Answer:
Decane> 2-octanone> acetaldehyde> acetic acid> 1-butanol
Explanation:
One of the most important factors that affect the retention factor (Rf) in chromatography is the polarity of the mobile phase.
We have to recall that effective separation in chromatography depends on the movement of the mobile phase over a stationary phase.
The more polar the mobile phase, the more it interacts with the solute and the more the distance it travels from the baseline. Thus high solvent polarity leads to higher Rf value.
Therefore, the order of increasing Rf values for the solvents in the question is; Decane> 2-octanone> acetaldehyde> acetic acid> 1-butanol
What is the standard enthalpy change for the decomposition of one mole of SO3?
Answer:
see explanation
Explanation:
SO₃(g) + 395.77 Kj/mole => S°(s) + 3/2O₂(g)
The standard heat of formation for SO₃(g) is given by the following rxn:
S°(s) + 3/2O₂(g) => SO₃(g) + 395.77 kJ/mole. Reversing this reaction is the decomposition of SO₃(g) into its basic elements in their standard state (25°C, 1atm) and is endothermic with +295.77Kj/mole.
Calculate the new boiling point of a solution if 10.00 g of a non-ionizing compound (C3H5(OH)3) is dissolved in 90.00 g of H2O. Molar Mass of C3H5(OH)3 = 92.09 g/mol Kb = 0.51 oC/m (Answer must be in 4 sig fig. Do not include units in your answer).
Answer:
Boiling T° of solution = 100.6
Explanation:
Formula for elevation of boiling point is:
ΔT = Kb . m . i
where ΔT means Boiling T° of solution - Boiling T° of pure solvent
Our solute is a non ionizing compound.
i = 1, because it is a non ionizing compound. i, indicates the ions dissolved in solution.
m = molality (moles of solute dissolved in 1 kg of solvent)
90 g of solvent = 0.09 kg of solvent
We convert mass of solute to moles (by the molar mass):
10 g . 1 mol /92.09 g = 0.108 moles
m = 0.108 mol /0.09 kg = 1.21 m
Let's replace data: Boiling T° of solution - 100°C = 0.51 °C/m . 1.21 m . 1
Boiling T° of solution = 0.51 °C/m . 1.21 m . 1 + 100°C
Boiling T° of solution = 100.6
Which of the following equations is not balanced?
A.
C5H12(1)
+
802(8)
5CO2(8)
+ 6H2O(g)
B.
H3PO4(aq)
+ 3NaOH(aq)
Na3PO4(aq) + 3H20(1)
C.
PC15(8)
+ 3H2O(g)
H3PO46
+ 5HCl(aq)
D.
Pb(NO3)2(aq)
+ 2HCl(aq)
PbCl2(s)
+ 2HNO3(aq)
Answer:
The answer is C.
Explanation:
In the chemical equation in C, there are 15 Carbon atoms and 1 Phosphorus atom in PC15, and 3 Hydrogen atoms and 6 Oxygen atoms in 3H2O on the left hand side. On the right hand side, there are 3 Hydrogen atoms, 1 Phosphorus atom and 46 Oxygen atoms in H3PO46, and 5 Hydrogen atoms and 5 Chlorine atoms in 5HCl.
Therefore, tallying up, there are 6 Oxygen atoms on the left hand side and 46 Oxygen atoms on the right hand side. This means the chemical equation in C is unbalanced.
Hope this helped!
What is the IUPAC name of the following compound?
OH
s
Answer:
2-isopropyl-4-methylphenol.
Explanation:
Hey there!
In this case, according to the given information, it turns out possible for us to assign the appropriate IUPAC name of the given compound, by considering that the phenol stands for the parent chain and we have isopropyl methyl radicals which the former is called first due to the alphabet consideration.
In such a way, the name would be 2-isopropyl-4-methylphenol.
Regards!
Give the balanced equation for the neutralisation of the excess NaOH with HCI.
1 NaCl + 1 HCl ➡️ 1 NaCl + Water (H2O) .
The three parts of quality assurance are determining use objectives, setting specifications, and assessment of results. Classify the actions taken during quality assurance by the part of quality assurance in which they should be taken.
a. Document procedures and keep suitable records.
b. Use quality control samples to monitor performance.
c. Compare data and results with specifications.
d. Consider the accuracy and precision needed.
e. Determine the sampling requirements.
f. Follow standard operating procedures.
The actions taken during quality assurance by the part of quality assurance in which they should be taken is to document procedures and keep suitable records. The correct option is a.
What is quality assurance?Quality assurance is checking the quality of objects and services. They are assured in the companies and factories and other places to check the quality of the products.
The different type of quality assurance is: There are different types of quality assurance.
control.acceptance sampling. control charts.product quality control.They work in the set quality and set requirements. They maintain the quality and develop those sets. Furthermore, they manage waste and quality.
Thus, the correct option is a. Document procedures and keep suitable records.
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Na lost an e- it shrank and the Cl+ grew in size. Does that say anything about the location of the electron before and after it was transferred
Answer:
48
Explanation:
Name the following ketone:
여 o
A. cyclohexyne
B. cyclohexanal
C. cyclohexanol
D. cyclohexanone
Answer:
It is D).cyclohexanone ( in acellus)
Explanation:
. Diatomic chlorine and sodium hydroxide (lye) are commodity chemicals produced in large quantities, along with diatomic hydrogen, via the electrolysis of brine, according to the following unbalanced equation:
NaCl(aq)+ H2O(l) → NaOH (aq) + H2(g) + Cl2(g)
Write balanced molecular, complete ionic, and net ionic equations for this process.
Answer:
See explanation
Explanation:
The molecular equation is depicted below;
The balanced molecular reaction equation is;
2 NaCl (aq) + 2 H2O (l) → 2 NaOH (aq) + Cl2 (g) + H2 (g)
The complete ionic equation is;
2Na^+(aq) + 2Cl^-(aq) + 2H2O(l) ---> 2Na^+(aq) + 2OH^-(aq) + Cl2(g) + H2(g)
Hence, net Ionic equation;
2Cl^-(aq) + 2H2O(l) ---> 2OH^-(aq) + Cl2(g) + H2(g)
When 10.0 g of sulfur is combined with 10.0 g of oxygen, 20.0 g of sulfur dioxide is formed. What mass of oxygen would be required to convert 10.0 g of sulfur into sulfur trioxide?
Answer:
Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.
2 S + 3 O₂ → 2 SO₃
The stoichiometric calculations is as follows:
6 g S * 1 mol/32.06 g S = 0.187 mol S
Moles O₂ needed = 0.187 mol S * 3 mol O₂/2 mol S = 0.2805 mol O₂
Since the molar mas of O₂ is 32 g/mol,
Mass of O₂ needed = 0.2805 mol O₂ * 32 g/mol = 8.976 g O₂
Potassium Chlorate decomposes according to the reaction below.
2KClO3(s) 2KCl(s) + 3O2(g)
A 4.35 g sample of KClO3 is heated and the O2 gas produced by the reaction is collected in an evacuated flask. What is the volume of the O2 gas if the pressure of the flask is 0.75 atm and the gas temperature is 27oC? R=0.0821 (L*atm)/(mol*K)
Answer:
1.75L
Explanation:
Reaction of decomposition is:
2KClO₃(s) → 2KCl(s) + 3O₂(g)
We determine moles of salt:
4.35 g . 1 mol /122.55 g = 0.0355 moles
Ratio is 2:3. 2 moles of salt can produce 3 moles of oxygen
Then, our 0.0355 moles of chlorate may produce (0.0355 . 3)/ 2 = 0.0532 moles.
We have determined, moles of gas and we have data of pressure and temperature. To find out the volume, we apply the Ideal Gases Law:
We convert T° from °C to K → 27°C + 273 = 300K
P . V = n . R . T
0.75 atm . V = 0.0532 mol . 0.0821 L.atm/mol.K . 300K
V = (0.0532 mol . 0.0821 L.atm/mol.K . 300K) / 0.75 atm
V = 1.75 Liters
One of the most common causes of inaccurate melting point ranges is rapid heating of the compound. Under these circumstances, how will the observed MP range compare to the true MP range
Answer:
INCREASE in the difference between the melting point measured and the true melting temperature.
Explanation:
Melting point of a compound is defined as the temperature at which the soils compound changes into liquid at the atmospheric pressure. There are different circumstances that can lead to inaccurate melting point. These include:
--> presence of impurities in the compound,
--> Molecular composition,
--> Force of attraction, and
--> Rapid heating of the compound.
Under the circumstances of rapid heating of the compound, there would be an increase in the melting point range when compared with the true melting point range of the compound.
The higher the heating rate, the more rapid the rise in oven temperature, increasing the difference between the melting point measured and the true melting temperature.
describe how lyophobic sols are synthesize by dispersion method
Explanation:
For preparing lyophobic sol, the substance in bulk is broken down into particles of colloidal dimensions (Dispersion) or aggregating smaller particles into particles of colloidal dimensions (condensation).
Using the following equation how many grams of water you would get from 886 g of glucose:
C6H12O6 + 6O2 → 6CO2 + 6H2O + energy
Answer:
531.6g
Explanation:
Total moles of glucose in this case is: 886/180= 4.922 (mole)
For every 1 mole glucose we get 6 mole water
-> Mole of water is: 4.922 * 6= 29.533 (mole)
weight of water is 18. Therefore, total weight of water that we will have from 886g of glucose are: 25.933*18= 531.6g
Epinephrine (adrenaline) is a hormone secreted into the bloodstream in times of danger and stress. It is 59.0% carbon, 7.15% hydrogen, 26.20% oxygen, and 7.65% nitrogen by mass and has a molar mass of 183 g/mol. Determine the empirical formula for Epinephrine.
Answer: The empirical and molecular formula for the given organic compound is [tex]C_9H_{13}O_3N[/tex]
Explanation:
Let the mass of the compound be 100 g
Given values:
% of C = 59.0%
% of H = 7.15%
% of O = 26.20%
% of N = 7.65%
Mass of C = 59.0 g
Mass of H = 7.15 g
Mass of O = 26.20 g
Mass of N = 7.65 g
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Molar mass of C = 12 g/mol
Molar mass of H = 1 g/mol
Molar mass of O = 16 g/mol
Molar mass of N = 14 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of C}=\frac{59.0g}{12g/mol}=4.917 mol[/tex]
[tex]\text{Moles of H}=\frac{7.15g}{1g/mol}=7.15 mol[/tex]
[tex]\text{Moles of O}=\frac{26.20g}{16g/mol}=1.6375 mol[/tex]
[tex]\text{Moles of N}=\frac{7.65g}{14g/mol}=0.546 mol[/tex]
Step 2: Calculating the mole ratio of the given elements.Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.546 moles
[tex]\text{Mole fraction of C}=\frac{4.917}{0.546 }=9[/tex]
[tex]\text{Mole fraction of H}=\frac{7.15}{0.546 }=13[/tex]
[tex]\text{Mole fraction of O}=\frac{1.6375}{0.546 }=2.99\approx 3[/tex]
[tex]\text{Mole fraction of N}=\frac{0.546}{0.546 }=1[/tex]
Step 3: Taking the mole ratio as their subscripts.The ratio of C : H : O : N = 9 : 13 : 3 : 1
The empirical formula of the compound becomes [tex]C_9H_{13}O_3N_1=C_9H_{13}O_3N[/tex]
To calculate the molecular formula, the number of atoms of the empirical formula is multiplied by a factor known as valency that is represented by the symbol, 'n'.
[tex]n =\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex] .....(2)
We are given:
Mass of molecular formula = 183 g/mol
Mass of empirical formula = 183 g/mol
Putting values in equation 2, we get:
[tex]n=\frac{183g/mol}{183g/mol}=1[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_{1\times 9}H_{1\times 13}O_{1\times 3}N_{1\times 1}=C_9H_{13}O_3N[/tex]
Hence, the empirical and molecular formula for the given organic compound is [tex]C_9H_{13}O_3N[/tex]
Using Hess’s law, what is the standard enthalpy of formation of manganese (II) oxide, MnO(s)?
With the help of hess's law:
ΔHf(MnO)=−248.9 kJ−12(272.0) kJ=−384.9 kJ(per mole) Δ H f ( M n O ) = − 248.9 k J − 1 2 ( 272.0 ) k J = − 384.9 k J ( p e r m o l e )
What is Hess's law?Hess's law of constant heat summation, also known simply as Hess' law, is a relationship in physical chemistry named after Germain Hess, a Swiss-born Russian chemist and physician who published it in 1840.
Moreover, hess's Law of Constant Heat Summation (or just Hess's Law) states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.
Therefore, hess' law is based on the state function character of enthalpy and the first law of thermodynamics. Energy (enthalpy) of a system (molecule) is a state function.
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A solution has a [H3O+] of 1 × 10−5 M. What is the [OH−] of the solution?
A) 9 M
B) 14 M
C) 1 x 10^{-9}
D) 1 x 10^{-14}
What is the pH of a 1.0 x103 M KOH solution?
A. 11
B. 4.0
C. 10
D. 3.0
Answer:
Option A. 11
Explanation:
We'll begin by calculating the concentration of Hydroxide ion in the solution. This can be obtained as follow:
KOH (aq) —> K⁺(aq) + OH¯(aq)
From the balanced equation above,
1 mole of KOH produced 1 mole of OH¯.
Therefore, 1×10¯³ M KOH will also produce 1×10¯³ M OH¯.
Next, we shall determine the pOH of the solution. This can be obtained as follow:
Concentration of Hydroxide ion [OH¯] = 1×10¯³ M
pOH =?
pOH = –Log [OH¯]
pOH = –Log 1×10¯³
pOH = 3
Finally, we shall determine the pH of the solution. This can be obtained as follow:
pOH = 3
pH =?
pH + pOH = 14
pH + 3 = 14
Collect like terms
pH = 14 – 3
pH = 11
Therefore, the pH of the solution is 11
Which tasks can be used to start a descriptive investigation
Answer:
The tasks which form the preliminaries to a descriptive investigation are: Making careful objective observations. Asking the relevant scientific questions.
A sample of gas contains 0.1800 mol of CO(g) and 0.1800 mol of NO(g) and occupies a volume of 23.2 L. The following reaction takes place:
2CO(g) + 2NO(g 2Co2(g) +N2(g)
Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.
Answer:
The volume of the sample is 17.4L
Explanation:
The reaction that occurs requires the same amount of CO and NO. As the moles added of both reactants are the same you don't have any limiting reactant. The only thing we need is the reaction where 4 moles of gases (2mol CO + 2mol NO) produce 3 moles of gases (2mol CO2 + 1mol N2). The moles produced are:
0.1800mol + 0.1800mol reactants =
0.3600mol reactant * (3mol products / 4mol reactants) = 0.2700 moles products.
Using Avogadro's law (States the moles of a gas are directly proportional to its pressure under constant temperature and pressure) we can find the volume of the products:
V1n2 = V2n1
Where V is volume and n moles of 1, initial state and 2, final state of the gas
Replacing:
V1 = 23.2L
n2 = 0.2700 moles
V2 = ??
n1 = 0.3600 moles
23.2L*0.2700mol = V2*0.3600moles
17.4L = V2
The volume of the sample is 17.4LA molecular compound has the following empirical formula: CH2O. The molar mass of the empirical formula is g. Write your answer using 3 significant figures. If the molar mass of the molecular compound is 180.0 g/mol, write the molecular formula of the compound.
Answer:
Empirical formula has a molar mass of 30.01g/mol and molecular formula is C₆H₁₂O₆
Explanation:
Molar mass of a molecule is the sum of the molar mass of each atom. In CH2O we have:
1C = 1*12.01g/mol = 12.01g/mol
2H = 2*1g/mol = 2g/mol
1O = 1*16g/mol = 16g/mol
Empirical formula of CH2O is:
12.01g/mol + 2g/mol + 16g/mol = 30.01g/mol
As the molecular compound has a molar mass of 180.0g/mol the molecular formula is:
180.0g/mol / 30.01g/mol = 6 times the empirical formula. That is:
C₆H₁₂O₆(NH4)2SO4(aq)+SrCl2(aq)→
what is the theoretical yield of Mg(s) + O2(g) → MgO(s)
will give brainliest fakes will be reported
The constitutional isomer of ethanol, dimethyl ether (CH3OCH3), is a gas at room temperature. Suggest an explanation for this observation.
Answer:
Because of its weak intermolecular forces.
Explanation:
Hello there!
In this case, according to the given description, it turns out possible for us to recall the chemical structures of both ethanol and dimethyl ether as follows:
[tex]CH_3CH_2OH\\\\CH_3COCH_3[/tex]
Thus, we can see that ethanol have London dispersion forces (C-C bonds), dipole-dipole forces (C-O bonds) and also hydrogen bonds (O-H bonds) which make ethanol a liquid due to the strong hydrogen bonds. On the other hand, we can see that dimethyl ether has just London and dipole forces, which are by far weaker than hydrogen bonding, that makes it unstable when liquid and therefore it tends to vaporize quite readily.
Regards!
CAN SOMEONE PLEASE HELP ME I WILL MARK YOU AS BRAINLIEST
Explanation:
[tex]2H_2O_2 \rightarrow 2H_2O + O_2[/tex]
First convert the amount of water into moles:
360 g H2O × [tex]\left(\dfrac{1\:\text{mol}H_2O}{18.015\:\text{g}H_2O}\right)[/tex]
[tex] = 20. \:\text{mol}H_2O[/tex]
Now let's calculate the number of moles of O2 gas produced.
20 mol H2O × [tex]\left(\dfrac{1\:\text{mol}O_2}{2\:\text{mol}H_2O}\right)=10\:\text{mol}O_2[/tex]
The volume of gas at 10°C and 5 atm can be found using the ideal gas law:
[tex]PV=nRT[/tex]
[tex]V= \dfrac{nRT}{P}[/tex]
[tex]= \dfrac{(10)(0.082)(283)}{(5)}=46.4\:L[/tex]
Ch3-ch2-o-ch2-ch2-och3
Explanation:
ethoxypropane
Ch3-ch2-o-ch2-ch2-och3
A chemistry student must write down in her lab notebook the concentration of a solution of sodium thiosulfate. The concentration of a solution equals the mass of what's dissolved divided by the total volume of the solution.
Answer:
A chemistry student must write down in her lab notebook the concentration of a solution of sodium thiosulfate. The concentration of a solution equals the mass of what's dissolved divided by the total volume of the solution.
Explanation:
The concentration of a solution can be measured in terms of molarity.
The molarity of a solution can be defined as the number of moles of solute present in the total volume of the solution.
The number of moles of solute is the ratio of mass of solute to molar mass of solute.
Hence,
[tex]Molarity=\frac{mass of solute}{molar mass of solute} * \frac{1}{volume of solution in L.}[/tex]
Which one is the dependent variable in distance, force, or work
Answer:
Work
Explanation:
Formula for work is given by;
Work = force × distance.
It is clear from this formula that Work depends on force and distance.
This means that force and distance are independent of the workdone and so they are classified as independent variables while work will be the dependent variable.
Cellulose and starch are examples of:
Select one:
a. monosaccharides
b. disaccharides
c. lipids
d. polysaccharides
Answer:
The choose (d)
d. polysaccharides
An object with a mass of 0.255 kg and density of 2.89 g/cm^3 measures 34 mm in length and 46 mm in width. What is the height of the object?
1) 5.6 cm
2) 5.6•10^-2 cm
3) 7.2 •10^-4 cm
5.64 I think, I'm sorry if I'm wrong