The question is incomplete. The complete question is :
Hydrogen [tex](H_2)[/tex] gas and oxygen [tex](O_2)[/tex] gas react to form water vapor [tex](H_2O)[/tex]. Suppose you have 11.0 mol of [tex]H_2[/tex] and 13.0 mol of [tex]O_2[/tex] in a reactor. Calculate the largest amount of [tex]H_2O[/tex] that could be produced. Round your answer to the nearest 0.1 mol .
Solution :
The balanced reaction for reaction is :
[tex]$2H_2(g) \ \ \ \ + \ \ \ \ \ O_2(g)\ \ \ \rightarrow \ \ \ \ 2H_2O(g)$[/tex]
11.0 13.0
11/2 13/1 (dividing by the co-efficient)
6.5 mol 13 mol (minimum is limiting reagent as it is completely consumed during the reaction)
Therefore, [tex]H_2[/tex] is limiting reagent. It's stoichiometry decides the product formation amount from equation above it is clear that number of moles for [tex]H_2O[/tex] will be produced = number of moles of [tex]H_2[/tex]
= 11.0 mol
Sofia orders a spare part for her custom-built bike from Oregon Technologies Inc. The company makes use of a computer-aided design model to produce the spare part at its location closest to Sofia's home. In this case, which of the following technologies is used to produce the spare part?
a. Molding
b. Additive manufacturing
c. Lenticular printing
d. Tampography
Answer:
b. Additive manufacturing
Explanation:
Additive manufacturing is defined as that manufacturing process where light parts and components are being developed or manufactured in 3D form by adding materials to it.
It is a process of adding materials to produce the final product. It is also known as 3D printing.
In the context, Oregon Technologies Inc. uses computer-aided design model in order to manufacture a spare part required by Sofia for her custom made bike by using a process called additive manufacturing.
Thus the correct option is (b).
The target compound that you should synthesize is 3-chloro-1-butene. Again, this is an electrophilic alkene addition reaction.Examine the product to determine the location of the new functionality. Keep in mind the nature of the intermediate. The regioselectivity is controlled by the stability of this intermediate. Assume that only one equivalent of reagent is used.
Required:
State the starting agents, solvents, and products. What is the main reaction and mechanism? What are the TLC values?
Answer:
Attached below
Explanation:
The starting agents : attached below
There is no Solvent required to carry out this electrophilic alkene addition reaction
The products are : attached below ( Cl )
The TLC values can only be determined by carrying out the experiment in the laboratory ( i.e. it is an experimental observation )
Attached below is the Mechanism showing the starting agents and products
you want to remove as much CO2 gas as possible from a water solution. Which of the following treatments would be most effective?
Answer:
Aerate solution
Explanation:
aerate solution is the best way to remove CO2 from water (Carbon dioxide in the water that does not form bicarbonates is “uncombined” and can be removed by aeration).
Arrange the following compounds in order of increasing reactivity (least reactive first.) to electrophilic aromatic substitution:.
Bromobenzene Nitrobenzene Benzene Phenol
a. Bromobenzene < Nitrobenzene < Benzene < Phenol
b. Nitrobenzene < Bromobenzene < Benzene < Phenol
c. Phenol < Benzene < Bromobenzene < Nitrobenzene
d. Nitrobenzene < Benzene < Bromobenzene < Phenol
Answer:
Nitrobenzene < Bromobenzene < Benzene < Phenol
Explanation:
Aromatic compounds undergo electrophilic aromatic substitution reaction in the presence of relevant electrophiles. Certain substituents tend to increase or decrease the tendency of an aromatic compound towards electrophilic aromatic substitution reaction.
Substituents that increase the electron density around the ring such as in phenol tends to make the ring more reactive towards electrophilic substitution. Halogens such as bromine has a -I inductive effect as well as a +M mesomeric effect.
However the -I(electron withdrawing effect) of the halogens supersedes the +M electron donation due to mesomeric effect.
Putting all these together, the order of increasing reactivity of the compounds towards electrophilic aromatic substitution is;
Nitrobenzene < Bromobenzene < Benzene < Phenol
The shape of a molecule is determined by:
A. All of these
B. The number of electron clouds around the atom.
C. The number of bonds.
D. Mutual repulsion between electrons.
A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.60 mL of a 0.400 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
A beaker with 120mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1M. A student adds 6.60mL of a 0.300M HCl solution to the beaker. How much will the pH change?
The pKa of acetic acid is 4.76.
Chemistry Buffer Calculations
1 Answer
Stefan V.
May 8, 2016
Δ
pH
=
0.29
Explanation:
!! LONG ANSWER !!
The idea here is that you need to use the Henderson-Hasselbalch equation to determine the ratio that exists between the concentration of the weak acid and of its conjugate base in the buffer solution.
Once you know that, you can use the total molarity of the acid and of the conjugate base to find the number of moles of these two chemical species present in the buffer.
So, the Henderson-Hasselbalch equation looks like this
∣
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
pH
=
p
K
a
+
log
(
[
conjugate base
]
[
weak acid
]
)
a
a
∣
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
In your case, you have acetic acid,
CH
3
COOH
, as the weak acid and the acetate anion,
CH
3
COO
−
, as its conjugate base. The
p
K
a
of the acid is said to be equal to
4.76
, which means that you have
pH
=
4.76
+
log
(
[
CH
3
COO
−
]
[
CH
3
COOH
]
)
The pH is equal to
5
, and so
5.00
=
4.76
+
log
(
[
CH
3
COO
−
]
[
CH
3
COOH
]
)
log
(
[
CH
3
COO
−
]
[
CH
3
COOH
]
)
=
0.24
This will be equivalent to
10
log
(
[
CH
3
COO
−
]
[
CH
3
COOH
]
)
=
10
0.24
which will give you
[
CH
3
COO
−
]
[
CH
3
COOH
]
=
1.74
This means that your buffer contains
1.74
times more conjugate base than weak acid
[
CH
3
COO
−
]
=
1.74
×
[
CH
3
COOH
]
Now, because both chemical species share the same volume,
120 mL
, this can be rewritten as
n
C
H
3
C
O
O
−
120
⋅
10
−
3
L
=
1.74
×
n
C
H
3
C
O
O
H
120
⋅
10
−
3
L
which is
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
n
C
H
3
C
O
O
−
=
1.74
×
n
C
H
3
C
O
O
H
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
(
1
)
So, the buffer contains
1.74
times more moles of acetate anions that of acetic acid.
Now, the total molarity of the buffer is said to be equal to
0.1 M
. You thus have
[
CH
3
COOH
]
+
[
CH
3
COO
−
]
=
0.10 M
Once again, use the volume of the buffer to write
n
C
H
3
C
O
O
H
120
⋅
10
−
3
L
+
n
C
H
3
C
O
O
−
120
⋅
10
−
3
L
=
0.1
moles
L
This will be equivalent to
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
n
C
H
3
C
O
O
−
+
n
C
H
3
C
O
O
H
=
0.012
a
a
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
(
2
)
Use equations
(
1
)
and
(
2
)
to find how many moles of acetate ions you have in the buffer
1.74
⋅
n
C
H
3
C
O
O
H
+
n
C
H
3
C
O
O
H
=
0.012
n
C
H
3
C
O
O
H
=
0.012
1.74
+
1
=
0.004380 moles CH
3
COOH
This means that you have
n
C
H
3
C
O
O
−
=
1.74
⋅
0.004380 moles
n
C
H
3
C
O
O
−
=
0.007621 moles CH
3
COO
−
Now, hydrochloric acid,
HCl
, will react with the acetate anions to form acetic acid and chloride anions,
Cl
−
H
Cl
(
a
q
)
+
CH
3
COO
−
(
a
q
)
→
CH
3
COO
H
(
a
q
)
+
Cl
−
(
a
q
)
Notice that the reaction consumes hydrochloric acid and acetate ions in a
1
:
1
mole ratio, and produces acetic acid in a
1
:
1
mole ratio.
Use the molarity and volume of the hydrochloric acid solution to determine how many moles of strong acid you have
∣
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
c
=
n
solute
V
solution
⇒
n
solute
=
c
⋅
V
solution
a
a
∣
∣
∣
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
In your case, this gets you
n
H
C
l
=
0.300 mol
L
−
1
⋅
volume in liters
6.60
⋅
10
−
3
L
n
H
C
l
=
0.001980 moles HCl
The hydrochloric acid will be completely consumed by the reaction, and the resulting solution will contain
n
H
C
l
=
0 moles
→
completely consumed
n
C
H
3
C
O
O
−
=
0.007621 moles
−
0.001980 moles
=
0.005641 moles CH
3
COO
−
n
C
H
3
C
O
O
H
=
0.004380 moles
+
0.001980 moles
=
0.006360 moles CH
3
COOH
The total volume of the solution will now be
V
total
=
120 mL
+
6.60 mL
=
126.6 mL
The concentrations of acetic acid and acetate ions will be
[
CH
3
COOH
]
=
0.006360 moles
126.6
⋅
10
−
3
L
=
0.05024 M
[
CH
3
COO
−
]
=
0.005641 moles
126.6
⋅
10
−
3
L
=
0.04456 M
Use the Henderson-Hasselbalch equation to find the new pH of the solution
pH
=
4.76
+
log
(
0.04456
M
0.05024
M
)
pH
=
4.71
Therefore, the pH of the solution decreased by
Δ
pH
=
|
4.71
−
5.00
|
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
0.29 units
a
a
∣
∣
−−−−−−−−−−−−−
Answer link
Related topic
Buffer Calculations
Questions
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What is the volume of the fluid in the graduated cylinder measured to the correct degree of precision?
37.22 mL
38.05 mL
37.0 ml
37.8 ml
Answer:
37.0. gsgggsgsghddhhdd
Low-density polyethylene is formed because _______ polymerization is very unpredictable and difficult to control.
dehydration-condensation
anionic-initiated
radical-initiated
esterification
Answer:
radical-initiated
Explanation:
Radical-initiated polymerization is unpredictable and difficult to control. The reaction proceeds indiscriminately and produces shortened chains, loops, and branches that create holes in the polymer. This reduces its mass to volume ratio.
What volume (in liters) of a solution contains 0.14 mol of KCl?
1.8 M KCl
Express your answer using two significant figures.
Answer:
[tex]\boxed {\boxed {\sf 0.078 \ L }}[/tex]
Explanation:
We are asked to find the volume of a solution given the moles of solute and molarity.
Molarity is a measure of concentration in moles per liter. It is calculated using the following formula:
[tex]molarity= \frac{moles \ of \ solute}{liters \ of \ solution}[/tex]
We know there are 0.14 moles of potassium chloride (KCl), which is the solute. The molarity of the solution is 1.8 molar or 1.8 moles of potassium chloride per liter.
moles of solute = 0.14 mol KCl molarity= 1.8 mol KCl/ Lliters of solution=xSubstitute these values/variables into the formula.
[tex]1.8 \ mol \ KCl/ L = \frac { 0.14 \ mol \ KCl}{x}[/tex]
We are solving for x, so we must isolate the variable. First, cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.
[tex]\frac {1.8 \ mol \ KCl/L}{1} = \frac{0.14 \ mol \ KCl}{x}[/tex]
[tex]1.8 \ mol \ KCl/ L *x = 1*0.14 \ mol \ KCl[/tex]
[tex]1.8 \ mol \ KCl/ L *x = 0.14 \ mol \ KCl[/tex]
Now x is being multiplied by 1.8 moles of potassium chloride per liter. The inverse operation of multiplication is division, so we divide both sides by 1.8 mol KCl/L.
[tex]\frac {1.8 \ mol \ KCl/ L *x}{1.8 \ mol \ KCl/L} = \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}[/tex]
[tex]x= \frac{0.14 \ mol \ KCl}{1.8 \ mol \ KCl/L}[/tex]
The units of moles of potassium chloride cancel.
[tex]x= \frac{0.14 }{1.8 L}[/tex]
[tex]x=0.07777777778 \ L[/tex]
The original measurements of moles and molarity have 2 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in the ten-thousandth place tells us to round the 7 up to a 8.
[tex]x \approx 0.078 \ L[/tex]
There are approximately 0.078 liters of solution.
Different steps of the oxidative decarboxylation of pyruvate by the pyruvate dehydrogenase PDH complex are given. Place these five steps in the correct order. Note that thiamine pyrophosphate, TPP, is sometimes called thiamine diphosphate, TDP.
1. FADH2 is reoxidized to FAD reducing NAD* to NADH.
2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.
3. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.
4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.
5. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.
Answer:
1. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.
2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.
3. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.
4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.
5. FADH2 is reoxidized to FAD reducing NAD+ to NADH.
Explanation:
The oxidation of pyruvate to AcetylCoA is catalyzed by the pyruvate dehydrogenase complex. The reaction is an irreversible oxidative decarboxylation process in which the carboxyl group of pyruvate is removed as a molecule of carbon dioxide, CO₂, while the remaining two carbons are attached to a CoASH molecule to form acetylCoA.
The pyruvate dehydrogenase complex contains three enzymes - Pyruvate dehydrogenase known as E₁, dihydrolipoyl transacetylase known as E₂, and dihydrolipoyl dehydrogenase known as E₃. It also requires five coenzymes namely: thiamine pyrophosphate (TPP), flavine adenine dinucleotide (NAD), coenzyme A (CoA-SH), nicotinamide adenine dinucleotide (NAD) and lipoate.
Oxidative decorbyxylation of pyruvate takes place in the pyruvate dehydrogenase complex in five steps:
1. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.
2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.
3. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.
4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.
5. FADH2 is reoxidized to FAD reducing NAD+ to NADH.
0. When measuring tert-butyl alcohol for this experiment, a student first weighs an empty graduated cylinder, then pours 15 mL of the alcohol into the graduated cylinder and weighs the cylinder again. He records the amount of alcohol used as the difference in these two masses. What is wrong with this method
Answer:
Both have solutions in the graduated cylinder.
Explanation:
Recording the amount of alcohol used as the difference between two masses is the wrong method used for measuring tert-butyl alcohol for the experiment. For measuring tert-butyl alcohol for this experiment, the student has to measure the two masses when both the graduated cylinders has solution of tert-butyl alcohol not when one of it is empty (having no tert-butyl alcohol ).
The wrong aspect is that the liquid didn't need to be weighed. Instead the volume should have been recorded with the aid of the graduated cylinder.
What is a Graduated cylinder?This is a cylinder with marked readings and is used to measure the volume of liquids in the laboratory.
The graduated cylinder will accurately measure the amount of alcohol used due to it being volatile and the mass fluctuating during the measurement.
Read more about Graduated cylinder here https://brainly.com/question/24869562
Nicotine is a toxic substance present in tobacco leaves. There are two lone pairs in the structure of nicotine. In general, localized lone pairs are much more reactive than delocalized lone pairs. With this information in mind, do you expect both lone pairs in nicotine to be reactive?
A. Both lone pairs are delocalized and, therefore, both are expected to have the same reactivity.
B.Lone pair in pyrrolidine ring is localized and, therefore, is expected to be more reactive.
C. Both lone pairs are localized and, therefore, both are expected to be reactive.
D. Lone pair in pyridine ring is localized and, therefore, is expected to be more reactive.
Answer:
B.Lone pair in pyrrolidine ring is localized and, therefore, is expected to be more reactive.
Explanation:
There are two nitrogen atoms bearing lone pairs of electrons in the structure of nicotine as shown in the image attached.
One nitrogen atom is found in the pyrrolidine ring. The lone pair on this nitrogen atom is localized hence it is more reactive than the lone pair of electrons found on the nitrogen atom in the pyridine ring which is delocalized a shown in the image attached to this answer.
if an element has an atomic number of 9 what is the electronic structure of the same element
9 is the element Florine
Florine has 9 electrons as well as the 9 protons that determine its atomic number.
The ground state configuration is the lowest energy configuration.
lution: What is the molarity of 245 g of H, SO4 dissolved in 1.00 L of solution?
[tex]\\ \large\sf\longmapsto H_2SO_4[/tex]
[tex]\\ \large\sf\longmapsto 2(1u)+32u+4(16u)[/tex]
[tex]\\ \large\sf\longmapsto 2u+32u+64u[/tex]
[tex]\\ \large\sf\longmapsto 98u[/tex]
[tex]\\ \large\sf\longmapsto 98g/mol[/tex]
Given mass=245g[tex]\boxed{\sf No\:of\:moles=\dfrac{Given\:mass}{Molar\:Mass}}[/tex]
[tex]\\ \large\sf\longmapsto No\:of\:moles=\dfrac{245}{98}[/tex]
[tex]\\ \large\sf\longmapsto No\:of\:moles=2.5mol[/tex]
Now
[tex]\boxed{\sf Molarity=\dfrac{Moles\:of\:solute}{Volume\:of\:solution\;in\;L}}[/tex]
[tex]\\ \large\sf\longmapsto Molarity=\dfrac{2.5}{1}[/tex]
[tex]\\ \large\sf\longmapsto Molarity=2.5M[/tex]
When a 1:1 mixture of ethyl propanoate and ethyl butanoate is treated with sodium ethoxide, four Claisen condensation products are possible. Draw the structure(s) of the product(s) that have an ethyl group on the chiral center
Answer:
attached below
Explanation:
The Four Claisen condensation are grouped into :
Self Claisen condensation reaction Cross Claisen condensation reactionSelf Claisen condensation is when R = R'
Cross Claisen condensation is when R ≠ R'
attached below are the four Claisen condensation
A student dissolves 12.6g of amonium nitrate(NH4NO3) in 250.g of water in a well-insulated open cup. She then observed the temperature of the water fall from 23.0°C to 18°C over the course of 6.1 minutes.
NH4NO3 â NH4+ (aq) + NO3^-(aq)
a. Is this reaction exothermic, endothermic, or neither?
b. If you said the reaction was exothermic or calculate the amount of heat that was released or absorbed by the reaction in this case.
c. Calculate the reaction enthalpy ÎHrxn per mole of NH4NO3.
Answer:
a. Endothermic.
b. [tex]Q_{rxn}=5493.6J[/tex]
c. [tex]\Delta H_{rxn}=35.0kJ/mol[/tex]
Explanation:
Hello there!
In this case, according to the given information for this calorimetry problem, it turns out possible for us to proceed as follows:
a. Due to the fact that the temperature of water goes from 23.0 °C to 18.0 °C, we infer this reaction is endothermic as the ammonium nitrate absorbed heat from the water.
b. Here, we consider the following heat equation:
[tex]Q_{rxn}=-Q_{water}[/tex]
Whereas we solve for the heat of reaction by means of the mass of the solution (both water and ammonium nitrate), the specific heat of the solution (we assume it is equal to that of water) and the temperature change:
[tex]Q_{rxn}=-m_{solution}C_{solution}(T_f-T_i)\\\\Q_{rxn}=-(12.6g+250.g)(4.184\frac{J}{g\°C} )(18.0\°C-23.0\°C)\\\\Q_{rxn}=5493.6J[/tex]
c. Here, we divide the previously calculated heat by the moles of ammonium nitrate (molar mass = 80.043 g/mol) to obtain the enthalpy of reaction per mole of this compound:
[tex]n_{NH_4NO_3}=12.6g*\frac{1mol}{80.043 g}=0.157mol\\\\\Delta H_{rxn}=\frac{5493.6J}{0.157mol} =34898.7J/mol\\\\\Delta H_{rxn}=35.0kJ/mol[/tex]
Regards!
According to the EPA Lead and Copper Rule (LCR), the action level for Pb in drinking water (the level at which threat to human health requires public notification and action towards mitigation) is 15 ppb. If you were to add enough phosphate to the system
saturated with respect to Pb3(PO4)2(s), would the [Pb2+] be below the action limit?
Answer:
The right answer is "105.17 ppb".
Explanation:
According to the question,
The amount of [tex]Pb^{2+}[/tex] in ppb will be:
= [tex]0.5076\times 10^{-6}\times 207.2\times 106[/tex]
= [tex]105.17 \ ppb[/tex]
Thus, the amount of [tex]Pb^{2+}[/tex] is above action limit.
A substance is tested and has a pH of 7.0. How would you classify it?
#19.
An unknown sample weighs 45.2 g and takes 58.2 kJ to vaporize. What is
its heat of vaporization?
what characterizes a homogeneous mixture?
Answer:
a mixture that doesn't really show the ingredients or things put into the material or food.
Write the chemical formula for the following:
a. The conjugate acid of amide ion, NH₂-
b. The conjugate base of nitric acid, HNO₃
Follow the rules of Bronsted Lowry theory
Acids take a HBases donate a HSo
#a
NH_2-
Add a H
Conjugate acid is NH_3#b
HnO_3
Take a H
Conjugate base is NO_3-#1
Conjugate acid means one H+ is added
NH_2+H+=NH_3sw
#2
Conjugate base means donate a H+
HNO_3-H=NO_3-Part A
3.75 mol of LiCl in 3.36 L of solution
Express the molarity in moles per liter to three significant figures
Answer:
1.12 mol/L.
Explanation:
From the question given above, the following data were obtained:
Mole of LiCl = 3.75 moles
Volume = 3.36 L
Molarity =?
Molarity is simply defined as the mole of solute per unit litre of the solution. Mathematically, it is expressed as:
Molarity = mole / Volume
With the above formula, we can obtain the molarity of the solution as follow:
Mole of LiCl = 3.75 moles
Volume = 3.36 L
Molarity =?
Molarity = mole /Volume
Molarity = 3.75 / 3.36
Molarity = 1.12 mol/L
Thus, the molarity of the solution is 1.12 mol/L
A sample of an ideal gas is slowly compressed to one-half its original volume with no change in temperature. What happens to the average speed of the molecules in the sample
Answer:
See explanation
Explanation:
The average speed of the molecules of a gas depends on the temperature of the gas and its molar mass and not on the volume of the gas.
The average velocity of a gas is given by; vrms=√3RTM
R= gas constant
T= Absolute temperature
M= molar mass of the gas
Where the temperature of the gas is held constant, the average velocity of gas molecules depends on the molar mass of the gas. Hence, if a sample of gas is slowly compressed to one-half of its original volume with no change in temperature, the average speed of the molecules in the sample of gas remains the same.
he FAA restricts how much lithium to carry on an airplane. The rule-of-thumb is a battery has 0.3 g of lithium per Ampere-hour (Ah). A laptop battery is rated as 5 Ah per cell and contains 4 cells. Find the lithium content in grams. (compare with the FAA limit of 8 grams)
Answer:
The right answer is "6 gm".
Explanation:
Given:
Amount of Li,
= 0.3 gm
Battery rated,
= 5 Ah per cell
Total number of cells,
= 4
The total power limit will be:
= [tex]5\times 4[/tex]
= [tex]20 \ Ah[/tex]
hence,
The amount of Li in battery will be:
= [tex](0.3)\times 20[/tex]
= [tex]6 \ gm[/tex]
(Allowed for transportation).
Given the following reaction:
CO (g) + 2 H2(g) <==> CH3OH (g)
In an experiment, 0.42 mol of CO and 0.42 mol of H2 were placed in a 1.00-L reaction vessel. At equilibrium, there were 0.29 mol of CO remaining. Keq at the temperature of the experiment is ________.
A) 2.80
B) 0.357
C) 14.5
D) 17.5
E) none of the above
Answer:
Option D. 17.5
Explanation:
Equiibrium is: CO + 2H₂ ⇄ CH₃OH
1 mol of CO is in equibrium with 2 moles of hydrogen in order to make, methanol.
Initially we have 0.42 moles of CO and 0.42 moles of H₂
If 0.29 moles of CO remained, (0.42 - 0.29) = 0.13 moles have reacted.
So in the equilibrium we may have:
0.29 moles of CO, and (0.42 - 0.13 . 2) = 0.16 moles of H₂
Ratio is 1:2, if 0.13 moles of CO haved reacted, (0.13 . 2) moles have reacted of hydrogen
Finally 0.13 moles of methanol, are found after the equilibrium reach the end.
Let's make expression for KC: [Methanol] / [CO] . [Hydrogen]²
0.13 / (0.29 . 0.16²)
Kc = 17.5
Let's assume you were given 2.0 g benzil, 2.2 g dibenzyl ketone, 50 mL 95% ethanol and 0.3 g potassium hydroxide to synthesize tetraphenylcyclopentadienone. You isolated 3.0 g of tetraphenylcyclopentadienone. What is the % yield
Answer:
the % yield is 82%
Explanation:
Given the data in the question,
we know that;
Molar mass of benzil is 210.23 g·mol−1
Molar mass of dibenzyl ketone is 210.27 g·mol−1
Molar mass of tetraphenylcyclopentadienone is 384.5 g·mol−1
Now,
2.0 g benzil = 2 g / 210.23 g·mol−1 = 0.0095 mole
2.2 g dibenzyl ketone = 2.2 / 210.27 = 0.0105 mole
3.0 g of tetraphenylcyclopentadienone = 3 / 384.5 = 0.0078 mole
Now, the limiting reagent is benzil. 0.0095 mole can reacts wiyh 0.0095 mole of dibenzyl ketone
percentage yield = ( 0.0078 mole / 0.0095 mole ) × 100%
= 0.82 × 100%
= 82%
Therefore, the % yield is 82%
Electrophilic addition reaction of conjugated dienes that occur at high temperature and/or long reaction times (reversible conditions) are said to be under kinetic control. Group of answer choices True False
Answer:
False
Explanation:
Electrophilic addition reactions may be under kinetic or thermodynamic control. Whether the reaction is under kinetic or thermodynamic control is easily deducible from the reaction time.
Shorter reaction time often reflect kinetic control while longer reaction reaction times favour thermodynamic control.
Hence, electrophilic addition reaction of conjugated dienes that occur at high temperature and/or long reaction times (reversible conditions) are said to be under thermodynamic and not kinetic control.
Please help answering 11)
Answer:
the answer is C
Explanation:
6. In a particular atom, an electron moves from n = 3 to the ground state (n = 1), emitting a photon with frequency 5.2 x 1015 Hz as it does so. What is the difference in energy between n = 3 and n = 1 in this atom? g
Answer: The question wants you to determine the energy that the incoming photon must have in order to allow the electron that absorbs it to jump from
n
i
=
2
to
n
f
=
6
.
A good starting point here will be to calculate the energy of the photon emitted when the electron falls from
n
i
=
6
to
n
f
=
2
by using the Rydberg equation.
1
λ
=
R
⋅
(
1
n
2
f
−
1
n
2
i
)
Here
λ
si the wavelength of the emittted photon
R
is the Rydberg constant, equal to
1.097
⋅
10
7
m
−
1
Plug in your values to find
1
λ
=
1.097
⋅
10
7
.
m
−
1
⋅
(
1
2
2
−
1
6
2
)
1
λ
=
2.4378
⋅
10
6
.
m
−
1
This means that you have
λ
=
4.10
⋅
10
−
7
.
m
So, you know that when an electron falls from
n
i
=
6
to
n
f
=
2
, a photon of wavelength
410 nm
is emitted. This implies that in order for the electron to jump from
n
i
=
2
to
n
f
=
6
, it must absorb a photon of the same wavelength.
To find the energy of this photon, you can use the Planck - Einstein relation, which looks like this
E
=
h
⋅
c
λ
Here
E
is the energy of the photon
h
is Planck's constant, equal to
6.626
⋅
10
−
34
.
J s
c
is the speed of light in a vacuum, usually given as
3
⋅
10
8
.
m s
−
1
As you can see, this equation shows you that the energy of the photon is inversely proportional to its wavelength, which, of course, implies that it is directly proportional to its frequency.
Plug in the wavelength of the photon in meters to find its energy
E
=
6.626
⋅
10
−
34
.
J
s
⋅
3
⋅
10
8
m
s
−
1
4.10
⋅
10
−
7
m
E
=
4.85
⋅
10
−
19
.
J
−−−−−−−−−−−−−−−−−
I'll leave the answer rounded to three sig figs.
So, you can say that in a hydrogen atom, an electron located on
n
i
=
2
that absorbs a photon of energy
4.85
⋅
10
−
19
J
can make the jump to
n
f
=
6
.
Explanation:
Calculate the mass of isoborneol in 2.5 mmol of isoborneol and the theoretical yield (in grams) of camphor from that amount of isoborneol
isoborneol = 154.25 g mol?1
Camphor, Molar mass = 152.23 g/mol
Answer:
[tex]m_{isoborneol }=0.39g\\\\m_{Camphor}=0.38g\\[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to infer that the reaction whereby isoborneol goes to camphor occurs in a 1:1 mole ratio, that is why the theoretical yield of the latter is also 2.5 mmol (0.0025 mol) but the masses can be calculated as follows:
[tex]m_{isoborneol }=0.0025mol*\frac{154.25g}{1mol} =0.39g\\\\m_{Camphor}=0.0025mol*\frac{152.23 g}{1mol} =0.38g\\[/tex]
Because of the fact this is a rearrangement reaction whereas the number of atoms is not significantly modified.
Regards!
