Answer:
96 mmHg
[tex]h=96mmHg[/tex]
Explanation:
From this question,manometer end is closedw, So we can deduced that the height of the column will not be affected by the atmospheric pressure .
The difference of height of the mercury level is given as,
h=9.60cm
h=9.60(10mm/1cm)
[tex]h=96mm[/tex]
But it is obvious that in this closed end manometer.the pressure of the gas is equal to the height
P(gas)=h
P(gas)=96mmHg
This pressure is as a result of the presence of gas.
Therefore, the pressure of the argon gas in the container is 96mmHg.
The pressure of the argon in the container was 96mmHg.
We were told that the manometer has closed ends which means that the
height will not be affected by atmospheric pressure.
The height which is the difference in mercury level is
h=9.60cm
We can convert it to millimeter by multiplying it by 10
h=9.60 × 10 = 96mm
The pressure of the closed end manometer will be equal to the height
P(gas)=h
P(gas)=96mmHg
The pressure of the argon gas in the container is 96mmHg.
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Consider the following practical aspects of titration.
(a) how can you tell when nearing the end point in titration?
(b) What volume of NaOH is required to permanently change the indicator at the end point?
(c) If KHP sample #1 requires 19.90 mL of NaOH solution to reach an end point, what volume is required for samples #2 and #3?
(d) if vinegar sample #1 requires 29.05 mL of NaOH solution to reach an endpoint, what volume is required for samole #2 and #3?
Answer:
A) when the titration is nearing the end point in titration the color of the solution starts to change and the change in color does not disappear as fast as it was during the beginning of the titration
B) The volume of NaOH required to permanently change the indicator at the end point is a drop of NaOH
c) The volume required by samples #2 and #3 will be the same i.e 19.90 mL of NaOH, if the concentration of KHP base used for the samples are the same
D) The volume required by samples #2 and #3 will be the same i.e 29.05 mL OF NaOH, if the concentration of Vinegar and base for the samples are the same
Explanation:
A) when the titration is nearing the end point in titration the color of the solution starts to change and the change in color does not disappear as fast as it was during the beginning of the titration
B) The volume of NaOH required to permanently change the indicator at the end point is a drop of NaOH
c) The volume required by samples #2 and #3 will be the same i.e 19.90 mL of NaOH, if the concentration of KHP base used for the samples are the same
D) The volume required by samples #2 and #3 will be the same i.e 29.05 mL OF NaOH, if the concentration of Vinegar and base for the samples are the same
What is buffers and mention its importance?
Answer:
Buffer is the chemical substance that addition of acids and bases, maintaining constant environment,its called Buffer.
Explanation:
Buffers are use in the system to maintain the value of pH, and the contain the pH value is not to change.Buffer maintain the body of pH for the optimal activity,and they are solution of pH constant.Buffer in used in the lab and that to maintain growth of the micro tissues and the culture media.Buffer are used in maintain necessary optimal reaction activity,determine the indicator of solution with pH.Buffer capacity is that concentration to the buffering agent, is the very small increase,buffer capacity to the pH is 32% , of the maximum value of pH.Buffers in a acid regions to the desired of that value to the particular buffer agent.Buffers can be made from that a mixture of the base and acid, buffer can be a wide range of the obtained.Buffers that the pH calculation and they required to performed in the critic acid that the overlap over the buffer range.19. For the following unbalanced equation: S (s) O2 (g) H2O (l) ----> H2SO4 (aq) At what temperature (K) does O2 have to be if the volume of the gas is 5.01 L with a pressure of 0.860 atm to produce 17.55 grams of H2SO4
Answer:
195.5 K
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2S + 3O2 + 2H2O → 2H2SO4
Next, we shall determine the number of mole in 17.55 g of H2SO4. This can be obtained as follow:
Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 + 64 = 98 g/mol
Mass of H2SO4 = 17.55 g
Mole of H2SO4 =...?
Mole = mass /Molar mass
Mole of H2SO4 = 17.55/98
Mole of H2SO4 = 0.179 mole.
Next, we shall determine the number of mole of O2 needed for the reaction. This can be obtained as illustrated below:
From the balanced equation above,
3 moles of O2 reacted to produce 2 moles of H2SO4.
Therefore, Xmol of O2 will react to produce 0.179 moleof H2SO4 i.e
Xmol of O2 = (3 x 0.179) / 2
Xmol of O2 = 0.2685 mole
Therefore, 0.2685 mole of O2 is needed for the reaction.
Finally, we shall determine the temperature of O2. This can be obtained by using the ideal gas equation as follow:
Volume (V) of O2 = 5.01 L
Pressure (P) = 0.860 atm
Number of mole (n) of O2 = 0.2685 mole
Gas constant (R) = 0.0821 atm.L/Kmol
Temperature (T) =..?
PV =nRT
0.860 x 5.01 = 0.2685 x 0.0821 x T
Divide both side by 0.2685 x 0.0821
T = (0.860 x 5.01) /(0.2685 x 0.0821)
T = 195.5 K
Therefore, the temperature of O2 must be 195.5 K.
By December 31, 2003, concerns over arsenic contamination had prompted the manufacturers of pressure-treated lumber to voluntarily cease producing lumber treated with CCA (chromated copper arsenate) for residential use. CCA-treated lumber has a light greenish color and was widely used to build decks, sand boxes, and playground structures.
Required:
Draw the Lewis structure of the arsenate ion (ASO4^3-) that yields the most favorable formal charges.
Answer:
Explanation:
lewis structure can be defined as a process of how the valence shell electrons of a molecule is being arranged, the pattern of it arrangement and the relationship between the bonding atoms and the lone pairs present in the molecule.
In order to draw the Lewis structure for Arsenate ion [tex]\mathsf{AsO \ _4^{3-}}[/tex], first thing is to count the valence electrons in the molecule. Once we determine the valence electrons, then we distribute them around the central atom. The Arsenate ion structure is tetrahedral in nature with a bond angle of 109.5° and it is sp³ hybridized.
You have to prepare a pH 3.65 buffer, and you have the following 0.10M solutions available: HCOOH, CH3COOH, H3PO4, HCOONa, CH3COONa, and NaH2PO4. How many mL of HCOOH and HCOONa would you use to make approximately a liter of the buffer?
Answer:
550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M
Explanation:
It is possible to find the pH of a buffer by using H-H equation:
pH = pKa + log [A⁻]/[HA]
For the formic buffer (HCOOH/HCOONa):
pH = 3.74 + log [HCOONa]/[HCOOH]
As you need a buffer of pH 3.65:
pH = 3.74 + log [HCOONa]/[HCOOH]
3.65 = 3.74 + log [HCOONa]/[HCOOH]
0.81283 = [HCOONa]/[HCOOH] (1)Where [HCOONa]/[HCOOH] can be taken as the moles of each specie.
As molarity of both solutions is 0.10M (0.10mol / L) and you need 1L of solution, total moles of the buffer are:
0.10 moles = [HCOONa] + [HCOOH] (2)Replacing (2) in (1):
0.81283 = 0.10 - [HCOOH] /[HCOOH]
0.81283[HCOOH] = 0.10 - [HCOOH]
1.81283[HCOOH] = 0.10
[HCOOH] = 0.055 molesAnd moles of HCOONa are:
[HCOONa] = 0.1 mol - 0.055mol =
[HCOONa] = 0.045 molesAs concentration of the solutions is 0.1M, the volume you need to add of both solutions is:
HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M
HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M
The number should be considered like 550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M.
Calculation of mL:Here we used the H-H equation:
pH = pKa + log [A⁻]/[HA]
Now
For the formic buffer (HCOOH/HCOONa):
So,
pH = 3.74 + log [HCOONa]/[HCOOH]
Now
need a buffer of pH 3.65:
So,
pH = 3.74 + log [HCOONa]/[HCOOH]
3.65 = 3.74 + log [HCOONa]/[HCOOH]
0.81283 = [HCOONa]/[HCOOH] (1)
here [HCOONa]/[HCOOH] can be considered as the moles of each specie.
Now the total moles should be
0.10 moles = [HCOONa] + [HCOOH] (2)
Now
0.81283 = 0.10 - [HCOOH] /[HCOOH]
0.81283[HCOOH] = 0.10 - [HCOOH]
1.81283[HCOOH] = 0.10
[HCOOH] = 0.055 moles
And moles of HCOONa should be
[HCOONa] = 0.1 mol - 0.055mol =
[HCOONa] = 0.045 moles
Now
HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M
HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M
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What's the name for the part of Earth made of rock?
A. Geosphere
B. Atmosphere
C. Hydrosphere
D. Biosphere
SUBMIT
Answer:I think it's Geosphere
Explanation:
Answer:
A
Explanation:
Geo means rock, or earth. Hydro means water, Atmosphere is space, and Bio global ecosystem composed of living organisms
What type of chemist exclusively studies most carbon compounds?
-biochemist
-physical chemist
-organic chemist
-inorganic chemist
Answer:
Organic chemist? I do not know.
Explanation:
Thanks you.
The type of chemist exclusively studies most carbon compounds are organic chemist. Therefore, option C is correct.
What is an organic chemist ?The structure, characteristics, and reactivity of compounds containing carbon are studied by organic chemists. Additionally, they create novel organic materials with distinct features and uses.
Analytical capabilities, communication skills, and numeracy skills are three of the most important soft skills for an organic chemist.
Organic chemists often work in research and development in labs at universities, pharmaceutical, industrial, and biotechnology businesses, as well as government agencies, according to the American Chemical Society.
According to one assessment, organic chemistry is the hardest college course. According to certain statistics, almost one out of every two students in organic chemistry leave the course. The hopes of a medical career come tumbling down for those who fit this description. Organic chemistry is undoubtedly challenging.
Thus, option C is correct.
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The solubility of calcium oxalate, CaC2O4, in pure water is 4.8 × 10‑5 moles per liter. Calculate the value of Ksp for silver carbonate from this data.
Answer:
2.3 × 10⁻⁹
Explanation:
Step 1: Write the reaction for the solution of calcium oxalate
CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)
Step 2: Make an ICE chart
We can relate the molar solubility (S) with the solubility product constant (Ksp) through an ICE chart.
CaC₂O₄(s) ⇄ Ca²⁺(aq) + C₂O₄²⁻(aq)
I 0 0
C +S +S
E S S
The solubility product constant is:
Ksp = [Ca²⁺] × [C₂O₄²⁻] = S² = (4.8 × 10⁻⁵)² = 2.3 × 10⁻⁹
In the pictured cell, the side containing zinc is the Choose... and the side containing copper is the Choose... . The purpose of the N a 2 S O 4 NaX2SOX4 is to
Answer:
Zinc- anode
Copper- cathode
Sodium sulphate- salt bridge
Explanation:
A galvanic cell is an electrochemical cell in which electrical energy is produced by a spontaneous chemical reaction.
In the pictured galvanic cell, zinc is the anode since it looses electrons according to the reaction; Zn(s) -----> Zn^2+(aq) + 2e
Copper is the cathode as shown here; Cu^2+(aq) + 2e ----> Cu(s)
Sodium sulphate functions as the salt bridge. It keeps the both solutions neutral by ensuring charge balance in the both half cells.
Answer:
zinc=anode
copper=cathode
Explanation:
Predict the most likely bond type for the following.
a. Cu (Copper)
b. KCl (Potassium Chloride)
c. Si (Silicon)
d. CdTe (Cadmium Telluride)
e. ZnTe (Zinc Telluride)
Answer:
The three major types of bond are ionic, polar covalent, and covalent bonds. Ionic occurs majorly between metals and non-metals, which allows sharing of electrons to form an ionic compound. Whereas covalent bonding calls for complete transfer of electrons between atoms. Polar covalent bonds have unequaly shared electron-pair between two atoms.
Explanation:
a. Cu (Copper)- ionic bonding
b. KCl (Potassium Chloride) - ionic bonding
c. Si (Silicon) - covalent bonding
d. CdTe (Cadmium Telluride) - polar covalent bonding
e. ZnTe (Zinc Telluride)- polar covalent bonding
At 25 °C, only 0.0990 mol of the generic salt AB3 is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C? AB3(s)↽−−⇀A3+(aq)+3B−(aq)
Answer:
[tex]Ksp=2.59x10^{-3}[/tex]
Explanation:
Hello,
In this case, given the 0.0990 moles of the salt are soluble in 1.00 L of water only, we can infer that the molar solubility is 0.099 M. Next, since the dissociation of the salt is:
[tex]AB_3\rightleftharpoons A^{3+}+3B^-[/tex]
The concentrations of the A and B ions in the solution are:
[tex][A]=0.099 \frac{molAB_3}{L}*\frac{1molA}{1molAB_3} =0.0099M[/tex]
[tex][B]=0.099 \frac{molAB_3}{L}*\frac{3molB}{1molAB_3} =0.000.297M[/tex]
Then, as the solubility product is defined as:
[tex]Ksp=[A][B]^3[/tex]
Due to the given dissociation, it turns out:
[tex]Ksp=[0.099M][0.297M]^3\\\\Ksp=2.59x10^{-3}[/tex]
Regards.
Which of the following compounds is more soluble in a 0.10 M NaCN solution than in pure neutral water? Ca3(PO4)2 AgBr CaCO3 Mg(OH)2 NH4ClO4
Answer:
AgBr
Explanation:
Silver bromide has a very low solubility product constant of about 7.7 ×10^-13 in pure water hence it is not quite soluble in pure water.
However, with NaCN, the AgBr forms the complex [Ag(CN)2]^2- which has a formation constant of about 5.6 ×10^8. This very high formation constant implies that the complex is easily formed leading to the dissolution of AgBr in NaCN.
The equation for the dissolution of AgBr in cyanide is shown below;
AgBr(s) + 2CN^-(aq) ----> [Ag(CN)2]^2-(aq) + Br^-(aq)
The mole is a counting number that allows scientists to describe how individual molecules and atoms react. If one mole of atoms or molecules is equal to 6.022 x 10^32 atoms or molecules, how many molecules are in 23.45 g sample of copper (II) hydroxide, Cu(OH)2? Express your answer to the correct number of significant figures. (MM of Cu(OH)2 is 97.562g/mol. Be sure to show all steps completed to arrive at the answer.
Answer:
[tex]\large \boxed{1.503 \times 10^{23}\text{ molecules of Cu(OH)}_{2}}$}[/tex]
Explanation:
You must calculate the moles of Cu(OH)₂, then convert to molecules of Cu(OH)₂.
1. Moles of Cu(OH)₂[tex]\text{Moles of Cu(OH)}_{2} = \text{24.35 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \text{0.2496 mol Cu(OH)}_{2}[/tex]
2. Molecules of Cu(OH)₂[tex]\text{No. of molecules} = \text{0.2496 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= \mathbf{1.503 \times 10^{23}}\textbf{ molecules Cu(OH)}_{2}\\\text{There are $\large \boxed{\mathbf{1.503 \times 10^{23}}\textbf{ molecules of Cu(OH)}_{2}}$}[/tex]
The number of molecules of copper (II) hydroxide in 23.45 g sample has been [tex]\rm \bold {1.45\;\times\;10^2^3}[/tex].
According to the Avogadro number, the number of molecules in a mole of atom has been equivalent to the Avogadro constant. The value of Avogadro constant has been [tex]\rm 6.023\;\times\;10^2^3[/tex].
The moles of a compound has been given as:
[tex]\rm Moles=\dfrac{Mass}{Molar\;mass}[/tex]
The moles in 23.45 g copper (II) hydroxide has been:
[tex]\rm Moles=\dfrac{23.45}{97.562} \\Moles=0.24\;mol[/tex]
The moles of copper (II) hydroxide has been 0.24 mol.
The number of molecules in 0.24 mol sample has been driven by:
[tex]\rm 1\;mol=6.023\;\times\;10^2^3\;molecules\\0.24\;mol=0.24\;\times\;10^2^3\;molecules\\0.24\;mol=1.45\;\times\;10^2^3\;molecules[/tex]
The number of molecules of copper (II) hydroxide in 23.45 g sample has been [tex]\rm \bold {1.45\;\times\;10^2^3}[/tex].
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A compound (C_9H_9BrO_2) gives the following NMR data. Draw the structure of the compound.
'1^H-NMR: 1.39 ppm, t(3H); 4.38 ppm, q(2H); 7.57 ppm, d(2H); 7.90 ppm, d(2H)
13^C-NMR: 165.73; 131.56; 131.01; 129.84; 127.81; 61.18; 14.18
You do not have to consider stereo chemistry.
You do not have to explicitly draw H atoms.
Do not include lone pairs in your answer.
Answer:
ethyl 4-bromobenzoate
Explanation:
In this question, we can start with the Index of Hydrogen Deficiency (I.H.D):
[tex]I.H.D=\frac{2C+2+N-H-X}{2}=\frac{(2*9)+2+0-9-1}{2}~=~5[/tex]
This indicates, that we can have a benzene ring (I.H.D = 4) and a carbonyl group (I.H.D = 1), for a total of 5.
Additionally, in the 1H-NMR info, we have a triplet 1.39 (3H) followed by a doublet 4.38 (2H), this indicates the presence of an ethyl group ([tex]CH_3-CH_2-[/tex]). Also, in the formula, we have 2 oxygens if we have carbonyl group with 2 oxygens we have a high probability to have an ester group.
[tex]O=C-O-CH_2-CH_3[/tex]
Now, if we add this to the benzene ring and the "Br" atom that we have in the formula, we will have ethyl 4-bromobenzoate.
See figures 1 and 2 to further explanations.
I hope it helps!
How many milliliters of 7.10 M hydrobromic acid solution should be used to prepare 5.50 L of 0.400 M HBr
Answer:
310 mL
Explanation:
Step 1: Given data
Initial concentration (C₁): 7.10 MInitial volume (V₁): ?Final concentration (C₂): 0.400 MFinal volume (V₂): 5.50 LStep 2: Calculate the initial volume
We have a concentrated HBr solution and we want to prepare a diluted one. We can do so using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 0.400 M × 5.50 L / 7.10 M
V₁ = 0.310 L = 310 mL
A scientist observes that the electrical resistance of a superconducting material drops to zero when the material is cooled to very low temperatures. Which of the following statements best describes what the scientist is observing?
The scientist is observing the electrical power of a superconductor.
The scientist is observing the temperature of a superconductor.
The scientist is observing an intensive property of a superconductor.
The scientist is observing an extensive property of a superconductor
Answer:
The scientist is observing an intensive property of a superconductor.
Explanation:
An intensive property is a bulk property of matter. This means that an intensive property does not depend on the amount of substance present in the material under study. Typical examples of intensive properties include; conductivity, resistivity, density, hardness, etc.
An extensive property is a property that depends on the amount of substance present in a sample. Extensive properties depend on the quantity of matter present in the sample under study. Examples of extensive properties include, mass and volume.
Resistance of a superconducting material has nothing to do with the amount of the material present hence it is an intensive property of the superconductor.
Answer:
The scientist is observing an intensive property of a superconductor.
Explanation:
its the only one that makes logical sense
Determine the molar solubility of AgBr in a solution containing 0.150 M NaBr. Ksp (AgBr) = 7.7 × 10-13.
Answer:
Molar solubility of AgBr = 51.33 × 10⁻¹³
Explanation:
Given:
Amount of NaBr = 0.150 M
Ksp (AgBr) = 7.7 × 10⁻¹³
Find:
Molar solubility of AgBr
Computation:
Molar solubility of AgBr = Ksp (AgBr) / Amount of NaBr
Molar solubility of AgBr = 7.7 × 10⁻¹³ / 0.150
Molar solubility of AgBr = 51.33 × 10⁻¹³
When The Molar solubility of AgBr is = 51.33 × 10⁻¹³
Calculation of Solubility of AgBr
Given as per question:
The Amount of NaBr is = 0.150 M
Then Ksp (AgBr) is = 7.7 × 10⁻¹³
Now we Find:
The Molar solubility of AgBr
The we Computation is:
The Molar solubility of AgBr is = Ksp (AgBr) / Amount of NaBr
After that Molar solubility of AgBr is = 7.7 × 10⁻¹³ / 0.150
Therefore, Molar solubility of AgBr is = 51.33 × 10⁻¹³
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Consider this synthesis of isoamyl acetate based on this week's experimental methods, after refluxing the reaction mixture for 25 minutes, what is likely present in solution
Answer:
acetic acid and phosphoric acid
Explanation:
After refluxing the reaction mixture ( synthesis of isoamyl acetate ) what is likely present in the solution is acetic acid and phosphoric acid, this due to the fact that if the reaction time between the reactants was less than the refluxing time which is 25 minutes,
there will be no reactant ( 3-methylbutanol )left in the reaction mixture
What is the initial temperature (°C) of a system that has the pressure decreased by 10 times while the volume increased by 5 times with a final temperature of -123°C?
Answer:
27°C or 300K
Explanation
We were told that the pressureof the system decreased by 10 times implies that P2= P1/10
Where P2=final pressure
P1= initial pressure
Wew were also told that the volume of the system increased by 5 times this implies that V2= 5×V1
Where T2= final temperature =-123C= 273+(-123C)=150K
T1= initial temperature
But from gas law
PV=nRT
As n and R are constant
P1V1/T1 = P2V2/T2
T1= P1V1T2/P2V2
T1=2×T2
T1=2×150
T1=300K
=300-273
=27°C
the initial temperature (°C) of a system is 27°C
Please tell the answer
Answer:
see the photo
Explanation:
it was the answer
What is the atomic mass for Helium (He)? Question 5 options: 8 2 3 4
Complete the sentences describing the cell.
a. In the nickel-aluminum galvanic cell, the cathode is ____ .
b. Therefore electrons flow from___ to ____.
c. The ____ electrode loses mass, while the ____ electrode gains mass.
Answer:
a. In the nickel-aluminum galvanic cell, the cathode is nickel electrode.
b. Therefore electrons flow from the aluminium electrode to the nickel electrode.
c. The aluminium electrode loses mass, while the nickel electrode electrode gains mass.
Explanation:
Voltaic or galvanic cells are electrochemical cells in which spontaneous oxidation-reduction reactions. The two halves of the redox reaction are separate and electron transfer is required to occur through an external circuit for the redox reaction to take place. That is, one of the metals in one of the half cells is oxidized while the metal of the other half cell is reduced, producing an exchange of electrons through an external circuit. This makes it possible to take advantage of the electric current.
Given:
E ⁰N i ⁺² = − 0.23 V is the standard reduction potential for the nickel ion
E ⁰ A l ⁺³ = − 1.66 V is the standard reduction potential for the aluminum ion
The most negative potentials correspond to more reducing substances. In this case, the aluminum ion is the reducing agent, where oxidation takes place. In the anodic half cell oxidations occur, while in the cathode half cell reductions occur. So the aluminum cell acts as the anode while the nickel cell acts as the cathode.
So a. In the nickel-aluminum galvanic cell, the cathode is nickel electrode.
The metal that is oxidized gives electrons to the metal that is reduced through the outer conductor. Then the electrons flow spontaneously from the anode to the cathode.
Then b. Therefore electrons flow from the aluminium electrode to the nickel electrode.
Ni⁺², being the cathode, accepts electrons, becoming Ni (s) and depositing on the Ni electrodes.
So, c. The aluminium electrode loses mass, while the nickel electrode electrode gains mass.
A 2.87g sample of carbon reacts with hydrogen to form 3.41g of car fuel. What is the empirical formula of the car fuel?
Answer:
The empirical formulae for the car fuel is C4H9
Al diluir 25 g de sal de mesa en 250ml de agua, ¿En cuántos °C aumenta el punto de ebullición de la disolución formada? ( Ke = 0,52 °C/molal , PM NaCl = 58,44 g/mol)
Answer:
ΔT=[tex]0.87^{\circ}C[/tex]
Explanation:
Para esta pregunta debemos recordar la ecuación que nos permite calcular el aumento ebulliscopico (aumento del punto de ebullición):
ΔT=[tex]Kb*m[/tex]
Donde ΔT es el valor del aumento del punto de ebullición. Kb es la constante ebulloscopica para el agua ([tex]0.512\frac{Kg~^{\circ}C}{mol}[/tex]) y m es la molalidad ([tex]m=\frac{mol}{Kg~ of~ solvente}[/tex]).
Por lo tanto el primer paso es calcular la molalidad de la solución. Para lo cual tendremos que calcular las moles de sal en los 25 g. Si queremos hacer esto debemos recordar que la formula de sal de mesa es NaCl y que la masa molar de NaCl es 58.44 g/mol. Por lo tanto:
[tex]25~g~NaCl\frac{1~mol~NaCl}{58.44~g~NaCl}~=~0.42~mol~NaCl[/tex]
Ahora bien, también debemos saber los Kg de agua en la solución. Por lo que podemos usar la densidad del agua (1 g/mL) para convertir de mL a g y luego hacer la conversión a Kg:
[tex]250~mL\frac{1~g}{1~mL}\frac{1~Kg}{1000~g}~=~0.25~Kg[/tex]
Finalmente para el calculo de la molalidad podemos dividir los dos valores:
[tex]m=\frac{0.42~mol}{0.25~Kg}=1.68[/tex]
Con el valor de la molalidad se puede calcular ΔT al reemplazar los valores:
ΔT=[tex]1.68~\frac{mol}{Kg}*0.52\frac{Kg~^{\circ}C}{mol}=0.87^{\circ}C[/tex]
Si la temperatura de ebullición normal del agua es 100 ºC. Podemos calcular la temperatura final si adicionamos ΔT:
Temperatura final = 100 + 0.87 = 100.87 ºC
Espero sea de ayuda!
Emission of which one of the following leaves both atomic number and mass number unchanged?
(a) positron
(b) neutron
(c) alpha particle
(d) gamma radiation
(e) beta particle
Answer: Gamma Radiation
Explanation:
The emission of Gamma rays does not cause a change in both the atomic and mass number. They are electromagnetic radiation.
The radiations that leaves without changing the atomic mass and atomic number of the particle have been gamma radiations. Thus, option D is correct.
Radiations have been the energy that has been evolved by the particles during energy transitions. The nuclear decay results with the release of the energy from the particle resulting in the change in the atomic mass.
The electromagnetic radiations have been capable of emitting the radiation without changing the mass and atomic number of the element. The gamma radiations have been the electromagnetic radiations. Thus, option D is correct.
For more information about the emissions, refer to the link:
https://brainly.com/question/517329
For which of the following elements (in their normal, stable, forms) would it be correct to describe the bonding as involving "a sea of electrons"?
a. hydrogen
b. nellum
c. sulfur
d. Iodine
e. Ethium
Answer:
e. Lithium
Explanation:
Correct list of options!
a. hydrogen b. Helium c. sulfur d. Iodine e. Lithium
Sea of electrons generally refers to metal atoms. This is because of the delocalized nature of the electrons compared to non metals where the electrons are localized (fixed to a specific atoms).
Among all the elements in the options, the metal is option e. Lithium
what is a chemical that is safe to use in food but in small amounts?
Answer:
Toxins
Explanation:
H2S(g) 2H2O(l)3H2(g) SO2(g) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.60 moles of H2S(g) react at standard conditions.
Answer: [tex]\Delta S[/tex] = 473.92J/K.mol
Explanation: In physics, Entropy is defined as a degree of disorder in a system. Entropy change is given by the sum of all the products multiplied by their respective coeficients minus the sum of all the reagents multiplied by their respective coeficients:
[tex]\Delta S = m\Sigma product - n\Sigma reagent[/tex]
The balanced reaction:
[tex]H_{2}S_{(g)}+2H_{2}O_{(l)}=>3H_{2}_{(g)}+SO_{2}_{(g)}[/tex]
gives the proportion reagents react to form products, so, if 1.6 moles of [tex]H_{2}S_{(g)}[/tex]:
3.2 moles of water is used;
4.8 moles of hydrogen gas is formed;
1.6 moles of sulfur dioxide is also formed;
Calculating entropy change:
[tex]\Delta S = (4.8*131+1.6*248.8)-(1.6*205.6+3.2*70)[/tex]
[tex]\Delta S=628.8+398.08-328.96-224[/tex]
[tex]\Delta S[/tex] = 473.92J/K.mol
Entropy change for the given chemical reaction is [tex]\Delta S[/tex] = 473.92J/K.mol
A quantity of liquid methanol, CH 3OH, is introduced into a rigid 3.00-L vessel, the vessel is sealed, and the temperature is raised to 500K. At this temperature, the methanol vaporizes and decomposes according to the reaction CH 3OH(g) CO(g) + 2 H 2(g), K c= 6.90×10 –2. If the concentration of H 2 in the equilibrium mixture is 0.426M, what mass of methanol was initially introduced into the vessel?
Answer:
74.3g of methanol were introduced into the vessel
Explanation:
In the equilibrium:
CH₃OH(g) ⇄ CO(g) + 2H₂(g)
Kc is defined as the ratio between concentrations in equilibrium of :
Kc = 6.90x10⁻² = [CO] [H₂]² / [CH₃OH]
Some methanol added to the vessel will react producing H₂ and CO. And equilibrium concentrations must be:
[CH₃OH] = ? - X
[CO] = X
[H₂] = 2X
Where ? is the initial concentration of methanol
As [H₂] = 2X = 0.426M; X = 0.213M
[CH₃OH] = ? - 0.213M
[CO] = 0.213M
[H₂] = 0.426M
Replacing in Kc to solve equilibrium concentration of methanol:
6.90x10⁻² = [0.213] [0.426]² / [CH₃OH]
[CH₃OH] = 0.560
As:
[CH₃OH] = ? - 0.213M = 0.560M
? = 0.773M
0.7733M was the initial concentration of methanol. As volume of vessel is 3.00L, moles of methanol are:
3.00L * (0.773 mol / L) = 2.319 moles methanol.
Using molar mass of methanol (32.04g/mol), initial mass of methanol added was:
2.319 moles * (32.04g / mol) =
74.3g of methanol were introduced into the vesselThe recommended application for dicyclanil for an adult sheep is 65 mg/kg of body mass. If dicyclanil is supplied in a spray with a concentration of 50. mg/mL, how many milliliters of the spray are required to treat a 70.-kg adult sheep?
Answer:
91 millilitres
Explanation:
Recommended application = 65mg / Kg
This means 65 mg of dicyclanil per kg (1 kg of body mass).
Concentration = 50 mg / mL
How many millilitres required to treat 70kg adult?
If 65mg = 1 kg
x = 70 mg
x = 70 * 65 = 4550 mg
Concentration = Mass / Volume
50 mg/mL = 4550 / volume
volume = 4550 / 50 = 91 mL
