Answer:
which class are you please mention
Which of the given statements best represent what to do in the event of a spill of concentrated sulfuric acid.
A. First, rinse the affected area with copious amount of water.
B. First, rinse the affected area with copious amounts of sodium hydroxide.
C. Second, treat the area with aqueous sodium bicarbonate solution.
D. Second, add sand to absorb the remaining acid.
How did Kepler's discoveries contribute to astronomy?
O They supported the heliocentric model.
O They established the laws of planetary motion.
O They explained how the Sun rises and sets.
O They made astronomy accessible to people who spoke Italian.
They made astronomy accessible to people who spoke italian
Answer:
"They established the laws of planetary motion"
Explanation:
Mr. Kepler was the astronomer who came up with the "Laws of Planetary Motion."
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%
Use the reaction: 2AgNO3(aq) + H2SO4(aq) → Ag2SO4(s) + 2HNO3(aq) What volume (mL) of 0.568 M AgNO3(aq) is needed to form 0.21 g of Ag2SO4(s)
Answer:
The mole ratio of AgNO3 to Ag2SO4 IS 2:1 .0.657 g Ag2SO4 x 1 mol / 312 g = 0.00211 mol Ag2SO4.
0.00211 mol Ag2SO4 x 2 mol AgNO3 / 1 mol Ag2SO4 = 0.00421 mol AgNO3
0.00421 mol AgNO3 x 1 L / 0.123 mol AgNO3 = 0.0342 L = 34.2 mL of AgNO3 solution.Therefore,34.2ml of 0.123M AgNO3 will be required.
g An aqueous solution of nitric acid is standardized by titration with a 0.137 M solution of calcium hydroxide. If 19.0 mL of base are required to neutralize 21.8 mL of the acid, what is the molarity of the nitric acid solution
Answer:
M of HNO₃ is 0.119M
Explanation:
A basic concept of titration is that in equivalence point:
mmoles of acid = mmoles of base
We have data from base and we only have data from volume of acid.
In a case our titration is a strong acid against a strong base.
We apply formula:
M of acid . Vol of acid = M of base . Vol of base
M of acid . 21.8 mL = 0.137M . 19 mL
M of acid = (0.137M . 19 mL) / 21.8 mL
M of acid = 0.119 M
When we neutralize all the titrant we reach the equivalence point.
At this point, pH = 7
2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O
5. How many moles are present in 4.20x10^24 atoms of Pb
Explanation:
[tex]57816 \: moles[/tex]
are present in 4.20x10^24 atoms of Pb
Answer:
7 moles
Explanation:
(4.2*10^24)/(6*10^23)=7
· Acids are not safe to be used, but our stomach secretes hydrochloric acid. What would happen if the stomach does not carry out this task? Mark them brainlist
A sample of gas contains 0.1200 mol of H2(g) and 0.1200 mol of O2(g) and occupies a volume of 11.5 L. The following reaction
takes place:
H2(g) + O2(g)>H2O2(g)
Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.
L
Answer:
5.75L is the volume of the sample after the reaction
Explanation:
Based on the reaction, 1 mole of H2 reacts with 1 mole of O2 to produce 1 mole of H2O2.
As in the reaction, 0.1200 moles of H2 and 0.1200 moles of O2 are added, 0.1200 moles of H2O2 are produced.
Before the reaction, the moles of gas are 0.2400 moles and after the reaction the moles are 0.1200 moles of gas.
Based on Avogadro's law, the moles of a gas are directly proportional to the volume under temperatura and pressure constant. The equation is:
V1/n1 = V2/n2
Where V is volume and n are moles of 1, initial state and 2, final state.
Replacing:
V1 = 11.5L
n1 = 0.2400 moles
V2 = ?
n2 = 0.1200 moles
11.5L*0.1200 moles / 0.2400 moles = V2
V2 = 5.75L is the volume of the sample after the reaction
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.
Which chemical can remove color of red/Pink phenol and make it clear like water transparent?
CHEM 100Worksheet 6Summer2021Name:____________________(5pts each, 10 pts total) Complete the following multistep synthesis problems. Show all reagents and intermediates for full credit. You do not need to show the mechanisms.
Where are the questions?
what characterizes a homogeneous mixture?
Answer:
a mixture that doesn't really show the ingredients or things put into the material or food.
A compound made of elements A and B, has a cubic unit cell. There is an A atom at each corner of the cube and an A atom at the center of each face of the cube. There are four B atoms that lie entirely within the unit cell. Based on this information, the empirical formula for the compound is:
Answer:
A₅B₄
Explanation:
Since we have one atom of element A at the center of each face of the unit cell, since the unit cell is a cubic cell, we have 6 faces. Since the atom on the face of the unit cell is shared with another cell, we have half of it in the unit cell is shared So, the number of atoms per face is 1/2 atom/face × 6 faces = 4 atoms on the faces of the unit cell.
Also, we have 1 atom at each corner of the cubic unit cell. Since there are 8 corner in the cubic unit cell. Also, each atom at the corner is shared with 8 unit cells, so we have 1/8 atom per corner. So, the number of atoms per unit cell is 1/8 atom/corner × 8 corners = 1 atoms at the corners of the unit cell.
So, in total we have 4 + 1 = 5 atoms of element A in the unit cell.
Also, there are 4 atoms of element B in the unit cell.
So, the ratio of atoms of element A to element B is 5 : 4.
A:B = 5:4
So, the empirical formula of the compound containing elements A and B is A₅B₄
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.
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
Related questions
How are buffer solutions used?
How do you calculate buffer capacity?
How do buffer solutions maintain the pH of blood?
How do you buffer a solution with a pH of 12?
Why are buffer solutions used to calibrate pH?
What is the role of buffer solution in complexometric titrations?
How you would make 100.0 ml of a 1.00 mol/L buffer solution with a pH of 10.80 to be made using...
What is the Henderson-Hasselbalch equation?
What is an example of a pH buffer calculation problem?
Why is the bicarbonate buffering system important?
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Dung dich NaCl 0.9% có 0.9g NaCl trong 100 mL dung dịch
Answer:
Explanation: Độ thẩm thấu của NaCl 0.9% và glucose 5% lần lượt là 308 và 278 ... Dung dịch natri clorid sử dụng trong pha thuốc tiêm truyền thường dùng
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.
What is the mole of 98 mL of carbon dioxide gas at 36°C and 795 torr?
R = 0.0821 Latm/molk
Round to the thousandth place.
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]
What is the name of the compound shown below?
A. 2-pentene
B. 1-propene
C. 2-propene
D. 1-pentene
The name of the compound shown below is 1- pentene. The correct answer is option D.
A compound is a substance made up of two or more different elements chemically bonded together in a fixed ratio.
1-pentene is an unsaturated hydrocarbon with the chemical formula [tex]\rm C_5H_{10}[/tex]. It is an alkene, which means it contains a carbon-carbon double bond.
The structure of 1-pentene is characterized by a chain of five carbon atoms (pentane) with one double bond between the first and second carbon atoms. The double bond causes the molecule to have a planar structure, with all atoms lying in the same plane. The remaining three carbon atoms in the chain are each bonded to two hydrogen atoms.Therefore, option D. 1-pentene is the name of the compound shown.
Learn more about compound here:
https://brainly.com/question/14117795
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calculate the hydrogen ion concentration of a solution who's pH is 2.4
Answer:
I don't know sorry yyyyyyy6yyyyyyyyyyyyyyyyyyyyyyyyyyy
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!
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
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.
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
)Calculate the molar mass of glucose (C6H12O6)
Answer:
Molar mass = 180 g/mol
Explanation:
Relative Atomic Mass of C = 12
of H = 1
of O =16
Let Molar mass be mm
mm of C6H12O6 = 6(12) + 12(1) + 6(16)
= 72 + 12 + 96 = 180 g/mol
A substance which is made up of the same kind
of atom is known as?
Answer:
Element
Element : A pure substance composed of the same type of atom throughout. Compound : A substance made of two or more elements that are chemically combined in fixed amounts.
Explanation:
What did Millikan discover
Answer:
Robert Millikan was a physicist who discovered the elementary charge of an electron using the oil-drop experiment
Answer:
the mass of an electron using the Oil-Drop experiment.
Explanation:
What volume of water is produced when 38.5 g of ethanol reacts with oxygen at 500°C at 1.75 atm?
CH3CH2OH(g) + 3 O2(g)→ 2 CO2(g) + 3 H2O(g)
Answer:
90.99 or 91.0
Explanation:
Using the balanced equation, you convert 38.5g of ethanol to moles of water. From there, you plug the values into the Ideal Gas Equation: PV=nRT.
Answer: The volume of oxygen gas is 91.4 L.
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Given mass of ethanol = 38.5 g
Molar mass of ethanol = 46 g/mol
Plugging values in equation 1:
[tex]\text{Moles of ethanol}=\frac{38.5g}{46g/mol}=0.840 mol[/tex]
The given chemical equation follows:
[tex]CH_3CH_2OH(g)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(g)[/tex]
By stoichiometry of the reaction:
If 1 mole of ethanol produces 3 moles of water
So, 0.840 moles of ethanol will produce = [tex]\frac{3}{1}\times 0.840=2.52mol[/tex] of water
The ideal gas equation is given as:
[tex]PV=nRT[/tex] .......(2)
where
P = pressure = 1.75 atm
V = volume of oxygen gas = ?
n = number of moles= 2.52 moles
R = Gas constant = 0.0821 L.atm/mol.K
T = temperature of the tank = [tex]500^oC=[500+273]K=773K[/tex]
Putting values in equation 2, we get:
[tex]1.75 atm\times V=2.52mol\times 0.0821L.atm/mol.K\times 773K\\\\V=\frac{2.52\times 0.0821\times 773}{1.75}=91.4L[/tex]
Hence, the volume of oxygen gas is 91.4 L.
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).
