Answer:
[tex]\boxed {\boxed {\sf D. \ 0.25 \ M}}[/tex]
Explanation:
Molarity is a measure of concentration in moles per liter.
[tex]molarity= \frac{moles \ of \ solute}{ liters \ of \ solution}[/tex]
The solution contains 0.75 moles of sodium chloride and has a volume of 3.0 liters.
moles of solute = 0.75 mol NaCl liters of solution = 3.0 LSubstitute these values into the formula.
[tex]molarity= \frac{ 0.75 \ mol \ NaCl}{3.0 \ L}[/tex]
[tex]molarity= 0.25 \ mol \ NaCl/L[/tex]
Molarity has the molar (M) as its unit. 1 molar is equal to 1 mole per liter.
[tex]molarity= 0.25 \ M \[/tex]
The molarity of the solution is 0.25 Molar and Choice D is correct.
The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 75.0 mL of ethyl acetate (C4H8O2). This mixture is stored at 25.0 ∘C. The vapor pressure and the densities for the two pure components at 25.0 ∘C are given in the following table. What is the vapor pressure of the stored mixture?
Answer:
The answer is "170.9 mm Hg".
Explanation:
[tex]\text{Mass of acetone = volume} \times density[/tex]
[tex]= 70.0 \times 0.791\\\\ = 55.37\ g\\[/tex]
[tex]\text{Moles of acetone} = \frac{mass}{molar\ mass}\\\\[/tex]
[tex]=\frac{55.37}{58.08}\\\\ = 0.9533\ mol[/tex]
[tex]\text{Mass of ethyl acetate = volume} \times density[/tex]
[tex]= 73.0 \times 0.900\\\\ = 65.7\ g[/tex]
[tex]\text{Moles of ethyl acetate = mass} \times\ molar\ mass[/tex]
[tex]= \frac{65.7}{88.105} \\\\= 0.7457\ mol[/tex]
[tex]\text{Mole fraction of acetone x(acetone)} = \frac{0.9533}{(0.9533 + 0.7457)}\ = 0.5611\\\\[/tex] [tex]\text{Mole fraction of ethyl acetate x(ethyl acetate)} =\frac{0.7457}{(0.9533 + 0.7457) }= 0.4389[/tex]
Applying Raoult's law: [tex]\text{Vapor pressure = x(acetone)P(acetone) + x(ethyl acetate)P(ethyl acetate)}\\\\= 0.5611 \times 230.0 + 0.4389 \times 95.38\\\\ = 170.9\ mm \ Hg\\[/tex]
The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 75.0 mL of ethyl acetate (C4H8O2).
The vapor pressure of the stored mixture is: 170.03 mmHg
In the given information, there is some information that is still missing.
The parameters that we are being given include:
The volume of acetone = 70.0 mLThe volume of ethyl acetate = 75.0 mLThe standard temperature for the mixture = 25° CThe first step we need to take is to determine the mass and number of moles of each compound (i.e. for acetone and ethyl acetate)
For us to do that:
We need the density of acetone and ethyl acetate, which is not given:
Assuming that at a standard condition of vapour pressure:
230 mmHg of acetone has a density of 0.791 g/mL95.38 mmHg of ethyl acetate has a density of 0.900 g/mLThen;
Using the relation:
[tex]\mathbf{Density = \dfrac{Mass}{volume}}[/tex]
Mass of acetone = Density of acetone × volume of acetone
Mass of acetone = 0.791 g/mL × 70.0 mL
Mass of acetone = 55.37 g
Mass of ethyl acetate = Density of ethyl acetate × volume of ethyl acetate
Mass of ethyl acetate = 0.900 g/mL × 75.0 mL
Mass of ethyl acetate = 67.5 g
At standard conditions;
For acetone, molar mass = 58.08 g/molFor ethyl acetate, molar mass = 88.11 g/molNow, using the formula for calculating the numbers of moles which can be expressed as:
[tex]\mathbf{Number \ of \ moles = \dfrac{mass}{molar \ mass}}[/tex]
For acetone:
[tex]\mathbf{Number \ of \ moles = \dfrac{55.37 \ g}{58.08 \ g/mol}}[/tex]
[tex]\mathbf{Number \ of \ moles =0.95334 \ mol}[/tex]
For ethyl acetate:
[tex]\mathbf{Number \ of \ moles = \dfrac{67.5 \ g}{88.11 \ g/mol}}[/tex]
[tex]\mathbf{Number \ of \ moles =0.76609 \ mol}[/tex]
Now, we will determine the mole fraction of each compound.
The mole fraction describes the ratio a certain constituent of a mixture to the total amount of all the constitutent in the mixture.
Using the formula:
[tex]\mathbf{mole \ fraction = \dfrac{n_A}{n_A+n_B+...n_N}}[/tex]
For Acetone:
[tex]\mathbf{mole \ fraction = \dfrac{0.95334}{0.95334+0.76609}}[/tex]
[tex]\mathbf{mole \ fraction =0.5545 }[/tex]
For ethyl acetate:
[tex]\mathbf{mole \ fraction = \dfrac{0.76609}{0.76609+0.95334}}[/tex]
[tex]\mathbf{mole \ fraction =0.4455}[/tex]
Finally, we can compute determine the vapour pressure of the stored mixture using Raoult's Law.
Raoult's Law posits that the constituent of a partial pressure in a mixture of a liquid is proportional to the mole fraction of that constituent in the mixture provided the temperature is constant.
∴ For the stored mixture = Vapor pressure of acetone + vapour pressure of ethyl acetate.
where:
Vapour pressure of the solution = (mole fraction × vapor pressure) of solventFor acetone;
Vapor pressure = 0.5545 × 230 mmHg
Vapour pressure = 127.54 mmHg
For ethyl acetate:
Vapour pressure = 0.4455 × 95.38 mmHg
Vapour pressure ==42.49 mmHg
Thus, the vapor pressure of the stored mixture is
= (127.54 + 42.49 ) mmHg
= 170.03 mmHg
Therefore, we can conclude that the vapour pressure of the stored mixture is 170.03 mmHg
Learn more about Vapour pressure here:
https://brainly.com/question/16931217?referrer=searchResults
Calculate the molarity of a solution consisting of 65.5 g of K2S0 4 in 5.00 L of solution.
Answer:
Molarity is 0.075 M.
Explanation:
Moles:
[tex]{ \tt{ = \frac{65.5}{RFM} }}[/tex]
RFM of potassium sulphate :
[tex]{ \tt{ = (39 \times 2) + 32 + (16 \times 4)}} \\ = 174 \: g[/tex]
substitute:
[tex]{ \tt{moles = \frac{65.5}{174} = 0.376 \: moles}}[/tex]
In volume of 5.00 l:
[tex]{ \tt{5.00 \: l = 0.376 \: moles}} \\ { \tt{1 \: l = ( \frac{0.376}{5.00} ) \: moles}} \\ { \tt{molarity = 0.075 \: mol \: l {}^{ - 1} }}[/tex]
Hydrocarbons do not dissolve in concentrated sulfuric acid, but methyl benzoate does. Explain this difference and write an equation showing the ions that are produced.
Answer:
See explanation
Explanation:
For a substance to dissolve in another, there must be some sort of interaction between the substances.
Recall that like dissolves like. That is, polar substances dissolve polar substances and non polar substances dissolve nonpolar substances.
Hydrocarbons are nonpolar hence they do not dissolve in polar sulphuric acid. Methyl benzoate is polar hence it dissolve in polar sulphuric acid.
The equation showing the ions is depicted in the image attached to this answer.
Problem 7 (Diffusion due to viscosity) If the viscosity of a solution is quadrupled, the rms-average distance of a collection of diffusing molecules from their starting point would be _________ over the same amount of time.
Answer:
1/2 the distance
Explanation:
If the viscosity of a solution is quadrupled then the distance of collection of diffusing molecules would be half over the same amount of time. The viscosity of the molecules is dependent on density of the liquid. It is independent to the volume of the liquid.
Predict the products (if any) that will be formed by the reaction below. If no reaction occurs, write NR after the reaction arrow.
2HClO4(aq) + Co(s) -->
Answer:
The product is aqueous [tex]CO(HCl)_2[/tex] and [tex]O_2(g)[/tex].
Explanation:
Given:
⇒ [tex]2HClO_4(aq) +CO(s)[/tex]
then,
The reaction will be:
⇒ [tex]2HClO_4(aq)+CO(s) \rightarrow CO(HCl)_2 +O_2 (g)[/tex]
In the above reaction, we can see that
The products is:
aqueous [tex]CO(HCl)_2[/tex] and [tex]O_2(g)[/tex]
Thus the above is the correct answer.
Which statement describes the 3d, 4s, and 4p orbitals of Arsenic (As) based on its electronic configuration and position in the periodic table?
The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
The 3d orbital is completely filled, and the 4s and 4p orbitals are partially filled.
The 3d, 4s, and 4p orbitals are completely filled.
The 3d, 4s, and 4p orbitals are partially filled.
Answer:
The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
Explanation:
The correct answer is: The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
The d orbital contains 10 electrons, the s orbital takes 2 electrons and the p orbital takes six electrons.
The orbital in chemistry is defined as a region in space where there is a high probability of finding an electron. There are s, p, d, f orbitals in chemistry which correspond to sharp, principal, diffuse and fundamental.
The electronic configuration of arsenic is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3.
From this electronic configuration, we can see that the 4s and 3d orbitals are half filled while the 4p orbital is half filled.
For more about electronic configuration, see:
https://brainly.com/question/4949433
if an element has an atomic number of 9 what is the electronic structure of the same element
9 is the element Florine
Florine has 9 electrons as well as the 9 protons that determine its atomic number.
The ground state configuration is the lowest energy configuration.
Write the chemical formula for the following:
a. The conjugate acid of amide ion, NH₂-
b. The conjugate base of nitric acid, HNO₃
Follow the rules of Bronsted Lowry theory
Acids take a HBases donate a HSo
#a
NH_2-
Add a H
Conjugate acid is NH_3#b
HnO_3
Take a H
Conjugate base is NO_3-#1
Conjugate acid means one H+ is added
NH_2+H+=NH_3sw
#2
Conjugate base means donate a H+
HNO_3-H=NO_3-A substance is tested and has a pH of 7.0. How would you classify it?
Match the change to its definition.
Name of change Change
condenation gas to solid
freezing solid to liquid
melting gas to liquid
evaporation liquid to gas
sublimation solid to gas
deposition liquid to solid
Given the following reaction:
CO (g) + 2 H2(g) <==> CH3OH (g)
In an experiment, 0.42 mol of CO and 0.42 mol of H2 were placed in a 1.00-L reaction vessel. At equilibrium, there were 0.29 mol of CO remaining. Keq at the temperature of the experiment is ________.
A) 2.80
B) 0.357
C) 14.5
D) 17.5
E) none of the above
Answer:
Option D. 17.5
Explanation:
Equiibrium is: CO + 2H₂ ⇄ CH₃OH
1 mol of CO is in equibrium with 2 moles of hydrogen in order to make, methanol.
Initially we have 0.42 moles of CO and 0.42 moles of H₂
If 0.29 moles of CO remained, (0.42 - 0.29) = 0.13 moles have reacted.
So in the equilibrium we may have:
0.29 moles of CO, and (0.42 - 0.13 . 2) = 0.16 moles of H₂
Ratio is 1:2, if 0.13 moles of CO haved reacted, (0.13 . 2) moles have reacted of hydrogen
Finally 0.13 moles of methanol, are found after the equilibrium reach the end.
Let's make expression for KC: [Methanol] / [CO] . [Hydrogen]²
0.13 / (0.29 . 0.16²)
Kc = 17.5
A 25.00 gram sample of an unknown metal initially at 99.0 degrees Celcius is added to 50.00 grams of water initially at 10.55 degrees Celcius. The final temperature of the system is 20.15 degrees Celcius. Calculate the specific heat of the metal. (The specific heat of water is 4.184 J/g*C).
Answer:
1.0188 J/g*C
Explanation:
Using the formula; Q = m × c × ∆T
Q(water) = -Q(metal)
m × c × ∆T (water) = -{m × c × ∆T (metal)}
According to this question,
mass of metal = 25g
initial temp of metal = 99°C
mass of water = 50g
initial temp of water = 10.55°C
final temperature of the system = 20.15°C
c of water = 4.184 J/g*C
50 × 4.184 × (20.15 - 10.55) = 25 × c × (20.15 - 99)
209.2 × 9.6 = 25c × -78.85
2008.32 = -1971.25c
c = 2008.32 ÷ 1971.25
c of metal = 1.0188 J/g*C
The first step of electrophilic aromatic substitution involves the formation of the arenium ion intermediate.
a. True
b. Fasle
Answer:
True
Explanation:
Aromatic compounds undergo substitution rather than addition reactions because the aromatic structure is maintained.
Electrophilic aromatic substitution begins with attack of the electrophile on the aromatic ring to yield a delocalized intermediate called the arenium intermediate. Loss of hydrogen from this intermediate yields the final product.
Different steps of the oxidative decarboxylation of pyruvate by the pyruvate dehydrogenase PDH complex are given. Place these five steps in the correct order. Note that thiamine pyrophosphate, TPP, is sometimes called thiamine diphosphate, TDP.
1. FADH2 is reoxidized to FAD reducing NAD* to NADH.
2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.
3. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.
4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.
5. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.
Answer:
1. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.
2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.
3. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.
4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.
5. FADH2 is reoxidized to FAD reducing NAD+ to NADH.
Explanation:
The oxidation of pyruvate to AcetylCoA is catalyzed by the pyruvate dehydrogenase complex. The reaction is an irreversible oxidative decarboxylation process in which the carboxyl group of pyruvate is removed as a molecule of carbon dioxide, CO₂, while the remaining two carbons are attached to a CoASH molecule to form acetylCoA.
The pyruvate dehydrogenase complex contains three enzymes - Pyruvate dehydrogenase known as E₁, dihydrolipoyl transacetylase known as E₂, and dihydrolipoyl dehydrogenase known as E₃. It also requires five coenzymes namely: thiamine pyrophosphate (TPP), flavine adenine dinucleotide (NAD), coenzyme A (CoA-SH), nicotinamide adenine dinucleotide (NAD) and lipoate.
Oxidative decorbyxylation of pyruvate takes place in the pyruvate dehydrogenase complex in five steps:
1. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.
2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.
3. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.
4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.
5. FADH2 is reoxidized to FAD reducing NAD+ to NADH.
In the graphic, 195 represents the _______.
195 Pt
78
A. Atomic Mass
B. Atomic Number
C. Neutron Number
Answer:
ITS ANSWER IS
OPTION B. ATOMIC NUMBER
HI HAVE A NICE DAY
Four atoms and/or ions are sketched below in accordance with their relative atomic and/or ionic radii. Which of the following sets of species are compatible with the sketch?
Explain. (a) C,Ca2+,Cl−,Br−;
(b) Sr4, Cl,Br−,Na+
(d) Al,Ra2+,Zr2+
(c) Y,K,Ca,Na+, Mg2+;
e) Fe,Rb,Co,Cs
Answer:
Hence the correct option is an option (b) Sr4, Cl,Br−,Na+.
Explanation:
Bromine and chlorine belong to an equivalent group. As we go down the group the dimensions increases which too there's a charge on the bromine atom. therefore the size of the Br- is going to be larger in comparison to the chlorine atom.
Sr atom is within the second group, and also it's below the above-mentioned atoms.so Sr is going to be the larger one among all the atoms.
Sodium and chlorine belong to an equivalent period .size decrease from left to right. but due to the charge on sodium its size decreases and there's an opportunity that Na+ size could be adequate for Cl.
Here we finally assume that two atoms are of an equivalent size (Na+ and Cl) which are less in size compared to the opposite two(Sr and Br-) during which one is greater (Sr)and the opposite is smaller(Br-).
The mole fraction of NaCl in an
aqueous solution is 0.132. How
many moles of NaCl are present in
1 mole of this solution?
Molar Mass
NaCl: 58.44 g/mol
H2O: 18.016 g/mol
Answer:
Moles of water are 0.868
Explanation:
How many chromosomes do we not understand?
Answer:
we don't understand why humans have only 46 chromosomes
Answer:
46 chromosomes is what we don't understand
Nicotine is a toxic substance present in tobacco leaves. There are two lone pairs in the structure of nicotine. In general, localized lone pairs are much more reactive than delocalized lone pairs. With this information in mind, do you expect both lone pairs in nicotine to be reactive?
A. Both lone pairs are delocalized and, therefore, both are expected to have the same reactivity.
B.Lone pair in pyrrolidine ring is localized and, therefore, is expected to be more reactive.
C. Both lone pairs are localized and, therefore, both are expected to be reactive.
D. Lone pair in pyridine ring is localized and, therefore, is expected to be more reactive.
Answer:
B.Lone pair in pyrrolidine ring is localized and, therefore, is expected to be more reactive.
Explanation:
There are two nitrogen atoms bearing lone pairs of electrons in the structure of nicotine as shown in the image attached.
One nitrogen atom is found in the pyrrolidine ring. The lone pair on this nitrogen atom is localized hence it is more reactive than the lone pair of electrons found on the nitrogen atom in the pyridine ring which is delocalized a shown in the image attached to this answer.
Two common methods to generate an aldehyde is by oxidation of an alcohol and through ozonolysis.
a. True
b. False
Answer:
a. True.
Explanation:
Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.
weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C
take an example of ethanol:
[tex]{ \bf{CH _{3} CH_{2}OH \: \: \frac{Ag/O_{2} }{500 \degree C} > \: \:CH _{3} CHO}}[/tex]
[tex]{ \sf{CH _{3} CHO \: \: is \: ethanal}} [/tex]
By ozonolysis:
Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.
take an example of propanol:
if it undergoes ozonolysis, it gives ethanal and methanal.
Answer:
A. True
Explanation:
Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.
weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C
take an example of ethanol:
By ozonolysis:
Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.
take an example of propanol:
if it undergoes ozonolysis, it gives ethanal and methanal.
A student dissolves 12.6g of amonium nitrate(NH4NO3) in 250.g of water in a well-insulated open cup. She then observed the temperature of the water fall from 23.0°C to 18°C over the course of 6.1 minutes.
NH4NO3 â NH4+ (aq) + NO3^-(aq)
a. Is this reaction exothermic, endothermic, or neither?
b. If you said the reaction was exothermic or calculate the amount of heat that was released or absorbed by the reaction in this case.
c. Calculate the reaction enthalpy ÎHrxn per mole of NH4NO3.
Answer:
a. Endothermic.
b. [tex]Q_{rxn}=5493.6J[/tex]
c. [tex]\Delta H_{rxn}=35.0kJ/mol[/tex]
Explanation:
Hello there!
In this case, according to the given information for this calorimetry problem, it turns out possible for us to proceed as follows:
a. Due to the fact that the temperature of water goes from 23.0 °C to 18.0 °C, we infer this reaction is endothermic as the ammonium nitrate absorbed heat from the water.
b. Here, we consider the following heat equation:
[tex]Q_{rxn}=-Q_{water}[/tex]
Whereas we solve for the heat of reaction by means of the mass of the solution (both water and ammonium nitrate), the specific heat of the solution (we assume it is equal to that of water) and the temperature change:
[tex]Q_{rxn}=-m_{solution}C_{solution}(T_f-T_i)\\\\Q_{rxn}=-(12.6g+250.g)(4.184\frac{J}{g\°C} )(18.0\°C-23.0\°C)\\\\Q_{rxn}=5493.6J[/tex]
c. Here, we divide the previously calculated heat by the moles of ammonium nitrate (molar mass = 80.043 g/mol) to obtain the enthalpy of reaction per mole of this compound:
[tex]n_{NH_4NO_3}=12.6g*\frac{1mol}{80.043 g}=0.157mol\\\\\Delta H_{rxn}=\frac{5493.6J}{0.157mol} =34898.7J/mol\\\\\Delta H_{rxn}=35.0kJ/mol[/tex]
Regards!
When a 1:1 mixture of ethyl propanoate and ethyl butanoate is treated with sodium ethoxide, four Claisen condensation products are possible. Draw the structure(s) of the product(s) that have an ethyl group on the chiral center
Answer:
attached below
Explanation:
The Four Claisen condensation are grouped into :
Self Claisen condensation reaction Cross Claisen condensation reactionSelf Claisen condensation is when R = R'
Cross Claisen condensation is when R ≠ R'
attached below are the four Claisen condensation
The products of nuclear reaction usually have a different mass than the reactants why?
Answer:
Explanation:
The best way to explain this is to use an example
[tex]I\frac{125}{53} + e \frac{0}{-1} ====> Te\frac{125}{52}[/tex]
You have to understand what happened. A electron was shot into the nucleus of the Iodine. That electron change the entire composition of the nucleus resulting in 52 protons. The mass remained the same (125) but the nucleus was altered. The chemical became 125 52 Tellurium. But what is important is that it takes a tremendous amount of energy to disrupt a nucleus, and a new chemical is born from that disruption.
I performed an experiment and mixed copper nitrate and potassium iodide. When they reacted, they formed a precipitate, even though solubility rules would not lead one to predict so because potassium nitrate is obviously soluble and so should copper (II) iodide. One can deduce from the formation of a precipitate that copper is reduced. Write a proposed reaction for the oxidation reduction of copper (II) iodide. Justify the choice of the substance that reduces the copper based on experimental evidence. Also, justify the choice using the atomic structure of potassium ion and iodide ion.
Answer:
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Explanation:
The reaction performed in the experiment is;
2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2
The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.
The oxidation - reduction equation is as follows;
2Cu2^+ + 2e ----> 2Cu^+ reduction half equation
2I^- ----> I2 + 2e. Oxidation half equation
Balanced redox reaction equation;
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Using the molarity of vinegar, calculate the mass percent of acetic acid in the original sample. Assume the density of vinegar is 1.00 g/mL. (The formula for acetic acid is C2H4O2).
Answer:
5.37% w/w is the mass percent of vinegar assuming a molarity of 0.8935mol/L
Explanation:
Assuming the molarity of vinegar is 0.8935mol/L:
Mass percent is defined as 100 times the ratio between mass of solute (In this case, acetic acid), and the mass of the solution
To solve this question we need to find the mass of acetic acid from the moles using the molar mass and the mass of the solution from the volume in liters using the density:
Mass Acetic acid -Molar mass: 60.052g/mol-
0.8935mol * (60.052g / mol) = 53.656g Acetic Acid
Mass Solution:
1L = 1000mL * (1.00g/mL) = 1000g Solution
Mass Percent:
53.656g Acetic Acid / 1000g Solution * 100 =
5.37% w/w is the mass percent of vinegar assuming a molarity of 0.8935mol/LThe mass percent of acetic acid in the original sample of vinegar of molarity 0.8935mol/L is 5.37% w/w.
How do we calculate the mass percent?Mass percent of any solute present in any solution will be calculated as the:
Mass % of solute = (mass of solute / mass of solution) × 100
Let the molarity of vinegar = 0.8935mol/L
Means 0.8935 moles of vinegar present in the 1 liter of the solution.
Now we calculate mass from moles as:
n = W/M, where
W = required mass
M = molar mass = 60.052g /mol
W = (0.8935mol)(60.052g/mol) = 53.656g
Mass of solution = 1L = 1000mL×(1.00g/mL) = 1000g Solution
Then the mass % of acetic acid:
Mass % = (53.656g / 1000g) × 100 = 5.37% w/w
Hence the required % mass is 5.37% w/w.
To know more about mass percent, visit the below link:
https://brainly.com/question/26150306
what is meant by density
Answer:
The degree of compactness of a substance
What is the final volume, in L. of a balloon that was initially at 173.8 mL at 17.5°C and was then heated to 78.0*C?
Answer:
0.21 L.
Explanation:
From the question given above, the following data were obtained:
Initial temperature (T₁) = 17.5°C = 17.5°C + 273 = 290.5 K
Initial volume (V₁) = 173.8 mL
Final temperature (T₂) = 78 °C = 78 °C + 273 = 351 K
Final volume (V₂) =?
V₁/T₁ = V₂/T₂
173.8 / 290.5 = V₂ / 351
Cross multiply
290.5 × V₂ = 173.8 × 351
290.5 × V₂ = 61003.8
Divide both side by 290.5
V₂ = 61003.8 / 290.5
V₂ = 210 mL
Finally, we shall convert 210 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
210 mL = 210 mL × 1 L / 1000 mL
210 mL = 0.21 L
Thus, the final volume of the balloon is 0.21 L.
Answer:
[tex]\boxed {\boxed {\sf 0.775 \ L}}[/tex]
Explanation:
1. Calculated Final VolumeWe are asked to find the final volume of a balloon given a change in temperature. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula for this law is:
[tex]\frac{V_1}{T_1}= \frac{V_2}{T_2}[/tex]
The initial volume is 173.8 milliliters and the initial temperature is 17.5 degrees Celsius.
[tex]\frac {173.8 \ mL}{17.5 \textdegree C}= \frac{V_2}{T_2}[/tex]
The balloon is heated to a final temperature of 78.0 degrees Celsius, but the volume is unknown.
[tex]\frac {173.8 \ mL}{17.5 \textdegree C}= \frac{V_2}{78.0 \textdegree C}[/tex]
We are solving for the final volume, so we must isolate the variable V₂. It is being divided by 78.0 degrees Celsius. The inverse of division is multiplication, so we multiply both sides by 78.0 °C.
[tex]78.0 \textdegree C *\frac {173.8 \ mL}{17.5 \textdegree C}= \frac{V_2}{78.0 \textdegree C} * 78.0 \textdegree C[/tex]
[tex]78.0 \textdegree C *\frac {173.8 \ mL}{17.5 \textdegree C}=V_2[/tex]
The units of degrees Celsius cancel.
[tex]78.0 *\frac {173.8 \ mL}{17.5}=V_2[/tex]
[tex]78.0 *9.931428571 \ mL= V_2[/tex]
[tex]774.6514286 \ mL =V_2[/tex]
2. Convert to LitersWe are asked to give the volume in liters, so we must convert out units. Remember that 1 liter contains 1000 milliliters.
[tex]\frac { 1 \ L}{1000 \ mL}[/tex]
[tex]774.6514286 \ mL * \frac{ 1 \ L}{1000 \ mL}[/tex]
[tex]774.6514286 * \frac{ 1 \ L}{1000}[/tex]
[tex]0.7746514286 \ L[/tex]
3. RoundThe original values of volume and temperature have 3 and 4 significant figures. We always round our answer to the least number of sig figs, which is 3. This is the thousandths place for the number we calculated. The 6 in the ten-thousandths place tells us to round the 4 up to a 5.
[tex]0.775 \ L[/tex]
The final volume is approximately 0.775 liters.
#19.
An unknown sample weighs 45.2 g and takes 58.2 kJ to vaporize. What is
its heat of vaporization?
you want to remove as much CO2 gas as possible from a water solution. Which of the following treatments would be most effective?
Answer:
Aerate solution
Explanation:
aerate solution is the best way to remove CO2 from water (Carbon dioxide in the water that does not form bicarbonates is “uncombined” and can be removed by aeration).
Part A
3.75 mol of LiCl in 3.36 L of solution
Express the molarity in moles per liter to three significant figures
Answer:
1.12 mol/L.
Explanation:
From the question given above, the following data were obtained:
Mole of LiCl = 3.75 moles
Volume = 3.36 L
Molarity =?
Molarity is simply defined as the mole of solute per unit litre of the solution. Mathematically, it is expressed as:
Molarity = mole / Volume
With the above formula, we can obtain the molarity of the solution as follow:
Mole of LiCl = 3.75 moles
Volume = 3.36 L
Molarity =?
Molarity = mole /Volume
Molarity = 3.75 / 3.36
Molarity = 1.12 mol/L
Thus, the molarity of the solution is 1.12 mol/L
