg to g/L. Since the density of seawater is 1.02 g/mL, 1 kg of seawater is approximately equivalent to 1 L:
Thus concentration of Mg²⁺ becomes 1.09 g/kg = 1.09 g/L.
Calculate the number of moles of Mg²⁺:
The molar mass of Mg is 24.305 g/mol.
Moles of Mg²⁺ = 1.09 g / 24.305 g/mol = 0.044844 mol.
Since we have 1 liter of solution, the molarity is simply the number of moles per liter. Therefore,
Molarity of Mg²⁺ = 0.044844 M.
1) The PPF: Draw a Production Possibilities Frontier (PPF) for the production of two goods, machine tools (a capital good) and donuts (a consumer good). Put machine tools on the vertical axis and donuts on the horizontal axis. Briefly explain what the PPF curve represents.
Answer:
The diagram of the Possibilities Frontier (PPF) is shown on the first uploaded image
A Possibilities Frontier (PPF) can also be referred to as the production possibility curve and it can be defined as a curve that denotes the production of two types of goods in an economy at a given period of time in different proportion of combinations
When this curve shift to the right it means that there is technological advancement in that economy
If this curve a points lies within it then the resources are not fully utilized
Explanation:
An element, X, can form a chloride (XCl3) and an iodide (XI3). The chloride can be converted quantitatively into the iodide when heated and exposed to excess iodine. If 0.760 grams of XCl3 are heated with iodine, 1.610 g of XI3 are produced. What is the chemical symbol for this element
Answer:
La is the chemical symbol for this element.
Explanation:
[tex]2XCl_3+3I_2\rightarrow 2XI_3+3Cl_2[/tex]
Let the molar mass of [tex]XCl_3[/tex] be M.
Let the molar mass of [tex]XI_3[/tex] be M'.
Moles of [tex]XCl_3=\frac{0.760 g}{M}=n[/tex]
Moles of [tex]XI_3=\frac{1.610g}{M'}=n'[/tex]
According to reaction , 2 moles of [tex]XCl_3[/tex] gives 2 moles of [tex]XI_3[/tex], then n moles [tex]XCl_3[/tex] will give:
[tex]\frac{1}{1}\times n=n[/tex] moles of [tex]XI_3[/tex]
[tex]n=n'[/tex]
[tex]\frac{0.760 g}{M}=\frac{1.610g}{M'}[/tex]
Atomic mass of iodine = 127 g/mol
Atomic mass of chlorine = 35.5 g/mol
Atomic mass of X = x
[tex]\frac{0.760 g}{x+3\times 35.5 g/mol}=\frac{1.610g}{x+3\times 127 g/mol}[/tex]
Solving for x:
x = 138.9 g/mol
The value of atomic mass of X corresponds to compound named lanthanum.
La is the chemical symbol for this element.
Answer:
The chemical symbol for this element is La
Explanation:
Step 1: Data given
Mass of XCl3 = 0.760 grams
Mass of XI3 = 1.610 grams
Molar mass Cl = 35.45 g/mol
Step 2: The balanced equation
2XCl3 + 3I2 → 2XI3 + 3Cl2
The mol ratio for XCl3: XI3 = 2:2 or 1:1
Step 3: Calculate moles
Moles = mass / molar mass
Moles XCl3 = mass XCl3 / molar mass XCl3
Moles XCl3 = 0.760 grams / (X + 3*35.45 g/mol)
Moles XI3 = 1.610 grams / (X + 3*126.9 g/mol)
0.760 grams / (X + 3*35.45 g/mol) = 1.610 grams / (X + 3*126.9 g/mol)
0.760 / (X + 106.35) = 1.610 / (X + 380.7)
0.760 (X +380.7) = 1.610 ( X + 106.35)
0.760X + 289.3 = 1.610X + 171.2
118.1 = 0.85X
X = 138.9 g/mol
If we look for an element with atomic mass of 138.9 g/mol we find Lanthanum (La)
2LaCl3 + 3I2 → 2LaI3 + 3Cl2
Enter a balanced equation for the reaction between solid silicon dioxide and solid carbon that produces solid silicon carbide and carbon monoxide gas. Express your answer as a chemical equation. Identify all of the phases in your answer.
Answer:
SiO2(s) + 3C(s) ------> SiC(s) + 2CO(g)
Explanation:
The formula for silicon oxide is SiO2 and carbon is C. silicon carbide is SiC
and carbon monoxide is CO.
An arrow is always used to separate the reactants (left) and products (right).
A balanced equation must contain equal number of atoms in each side of the equation.
For example in the equation above, there are 1 atom of silicon appears on each side; 2 atoms of oxygen and three atoms of carbon.
Answer: SiO₂(s) + 2 C(s) → Si(s) + 2 CO(g)
Explanation:
There are 2 C in the reactants rather than 3. That way the equation is properly balanced.
A certain first-order reaction (A→productsA→products) has a rate constant of 9.90×10−3 s−1s−1 at 45 ∘C∘C. How many minutes does it take for the concentration of the reactant, [A][A] , to drop to 6.25%% of the original concentration? Express your answer with the appropriate units
Answer:
4.66667 minutes
Explanation:
Rate constant, k = 9.90×10−3 s−1
Time = ?
Initial concentration, [A]o = 100
Final concentration, [A] = 6.25
The integral rate law for first order reactions is given as;
ln[A] = ln[A]o − kt
kt = ln[A]o - ln[A]
t = ( ln[A]o - ln[A]) / k
t = [ln(100) - ln(6.25)] / 9.90×10−3
t = 2.77 / 9.90×10−3
t = 0.28006 ×103
t = 280 seconds
t = 4.66667 minutes (Upon conversion by dividing by 60)
A 15.0 g piece of copper wire is heated, and the temperature of the wire changes from 12.0oC to 79.0oC. The amount of heat absorbed is 775 cal. What is the specific heat of copper?
Answer:
The specific heat of the copper is 0.771 cal/ grams °C
Explanation:
Step 1: Data given
Mass of the piece of copper = 15.0 grams
The temperature of the wire changes from 12.0 °C to 79.0 °C
The amount of heat absorbed is 775 cal
Step 2: Calculate the specific heat of copper
Q = m*c*ΔT
⇒with Q = the heat absorbed = 775 cal
⇒with m = the mass of the copper = 15.0 grams
⇒with c = the specific heat of copper = TO BE DETERMINED
⇒with ΔT = The change in temperature = 79.0 °C - 12.0 °C = 67.0 °C
775 cal = 15.0 grams * c * 67.0 °C
c = 0.771 cal/gm °C
The specific heat of the copper is 0.771 cal/ grams °C