Answer:
Butanoic acid
Explanation:
The IUPAC name of CH3CH2CH2COOH is:
The IUPAC name for a given compound is Butanoic acid.
Compound X has the same molecular formula as butane but has a different boiling point and melting point. What can be concluded about Compound X?
A It is a four-carbon alkene or alkyne.
B It is an optical isomer of butane.
C It is a geometric isomer of butane.
D It is a structural isomer of butane.
need this for gradpoint:)
Answer:
d
Explanation:
Determine whether methylcyclohexane contains a chiral center. _____ How many different substituents do you count at atom
Answer:
methylcyclohexane does not possess a chiral center
Explanation:
A molecule is said to possess a chiral center if one or more of the atoms in the molecule is bonded to four different atoms or groups.
If we look at methylcyclohexane, none of the carbon atoms in the molecule is bonded to four different atoms or groups hence the molecule contains no chiral centers and will not exhibit optical isomerism.
So, methylcyclohexane has a plane of symmetry thus it is an achiral molecule.
The number of mol of oxygen in 2.5 mol of caffeine
Answer:
5.0molO
Explanation:
To find the moles of oxygen in 2.5 moles of caffeine, we will first research caffeine's molecular formula: C8H10N4O2. From the molecular formula, we can see there are 2 oxygen atoms in every 1 molecule of C8H10N4O2.We can therefore multiply by the following mole ratio to get the moles of oxygen.
2.5molC8H10N4O2×2molO/1molC8H10N4O2 = 5.0molO.
Calculate the mass of butane needed to produce 80.0g of carbon dioxide
Answer:
Multiply the number of moles of butane by its molar mass, 58.12g/mol, to produce the mass of butane. Mass of butane = 18.8g.
Diethyl ether (C2H5 )2O vaporizes at room temperature. If the vapor exerts a pressure of 233 mm Hg in a flask at 25 °C, what is the density of the vapor?
Answer: The density of the given vapor is 0.939 g/L.
Explanation:
Given: Pressure = 233 mm Hg (1 mm Hg = 0.00131579 atm) = 0.31 atm
Temperature = [tex]25^{o}C[/tex] = (25 + 273) K = 298 K
According to the ideal gas equation,
[tex]PV = \frac{m}{M}RT[/tex]
where,
P = pressure
V = volume
m = mass
M = molar mass
R = gas constant = 0.0821 L atm/mol K
T = temperature
This formula can be re-written as follows.
[tex]PM = \frac{m}{V}RT[/tex] (where, [tex]Density = \frac{mass (m)}{Volume (V)}[/tex] )
Hence, formula used to calculate density of diethy ether (molar mass = 74.12 g/mol) vapor is as follows.
[tex]d = \frac{PM}{RT}[/tex]
Substitute values into the above formula as follows.
[tex]d = \frac{PM}{RT}\\= \frac{0.31 atm \times 74.12 g/mol}{0.0821 L atm/mol K \times 298 K}\\= \frac{22.9772}{24.4658}\\= 0.939 g/L[/tex]
Thus, we can conclude that the density of the given vapor is 0.939 g/L.
What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:
2 NaN3 is) 2Na (s) + 3N2 (8)
Answer:
108.6 g
Explanation:
2NaN₃(s) → 2Na(s) + 3N₂(g)First we use the PV=nRT formula to calculate the number of nitrogen moles:
P = 1.00 atmV = 56.0 Ln = ?R = 0.082 atm·L·mol⁻¹·K⁻¹T = 0 °C ⇒ 0 + 273.2 = 273.2 KInputting the data:
1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 Kn = 2.5 molThen we convert 2.5 moles of N₂ into moles of NaN₃, using the stoichiometric coefficients of the balanced reaction:
2.5 mol N₂ * [tex]\frac{2molNaN_3}{3molN_2}[/tex] = 1.67 mol NaN₃Finally we convert 1.67 moles of NaN₃ into grams, using its molar mass:
1.67 mol * 65 g/mol = 108.6 g1. A 225-L barrel of white wine has an initial free SO2 concentration of 22 ppm and a pH of 3.70. How much SO2 (in grams) should be added to the barrel to result in the required SO2 level
Answer:
The appropriate answer is "9.225 g".
Explanation:
Given:
Required level,
= 63 ppm
Initial concentration,
= 22 ppm
Now,
The amount of free SO₂ will be:
= [tex]Required \ level -Initial \ concentration[/tex]
= [tex]63-22[/tex]
= [tex]41 \ ppm[/tex]
The amount of free SO₂ to be added will be:
= [tex]41\times 225[/tex]
= [tex]9225 \ mg[/tex]
∵ 1000 mg = 1 g
So,
= [tex]9225\times \frac{1}{1000}[/tex]
= [tex]9.225[/tex]
Thus,
"9.225 g" should be added.
What is the phase of water at 0.25 atm and 0°C?
Water
(liquid)
Pressure (atm)
0.5-
0.25
Ice
(solid)
Water vapor
(gas)
0
000
Temperature (°C)
O A. Gas
O B. Solid and gas
O C. Solid and liquid
D. Solid
Water is in the solid phase at 0.25 atm and 0°C.
In what phase is water at 25?A pressure of 50 kPa and a temperature of 50 °C correspond to the “water” region—here, water exists only as a liquid. At 25 kPa and 200 °C, water exists only in the gaseous state.
What phase is water in at 0 C?Under standard atmospheric conditions, water exists as a liquid. But if we lower the temperature below 0 degrees Celsius, or 32 degrees Fahrenheit, water changes its phase into a solid called ice.
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If the volume of the gas is increased to 9.6 L , what will the pressure be?
Using the following reaction:
H2SO4 (aq) + 2NaOH (aq) → Na2SO4 (aq) + 2H2O (l)
Calculate the molarity of the H2SO4 solution if 14.92 mL of NaOH was necessary to reach the endpoint of a titration. The molarity of the NaOH solution was 0.83 M and 25.18 mL of H2SO4 was added to the Erlenmeyer flask.
Answer:
The molar concentration of the H₂SO₄ solution is 0.28 M
Explanation:
Molar concentration = number of moles / volume in litres
Number of moles = molar concentration × volume
From the equation of reaction, molar ratio of acid to base = 1 : 2
Using the formula; Na/Nb = CaVa/CbVb
Where Na is the number of moles of acid; Nb = number of moles of base; Ca = concentration of acid; Va = volume of acid; Cb = concentration of base; Vb = volume of base; Na/Nb = mole ratio of acid to base
Substituting the given values in the equation:
1/2 = Ca × 25.18 / 0.93 × 14.92
Ca = 0.93 × 14.92/ 25.18 × 2
Ca = 0.28M
Therefore, the molar concentration of the H₂SO₄ solution is 0.28 M
Which 2 resonance forms destablize the carbocation intermediate if bezonitrile undergoes chlronation at the ortho or para positions
The question is incomplete, the complete question is shown in the image attached
Answer:
A and B
Explanation:
The electrophilic substitution of arenes yields a cation intermediate. The positive charge of the cation is delocalized over the entire ring.
The -CN group directs incoming electrophiles to the ortho/para position. The resonance structures for the chlorination of benzonitrile are shown in the question.
Recall that -CN is an electron withdrawing group. The resonance forms that destablize the carbocation intermediate are those in which the -CN group is directly attached to the carbon atom bearing the positive charge as in structures A and B.
The following statements either define or are closely related to the terms periodic law, period, or group. Match the terms to the appropriate statements.1. This is a vertical arrangement of elements in the periodic table. 2. This is a horizontal arrangement of elements in the periodic table. 3. The properties of the elements repeat in a regular way as the atomic numbers increase. 4. Element 19 begins this arrangement in the periodic table 5. The chemical properties of elements 12, 20, and 38 demonstrate this principle.
Answer:
Please find the complete solution in attached file.
Explanation:
Periodic law is the the properties of the elements that repeat in a regular way as the atomic numbers increase.
What is periodic table?Periodic chart organized collection of all chemical elements arranged roughly according to increasing atomic weight. The periodic recurrence of several features in the elements was originally identified by Dmitry I. Mendeleyev in 1869.
1. This is a vertical arrangement of elements in the periodic table group
2. This is a horizontal arrangement of elements in the periodic table period
3. The properties of the elements repeat in a regular way as the atomic numbers increase periodic law
4. Element 19 begins this arrangement in the periodic table period
5. The chemical properties of elements 12, 20, and 38 demonstrate this principle. periodic law
Therefore, periodic law is the the properties of the elements that repeat in a regular way as the atomic numbers increase.
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3)O que são políticas públicas?
Answer:
azertyuiopazertyuiiop
Suppose 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 14.2 mL of 0.0500 M NaOH to reach the equivalence point. What is the concentration of the unknown weak acid solution
Answer:
0.1775 M
Explanation:
The reaction that takes place is:
HA + NaOH → NaA + H₂OWhere HA is the unknown weak acid.
At the equivalence point all HA moles are converted by NaOH. First we calculate how many NaOH moles reacted, using the given concentration and volume:
0.0500 M NaOH * 14.2 mL = 0.71 mmol NaOHThat means that in 4.00 mL of the weak acid solution, there were 0.71 weak acid mmoles. With that in mind we can now calculate the concentration:
0.71 mmol HA / 4.00 mL = 0.1775 MWhat volume (mL) of the sweetened tea described in Example 1 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example
The question is incomplete, the complete question is:
What volume (mL) of the sweetened tea described in Example 3.14 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example. The example is attached below.
Answer: 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink
Explanation:
We first calculate the number of moles of soft drink in a volume of 10 mL
The formula used to calculate molarity:
[tex]\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}[/tex] .....(1)
Taking the concentration of soft drink from the example be = 0.375 M
Volume of solution = 10 mL
Putting values in equation 1, we get:
[tex]0.375=\frac{\text{Moles of sugar in soft drink}\times 1000}{10}\\\\\text{Moles of sugar in soft drink}=\frac{0.375\times 10}{1000}=0.00375mol[/tex]
Calculating volume of sweetened tea:
Moles of sugar = 0.00375 mol
Molarity of sweetened tea = 0.05 M
Putting values in equation 1, we get:
[tex]0.05=\frac{0.00375\times 1000}{\text{Volume of sweetened tea}}\\\\\text{Volume of sweetened tea}=\frac{0.00375\times 1000}{0.05}=75mL[/tex]
Hence, 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink
Draw bond-line formulas of all dichloro derivatives that might be formed when 1-chloro-2,2,3,3,-tetramethylpentane is allowed to react with Cl 2 under UV irradiation. For each structure, indicate, with an asterisk, any stereocenters that might be present. Predcit the percentage of each product using the relative reactivities 3 0 = 5.3, 2 0 = 3.6, 1 0 = 1
Answer:
Explanation:
This is a halogenation reaction i.e substitution or replacement of a single or more than a single hydrogen atom in the organic alkane compound with the halogen(here it is chlorine).
The chlorination of 1-chloro-2,2,3,3-tetramethylpentane under UV light resulted in the formation of five (5) dichloro derivatives which are shown in the image attached below.
Also, the compounds containing a stereocenter (i.e a location within the compound composing of various substituents in which the interchangeability of these substituents has the tendency of resulting into a stereoisomer) are indicated with an asterisk in the image below.
From the image below:
compound 1 ⇒ 1,1-dichloro-2,2,3,3-tetramethylpentane = 2° C
∴
The given relative reactivity rate for 2° = 3.6x
For compound 2 ⇒ 1,4-dichloro-2,2,3,3-tetramethylpentane = 2° = 3.6x
For compound 3 ⇒ 1,5-dichloro-2,2,3,3-tetramethylpentane = 1° = 1x
For compound 4 ⇒ 1-chloro-2-chloromethyl-2,3,3-trimethylpentane
= 1° = 1x
For compound 5 ⇒ 1-chloro-3-chloromethyl-2,2,3-trimethylpentane
= 1° = 1x
As such, we have:
2(3.6x) + 3(1x) = 100
7.2x + 3x = 100
10.2x = 100
x = 100/10.2
x = 9.803°
∴
For compound (1) = 3.6(9.803) = 35.3%
For compound (2) = 3.6(9.803) = 35.3%
For compound (3) = 1(9.803) = 9.803°%
For compound (4) = 1(9.803) = 9.803°%
For compound (5) = 1(9.803) = 9.803°%
The element antimony has two stable isotopes, antimony-121 with a mass of 120.90 amu and antimony-123 with a mass of 122.90 amu. From the atomic weight of Sb = 121.76 one can conclude that: ________
antimony-123 has the highest percent natural abundance
most antimony atoms have a mass of 121.76 amu
antimony-121 has the highest percent natural abundance
both isotopes have the same percent natural abundance
Answer:
antimony-121 has the highest percent natural abundance
Explanation:
percent natural abundance;
121.76 = 120.90 x + 122.90 (1 - x)
121.76 = 120.90 x + 122.90 - 122.90x
121.76 = -2x + 122.90
121.76 - 122.90 = -2x
x= 121.76 - 122.90/ -2
x= 0.57
Where x and 1 - x refers to the relative abundance of each of the isotopes
Percent natural abundance of antimony-121 = 57 %
Percent natural abundance of antimony-123 = (1 - 0.57) = 43%
Let us remember that isotopy refers to a phenomenon in which atoms of the same element have the same atomic number but different mass numbers. This results from differences in the number of neutrons in atoms of the same element.
We can clearly see that antimony-121 has the highest percent natural abundance.
Methanoic acid is the simplest carboxylic acid molecule. It has one carbon atom. Draw the structural model for methanoic acid (using C and H).
ANSWER IS ABOVE
THE METHANOIC ACID
A gas bottle contains 0.650 mol of gas at 730. mmHg pressure. If the final pressure is 1.15 atm, how many moles of gas were added to the bottle
Answer: There are 0.779 moles of gas were added to the bottle.
Explanation:
Given: [tex]n_{1}[/tex] = 0.650 mol, [tex]P_{1}[/tex] = 730 mm Hg (1 mm Hg = 0.00131579 atm) = 0.96 atm
[tex]n_{2}[/tex] = ?, [tex]P_{2}[/tex] = 1.15 atm
Formula used is as follows.
[tex]\frac{P_{1}}{n_{1}} = \frac{P_{2}}{n_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{P_{1}}{n_{1}} = \frac{P_{2}}{n_{2}}\\\frac{0.96 atm}{0.650 mol} = \frac{1.15 atm}{n_{2}}\\n_{2} = 0.779 mol[/tex]
Thus, we can conclude that there are 0.779 moles of gas were added to the bottle.
Calculate the moles of H3PO4 that reacted (8). Consult the coefficients in the balance chemical reaction to obtain the mole ratio. Show your calculation here.
Answer:
The number of moles of H₃PO₄ that reacted is 0.000343 moles
Note: Some data is missing. Data from the attachment is used for the calculationsinnthe explanation below.
Explanation:
The reaction is a neutralization reaction between NaOH and H₃PO₄. The equation of the reaction is given as follows:
3 NaOH + H₃PO₄ ---> Na₃PO₄ + 3 H₂O
The molarity of the NaOH solution is 0.238 mol/L.
Average volume of NaOH used during the titration to arrive to endpoint = (4.6 + 3.9 + 4.5) mL / 3 = 4.33 mL
Molarity is defined ratio of the number of moles of solute to the volume of solution. Mathematically, molarity = number of moles/volume in Litres
Number of moles of NaOH reacted = 0.238 mol/L × (4.33mL × 1 L/1000 mL)
Number of moles of NaOH = 0.00103 moles
From the equation of the reaction, 3 moles of NaOH reacts with 1 mole of H₃PO₄
0.00103 moles of NaOH will react with 0.00103 x 1/3 moles of H₃PO₄ = 0.000343 moles of H₃PO₄.
Therefore, number of moles of H₃PO₄ that reacted is 0.000343 moles
43 mg = [?]g
A. 0.043 g
B. 4.3 g
C. 4300 g
D. 43,000 g
Answer:
Option A (0.043 g) is the correct answer.
Explanation:
Given:
= 43 mg
As we know,
[tex]1 \ mg = \frac{1}{1000} \ g[/tex]
then,
⇒ [tex]43 \ mg = \frac{43}{1000} \ g[/tex]
[tex]= 0.043 \ g[/tex]
Thus, the above is the correct alternative.
Calculate the mass percent of carbon in the following molecule: Mn3[Mn(CO)4]3
Answer:
21.63 %
Explanation:
The molar mass of Mn₃[Mn(CO)₄]₃ is 665.64 g/mol.
Let's assume we have 1 mol of Mn₃[Mn(CO)₄]₃, if that were the case then we would have 665.64 grams.
There are 12 C moles per Mn₃[Mn(CO)₄]₃, with that in mind we calculate the weight of 12 C moles:
12 mol C * 12 g/mol = 144 gFinally we calculate the mass percent of carbon:
144 g / 665.64 g * 100% = 21.63 %The phrases or terms describe different fundamental processes of nucleic acids. Classify each phrase or term as relating to replication, transcription, or translation.
a. Single DNA strand used to produce mRNA
b. Requires tRNA
c. Ribosome
d. DNA polymerase
e. Both DNA strands are duplicate
f. Described as semi-conservative
g. Amino acids added to peptide chain
Answer:
I don't know what to do
Explanation:
bye
Answer:
a. Single DNA strand used to produce mRNA ⇒ Transcription
b. Requires tRNA ⇒ Translation
c. Ribosome ⇒ Translation
d. DNA polymerase ⇒ Replication
e. Both DNA strands are duplicate ⇒ Replication
f. Described as semi-conservative ⇒ Replication
g. Amino acids added to peptide chain ⇒ Translation
Explanation:
Replication: the double-strand DNA is separated into two strands. Each strand is used as a template by DNA polymerase to produce the other strand. The leading strand is read by DNA polymerase in a continuous form (3' - 5') and the lagging strand is read in a discontinuous form (5'-3'). In this way, both strands are duplicated. The process is semi-conservative because the DNA molecule produced conserves 1 original strand and the other strand is the new synthesized one. The corresponding options are: d, e, f.
Transcription: is a process in which the genetic code of DNA is copied into a molecule called messenger RNA (mRNA). The double-strand DNA is opened and one strand is read. The enzyme involved is RNA polymerase, which binds the DNA (in a sequence called promoter) and uses the nucleotide code of DNA as a template to produce a molecule of RNA (the mRNA). Therefore, the correct option for this process is a.
Translation: is the process in which the mRNA sequence (copied from a DNA molecule) is translated into an amino acid sequence to produce a protein. This process is carried out within the cell ribosome. The mRNA is read in groups of three nucleotides (a codon) that codifies amino acids. The translation between codon and amino acid is assisted by molecules called transference RNA (tRNA). As each codon is decoded, an amino acid is added and the new polypeptide sequence is synthesized. Therefore, the correct options are: b, c, g.
A sample of calcium fluoride was decomposed into the constituent elements. Write a balanced chemical equation for the decomposition reaction. If the sample produced 294 mg of calcium, how many g of fluorine were formed
Answer:
A sample of calcium fluoride was decomposed into the constituent elements. Write a balanced chemical equation for the decomposition reaction. If the sample produced 294 mg of calcium, how many g of fluorine was formed
Explanation:
The balanced chemical equation for the decomposition of calcium fluoride is shown below:
[tex]CaF_2(s)->Ca(s)+F_2(g)[/tex]
The sample produced 294 g of calcium then, how many grams of fluorine is formed?
From the balanced chemical equation,
1 mol of CaF2 forms 1mol of calcium and 1 mol of fluorine.
That is:
40g of calcium and 38.0 g of fluorine are formed.
then,
If 294 g of calcium is formed then how many grams of fluorine is formed?
[tex]294g Ca * 38g F2 / 40g Ca\\=279.3 g F_2[/tex]
Hence, 279.3 g of fluorine will be formed.
In a single displacement reaction Zinc can displace ALL but…
Iron
Nickel
Calcium
Lead
Answer:
Calcium
Explanation:
Zinc cannot displace Ca because calcium is above it in the reactivity series
A 0.50 mol sample of COBr2 is transferred to a 9.50-L flask and heated until equilibrium is attained. Calculate the equilibrium concentrations of each species.
Answer:
Equlibrium concentration for each species ae as follows:
[CO] = 0.043 mol/L
[Br₂] = 0.043 mol/L
[COBr₂] = 0.01 mol/L
Explanation:
Let take a look at the chemical equation taking place at equilibrium
COBr2(g) ⇄ CO(g) + Br2(g)
The concentration of COBr2 i.e.
[COBr2] = no of moles/volume
= 0.50 mol/9.50 L
[COBr2] = 0.0530 mol/L
At standard conditions
Kc for COBr2 = 0.190
Now, the ICE table for the above reaction can be computed as follows:
COBr2(g) ⇄ CO(g) + Br2(g)
Initial 0.053 0 0
Change -x +x +x
Equilibrium (0.053 - x) x x
[tex]\mathsf{K_c = \dfrac{[CO][Br_2]}{[COBr_2]}}[/tex]
[tex]K_c = \dfrac{(x) (x)}{(0,053 -x)}[/tex]
[tex]0.190= \dfrac{x^2}{(0.053 -x)}[/tex]
x² = 0.190(0.053 - x)
x² = 0.01007 - 0.190x
x² + 0.190x - 0.01007 = 0
Using quadratic formula:
x ≅ 0.043 mol/L
SInce: x = [CO][Br₂] = 0.043 mol/L
[COBr₂] = 0.053 - x
[COBr₂] = 0.053 - 0.043 mol/L
[COBr₂] = 0.01 mol/L
Which compound contains both sigma and pi bonds... HCCl3, H2CO, H2S, or HBr?
Answer:
H2CO
Explanation:
Becuase it has 2 sigma bonds plus one pi bond and one sigma bond that consitute the double bond between C and O.
Answer:
B. H2CO
Explanation:
Determine whether each of the properties described applies to volumetric or graduated glassware. 1. Used for applications in which great accuracy is needed_____.a. volumetricb. graduated 2. Capable of measuring a range of volumes of liquid_____.a. volumetric
b. graduated 3. Designed to measure one specific volume of liquid_____.a. volumetric
b. graduated
Answer:
1) volumetric
2) graduated
3) volumetric
Explanation:
A volumetric glassware is a glassware that is marked at a particular point. A typical example of a volumetric glassware is the volumetric flask. A volumetric glassware is capable of measuring only a specific volume of a liquid.
On the other hand, graduated glassware can measure a range of volumes of liquid. However, a volumetric glassware is still required where a high degree of accuracy is important.
The following data were obtained using the same procedure as in your experiment. Using these data, calculate the percent malachite in the sample.
FW information (in g/mol): malachite = 221.11; CO2 = 44.01; H2O = 18.02
Mass of crucible: 23.8839 g Mass of crucible + sample: 26.0496 g
Mass of crucible + decomposed sample: 25.7013 g
One molecule of malachite decomposes to make:_______.
___________molecule(s) of copper oxide
___________molecule(s) of carbon dioxide
____________molecule(s) of water
a. 1, 2,2
b. 2, 1, 1
c. 1, 1, 1
d. 1,0.5, 0.5
Answer:
c
Explanation:
9. Which property is not important when selecting a material to use as a light bulb filament?
d
high melting point
high vapor pressure
high ductility
Answer:
high melting point
Explanation:
The filament of a bulb is often heated to very high temperatures as the bulb is in operation.
Many times, electric bulbs may have to be on for a whole day and they may reach temperatures that are outrageously high in the process.
The material of the filament must have a very high melting point so that it doesn't melt while the bulb is still in operation.
Answer:
Actually the answer is High Vapor pressure
Explanation:
