Answer:
the change in the internal energy of the system is 3,752.67 J
Explanation:
Given;
initial volume of the gas, V₁ = 30.6 L
final volume of the gas, V₂ = 1.8 L
constant pressure of the gas, P = 1.8 atm
Energy released by the system, Q = 1.5 kJ = 1,500 J
Apply pressure-volume work equation, to determine the work done on the gas;
w = -PΔV
w = -P(V₂ - V₁)
w = - 1.8 atm(1.8 L - 30.6 L)
w = 51.84 L.atm
w = 51.84 L.atm x 101.325 J/L.atm
w = 5,252.67 J
The change in the internal energy of the system is calculated as;
ΔU = Q + w
Since the heat is given out, Q = - 1,500 J
ΔU = -1,500 J + 5,252.67 J
ΔU = 3,752.67 J
Therefore, the change in the internal energy of the system is 3,752.67 J
Glass tubes of the following diameters are arranged in a water trough. Rank the following glass tubes in decreasing order of their capillary rise.
Rank from highest to lowest capillary rise. To rank items as equivalent, overlap them.
1.25 mm, 1 mm, 0.75 mm, 0.25 mm, 0.50 mm
Answer:
0.25 mm> 0.50 mm > 0.75 mm> 1mm> 1.25 mm
Explanation:
Capillary movement involves the movement of liquid up through a capillary tube.
Looking at the question, we can see all the various diameters of the capillary tube are involved.
Note that, that the narrower the diameter of the capillary tube, the better the capillary action of the liquid (the liquid rises more now than then).
Draw structures corresponding to the following IUPAC names:(a) (Z)-2-Ethyl-2-buten-1-ol (b) 3-Cyclohexen-1-ol(c) trans-3-Chlorocycloheptanol (d) 1,4-Pentanediol(e) 2,6-Dimethylphenol (f ) o-(2-Hydroxyethyl)phenol
Answer:
Draw structures corresponding to the following IUPAC names:(a) (Z)-2-Ethyl-2-buten-1-ol (b) 3-Cyclohexen-1-ol(c) trans-3-Chlorocycloheptanol (d) 1,4-Pentanediol(e) 2,6-Dimethylphenol (f ) o-(2-Hydroxyethyl)phenol
Explanation:
According to IUPAC rules, the name of a compound is:
Prefix+root word+suffix
1) Select the longest carbon chain and it gives the root word.
2) The substituents give the prefix.
3) The functional group gives the secondary suffix and the type of carbon chain gives the primary suffix.
The structure of the given compounds are shown below:
You are asked to prepare a buffer solution with a pH of 3.50. The following solutions, all 0.100 M, are available to you: HCOOH, CH3COOH, H3PO4 , NaCHOO, NaCH3COO, and NaH2PO4. What would be the best combination to make the required buffer solution? Select one:
a. NaH2PO4 and NaCHOO
b. H3PO4 and NaH2PO4
c. NaH2PO4 and HCOOH
d. CH3COOH and NaCH3COO e. HCOOH and NaCHOO
can someone helo me with this
Answer:
e. HCOOH and NaCHOO
Explanation:
For a buffer solution, both an acid and its conjugate base are required.
With the information above in mind, we can discard options a) and c), as those combinations are not of an acid and its conjugate base.
Now it is a matter of comparing the pKa (found in literature tables) of the acids of the remaining three acids:
H₃PO₄ pKa = 2.12CH₃COOH pKa = 2.8HCOOH pKa = 3.74The acid with the pKa closest to the desired pH is HCOOH, so the correct answer is e. HCOOH and NaCHOO
Suppose you ran this reaction without triethylamine and simply used an excess of reactant 1. At the end of the reaction, your methylene chloride solution would contain mostly reactant 1 and the product. What would you do to remove reactant 1 from the solution
ummm is that chemistry?
Answer:
is this chem
Explanation:
Aluminum hydroxide, with heat, creates____
Answer:
Water and Aluminium oxide
Explanation:
Have a nice day.
A sample of gas is placed into an enclosed cylinder and fitted with a movable piston. Calculate the work (in joules) done by the gas if it expands from 5.33 L to 11.05 L against a pressure of 1.50 atm.
Explanation:
here is the answer. Feel free to ask for more chem help
How many grams of solute are present in 635mL of 0.450 M KBr?
Answer:
34.03 g
Explanation:
We'll begin by converting 635 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
635 mL = 635 mL × 1 L / 1000 mL
635 mL = 0.635 L
Thus, 635 mL is equivalent to 0.635 L
Next, we shall determine the number of mole of the solute (KBr) in the solution. This can be obtained as follow:
Volume = 0.635 L
Molarity = 0.450 M
Mole of KBr =?
Mole = Molarity × Volume
Mole of KBr = 0.450 × 0.635
Mole of KBr = 0.286 mole
Finally, we shall determine the mass of 0.286 mole of KBr. This can be obtained as follow:
Mole of KBr = 0.286 mole
Molar mass of KBr = 39 + 80
= 119 g/mol
Mass of KBr =?
Mass = mole × molar mass
Mass of KBr = 0.286 × 119
Mass of KBr = 34.03 g
Thus, the mass of the solute (KBr) in the solution is 34.03 g
Consider the balanced chemical equation below.
2 A ⟶ C + 4 D
How many moles of D would be produced if 6 moles of A were used?
Explanation:
[tex]here \: is \: your \: explanation : - \\ \\ given \: balanced \:equation \: = > \\ \\ 2 A=>C \: + \: 4D \\ \\ by \: this \: equation \: we \: get \: \\ \\ 2 \: moles \: of \: A \: produce \: 4 \: moles \: \\ \\ of \: D \\ \\ hence \: . \: 1 \: mole \: can \: produce \: = 4 \div 2 \\ \\ = > 2 \: moles \: \\ \\ so \: if \: 6 \: moles \: of \: A \: used \: then \: \\ \\ amount \: of \: D \: produced \: = (6 \times 2) \\ \\ = > 12 \: moles \: of \: D \\ \\ \mathcal\blue{ Hope \: it \: helps \: you \: (. ❛ ᴗ ❛.) }[/tex]
ype the correct answer in the box.
Calculate the density of the substance.
A sample of a substance has a mass of 4.2 grams and a volume of 6 milliliters. The density of this substance is grams/milliliter.
Reset Next
Explanation:
here is your answer. Hope it helps
An unidentified gas is determined to be 24.0% carbon and 76% fluorine by mass. What is the empirical formula of this gas
Answer:
CF₂
Explanation:
Let's assume we have 100 g of the gas. If that were the case we'd have
24 g of C76 g of FNow we convert both masses into moles, using their respective molar mass:
24 g C ÷ 12 g/mol = 2 mol C76 g F ÷ 19 g/mol = 4 mol FWe can express those results as C₂F₄.
To determine the empirical formula we reduce those coefficients to the lowest possible integers, leaving us with CF₂.
Which diagram correctly depicts the trend in electronegativity?
a.
b.
c.
d.
The electronegativity increases across the period and decreases down the group. Thus, option B is correct.
Electronegativity can be defined as the tendency of an atom to gain or attract an electron. The electronegativity has been dependent on the size of the atom, as well as the atomic number and valence electrons.
The atom with the requirement of a less number of atoms to complete its octet can easily gain the electron and thereby have high electronegativity. The atomic size also plays a role in the electronegativity of the atom.
The atom with a bigger size has the lesser force of attraction from the nucleus and thus has difficulty attracting the electron, however, the smaller size atom can easily attract the electron with the attraction force from the nucleus.
Thus, the elements with smaller sizes and a high number of valence electrons are more electronegative. In the periodic table, on moving from left to right the valence electrons increase, thus the electronegativity increases.
On moving down the group, the element size increase, thus the electronegativity decreases down the group.
The electronegativity increases across the period and decreases down the group. Thus, option B is correct.
For more information about electronegativity, refer to the link:
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A compound, C7H13Cl, is reacted with sodium ethoxide and gives a single elimination product, C7H12. Treatment with ozone followed by zinc and water gives the compound below. Identify the original compound.
A. 2-chloro-1,1-dimethylcyclopentane.
B. 1-chloro-1,2-dimethylcyclopentane.
C. 4-chloro-1,2-dimethylcyclopentane.
D. 2-chloro-1,3-dimethylcyclopentane.
Answer:2-chloro-1,3-dimethylcyclopentane
Explanation
A dilute solution of an unknown ionic compound must be identified. Several experiments are run on solutions that contain the possible ions with the results tabulated below. Cation Addition of HCl Heated in hot water Ag+ White precipitate is formed No observable change Pb+2 White precipitate is formed Precipitate dissolves Zn+2 No observable change No observable change Anion Addition of Ca(NO3)2 Addition of HNO3 CO3–2 White precipitate is formed Precipitate dissolves and bubbles form PO4–3 White precipitate is formed Precipitate dissolves A white precipitate is formed when HCl is added to the unknown compound. When the white precipitate is heated in hot water, it does not dissolve. A white precipitate is formed when Ca(NO3)2 is added to the unknown compound. When HNO3 is added to the precipitate, it dissolves and bubbles are formed. What is the chemical formula of the unknown ionic compound?
Answer:
The chemical formula of the unknown ionic compound is Ag₂CO₃.
Explanation:
The method used to identify the ions present in a given solution of an unknown compound is known as qualitative analysis. Specific tests are carried based on the chemical reaction of these ions and the results they give in these reactions, the ions present innthe compound are identified. Some of the results or observations include formation of precipitates on addition of certain reagents, evolution of gases, colour changes, as well as the color changes that are observed in a flame test of a solutionmof these compounds.
In the given question, when testing for cations, a white precipitate is formed when HCl is added to the unknown compound. When the white precipitate is heated in hot water, it does not dissolve.
From the table of observations, both Ag+ and Pb²+ give a white precipitate on addition of HCl but Zn²+ does not. So the cation present is either Ag+ or Pb²+. However, when heated in hot water, the white precipitate does not dissolve as will be observed if Ag+ were the cation present. Thus, the cation is Ag+.
When testing for anions, a white precipitate is formed when Ca(NO3)2 is added to the unknown compound and when HNO3 is added to the precipitate, it dissolves and bubbles are formed.
Now comparing these results with that as observed with CO₃²- and PO₄³-, both form white precipitates on addition of Ca(NO₃)₂, and the precipitate dissolves on addition of HNO₃. However, only CO₃²- dissolves with the production of bubbles when HNO₃ is added. Thus, the anion is CO₃²-.
Therefore, the compound is Ag₂CO₃ which is formed from the cation Ag+ and CO₃²-.
An element of the environment that causes a population the decrease is
A) the birthrate minus the death rate
B) the carrying capacity
C) a limiting factor
D) the death rate
Consider the arrangement of gases shown below. If the value between the gases is opened and the temperature is held constant, determine the following.
Answer:
I don't know what to say . just for points
How many molecules in each sample?
64.7 g N2
83 g CCl4
19 g C6H12O6
Answer:
1.39x10²⁴ molecules N₂.25x10²³ molecules CCl₄6.38x10²² molecules C₆H₁₂O₆Explanation:
First we convert the given masses into moles, using the compounds' respective molar mass:
64.7 g N₂ ÷ 28 g/mol = 2.31 mol N₂83 g CCl₄ ÷ 153.82 g/mol = 0.540 mol CCl₄19 g C₆H₁₂O₆ ÷ 180 g/mol = 0.106 mol C₆H₁₂O₆Then we multiply each amount by Avogadro's number, to calculate the number of molecules:
2.31 mol N₂ * 6.023x10²³ molecules/mol = 1.39x10²⁴ molecules0.540 mol CCl₄ * 6.023x10²³ molecules/mol = 3.25x10²³ molecules0.106 mol C₆H₁₂O₆ * 6.023x10²³ molecules/mol = 6.38x10²² molecules. The nucleophile in the reaction is _______ b. The Lewis acid catalyst in the reaction is ______ c. This reaction proceeds___________(faster or slower) than benzene. d. Draw the structure of product D
Answer:
a. eletrophile
b. able to impose regioselectivity and stereo selectivity.
c. faster
Explanation:
Necleophile reaction is chemical reaction in which electron rich chemical specie replaces functional group with another electron deficient molecule. Lewis acid catalyst is organic chemical reaction which lewis acid act as electron pair acceptor. Nucleophile reaction proceeds about 25 times more faster than benzene.
In a sample of oxygen gas at room temperature, the average kinetic energy of all the balls stays constant. Which postulate of kinetic molecular theory best explains how this is possible?
A. Attractive forces between gas particles are negligible because the particles of an ideal gas are moving so quickly.
B. Collisions between gas particles are elastic; there is no net gain or loss of kinetic energy.
C. Gases consist of a large number of small particles, with a lot of space between the particles.
D. Gas particles are in constant, random motion, and higher kinetic energy means faster movement.
Answer:
Collisions between gas particles are elastic; there is no net gain or loss of kinetic energy.
Explanation:
When a gas is paced in a container, the molecules of the gas have little or no intermolecular interaction between them. There is a lot of space between the molecules of the gas.
The gas molecules move at very high speed and collide with each other and with the walls of container.
The collision of these particles with each other is perfectly elastic hence the kinetic energy of the colliding gas particles do not change.
What is unit? Write down the units of mass, temperature and power
Explanation:
a single thing, person, or group forming part of a whole There are 36 units in my apartment building. the least whole number : one. a fixed quantity (as of length, time, or value) used as a standard of measurement An inch is a unit of length.mass=kilogram (kg)
temperature=kelvin
power=watt
hope it helps
stay safe healthy and happy..how many lone pair electrons are on the central oxygen atom in the Lewis structure for dinitrogen pentoxide 9
Answer:
Two
Explanation:
Lone pairs are electron pairs on an atom that resides only with one of the atoms in a molecule.
Dinitrogen pentaoxide is shown in the image attached. There are five oxygen atoms and two nitrogen atoms in the molecule. The molecule has a total of 40 valence electrons.
There are two electrons present on the central oxygen atom in the Lewis structure of dinitrogen pentaoxide as shown in the image attached.
if salt and sand is mixed with distilled water, what will be the residue? and, what will be the filtrate?
Answer:
Filtration is a technique used as a remedy to separate mixes
Explanation:
If you have a sodium, then you will explore that salt dissolves but the sand is still the same.
If the salt in the resin water solution scanners, the sand remains the residue and passes through the filter paper.
All you have to do now is pleasant the salty water so that the water can evaporate, leaving the salt behind.
chemistry help!
This mysterious gas has a volume of 4.35 L and a pressure of 1.20 atm. If the pressure is changed to 0.95 atm, what will the new volume be? How do i set up this problem and solve it?
Refer to the attachment.
Hope this helps you...
In some sheep, the presence of horns is produced by an autosomal allele that is dominant in males and recessive in females.A horned female is crossed with a hornless male. One of the resulting F1 females is crossed with a hornless male. What proportion of the male and female progeny from this cross will have horns?(5 marks)
Answer:
1/2 f1 will cross
Explanation:
answer it
If 12.3 g of Cu is deposited at the cathode of an electrolytic cell after 5.50 h, what was the current used?
Answer:
1.88 A
Explanation:
Let's consider the reduction of copper in an electrolytic cell.
Cu²⁺ + 2 e⁻ ⇒ Cu
We can calculate the charge used to deposit 12.3 g of Cu using the following relations.
The molar mass of Cu is 63.55 g/mol.1 mole of Cu is deposited when 2 moles of electrons circulate.1 mole of electrons has a charge of 96486 C (Faraday's constant).The charge used is:
[tex]12.3 g \times \frac{1 molCu}{63.55gCu} \times \frac{2molElectron}{1molCu} \times \frac{96486C}{1molElectron} = 3.73 \times 10^{4} C[/tex]
We can convert 5.50 h to seconds using the conversion factor 1 h = 3600 s.
5.50 h × 3600 s/1 h = 1.98 × 10⁴ s
The current used is:
I = q/t = 3.73 × 10⁴ C/1.98 × 10⁴ s = 1.88 A
An acetic acid buffer containing 0.50 M acetic acid (CH3COOH) and 0.50 M sodium acetate (CH3COONa) has a pH of 4.74. What will the pH be after 0.0020 mol of HCl has been added to 100.0 mL of the buffer
Answer:
pH = 4.71
Explanation:
We can find the pH of a buffer (Mixture of weak acid: CH3COOH, and its conjugate base: CH3COONa) using H-H equation:
pH = pKa + log [CH3COONa] / [CH3COOH]
Where pH is the pH of the buffere = 4.74, pKa the pka of the buffer and [] could be taken as the moles of each reactant.
As initially [CH3COONa] = [CH3COOH], [CH3COONa] / [CH3COOH] = 1:
pH = pKa + log 1
4.74 = pKa
To solve this question we need to find the initial moles of each species, The CH3COONa reacts with HCl to produce CH3COOH. That means the moles of CH3COOH after the reaction are: Initial CH3COOH + Moles HCl
Moles CH3COONa: Initial CH3COONa - Moles HCl.
Moles CH3COOH:
0.100L * (0.50mol / L) = 0.050 moles CH3COOH + 0.0020 moles HCl =
0.052 moles CH3COOH
Moles CH3COONa:
0.100L * (0.50mol / L) = 0.050 moles CH3COONa - 0.0020 moles HCl =
0.048 moles CH3COONa
Using H-H equation:
pH = 4.74 + log [0.048 moles] / [0.052 moles]
pH = 4.71The pH be after 0.0020 mol of HCl has been added to 100.0 mL of the buffer is 4.71.
What is buffer solution?Buffer solution is a mixture of weak acid and its conjugate base or vice versa.
We can calculate the pH of buffer solution by using Henderson - Hasselbalch Equation:
pH = pKa + log[CH₃COONa] / [CH₃COOH]
Initially concentration of CH₃COONa is equal to the concentration of CH₃COOH and equal becomes:
4.74 = pKa + log(1)
pKa = 4.74
Given moles of added HCl moles = 0.002 mole
Given molarity of each CH₃COOH & CH₃COONa = 0.50M
Given Volume = 100mL
We can calculate the moles by using the formula:
n = M × V
Moles of CH₃COOH & CH₃COONa = 0.100 × 0.50 = 0.050 moles
Moles of CH₃COOH = 0.050moles CH₃COOH + 0.0020moles HCl =
0.052moles CH₃COOH
Moles of CH₃COONa = 0.050moles CH3COONa - 0.0020moles HCl =
0.048moles CH₃COONa
Now, resultant pH will be:
pH = 4.74 + log [0.048 moles] / [0.052 moles]
pH = 4.71
Hence, pH of resultant solution is 4.71.
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A soluble unknown has contaminated your sample. It absorbs the same wavelength as your analyte, Allura Red dye. How will this affect your results, and what type of error is introduced
Answer:
The explanation as per the given query is presented in the following paragraph.
Explanation:
The Allura Reds Pigment or coloring increases the absorption by absorbing at around a very similar wavelength as that of the analysis.Therefore, the approximated analysis concentration or its intensity seemed substantially higher than that of the authentic one. Some positive mistake is going to exist.Thus the above is the correct explanation.
The measured absorbance will be higher, thereby creating systematic error.
Absorbance can be defined as the value or quantity of light absorbed by a given sample.The presence of impurity with a similar absorption wavelength will increase the absorbance. Moreover, systematic error is a type of error that is repeated along with different processes of measurement.In conclusion, the measured absorbance will be higher, thereby creating systematic error.
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howtocalculatethevolumeofcarbondioxideproducedwhen400gofmarblewereats.t.p
Answer:
so 0.15 moles X 22.4 dm3/mole=3.36 dm3. Next we find the moles of hexane combusted, and then the moles of CO2. Finally, we find the volume of CO2 using the fact that at STP, 1 mole of gas = 22.4 dm3.
A sample of Br2(g) takes 12.0 min to effuse through a membrane. How long would it take the same number of moles of Ar(g) to effuse through the same membrane
Answer:
6 mins
Explanation:
The time taken for Ar to effuse can be obtained as follow:
Time for Br₂ (t₁) = 12 mins
Molar mass of Br₂ (M₁) = 2 × 80 = 160 g/mol
Molar mass of Ar (M₂) = 40 g/mol
Time for Ar (t₂) =?
t₂/t₁= √(M₂/M₁)
t₂ / 12 = √(40/160)
Cross multiply
t₂ = 12 × √(40/160)
t₂ = 12 × 0.5
t₂ = 6 mins
Therefore, it will take 6 mins for the same amount of Ar to effused out.
The energy needed to collapse the H-bonding of ice is _____.
it depends on the widths of the recesses, and if it is causing the shelf to fracture and collapse into the sea, then a massive iceberg could be called from the life she,f and the ice shelf are way more important because it holds it up
What is the phase change from solid to gas?
O A. Condensation
O B. Sublimation
O C. Freezing
O D. Vaporization
Answer:
The answer is B, sublimation.
Answer:
The correct answer
B . Sublimation
