1. 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.
atomaticity of chlorine 1) 2, 2)1, 3) 32 , 4) 4.
Answer:
ATOMICITY OF CHLORINE IS 2Explanation:
Atomicity is defined as the total number of atoms present in a molecule.
What effect does a high carbon level have on a deep ocean
Explanation:
High carbon concentration in the deep ocean means increased absorption of carbon to the atmosphere resulting to even greater and harmful amounts of carbon in the atmosphere. Therefore we need to keep a close eye of the deep ocean in the quest to monitor and pump out excess carbon from this part of marine life.
Given the amount of camphor (200mg) we are using in this experiment, please determine how many mg of sodium borohydride to use in this reaction. We would like you to use 5.2 molar equivalents of this reagent. This means 5.2 times the mmol of camphor we are using. As an example: for 110.0 mg of camphor,142 mg of NaBH4 would be used (see if you can confirm this result). For complete credit, your work needs to be clearly drawn out!
Answer:
Explanation:
From the given information:
Camphor may be reduced as readily in the presence of sodium borohydride(NaHB4). The resulting compound which is stereoselective requires 1 mole of sodium borohydride (NaHB4) to reduce 1 mole of camphor in this reaction. The reaction is shown below.
Through the reduction process of camphor, the reducing agent can reach the carbonyl face with a one-carbon linkage. The product stereoisomer is known as borneol.
If the molecular weight of camphor = 152.24 g/mol
and it mass = 200 mg
The its no of moles = 200 mg/ 152.24 g/mol
= 1.3137 mmol
Now the amount of the required mmol for NaBH4 to be consumed in the reaction = 5.2 × 1.3137 mmol
= 6.831 mmol
since the molar mass of NaBH4 = 37.83 g/mol
Then, using the same formula:
No of moles = mass/molar mass
mass = No of moles × molar mass
mass = 6.831 mmol × 37.83 g/mol
mass of NaBH4 used = 258.42 mg
Calculate the mass of water produced when 1.57g of butane reacts with excess oxygen
Explanation:
So, first you will want to write the balanced chemical equation for this reaction.
Butane = [tex]C_4H_{10}[/tex]
[tex]2C_4H_{10}+13O_2=>10H_2O+8CO_2[/tex]
^ This ends up being your balanced chemical equation. Now, you can do the math!
[tex]1.57gC_4H_{10}*\frac{1molC_4H_{10}}{58.12gC_4H_{10}}*\frac{10molH_2O}{2molC_4H_{10}}*\frac{18gH_2O}{1molH_2O}[/tex]
After plugging this into a calculator, your final mass of water should be:
2.43gH2O
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.
What 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
15. You are interested in separating 4-methylbenzoic acid from 1,4-dimethoxybenzene using a procedure similar to the extraction procedure we used in lab. You plan to use sodium bicarbonate instead of sodium hydroxide. a) Show the reaction between salicylic acid and sodium bicarbonate. Label the acid, base, conjugate acid, conjugate base. b) Give the pKa values of the acid and conjugate acid. c) Which base will work better, sodium hydroxide or sodium bicarbonate
Solution :
a). The separation of 4-methylbenzoic acid from 1,4-dimethoxybenzene will work but it will result in lower recovery.
In the reaction of acid-base to form a sodium 4 - methoxy benzoate, that is soluble in the water, 4-methoxy benzoic acid reacts with the sodium bicarbonate to give sodium 4-methoxybenzoate as well as carbonic acid.
b). The pKa for the 4-methoxybenzoic acid is [tex]4.46[/tex], and that of carbonic acid is [tex]6.37[/tex]
c). The Keq for the reaction is [tex]10(6.37 - 4.46) = 101.91[/tex]
Therefore, the equilibrium lies to the right and also the reaction favors the products and the separation works.
But the recovery will be low when compared to the extraction with Sodium hydroxide as the strong base will drive the equilibrium further to the right position, thus neutralizing all the acids virtually. And the weak base will partially neutralize the 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:
How many molecules of C 2H 5Br will be present if you had 4.52 g of this compound?
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.
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°%
If the volume of the gas is increased to 9.6 L , what will the pressure be?
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 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.
Name two natural sources of esters.
Answer:
"Esters are widespread in nature and are widely used in industry. In nature, fats are in general triesters derived from glycerol and fatty acids. Esters are responsible for the aroma of many fruits, including apples, durians, pears, bananas, pineapples, and strawberries."
Explanation:
Under certain conditions, the substance mercury(II) oxide can be broken down to form mercury and oxygen. If 32.2 grams of mercury(II) oxide react to form 29.8 grams of mercury, how many grams of oxygen must simultaneously be formed
Explanation:
This is a decomposition reaction. Firstly, you will want to write the chemical equation out and balance it.
[tex]2Hg_2O->4Hg+O_2[/tex] (The -> is supposed to be an arrow, sorry!)
We see that there's only 1mol of Oxygen made in the products, we can do some simple math to solve for the amount of grams of Oxygen produced according to the amount of the reactant (Hg2O).
[tex]32.2gHg_2O*\frac{1molHg_2O}{417.18gHg_2O}*\frac{1molO_2}{2molHg_2O}*\frac{32gO_2}{1molO_2}[/tex]
I want to break this down, just in case:
The 417.18gHg2O is the molecular mass of the molecule (so I doubled Hg and added 16 to it to get this number).
As we can see in the chemical equation, 1mol Hg2O produces 2mol O because Oxygen is a diatomic molecule (so there will always be two of it when it's by itself).
And finally, in 1mol O2 there are 32g of O2.
** When you do math like this, always make sure that all of your units cancel out except for the units you're looking for. For example, here we're looking for the grams of Oxygen, so after everything else cancels out, we should only have grams O2.
So, 1.23gO2 should be your answer.
According to Newton's law of universal gravitation, how do the masses of two
objects relate to the gravitational force between them?
A. As either mass increases, the gravitational force between them
increases.
B. As either mass increases, the gravitational force between them
decreases.
C. Gravitational force increases only when both masses increase.
D. Gravitational force increases only when both masses decrease.
Answer:
As either mass increases, the gravitational force between them
increases.
Explanation:
According to Newton's law of universal gravitation;
F α m1m2/r^2
That is, the force between two masses in a gravitational field is directly proportional to the product of the two masses and inversely proportional to their distance apart.
Hence, as either of the masses increase, the force of gravitation between the two masses increases. Hence the answer.
Wine goes bad soon after opening because the ethanol dissolved in it reacts with oxygen gas to form water and aqueous acetic acid , the main ingredient in vinegar. Calculate the moles of acetic acid produced by the reaction of of ethanol. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.
Answer:
The answer is "It takes 1,70 mol of ethanol".
Explanation:
To make acetic acid, we must first write the balanced reaction that occurs of ethanol with oxygen
The response is balanced:
[tex]CH_3CH_2OH+O_2\to CH_3COOH+H_2O[/tex]
1 mol of ethanol creates 1 mol of According the equilibrium Ethanol moles, therefore, required 1.70 mol of water = 1.70 mol
g Arrange the following compounds in order of acidity (highest to lowest): H2O, H3O , HCl A. CH3COOH > HCl > H2O B. H2O > CH3COOH > HCl C. HCl > H2O > CH3COOH D. HCl > CH3COOH > H2O
Answer:
Arrange the following compounds in order of acidity (highest to lowest): H2O, CH3COOH , HCl
A. CH3COOH > HCl > H2O
B. H2O > CH3COOH > HCl
C. HCl > H2O > CH3COOH
D. HCl > CH3COOH > H2O
Explanation:
The given substances are acetic acid, hydrochloric acid, and water.
Since HCl is a strong acid and it undergoes complete ionization.
CH3COOH acetic acid is a weak acid and it undergoes partial dissociation in water.
Pure water is a neutral substance.
Hence, the order of acidity is shown below:
HCl > CH3COOH > H2O.
Among the given options, option D is the correct answer.
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.
We can use bond-line formulas to represent alkenes in much the same way that we use them to represent alkanes. Consider the following alkene: h5ch5e4 How many carbon atoms are sp2−hybridized in this alkene?
Answer:
2
Explanation:
The number of carbon atoms that are sp²-hybridized in this alkene is 2
Because all the single bonded carbon atoms in the alkene are sp²-hybridized
There are three(3) single formed via sp² orbitals and one ( 1 ) PI bond formed via Pure-P-orbital
attached below is the some part of the solution
I don’t want to fail help
I need correct answer if u don’t know I will report
When the researcher compiled information which research method did they most likely utilize?
a) documentary
b) survey
c) participant observation
d) case study
Answer:
a
Explanation:
documentary is best researcher!.
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.
g 0.500 L of a solution with a concentration of 0.25 M is needed. To prepare this solution, a stock solution with a concentration of 1.25 M is prepared. What volume of the stock solution is needed to create the desired solution
Answer:
0.1 L
Explanation:
From the question given above, the following data were obtained:
Concentration of stock solution (C₁) = 1.25 M
Volume of diluted solution (V₂) = 0.5 L
Concentration of diluted solution (C₂) = 0.25 M
Volume of stock solution needed solution (V₁) =?
The volume of the stock solution needed can be obtained as follow:
C₁V₁ = C₂V₂
1.25 × V₁ = 0.25 × 0.5
1.25 × V₁ = 0.125
Divide both side by 1.25
V₁ = 0.125 / 1.25
V₁ = 0.1 L
Therefore, the volume of the stock solution needed is 0.1 L
compound of aspartame is a dipeptide that is often used as a sugar substitute which functional groups are present
Answer:
Carboxyl, primary amine, amide, ester, and phenyl.
Explanation:
The functional groups present in the compound of aspartame are carboxyl, primary amine, amide, ester, and phenyl. Aspartame is an artificial non-saccharide sweetener which is 200 times sweeter than sucrose. This aspartame is commonly used as a sugar substitute in many foods and beverages. It has the trade names such as NutraSweet, Equal, and Canderel.
The volume of a single tantalum atom is 1.20×10-23 cm3. What is the volume of a tantalum atom in microliters?
Answer:
1.20x10⁻²⁰μL
Explanation:
1cm³ is equal to 1milliliter. As we must know, 1milliliter = 1000 microliters, 1000μL. To convert the 1.20x10⁻²³mL we need to use the conversion factor: 1mL = 1000μL.
The volume of tantalum in μL is:
1.20x10⁻²³mL * (1000μL /1L) = 1.20x10⁻²⁰μL
A sample of 10.6 g of KNO3 was dissolved in 251.0 g of water at 25 oC in a calorimeter. The final temperature of the solution was 21.5 oC. What is the molar heat of solution of KNO3
Answer:
36.55kJ/mol
Explanation:
The heat of solution is the change in heat when the KNO3 dissolves in water:
KNO3(aq) → K+(aq) + NO3-(aq)
As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.
To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:
Moles KNO3 -Molar mass: 101.1032g/mol-
10.6g * (1mol/101.1032g) = 0.1048 moles KNO3
Change in heat:
q = m*S*ΔT
Where q is heat in J,
m is the mass of the solution: 10.6g + 251.0g = 261.6g
S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-
And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C
q = 261.6g*4.184J/g°C*3.5°C
q = 3830.87J
Molar heat of solution:
3830.87J/0.1048 moles KNO3 =
36554J/mol =
36.55kJ/mol
3)O que são políticas públicas?
Answer:
azertyuiopazertyuiiop
Name the following aldehyde PLEASE PLEASE HELP
Answer:
Explanation:
Answer is D 2,5-dimethylheptanal
You should accern the lowest possible number close to the parent name