Answer:
CO2
Explanation:
Carbon dioxide is a chemical compound that consists of two oxygen atoms and one carbon atom
The titration of a weak acid with a strong base is a common exercise in chemistry labs. We have looked at all of the individual types of calculations needed to determine the pH at any point of a titration. Consider the titration of 0.100 L of 0.300 M acetic acid with 2.00 M NaOH. What type of calculation will we be performing at the following points of the titration curve
Answer:
1. Weak acid calculation
2. Strong base calculation
3. Strong base calculation
4. Strong base calculation
Explanation:
1. Zero equivalents of base added
2. 0.15 equivalents of base added
3. One equivalent of base added
4. 1.75 equivalents of base added
The equivalents of acetic acid are:
0.100L * (0.300mol/L) = 0.0300 moles of acetic acid = 0.0300Eq.
1. With Zero equivalents of base added you will calculate pH using weak acid calculation because you have only acetic acid (Weak acid) in the solution.
2. When 0.15 equivalents of base are added all the acid reacted and 0.15-0.03 = 0.12 equivalents of Strong base are in excess. With these equivalents and the volume we can find pH using strong base calculation.
3 and 4. As with 0.15 eq. added the 1 eq. and 1.75 eq. produce an excess of strong base and the pH must be calculated by using strong base calculation
The main importance of the titration curve is that it is used show how the pH of the solution changes as a titrant is added to the solution.
What is a titration curve?A titration curve is a plot of pH of the system against the volume of titrant added. The question is incomplete but I will try to explain what a titration curve is.
The main importance of the titration curve is that it is used show how the pH of the solution changes as a titrant is added to the solution.
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Perform the following operation and express the answer in scientific notation.
7.296×10² ÷ 9.6×10^-9
Answer:
7.6×10¹⁰
Explanation:
7.296×10²÷9.6×10⁻⁹
To solve such problem,
We group the whole number ans solved seperately and also group the indices and solve the seperately
Step1 : 7.296/9.6 = 0.76
Step 2: applying the law of indices,
10²÷10⁻⁹ = 10⁽²⁺⁹⁾ = 10¹¹
Therefore,
7.296×10²÷9.6×10⁻⁹ = 0.76×10¹¹ = 7.6×10¹⁰
Which one of the following reactions is NOT balanced?
2 CO + O2 + 2 CO2
2 SO2 + O2 +2 SO3
2 KNO3 + 10 K 5 K20 + N2
SF4 + 3 H2O → H2SO3 + 4HF
Answer:
co+ o2+ 2co2 is not balanced reaction
4. If matter goes through a chemical change, are the physical properties likely to stay the same?
•
Please say yes or no and explain why.
Answer: If matter goes through a chemical change then the physical properties are not likely to stay the same.
Explanation:
When chemical composition of a substance changes during a chemical reaction then it is called a chemical change.
Chemical change always leads to the formation of new substances. Properties like chemical reactivity, combustion, rusting etc are chemical changes.
For example, [tex]N_{2} + 3H_{2} \rightarrow 2NH_{3}[/tex]
Here, [tex]NH_{3}[/tex] will have different chemical as well as physical properties as compared to [tex]N_{2}[/tex] and [tex]H_{2}[/tex].
As physical properties are the properties that cause change in state of a substance.
Properties like boiling point, state of substance etc are physical properties.
Thus, we can conclude that if matter goes through a chemical change then the physical properties are not likely to stay the same.
Define mass number of an atom.
Predict the products from reaction of 2-hexyne with the following reagents: (a) 2 equiv Br2 (b) 1 equiv HBr (c) Excess HBr (d) Li in NH3 (e) H2O, H2SO4, HgSO4
Answer:
See explanation and image attached
Explanation:
The reactions of the alkynes involved are shown in the image attached to this answer.
First of all, the reaction of two equivalents of bromine with the alkyne converts it to a saturated compound as shown. One equivalent of HBr converts the alkyne to alkene while excess HBr completely reduces the compound to a saturated compound.
Li/NH3 reduces the alkyne to an alkeneby anti addition to the triple bond.
Reaction of the alkyne with H2O, H2SO4, HgSO4 converts it to an aldehyde as shown.
(S)-Pentan-2-ol was treated sequentially with methanesulfonyl chloride (CH3SO2Cl) and then potassium iodide. What is the final product that forms
Answer:
(S)-Pentan-2-ol was treated sequentially with methanesulfonyl chloride (CH3SO2Cl) and then potassium iodide. What is the final product that forms
Explanation:
Alcohols are poor leaving groups.
To make -OH group a better-leaving group, it should be treated with sulfonyl chlorides.
Then, methane sulfonyl group makes will be substituted on the -OH group and forms sulfonyl esters and makes it a better leaving group.
After that treating with KI proceeds through nucleophilic bimolecular substitution and the final product formed is shown below:.
After standardizing a NaOH solution, you use it to titrate an HCl solution known to have a concentration of 0.203 M. You perform five titrations and obtain the following results: 0.210, 0.204, 0.201, 0.202, and 0.197 M. a) What is the mean
Answer:
0.203 is the mean of the concentration of the HCl solution
Explanation:
You have 5 concentrations. The most appropiate result is the mean of those results. The mean is a statistical defined as the sum of each result divided by the total amount of results. For the results of the problem, the mean is:
0.210 + 0.204 + 0.201 + 0.202 + 0.197 = 1.014 / 5 =
0.203 is the mean of the concentration of the HCl solutionSuppose that you move from a Suppose that you move from a town near the ocean to a town in the mountains. To what atmospheric changes would your body need to adjust? town near the ocean to a town in the mountains. To what atmospheric changes would your body need to adjust?
Answer:
all I can say is town near the ocean atmospheric changes will be cooler, warm, sea breeze, and fresh healthy air. Then when it comes to the mountain lot of change firstly there's a dry air
g The theoretical yield of a certain reaction is 123 g of Al2O3. If the actual yield when the experiment is performed is 0.209 mol Al2O3, what is the percent yield
Answer:
Percent yield = 17.3%
Explanation:
The percent yield is defined as 100 times the ratio between actual yield in grams and theoretical yield in grams (123g).
Percent Yield = Actual Yield / Theoretical yield (123g) * 100
To find actual yield we need to convert the 0.209moles to grams:
Actual yield -Molar mass Al2O3: 101.96g/mol-
0.209mol * (101.96g/mol) = 21.3g
Percent yield = 21.3g / 123g * 100
Percent yield = 17.3%
Question 4 of 30
Scientific research shows that Earth's climate is changing due to human
activities. How can scientific research on climate change help society?
A. It can help us stop storms before they occur.
B. It can help us find a new way to make more water.
C. It can help us track how quickly elements of the climate are
changing
D. It can help us reverse the effects of climate change.
th
C. It can help us track how quickly element of the climate are changing
Calculate the average atomic mass for X
Answer:
39.0229 amu
Explanation:
Hello there!
In this case, according to given information, the idea here is to multiply the percent abundance by the mass number of each isotope and then add them all together as shown below:
[tex]=0.0967*38+0.7868*39+0.1134*40+0.0031*41\\\\=3.6746+30.6852+4.536+0.1271\\\\=39.0229amu[/tex]
Regards!
7.23 One equivalent of sodium methanethiolate is added to an electrophile that has two leaving groups. Which product will be major
The question is incomplete, the complete question is shown in the image attached to this answer
Answer:
A
Explanation:
We can see from the conditions of the reaction that the incoming nucleophile is -SCH3 and there are two possible leaving groups in the substrate.
First of all, we have to look at the conditions of the reaction. We can see that the reaction is taking place in DMF, a polar aprotoc solvent. This condition favours the SN2 synchronous mechanism over the SN1 ionic mechanism.
Hence, the nucleophile at the 1-position is preferentially substituted owing to the conditions of the reaction.
Thus, option A is the major product of the reaction.
Methyl orange can change color by transitioning from one chromophore to another. When added to a clear solution and the solution turns red, it is determined to be a(n) __________ in its __________ stable form.
Answer:
acidic titration in its stable form
Explanation:
Methyl orange can change its color in titration solution. The yellow color is towards alkaline solution and red color is towards acidic solution. The Ph value of solution will change during this chemical process.
Pls pls pls help me
Does Every Chemical Have a formula?
Yes.
Each chemical substance has a specific chemical composition, so these chemical substances have their own chemical formula.
I hope this helps!
How much energy is required to melt 2 kg of gold? Use the table below and this equation.
a. 125.6 kJ
b. 1729 kJ
c. 10.4 kJ
d. 3440kJ
The equation for the energy required to melt 2 kg of gold is 3440 kJ.
What is energy?Energy is the ability to do work or cause change. It is an essential part of everyday life and is present in many forms, such as thermal energy, electrical energy, chemical energy, and mechanical energy. Energy can be converted from one form to another in order to do work.
The equation for calculating the energy required to melt a certain mass of material is Q = m x Lf, where Q is the energy required (in joules), m is the mass of the material (in kilograms), and Lf is the latent heat of fusion (in joules per kilogram).
Using the table below, we can see that the latent heat of fusion for gold is 1760 kJ/kg. Therefore, the equation for the energy required to melt 2 kg of gold is: Q = 2 kg x 1760 kJ/kg = 3440 kJ.
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14 protons,14 electrons and 14 neutrons
the answer is silicon!!
Consider the titration of 30 mL of 0.030 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added: a) 0 mL; b) 10 mL; c) 20 mL; d)35 mL; e) 36 mL; f) 37 mL.
Answer:
a)10.87
b)9.66
c)9.15
d)7.71
e) 5.56
f) 3.43
Explanation:
tep 1: Data given
Volume of 0.030 M NH3 solution = 30 mL = 0.030 L
Molarity of the HCl solution = 0.025 M
Step 2: Adding 0 mL of HCl
The reaction: NH3 + H2O ⇔ NH4+ + OH-
The initial concentration:
[NH3] = 0.030M [NH4+] = 0M [OH-] = OM
The concentration at the equilibrium:
[NH3] = 0.030 - XM
[NH4+] = [OH-] = XM
Kb = ([NH4+][OH-])/[NH3]
1.8*10^-5 = x² / 0.030-x
1.8*10^-5 = x² / 0.030
x = 7.35 * 10^-4 = [OH-]
pOH = -log [7.35 * 10^-4]
pOH = 3.13
pH = 14-3.13 = 10.87
Step 3: After adding 10 mL of HCl
The reaction:
NH3 + HCl ⇔ NH4+ + Cl-
NH3 + H3O+ ⇔ NH4+ + H2O
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.010 L = 0.00025 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.00025 =0.00065 moles
Moles HCl = 0
Moles NH4+ = 0.00025 moles
Concentration at the equilibrium:
[NH3]= 0.00065 moles / 0.040 L = 0.01625M
[NH4+] = 0.00625 M
pOH = pKb + log [NH4+]/[NH3]
pOH = 4.75 + log (0.00625/0.01625)
pOH = 4.34
pH = 9.66
Step 3: Adding 20 mL of HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.020 L = 0.00050 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.00050 =0.00040 moles
Moles HCl = 0
Moles NH4+ = 0.00050 moles
Concentration at the equilibrium:
[NH3]= 0.00040 moles / 0.050 L = 0.008M
[NH4+] = 0.01 M
pOH = pKb + log [NH4+]/[NH3]
pOH = 4.75 + log (0.01/0.008)
pOH = 4.85
pH = 14 - 4.85 = 9.15
Step 4: Adding 35 mL of HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.035 L = 0.000875 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.000875 =0.000025 moles
Moles HCl = 0
Moles NH4+ = 0.000875 moles
Concentration at the equilibrium:
[NH3]= 0.000025 moles / 0.065 L = 3.85*10^-4M
[NH4+] = 0.000875 M / 0.065 L = 0.0135 M
pOH = pKb + log [NH4+]/[NH3]
pOH = 4.75 + log (0.0135/3.85*10^-4)
pOH = 6.29
pH = 14 - 6.29 = 7.71
Step 5: adding 36 mL HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.036 L = 0.0009 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.0009 =0 moles
Moles HCl = 0
Moles NH4+ = 0.0009 moles
[NH4+] = 0.0009 moles / 0.066 L = 0.0136 M
Kw = Ka * Kb
Ka = 10^-14 / 1.8*10^-5
Ka = 5.6 * 10^-10
Ka = [NH3][H3O+] / [NH4+]
Ka =5.6 * 10^-10 = x² / 0.0136
x = 2.76 * 10^-6 = [H3O+]
pH = -log(2.76 * 10^-6)
pH = 5.56
Step 6: Adding 37 mL of HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.037 L = 0.000925 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.000925 =0 moles
Moles HCl = 0.000025 moles
Concentration of HCl = 0.000025 moles / 0.067 L = 3.73 * 10^-4 M
pH = -log 3.73*10^-4= 3.43
The pH of the solution in the titration of 30 mL of 0.030 M NH₃ with 0.025 M HCl, is:
a) pH = 10.86
b) pH = 9.66
c) pH = 9.15
d) pH = 7.70
e) pH = 5.56
f) pH = 3.43
Calculating the pH a) 0 mL
Initially, the pH of the solution is given by the dissociation of NH₃ in water.
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (1)
The constant of the above reaction is:
[tex] Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]} = 1.76\cdot 10^{-5} [/tex] (2)
At the equilibrium, we have:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (3)
0.030 M - x x x
[tex] 1.76\cdot 10^{-5}*(0.030 - x) - x^{2} = 0 [/tex]
After solving for x and taking the positive value:
x = 7.18x10⁻⁴ = [OH⁻]
Now, we can calculate the pH of the solution as follows:
[tex] pH = 14 - pOH = 14 + log(7.18\cdot 10^{-4}) = 10.86 [/tex]
Hence, the initial pH is 10.86.
b) 10 mL
After the addition of HCl, the following reaction takes place:
NH₃ + HCl ⇄ NH₄⁺ + Cl⁻ (4)
We can calculate the pH of the solution from the equilibrium reaction (3).
[tex] 1.76\cdot 10^{-5}(Cb - x) - (Ca + x)*x = 0 [/tex] (5)
Finding the number of moles of NH₃ and NH₄⁺
The number of moles of NH₃ (nb) and NH₄⁺ (na) are given by:
[tex] n_{b} = n_{i} - n_{HCl} [/tex] (6)
[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.010 L = 6.5\cdot 10^{-4} moles [/tex]
[tex] n_{a} = n_{HCl} [/tex] (7)
[tex] n_{a} = 0.025 mol/L*0.010 L = 2.5 \cdot 10^{-4} moles [/tex]
Calculating the concentrations of NH₃ and NH₄⁺The concentrations are given by:
[tex] Cb = \frac{6.5\cdot 10^{-4} moles}{(0.030 L + 0.010 L)} = 0.0163 M [/tex] (8)
[tex] Ca = \frac{2.5 \cdot 10^{-4} mole}{(0.030 L + 0.010 L)} = 6.25 \cdot 10^{-3} M [/tex] (9)
Calculating the pHAfter entering the values of Ca and Cb into equation (5) and solving for x, we have:
[tex] 1.76\cdot 10^{-5}(0.0163 - x) - (6.25 \cdot 10^{-3} + x)*x = 0 [/tex]
x = 4.54x10⁻⁵ = [OH⁻]
Then, the pH is:
[tex] pH = 14 + log(4.54\cdot 10^{-5}) = 9.66 [/tex]
Hence, the pH is 9.66.
c) 20 mLWe can find the pH of the solution from the reaction of equilibrium (3).
Calculating the concentrations of NH₃ and NH₄⁺The concentrations are (eq 8 and 9):
[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 8.0\cdot 10^{-3} M [/tex]
[tex] Ca = \frac{0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 0.01 M [/tex]
Calculating the pHAfter solving the equation (5) for x, we have:
[tex] 1.76\cdot 10^{-5}(8.0\cdot 10^{-3} - x) - (0.01 + x)*x = 0 [/tex]
x = 1.40x10⁻⁵ = [OH⁻]
Then, the pH is:
[tex] pH = 14 + log(1.40\cdot 10^{-5}) = 9.15 [/tex]
So, the pH is 9.15.
d) 35 mLWe can find the pH of the solution from reaction (3).
Calculating the concentrations of NH₃ and NH₄⁺[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 3.85\cdot 10^{-4} M [/tex]
[tex] Ca = \frac{0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 0.0135 M [/tex]
Calculating the pHAfter solving the equation (5) for x, we have:
[tex] 1.76\cdot 10^{-5}(3.85\cdot 10^{-4} - x) - (0.0135 + x)*x = 0 [/tex]
x = 5.013x10⁻⁷ = [OH⁻]
Then, the pH is:
[tex] pH = 14 + log(5.013\cdot 10^{-7}) = 7.70 [/tex]
So, the pH is 7.70.
e) 36 mL Finding the number of moles of NH₃ and NH₄⁺[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.036 L = 0 [/tex]
[tex] n_{a} = 0.025 mol/L*0.036 L = 9.0 \cdot 10^{-4} moles [/tex]
Since all the NH₃ reacts with the HCl added, the pH of the solution is given by the dissociation reaction of the NH₄⁺ produced in water.
At the equilibrium, we have:
NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺
Ca - x x x
[tex] Ka = \frac{x^{2}}{Ca - x} [/tex]
[tex] Ka(Ca - x) - x^{2} = 0 [/tex] (10)
Calculating the acid constant of NH₄⁺
We can find the acid constant as follows:
[tex] Kw = Ka*Kb [/tex]
Where Kw is the constant of water = 10⁻¹⁴
[tex] Ka = \frac{1\cdot 10^{-14}}{1.76 \cdot 10^{-5}} = 5.68 \cdot 10^{-10} [/tex]
Calculating the pH
The concentration of NH₄⁺ is:
[tex] Ca = \frac{9.0 \cdot 10^{-4} moles}{(0.030 L + 0.036 L)} = 0.0136 M [/tex]
After solving the equation (10) for x, we have:
x = 2.78x10⁻⁶ = [H₃O⁺]
Then, the pH is:
[tex] pH = -log(H_{3}O^{+}) = -log(2.78\cdot 10^{-6}) = 5.56 [/tex]
Hence, the pH is 5.56.
f) 37 mLNow, the pH is given by the concentration of HCl that remain in solution after reacting with NH₃ (HCl is in excess).
Calculating the concentration of HCl
[tex] C_{HCl} = \frac{0.025 mol/L*0.037 L - 0.030 mol/L*0.030 L}{(0.030 L + 0.037 L)} = 3.73 \cdot 10^{-4} M = [H_{3}O^{+}] [/tex]
Calculating the pH
[tex] pH = -log(H_{3}O^{+}) = -log(3.73 \cdot 10^{-4}) = 3.43 [/tex]
Therefore, the pH is 3.43.
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In a closed system, If a gas is transported to a container with double the volume of the previous container, the gas was held in, what is the gases' new volume?
The volume of the gas is fixed and will not change.
The volume of the gas will be half the original volume.
The volume of the gas will be the original volume squared.
The volume of the gas will be double the original volume.
Answer:
The volume of the gas is fixed and will not change.
Explanation:
The volume of the gas will not change because there is no change in temperature. Temperature increases the volume of gases enclosed in a container.
Determine the theoretical yield and the percent yield if 21.8 g of K2CO3 is produced from reacting 27.9 g KO2 with 57.0 g CO2. The molar mass of KO2
Answer:
26.9 g
81%
Explanation:
The equation of the reaction is;
4 KO2(s) + 2 CO2(g) → 3 O2(g) + 2 K2CO3(s)
Number of moles of KO2= 27.9g/71.1 g/mol = 0.39 moles
4 moles of KO2 yields 2 moles of K2CO3
0.39 moles of KO2 yields 0.39 × 2/4 = 0.195 moles of K2CO3
Number of moles of CO2 = 57g/ 44.01 g/mol = 1.295 moles
2 moles of CO2 yields 2 moles of K2CO3
1.295 moles of CO2 yields 1.295 × 2/2 = 1.295 moles of K2CO3
Hence the limiting reactant is KO2
Theoretical yield = 0.195 moles of K2CO3 × 138.205 g/mol = 26.9 g
Percent yield = actual yield/theoretical yield × 100
Percent yield = 21.8/26.9 × 100
Percent yield = 81%
What intermolecular force or bond is primarily responsible for the solubility of chlorine (Cl2) in water
The question is incomplete, the common question is;
What intermolecular attractive force is primarily responsible for the solubility of chlorine, Cl2, in water?
a. dipole - dipole
b. hydrogen bonding
c. dipole-induced dipole
d. ion-dipole
e. ion-induced dipole
Answer:
dipole-induced dipole
Explanation:
We have to remember that water is a polar molecule hence it possesses a dipole moment. Its dipole moment is responsible for the ability of water to dissolve many substances.
On the other hand, Cl2 is a nonpolar molecule bound together by only weak dispersion forces.
Recall that dispersion forces involve transient appearance of a dipole in a molecule.
Water molecules can induce a dipole in Cl2 thereby causing the both molecules to interact and Cl2 dissolves in water.
EXTRACTION OF CAFFEINE 1.Explain the reason each step of the separation is performed with three portions of the solvent rather than with a single portion of solvent. (e.g. Done in triplicate.) (0.5 pt) 2.Why should a separatory funnel not be vigorously shaken
Answer:
Throughout the explanations section below you will find a description of the question.
Explanation:
(1)
Whether a solution would be positioned inside a separative funnel, combined water, as well as solvent, disintegrate particulate caffeine. In every stage, the caffeine content incorporated relies upon the coefficient of caffeine partitioning throughout the combination of water as well as fluid.Thus, increasingly caffeine is taken from the solvent whenever the moment you bring additional solvent. Consequently, we separate the solvent from the single component.(2)
For compounds to be mixed thoroughly and separated into different layers, a shuddering mixture within the dividing funnel would be essential.However, it vibrates the separation funnel forcefully, restricts airflow within the funnel, which can also induce the fluid under it to burst or causing fluid to fire.Using the balanced equation for the combustion of ethane: 2C2H6 + 7O2 → 4CO2 + 6H2O, how many moles of O2 needed to produce 12 moles of H2O?
Answer:
14 moles of oxygen needed to produce 12 moles of H2O.
Explanation:
We are given that balance eqaution
[tex]2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O[/tex]
We have to find number of moles of O2 needed to produce 12 moles of H2O.
From given equation
We can see that
6 moles of H2O produced by Oxygen =7 moles
1 mole of H2O produced by Oxygen=[tex]\frac{7}{6}[/tex]moles
12 moles of H2O produced by Oxygen=[tex]\frac{7}{6}\times 12[/tex]moles
12 moles of H2O produced by Oxygen=[tex]7\times 2[/tex]moles
12 moles of H2O produced by Oxygen=14 moles
Hence, 14 moles of oxygen needed to produce 12 moles of H2O.
The amount of oxygen required for the combustion of ethane to produce 12 moles of water is 14 moles.
How are the moles produced in reaction calculated?The moles of oxygen produced in the reaction can be given from the stoichiometric law of the balanced chemical equation.
The balanced chemical equation for the combustion of ethane is:
[tex]\rm 2\;C_2H_6\;+\;7\;O_2\;\rightarrow\;4\;CO_2\;+\;6\;H_2O[/tex]
The 6 moles of water are produced from 7 moles of oxygen. The moles of oxygen required to produce 12 moles of water are:
[tex]\rm 6\;mol\;H_2O=7\;mol\;Oxygen\\12\;mol\;H_2O=\dfrac{7}{6}\;\times\;12\;mol\;O_2\\ 12\;mol\;H_2O=14\;mol\;O_2[/tex]
The moles of oxygen required to produce 12 moles of water are 14 moles.
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The pKa of an acid can be determined through _____ (reduction, titration, filtration) with a strong base.
Gradually increase the volume of the base, stopping _____ (before, as, after) the equivalence point is reached.
The pKa of the acid is equal to the pH at the _____ (equivalence point, midway volume to the equivalence point, maximum volume).
Answer:
The pKa of an acid can be determined through titration with a strong base.
Gradually increase the volume of the base, stopping before the equivalence point is reached.
The pKa of the acid is equal to the pH at the midway volume to the equivalence point.
Explanation:
An acid HA dissociates in water as follows:
HA ⇄ H⁺ + A⁻ Ka
So, it produces hydrogen ions (H⁺) and a conjugate base (A⁻). The concentrations of HA, H⁺ and A⁻ at equilibrium determine the constant Ka. The pKa is calculated as:
pKa = -log Ka
The relationship between the pH of the solution and the pKa of the acid is described by the Henderson-Hasselbalch equation:
pH = pKa + log ([A⁻]/[HA])
The pKa can be experimentally determined by acid-base titration, in which a strong base is added to the acid solution. As the base is added, the acid HA is neutralized and the conjugate base A⁻ is formed. Thus, the concentration of the acid ([HA]) increases and the concentration of the conjugate base ([A⁻] decreases. The equivalence point is reached when the total amount of acid is neutralized with the added base. Before reaching the equivalence point, at the halfway point, half of the acid is neutralized and converted into the conjugate base. Thus:
[A-] = [HA] ⇒ log [A-]/[HA] = log 1 = 0 ⇒ pH = pKa
We measure the pH at that point and it is equal to the pKa of the acid.
What quantity of heat is transferred when a 150.0g block of iron metal is heated from 25.0°C to 73.3°C? What is the direction of the heat flow?
The direction of heat flow is increased which means blocks temperature is higher and hotter than it was before
Na2CO3 reacts with dil.HCl to produce NaCl, H2O and CO2. If 21.2 g of pure Na2CO3 are added in a solution containing 21.9g HCl , a. Find the limiting reagent. (2) b. Calculate the number of moles of excess reagent left over.(2) c. Calculate the number of molecules of H2O formed.(1) d. Calculate volume of CO2 gas produced at 270C and 760mm Hg pressure.(2) e. Write significance of limiting reagent
Answer:
See explanation
Explanation:
Equation of the reaction;
Na2CO3(aq) + 2HCl(aq) -------> 2NaCl(aq) + H2O(l) + CO2(g)
Number of moles of Na2CO3 = 21.2g/106g/mol = 0.2 moles Na2CO3
Number of moles of HCl = 21.9g/36.5g/mol = 0.6 moles of HCl
1 mole of Na2CO3 reacts with 2 moles of HCl
0.2 moles of Na2CO3 reacts with 0.2 × 2/1 = 0.4 moles of HCl
Hence Na2CO3 is the limiting reactant
Since there is 0.6 moles of HCl present, the number of moles of excess reagent=
0.6 moles - 0.4 moles = 0.2 moles of HCl
1 mole of Na2CO3 forms 1 mole of water
0.2 moles of Na2CO3 forms 0.2 moles of water
Number of molecules of water formed = 0.2 moles × 6.02 × 10^23 = 1.2 × 10^23 molecules of water
1 mole of Na2CO3 yields 1 mole of CO2
0.2 moles of Na2CO3 yields 0.2 moles of CO2
1 mole of CO2 occupies 22.4 L
0.2 moles of CO2 occupies 0.2 × 22.4 = 4.48 L at STP
Hence;
V1=4.48 L
T1 = 273 K
P1= 760 mmHg
T2 = 27°C + 273 = 300 K
P2 = 760 mmHg
V2 =
P1V1/T1 = P2V2/T2
P1V1T2 = P2V2T1
V2 = P1V1T2/P2T1
V2 = 760 × 4.48 × 300/760 × 273
V2= 4.9 L
The limiting reactant is the reactant that determines the amount of product formed in a reaction. When the limiting reactant is exhausted, the reaction stops.
Be sure to answer all parts.
Calculate the molarity of each of the following solutions:
(a) 28.5 g of ethanol (C2H5OH) in 4.50 x 102 mL of solution
M M
es
(b) 21.6 g of sucrose (C12H22011) in 67.0 mL of solution
M
(c) 6.65 g of sodium chloride (NaCl) in 96.2 mL of solution
M
Answer: The molarity of each of the given solutions is:
(a) 1.38 M
(b) 0.94 M
(c) 1.182 M
Explanation:
Molarity is the number of moles of a substance present in liter of a solution.
And, moles is the mass of a substance divided by its molar mass.
(a) Moles of ethanol (molar mass = 46 g/mol) is as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{28.5 g}{46 g/mol}\\= 0.619 mol[/tex]
Now, molarity of ethanol solution is as follows.
[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.619 mol}{4.50 \times 10^{2} \times 10^{-3}L}\\= 1.38 M[/tex]
(b) Moles of sucrose (molar mass = 342.3 g/mol) is as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{21.6 g}{342.3 g/mol}\\= 0.063 mol[/tex]
Now, molarity of sucrose solution is as follows.
[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.063 mol}{0.067 L} (1 mL = 0.001 L)\\= 0.94 M[/tex]
(c) Moles of sodium chloride (molar mass = 58.44 g/mol) are as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{6.65 g}{58.44 g/mol}\\= 0.114 mol[/tex]
Now, molarity of sodium chloride solution is as follows.
[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.114 mol}{0.0962 L}\\= 1.182 M[/tex]
Thus, we can conclude that the molarity of each of the given solutions is:
(a) 1.38 M
(b) 0.94 M
(c) 1.182 M
A student dropped a piece of nickel metal into a solution of HCl(aq). He observed the formation of gas bubbles and collected the gas into another test tube. The student performed a splint test and observed that the splint flared up when he placed the splint into the test tube of the gas. What can be said about the results of this students experiment?
a. The student performed the splint test incorrectly. He should of observed a popping sound when the splint was placed in the test tube.
b. The experiment was performed incorrectly. Nickel doesn't react with HCl. Therefore, the student picked up the wrong metal when conducting the experiment.
c. The student completed the experiment correctly and there were no errors in the experiment.
d. The student performed the splint test incorrectly. He should of observed the flame being extinguished when the splint was placed in the test tube.
Answer:
a. The student performed the splint test incorrectly. He should of observed a popping sound when the splint was placed in the test tube.
Explanation:
It is given that a student performed an experiment where he dropped a nickel metal in to HCl solution. He observed the reaction and performed a splint test in the test tube that is filled with a gas which is formed while Nickle is dropped into the solution of HCl.
But the experiment that the student performed was incorrect. He must have observed the popping sound when the splint was placed in the test tube.
When the splint was added to the gas splint flared up. The hydrogen gas pops out when exposed to the flame.
[tex]$Ni + HCl(aq) = NiCl + H_2$[/tex]
Thus the correct option is (a).
If 0.21J of heat cause a 0.308 degree C temperature change, what mass of water is present?
a 0.0702 g
b 0.00540 g
c 0.163 g
d 18.4 g
Answer:
The correct answer is Option c (0.163 g).
Explanation:
Given:
Heat energy,
Q = 0.21 J
Specific heat,
c = 4.184 J/g°c
Change in temperature,
ΔT = 0.308°C
As we know,
⇒ [tex]Q=mc \Delta T[/tex]
By substituting the values, we get
[tex]0.21=m\times 4.184\times 0.308[/tex]
[tex]m=\frac{0.21}{0.308\times 4.184}[/tex]
[tex]=\frac{0.21}{1.28867}[/tex]
[tex]=0.163 \ g[/tex]
