Answer:
The difference is due to the ion-pairing effect which effectively reduces the number of solute particles present in the solution.
Explanation:
Colligative properties are those properties that depend on the amount of solute present. Such properties include; boiling point elevation, freezing point depression etc.
Ion pairing causes the Van't Hoff factor to deviate from whole numbers. Ion pairing effect generally reduces the number of solute particles in solution thereby decreasing the experimental value of the Van't Hoff factor (i).
Hence, the reason why Rita determined the Van't Hoff factor as 1.9 and not the theoretical value of 2 is because of on-pairing effect which effectively reduces the number of solute particles present in the solution.
The difference between the theoretical and experimental is A. The difference is due to the ion-pairing effect which effectively reduces the number of solute particles present in the solution.
Colligative propertiesIt should be noted that colligative properties simply means the properties that depend on the amount of solute present.
The ion pairing causes the Van't Hoff factor to deviate from whole numbers. Therefore, they caused the difference between the theoretical and experimental values.
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What would you use to measure
to measure an amount of solid sodium chloride(NaCI)
crystals to add to a 30 mL solution?
A. Triple Beam Balance that measures to the nearest 0.1 grams.
) B. 10 ml graduated cylinder.
Scale that can weigh up to 250 lbs.
D. Spectrophotometer that measures between 100 and 900 nanometers.
You HAVE 1L of 2.0 mol/L HCI. You want to dilute the solution to a concentration of 0.1 mol/L. What volume of the new solution (V2)?
0.05 L
O 20L
0.2 L
M₁ = 2.0 mol/L
V₁ = 1 L
M₂ = 0.1 mol/L
Required:V₂
Solution:M₁V₁ = M₂V₂
V₂ = M₁V₁ / M₂
V₂ = (2.0 mol/L)(1 L) / (0.1 L)
V₂ = 20 L
Therefore, the volume of the new solution will be 20 L.
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M₁V₁ = M₂V₂
V₂ = M₁V₁ / M₂
V₂ = (2.0 mol/L)(1 L) / (0.1 L)
V₂ = 20 L
THE BEST ANSWER IS
O 20L
Which of the following ions is formed when an acid is dissolved in a solution? (5 points)
H+
O−
OH−
SO42+
Answer:
H +
Explanation:
Acid are those species which donates the H+ ion when dissolved in a solution. In other way acid are those chemical species which donate proton to other species as proton means H+ ion so H+ ions are formed as a result of dissolution ..
Identify the indicated protons in the following molecules as unrelated, homotopic, enantiotopic, or diastereotopic. a) Methyls a & b: _________ b) Ha & Hc: _________
Answer:
Identify the indicated protons in the following molecules as unrelated, homotopic, enantiotopic, or diastereotopic. a) Methyls a & b: _________ b) Ha & Hc: ________
Explanation:
Homotopic hydrogens:
Consider two hydrogens in the given molecule and replace one by one with a different atom say for example deuterium, then if the two molecules formed by replacing hydrogens are the same then the two hydrogens are called homotopic hydrogens.
After replacing the two hydrogens with a different atom then, enantiomers are formed then, the two hydrogens are called enantiotopic hydrogens.
After replacing the two hydrogens with a different atom then, diastereomers are formed then, the two hydrogens are called diastereotopic hydrogens.
In the methyl group, select two hydrogens and replace one hydrogen atom with a D-atom name the compound.
Again replace another hydrogen atom with D-atom.
Name it.
If both are the same then, the hydrogens are homotopic and they are shown below:
Hence, they are homotopic protons.
3. 3. A 90.-mL sample of juice was titrated with the I2(aq) solution described above using a sb buret. The initial reading of the sb buret was 58.3 mL. When the endpoint was reached, the reading on the sb buret was 26.9 mL. How many mg of Vitamin C were in the juice sample
Answer:
2765mg of Vitamin C assuming the concentration of I2 = 0.500M
Explanation:
Iodine, I2, reacts with vitamin C in a ratio of 1:1 (The moles of I2 added = Moles of Vitamin C in the sample). To solve this question we must find the difference in volume from the initial reading to the endpoint reading. Now, ASSUMING the molar concentration of the I2(aq) solution is 0.500M we can find the moles of I2 = Moles of Vitamin C. With its molar mass -176.12g/mol- we can find its mass:
Moles I2:
Volume = 58.3mL - 26.9mL = 31.4mL = 0.0314L * (0.500mol / L) = 0.0157 moles I2 = Moles Vitamin C
Mass Vitamin C:
0.0157 moles * (176.12g/mol) = 2.765g of Vitamin C * (1000mg/1g) =
2765mg of Vitamin C assuming the concentration of I2 = 0.500M
12.0: A
Mention three body fluids that are alkaline in nature
Whether or not the process is observed in nature, which of the following could account for the transformation of phosphorus-28 to silicon-28?
a. beta decay.
b. eletron capture
c. positron emission
d. alpha decay
Answer:
positron emission
Explanation:
Positron emission,involves the decrease in the nuclear positive charge thus the atomic number of the daughter nucleus decreases by one unit compared to that of the parent nucleus while the mass number of the daughter nucleus remains the same as that of the parent nucleus.
Thus, when phosphorus-28 is transformed into to silicon-28, a positron emission has taken place because the mass number of the daughter nucleus remained the same while the atomic number of the daughter nucleus was decreased by one unit.
To prepare a sample in a capillary tube for a melting point determination, gently tap the tube into the sample with the ___________ end of the tube down. Continue tapping until the sample. _____________ Then, with the __________ end of the tube down, tap the sample down slowly or __________ to move the sample down faster. Finally, make sure that you can see ___________ in the magnifier when placed in the melting point apparatus before turning on the heat.
Answer: Hello I was able to find the missing options related to your question online but I couldn't attach them hence I just used them to provide the relevant solution
answer:
Open Is a couple of millimeters highclosed Drop tube into a longer tube sample clearlyExplanation:
OPEN end of the tube down. Continue tapping until the sample. Is a couple of mm high Then, with the Closed end of the tube down, tap the sample down slowly or drop tube into a longer/larger tube to move the sample down faster. Finally, make sure that you can see sample clearly in the magnifier when placed in the melting point apparatus before turning on the heat.
Round 5578 L to three significant figures.
a. 557 L
b. 558 L
c. 5.58 x 10(3) L
d. 5.58 x 10(-3) L
Answer:
b
Explanation:
The figures of a number that are noteworthy in high accuracy or precision are known as significant figures. They are as follows:
Any digit that isn't zeroAs in 20012 or 64.60007, there are zeros between non-zero numbers.Only use leading zeros when a decimal point is present, such as in 5640.0 or 532.330.If the figure is less than 5, remove it and leave the rest of the number unaffected but if more than 5 add, round it up to 1, and add it to the next figure in the line.From the given information:
To (3) sig. fig: 5578 L = 558 L
A flexible vessel contains 65.8 L of gas at a pressure of 2.07 atm. Under the conditions of constant temperature and constant number of moles of gas, what is the pressure of the gas (in atm) when the volume of the vessel increased by a factor of 16.00
Answer: Pressure of the gas is 0.129375 atm when the volume of the vessel increased by a factor of 16.00.
Explanation:
The formula for ideal gas equation is as follows.
[tex]PV = Nk_{b}T[/tex]
where,
[tex]k_{b}[/tex] = Boltzmann constant
N = number of moles
That can also be written as:
[tex]\frac{PV}{T} = constant[/tex]
As pressure and volume are inversely proportional to each other. So, if one of the state variable is increased then the other one will decrease or vice-versa.
So, if volume of the vessel increased by a factor of 16.00 then it means pressure is decreased by a factor of 16.00
Therefore, final volume is as follows.
[tex]65.8 L \times 16.00\\= 1052.8 L[/tex]
Now, final pressure is as follows.
[tex]\frac{2.07}{16.00}\\= 0.129375 atm[/tex]
Initially the product of pressure and volume is as follows.
[tex]PV = 2.07 \times 65.8\\= 136.206[/tex]
Hence, if volume of the vessel increased by a factor of 16.00 and pressure is decreased by a factor of 16.00 then its product is as follows.
[tex]PV = 0.129375 \times 1052.8\\= 136.206[/tex]
Here, product of pressure and volume remains the same.
Thus, we can conclude that pressure of the gas is 0.129375 atm when the volume of the vessel increased by a factor of 16.00.
0.159 mol 2.25 M = A L of HCI
Answer:
hmmm?what?
Explanation:
ayusin mo flece:>
What waves are used to send signals because they can travel over very long distances and pass through buildings
Answer:
sound waves for used I think
What are the lengths of the diagonals of the kite?
The answer ( 13 and 8 )
x²=5²+12²
x²=25+144
x²=169
x=13
x²=5²+6²
x²=25+36
x²=61
x=7.8
x=8
Convert the following:
1) 367.5 mg = _______ g
2) 367 mL = _______ L
3) 28.59 in =______ cm
4) 8 0z =_______lb
5) 0.671 mm =_____m
Draw the predominant product(s) of the following reactions including stereochemistry when it is appropriate. CH3CH2 C C CH3 H2O/H2SO4/HgSO4
Answer:
Draw the predominant product(s) of the following reactions including stereochemistry when it is appropriate.
CH3CH2 C C CH3 H2O/H2SO4/HgSO4
Explanation:
The given compound is: pent-2-yne.
When it reacts with water, in presence of sulphuric acid and mercuric sulphate then a ketone is formed as shown below:
This reaction is an example of nucleophilic attack of water on carbon carbon triple bond.
The general mechanism of the reaction is hsown below:
Pent-2-yne reacts with water and form 3-pentanone.
The reaction is shown below:
Which state of matter is characterized by having an indefinite shape, but a definite volume?
solid
gas
liquid
Answer:
liquid is the right answer k
Answer:
liquid
Explanation:
What is the speed of a wave with a frequency of 2 Hz and a wavelength of 87m (subject is science) pls answer fast
Answer:
43.5
Explanation:
Hope that helps
Hydrogen and chlorine gases react to form HCl. You and a friend are on opposite sides of a long hallway, you with H2 and your friend with Cl2. You both want to form HCl in the middle of the room. Which of the following is true?
a) You should release the H2 first.
b) Your friend should release the Cl2 first.
c) You both should release the gases at the same time.
d) You need to know the length of the room to answer this question.
e) You need to know the temperature to answer this question.
Answer: It is true that Your friend should release the [tex]Cl_{2}[/tex] first.
Explanation:
The velocity of particles of a gas is inversely proportional to the mass of gas. This means that more is the mass of gas less will be its velocity.
Or, more will be the mass of gas more slowly it will move from one place to another.
The molar mass of chlorine gas is more than the molar mass of hydrogen gas. Therefore, chlorine gas will move slowly.
So, your friend should release the [tex]Cl_{2}[/tex] gas first and then according to the length of room you should release the [tex]H_{2}[/tex] gas.
Thus, we can conclude that it is true that Your friend should release the [tex]Cl_{2}[/tex] first.
once water reach its boiling point what does you notice about its temperature??
4. After reaching the final titration endpoint the solution will be cloudy white. As time goes on the solution will turn back to a cloudy dark purple color. Why does this occur if you have already reached the endpoint
Answer: hello some part of your question is missing below is the missing part
In an experiment to determine the % of ascorbic acid in Vitamin C Tablets by Titration with Potassium Bromate,
answer:
Oxidation half reaction of Vitamin C
Explanation:
The solution will turn cloudy dark purple even after reaching endpoint when allowed to settle with time. because of the Oxidation half reaction of Vitamin C. also during the Titration process few drops of starch solution will be added to help determine the endpoint of the experiment .
What is the specific heat value of brass?
Answer:
Nhiệt dung riêng của đồng là
380
J
/
k
g
.
K
Explanation:
According to hydrogen’s chemical properties, it belongs in the same group as:
1. alkali metals
2. alkaline earth metals
3 .halogens
4. both a and c
---
Select the following group of elements that contains elements with similar chemical properties
A.
N, As, Bi
B.
Rb, Sn, In
C.
Li, Mg, Ga
D.
W, Tc, Fe
---
2.67 Determine the density (g/mL) for each of the following:
a. A 20.0-mL sample of a salt solution has a mass of 24.0 g.
The density (g/mL) for a 20.0-mL sample of a salt solution has a mass of 24.0 g is 1.2 g/ml.
What is density?Density is the mass per unit volume. Density is a scalar quantity. It is denoted by d and the symbol for density is given as rho, a Greek symbol. Density is calculated as mass divided by volume.
Mass is the quantity of matter in a physical body. The product of the molar mass of the compound and the moles of the substance are defined as mass.
Volume is the space occupied by a three-dimensional object. Volume is calculated by dividing mass by density.
Given, the 20.0-mL sample of a salt solution, which is the volume.
The mass of the solution is 24.0 g
To calculate the density
mass/volume
24.0 / 20.0 = 1.2 g/ml
Thus, the density of the given salt solution is 1.2 g/ml.
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Helllppp asapppp
Which of the following is matter?
O A. Light
O
B. Sound
O C. Atoms
O D. Radio waves
A solution of KMnO4 has an absorbance of 0.539 when measured in the colorimeter. Determine the concentration of the KMnO4 given the following data for a calibration plot.
Concentration of KMNO4 (M) Absorbance
0.0150 0.081
0.0300 0.159
0.0450 0.260
0.0600 0.334
Answer:
Concentration of unknown solution is 0.0416 M
Explanation:
As we know
Absorbance is equal to the product of molar absorptivity of KMnO4 m, path length and concentration
From the given set of graphical data, it is clear that the absorbance vs concentration is a straight line.
From the graph, we can obtain-
Y = 5.73 X – 0.0065
Absorbance = 0.232
0.232 = 5.73 X – 0.0065
X = 0.0416
Concentration of unknown solution is 0.0416 M
calculate the relative molecular
mass of Zn (NO3)2
Answer:
Okyu
Explanation:
Mass of Zn = 65
(NO3)2 = 124
[tex]65 + (14 + 16 \times 2) \times 2 \\ 65 + 124 \\ = 184g mol[/tex]
Explanation:
Mass of Zn = 65
(NO3)2 = 124
\begin{gathered}65 + (14 + 16 \times 2) \times 2 \\ 65 + 124 \\ = 184g mol\end{gathered}
65+(14+16×2)×2
65+124
=184gmol
Lab 2: paper chromatography of organic dyes
Picture of questions below.
Answer:
The three primary colors used when mixing dyes or paints are red, yellow, and blue. Other colors are often a mixture of these three colors. Try running a chromatography test again with non-primary-color markers, like purple, brown, and orange.
Explanation:
Mixtures that are suitable for separation by chromatography include inks, dyes and colouring agents in food. ... As the solvent soaks up the paper, it carries the mixtures with it. Different components of the mixture will move at different rates. This separates the mixture out.
2.
A solution contains 50 g of KCl per 100 g of water at
90°C. Is the solution unsaturated, saturated or
supersaturated?
unsaturated, saturated or supersaturated?
张
ci Metal displaces 50cm of water
when completely immersed in water if the mos
of the metal is 35.0q, calculate its
density
Answer:
0.70 g/cm³
Explanation:
Step 1: Given data
Volume of water displaced by the metal (V): 50 cm³ (this is equal to the volume of the metal)Mass of the metal (m): 35.0 gStep 2: Calculate the density (ρ) of the metal
Density is an intrinsic property of matter. It is equal to the mass of the metal divided by its volume.
ρ = m/V
ρ = 35.0 g/50 cm³ = 0.70 g/cm³
The oxygen atom of a ketone (such as cyclohexanone) contains 2 lone pairs of electrons. These pairs of electrons most likely reside in what type of orbital
Answer: The given pairs of electrons most likely reside in [tex]sp^{2}[/tex] type of orbital.
Explanation:
As it is given that two lone pair of electrons are present on the oxygen atom of ketone (such as cyclohexanone).
Also, there will be one bond pair between carbon and oxygen atom.
Hence, total electrons present in the domain are as follows.
2 lone pairs + 1 bon pair of electron = 3 electron domains
This means that there will be [tex]sp^{2}[/tex] type of orbital present.
Thus, we can conclude that given pairs of electrons most likely reside in [tex]sp^{2}[/tex] type of orbital.