Answer:
0.1 M Nal
0.1M C2H5NH3Br
0.1M KClO
0.1M NaCN
Explanation:
The strongest acid is the one that has the higher Ka. Now, the weakest conjugate base is the conjugate base of the strongest acid and vice versa:
In the problem, we have only conjugate bases, as the HCN is the weakest acid, the strongest conjugate base is NaCN, then KClO and as last C2H5NH3Br and NaI (The conjugate base of a strong acid, HI).
The strongest base has the higher pH, that means. Thus, the rank in order of increasing pH is:
0.1 M Nal
0.1M C2H5NH3Br
0.1M KClO
0.1M NaCN
Given the chemical equation: KI +Pb(NO3)2—>KNO3 + Pbl2
Balance this chemical equation.
Indicate the type of reaction. How do you know?
Thoroughly discuss how your balanced chemical equation agrees with the law of conservation of mass.
Answer:
[tex]2KI +Pb(NO_3)_2\rightarrow 2KNO_3 + Pbl_2[/tex]
Double replacement reaction.
It is in agreement with the law of conservation of mass because we have two potassium atoms, two iodine atoms, one lead atom, two nitrogen atoms and six oxygen atoms on both sides of the chemical equation (count them).
Explanation:
Hello there!
In this case, according to the given information, it turns possible for us to solve this problem by firstly considering that this reaction occurs between potassium iodide and lead (II) nitrate to yield potassium nitrate and lead (II) iodide which is clearly not balanced since we have one iodine atom on the reactants and two on the products, that is why the balance implies the placement of a coefficient of 2 in front of both KI and KNO3 as shown below:
[tex]2KI +Pb(NO_3)_2\rightarrow 2KNO_3 + Pbl_2[/tex]
Thus, we infer this is a double replacement reaction due to the exchange of both cations, K and Pb with both anions, I and NO3. Moreover, we can tell this balanced reaction is in agreement with the law of conservation of mass because we have two potassium atoms, two iodine atoms, one lead atom, two nitrogen atoms and six oxygen atoms on both sides of the chemical equation (count them).
Regards!
Which does not result in deviations from linearity in a Beer's law plot of absorbance versus concentration?a. light losses at the cell interface b. all are sources of nonlinearity c. stray radiation d. equilibrium between different forms of the analyte e. a wide bandwidth relative to the width of the absorption band
Answer:
a
Explanation:
Beer-Lambert Law shows the relationship between the factors affecting the absorbance of a sample in relation to the concentration. These factors are:
the concentration c, path length (l), and the molar absorptivity (ε).
As a result, more radiation is assimilated as the concentration rises, and the absorbance rises as well. However, the longer the path length, the increase in the number of molecules and the higher the absorbance.
Thus, the straight-line equation for Beer-Lambert's law is:
A = εcl
From the above explanation, the option that doesn't relate to the deviations from linearity of Beer's law plot is in Option (a).
Identify each of the following as a covalent compound or ionic compound. Then provide
either the formula for compounds identified by name or the name for those identified by
formula. (1 point each)
a. Li2O
b. Dinitrogen trioxide:
c. PCI3
d. Manganese(III) oxide:
Answer:
Explanation:
a) Ionic
Lithium oxide
b) Covalent
[tex]$\ce{N_2O_3}$[/tex]
c) Covalent
Phosphorus trichloride
d) Ionic
[tex]Mn_2O_3[/tex]
Curium – 245 is an alpha emitter. Write the equation for the nuclear reaction and identify the product nucleus.
Answer:
Please find the complete solution in the attached file.
Explanation:
You have 10 pounds of egg whites. You need 6oz to make one serving of cosomme. How many servings can you make?
Answer:
I think you can make 26, hope this helped.
Explanation:
examples s name of thosse food items we can store for a month?
Answer:
1. Nuts
2. Canned meats and seafood
3. Dried grains
4. Dark chocolate
5. Protein powders
why beta carbon hydrogen is easily replaceable but not alpha carbon hydrogen
Answer:
Four common types of reactions involving carbonyl reactions: 1) nucleophilic addition; 2) nucleophilic acyl substitution; 3) alpha substitution; 4) carbonyl condensations. The first two were previously discussed and the second two involve the properties of the carbon directly adjacent to the carbonyls, α carbons.
Alpha-substitution reactions results in the replacement of an H attached to the alpha carbon with an electrophile.
The nucleophile in these reactions are new and called enols and enolates.
Explanation:
The carbon in the carbonyl is the reference point and the alpha carbon is adjacent to the carbonyl carbon.
Hydrogen atoms attached the these carbons denoted with Greek letters will have the same designation, so an alpha hydrogen is attached to an alpha carbon.
Aldehyde hydrogens not given Greek leters.
α hydrogens display unusual acidity, due to the resonance stabilization of the carbanion conjugate base, called an enolate.
Tautomers are readily interconverted constitutional isomers, usually distinguished by a different location for an atom or a group, which is different than resonance.
The tautomerization in this chapter focuses on the carbonyl group with alpha hydrogen, which undergo keto-enol tautomerism.
Keto refers to the tautomer containing the carbonyl while enol implies a double bond and a hydroxyl group present in the tautomer.
The keto-enol tautomerization equilibrium is dependent on stabilization factors of both the keto tautomer and the enol tautomer, though the keto form is typically favored for simple carbonyl compounds.
The 1,3 arrangement of two carbonyl groups can work synergistically to stabilize the enol tautomer, increasing the amount present at equilibrium.
The positioning of the carbonyl groups in the 1,3 arrangement allows for the formation of a stabilizing intramolecular hydrogen bond between the hydroxyl group of the enol and the carbonyl oxygen as well as the alkene group of the enol tautomer is also conjugated with the carbonyl double bond which provides additional stabilization.
Aromaticity can also stabilize the enol tautomer over the keto tautomer.
Under neutral conditions, the tautomerization is slow, but both acid and base catalysts can be utilized to speed the reaction up.
Biological enol forming reactions use isomerase enzymes to catalyze the shifting of a carbonyl group in sugar molecules, often converting between a ketose and an aldose in a process called carbonyl isomerization.
Choose the correct answer to make the statement true.
a. An exothermic reaction has a positive ΔH and absorbs heat from the surroundings.
b. An exothermic reaction feels warm to the touch. a positive ΔH and gives off heat to the surroundings.
c. An exothermic reaction feels warm to the touch. a negative ΔH and absorbs heat from the surroundings.
d. An exothermic reaction feels warm to the touch. a negative ΔH and gives off heat to the surroundings.
e. An exothermic reaction feels warm to the touch.
sự sắp xếp nguyên tử trong vật chất
Answer:
sosksjsjjs
Explanation:
even i know how to type şïllily
Part A
If the theoretical yield of a reaction is 23.5 g and the actual yield is 14.8 g, what is the percent yield?
Answer:
[tex]\boxed {\boxed {\sf 63.0 \%}}[/tex]
Explanation:
The percent yield is the ratio of the actual yield to the theoretical yield.
[tex]percent \ yield = \frac{actual \ yield}{theoretical \ yield} * 100[/tex]
The actual yield is the amount obtained from performing a chemical reaction. For this problem, it is 14.8 grams. The theoretical yield is the potential amount from performing a chemical reaction at maximum performance. For this problem, it is 23.5 grams.We can substitute the known values into the formula.
[tex]percent \ yield= \frac{ 14.8 \ g}{23.5 \ g}*100[/tex]
Divide.
[tex]percent \ yield = 0.629787234043*100[/tex]
Multiply.
[tex]percent \ yield = 62.9787234043[/tex]
The original measurements for the theoretical and actual yields have 3 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place.
The 7 to the right, in the hundredths place, tells us to round the 9 up to a 0. Since we rounded up to 0, we have to move to the next place to the left and round the 2 up to a 3.
[tex]percent \ yield \approx 63.0[/tex]
The percent yield is approximately 63.0 percent.
2. 27.8 mL of an unknown were added to a 50.0-mL flask that weighs 464.7 g. The total mass of the flask and the liquid is 552.4 g. Calculate the density of the liquid in Lbs/ in3.
Answer:
[tex]d=4.24x10^{-4}\frac{lb}{in^3}[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to set the equation for the calculation of density and mass divided by volume:
[tex]d=\frac{m}{V}[/tex]
Thus, we can find the mass of the unknown by subtracting the total mass of the liquid to the mass of the flask and the liquid:
[tex]m=552.4g-464.7g=87.7g[/tex]
So that we are now able to calculate the density in g/mL first:
[tex]d=\frac{87.7g}{27.8mL}=3.15g/mL[/tex]
Now, we proceed to the conversion to lb/in³ by using the following setup:
[tex]d=3.15\frac{g}{mL}*\frac{1lb}{453.6g}*\frac{1in^3}{16.3871mL}\\\\d=4.24x10^{-4}\frac{lb}{in^3}[/tex]
Regards!
How many molecules of Iron(II)oxide are present in 35.2*10^-23 g of Iron (II)oxide?
Answer:
R.F.M of Iron (II) oxide :
[tex]{ \tt{ = (56 \times 2) + (16 \times 3)}} \\ = 160 \: g[/tex]
Moles :
[tex]{ \tt{ \frac{35.2 \times {10}^{ - 23} }{160} }} \\ = 2.2 \times {10}^{ - 24} \: moles[/tex]
Molecules :
[tex]{ \tt{ = 2.2 \times {10}^{ - 24} \times 6.02 \times {10}^{23} }} \\ = 1.3244 \: molecules[/tex]
The number of molecules of Iron(II) oxide present in 35.2 ×10⁻²³ g of Iron(II) oxide is equal to 2.95.
What is Avogadro's number?Avogadro’s number can be described as the proportionality constant that is used to represent the number of entities or particles in one mole of any substance. Generally, it is used to count atoms, molecules, ions, electrons, or protons, depending upon the chemical reaction or reactant and product.
The value of Avogadro’s constant can be represented as numerically approximately equal to 6.022 × 10²³ mol⁻¹.
Given, the mass of the iron oxide = 35.2 ×10⁻²³ g
The molar mass of the Iron(II) oxide, FeO = 71.84 g/mol
71.84 g of Iron (II) oxide have molecules = 6.022 × 10²³
35.2 ×10⁻²³ g of FeO have molecules = 6.022 × 10²³ × (35.2 ×10⁻²³ /71.84)
The number of molecules of FeO in a given mass = 2.95 molecules
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Monomers that each contain a 5-carbon sugar, a phosphate group, and a nitrogenous base combine and form which type of polymer?
A. Amino acid
B. Carboxylic acid
C. Nucleic acid
D. Fatty acid
Answer:
The correct answer is C. Nucleic acid
Explanation:
Nucleic acids are biological polymers which play an important role in the storage and expresion of genetic information. There are two types of nucleic acids: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). Both are basically composed of:
- a 5-carbon sugar: deoxyribose in DNA and ribose in RNA
- phosphate group
- a nitrogenous base: adenine, cytosine, guanine and thymine in DNA; while RNA contains adenine, cytosine, guanine and uracil.
Write the formulas of all species in solution for the following ionic compounds by writing their dissolving equations:
(Use the lowest possible coefficients.)
1. Rubidium hydroxide: __--__+___
2. Sodium carbonate: __--__+__
3. Ammonium selenite:__--__+__
Answer:
1. RbOH(s) ⇒ Rb⁺(aq) + OH⁻(aq)
2. Na₂CO₃(s) ⇒ 2 Na⁺(aq) + CO₃²⁻(aq)
3. (NH₄)₂SeO₃(s) ⇒2 NH₄⁺(aq) + SeO₃²⁻(aq)
Explanation:
Let's consider the dissolving equations for the following compounds.
1. Rubidium hydroxide
RbOH(s) ⇒ Rb⁺(aq) + OH⁻(aq)
2. Sodium carbonate
Na₂CO₃(s) ⇒ 2 Na⁺(aq) + CO₃²⁻(aq)
3. Ammonium selenite
(NH₄)₂SeO₃(s) ⇒2 NH₄⁺(aq) + SeO₃²⁻(aq)
In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by
A) NaF
B) MgF₂
C) MgBr₂
D) AlF₃
E) AlBr₃
In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by AlF₃. Hence , Option (D) is correct
What factors affect the magnitude of energy of ionic crystalline solids ?For an ionic compound, there are two main terms that this magnitude depends upon: ion size and ion charge.
Ion size: the smaller the ionic radii, the shorter the internuclear distance and, therefore, the closer the ions. This factor makes lattice enthalpy increase
Ion charge: the greater the charge on ions, the greater the attractive forces between them and, therefore, the larger the lattice enthalpy.
The lattice enthalpy of AlF₃ (5215 kJ/mol) is indeed greater than that of other given solids
Therefore , In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by AlF₃. Hence , Option (D) is correct
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The compound chromium(II) chloride is a strong electrolyte. Write the transformation that occurs when solid chromium(II) chloride dissolves in water. Be sure to specify states such as (aq) or (s).
Answer:
CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)
Explanation:
Chromium (II) chloride is a strong electrolyte, that is, when dissolved in water, it completely dissociates into the ions. The cation is chromium (II) and the anion is chloride. The balanced equation for the solution of chromium (II) chloride is:
CrCl₂(s) ⇒ Cr²⁺(aq) + 2 Cl⁻(aq)
A quantity of 0.27 mole of neon is confined in a container at 2.50 atm and 298 Kand then allowed to expand adiabatically under two different conditions: (a) reversibly to 1.00 atm and (b) against a constant pressure of 1.00 atm. Calculate the final temperature in each case.
Answer:
a) Hence, T = 207 K.
b) Hence, T2 = 226 K.
Explanation:
Now the given,
n = 0.27 moles ; P = 2.5 atm ; T = 298 K
a) γ = 5/3 since Ne is a monoatomic gas.
[tex](1 - \gamma )/\gamma = -2/5\\T1 P1^{(1-\gamma)/\gamma}=T2 P2^{(1-\gamma)/\gamma}\\T2 = T1(P1/P2)^{(1 - \gamma)/\gamma}\\T2 = 298 (2.5/1)^{-2/5}= 207 K\\[/tex]
Hence, T = 207 K
b) We know that,[tex]U = W = n Cv (T2 - T1) = -P (V2 - V1)[/tex]
[tex]n(3/2)R(T2 - T1) = -P( n R T2/P2 - n R T1/P1)\\3/2(T2 - T1) = -P (T2/P2 - T1/P1)[/tex]
But P = P2
[tex]3/2(T2 - T1) = -P2(T2/P2 - T1/P1)\\3/2(T2 - T1) = -T2 + P2T1/P1[/tex]
This gives us:
[tex]T2 = 2/5(P2/P1 + 3/2)T1\\T2 = 2/5 x (1 /2.5 + 3/2)/(298)\\T2 = 19/25 x 298 = 226 K[/tex]
Hence, T2 = 226 K
How many atom in protons
Answer:
Its atomic number is 14 and its atomic mass is 28. The most common isotope of uranium has 92 protons and 146 neutrons. Its atomic number is 92 and its atomic mass is 238 (92 + 146).
When solid Ni metal is put into an aqueous solution of Pb(NO3)2, solid Pb metal and a solution of Ni(NO3)2 result. Write the net ionic equation for the reaction.
Answer:
[tex]Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to write the complete molecular equation as shown below:
[tex]Pb(NO_3)_2(aq)+Ni(s)\rightarrow Ni(NO_3)_2(aq)+Pb(s)[/tex]
Now, we can separate the nitrates in ions as they are aqueous to obtain:
[tex]Pb^{2+}(aq)+2(NO_3)^-(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+2(NO_3)^-(aq)+Pb(s)[/tex]
And then, we cancel out the nitrate ions as the spectator ones, for us to obtain the net ionic equation:
[tex]Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)[/tex]
Best regards!
2. For each of the ionic compounds in the table below, name the compound and explain the rule that you
used in formulating your name for the compound.
Name:
Rule for naming compound:
-PbF4
-NH4NO3
-Li2S
Answer:
2
Explanation:
Lead(|V) fluoride
Ammonium Nitrate
Lithium sulfide
For the rules, I don't know what you were taught. I just do it intuitively since I have done so much chemistry.
The first one the roman numerals represents the charge of the lead which much match the 4- charge from the 4 fluorides.
The second one is just two polyatomic ions which you just have to remember.
The last one is the typical ionic compound naming technique i guess.
Use dimensional analysis to solve the following problems. Pay attention to correct use of units and correct use of significant figures in calculations. Please show work!
3) Convert 0.250 moles of aluminum sulfate to grams.
4) Convert 2.70 grams of ammonia to moles.
Answer:
0.000731 grams aluminium sulfate
46.0 mols ammonia
Explanation:
ALS = shorthand for aluminium sulfate which has a molar mass of 342.15 g/mol
[tex]ALS: \frac{0.250mols}{1} *\frac{1g}{342.15mols} = \frac{0.250g}{342.15}=0.0007307 g[/tex]
NH3 has a molar mass of 17.031 g/mol
[tex]NH3: \frac{2.70g}{1} *\frac{17.031mols}{1g} = \frac{0.250g}{342.15}=45.9837 mols[/tex]
Aluminium sulphate (AlS) whose molar mass is= [tex]\sf{ 342.15\dfrac{g}{mol} }[/tex]
we have to find the 0.250 moles of aluminum sulphate.
[tex]\implies AlS=\dfrac{1g}{342.15~mole}×0.250~mole \\\\\implies AlS=\dfrac{0.250}{342.15}\\\\\implies \dfrac{\frac{250}{1000}}{\frac{34215}{100}}\\\\\implies \dfrac{250}{1000}×\dfrac{100}{34215}\\\\=0.00073067\approx{0.0007307~g}[/tex]
[tex]\\\\\\[/tex]
Ammonia(NH3) whose molar mass is =[tex]\sf{17.031\dfrac{mol}{g} }[/tex]
We have to find 2.70 grams of ammonia
[tex]\implies NH_{3}=\dfrac{17.031~mol}{1g}×2.70g\\\\ 17.031×2.70\\\\\dfrac{17031}{1000}×\dfrac{270}{100}\\\\ \dfrac{4598370}{100000}\\\\=45.9837\approx{46~mole}[/tex]
repining of fruits is which type of change
Answer:
irreversible.
I hope this will help you
Help naming this plzzzzzzzzzzzzz
Answer:
A. 3-chloro-1-methylcyclobutane.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to infer that the name of this compound is A. 3-chloro-1-methylcyclobutane because of the fact that the parent chain is a cyclobutane which starts by the methyl radical as it has the priority over the chlorine radical which is actually named first at the third carbon (clockwise).
Therefore the name is given in A, accordingly to the IUPAC rules of nomenclature.
Regards!
A second-order reaction has a half-life of 12 s when the initial concentration of reactant is 0.98 M. The rate constant for this reaction is ________ M-1s-1. A) 12
Answer: 0.085 (Ms)⁻¹
Explanation: Half life = 12 s
is the initial concentration = 0.98 M
Half life expression for second order kinetic is:
k = 0.085 (Ms)⁻¹
The rate constant for this reaction is 0.085 (Ms)⁻¹ .
what is Lewis acid and Lewis base? give examples
Explanation:
example is copper iron...........
Question 9 of 25
How many hydrogen atoms are in a molecule of table sugar (C12H,2011)?
O A. 12
B. 45
C. 11
D. 22
SUBMIT
D.22
is my answer than welcome
How many colors are there in a rainbow?
[tex]\boxed{\large{\bold{\blue{ANSWER~:) }}}}[/tex]
There are 7 colours in a rainbow The colours of the rainbow are Red, Orange, Yellow, Green, Blue, Indigo and Violet.Explanation:
there r seven colors in a rainbow.red, orange, yellow, green, blue, indigo and violet.hope it helps.stay safe healthy and happy..what is the hybridisation of the central carbon in CH3C triple bonded to N
Explanation:
the carbon would be sp3 hybridized, and it doesn't matter which carbon, since either of them have a full octet
You will observe a weak acid-strong base titration in this experiment. Select all statements that are true about weak acid-strong base titrations.
A. Weak acid-strong base titrations always start at a higher pH than strong acid-strong base titrations, no matter the initial concentration.
B. The pH is less than 7 at the equivalence point.
C. The pH is greater than 7 at the equivalence point.
D. Half way to the equivalence point, a buffer region is observed.
Answer:
The pH is greater than 7 at the equivalence point.
Explanation:
Equivalence point is the point where the acid reacts with the base as stipulated in the equation of the reaction.
When a weak acid and a strong base are titrated, the pH of the solution at equivalence point is actually found to be around about pH ~ 9.
Hence, for a weak acid and strong base titration, The pH is greater than 7 at the equivalence point.
A titration between a weak acid and a strong base yields a solution whose pH is greater than 7 at the equivalence point.
What are weak acids?Weak acids are acids which only ionize partially in aqueous solutions.
When weak acids are dissolved in water, they produce only few hydrogen ions.
A strong base on the other hand ionizes completely to produce hydroxide ions in aqueous solutions.
The titration of a weak acid and a strong base gives a solution whose pH is greater than 7 at equivalence point.
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A student attempts to separate 4.656 g of a sand/salt mixture just like you did in this lab. After carrying out the experiment, she recovers 2.775 g of sand and 0.852 g of salt.a. What was the percent composition of sand in the mixture according to the student's data? b. What was the percent recovery?
Answer:
Explanation:
a ) Total mixture = 4.656 g
Sand recovered = 2.775 g
percent composition of sand in the mixture
= (2.775 g / 4.656 g ) x 100
= 59.6 % .
b )
Total of sand and salt recovered = 2.775 g + .852 g = 3.627 g .
Total mixture = 4.656 g
percent recovery = (3.627 / 4.656 ) x 100
= 77.9 % .
