Answer:
Vanlency of hydrogen - 11
Electrons of hydrogen - 1
Answer:
The answer is: 1
Hope this helps :) <3
Explanation:
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!
Suppose that a certain atom possesses only four distinct energy levels. Assuming that all transitions between levels are possible, how many spectral lines will this atom exhibit
Answer:
Following are the response to the given question:
Explanation:
The number of shells
n = 4
Calculating the spectral line:
[tex]= \frac{n(n-1)}{2}\\\\ = \frac{4(4-1)}{2} \\\\= \frac{4\times 3}{2}\\\\ = \frac{12}{2}\\\\ = 6[/tex]
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.
Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction.
half-reaction identification
Cu+(aq)--->Cu2+(aq) + e- _________
I2(s) + 2e--->2I-(aq) _________
Answer:
Cu+(aq)--->Cu2+(aq) + e- : oxidation
reason: there is loss of electrons.
I2(s) + 2e--->2I-(aq) : reduction
reason: There is reduction of electrons.
What will be the equilibrium temperature when a 245 g block of lead at 300oC is placed in 150-g aluminum container containing 820 g of water at 12.0oC?
Answer:
The correct approach is "12.25°C".
Explanation:
Given:
Mass of lead,
mc = 245 g
Initial temperature,
tc = 300°C
Mass of Aluminum,
ma = 150 g
Initial temperature,
ta = 12.0°C
Mass of water,
mw = 820 g
Initial temperature,
tw = 12.0°C
Now,
The heat received in equivalent to heat given by copper.
The quantity of heat = [tex]m\times s\times t \ J[/tex]
then,
⇒ [tex]245\times .013\times (300-T) = 150\times .9\times (T-12.0) + 820\times 4.2\times (T-12.0)[/tex]
⇒ [tex]3.185(300-T) = 135(T-12.0) + 3444(T-12.0)[/tex]
⇒ [tex]955.5-3.185T=135T-1620+3444T-41328[/tex]
⇒ [tex]43903.5 = 3582.185 T[/tex]
⇒ [tex]T = 12.25^{\circ} C[/tex]
g A high altitude balloon is filled with 1.41 x 104 L of hydrogen gas (H2) at a temperature of 21oC and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is -48oC and the pressure is 63.1 torr
Answer:
1.27 × 10⁵ L
Explanation:
Step 1: Given data
Initial pressure (P₁): 745 TorrInitial volume (V₁): 1.41 × 10⁴ LInital temperature (T₁): 21 °CFinal pressure (P₂): 63.1 TorrFinal volume (V₂): ?Final temperature (T₂): -48 °CStep 2: Convert the temperatures to the Kelvin scale
We will use the following expression.
K = °C + 273.15
K = 21 °C + 273.15 = 294 K
K = -48 °C + 273.15 = 225 K
Step 3: Calculate the final volume of the balloon
We will use the combined gas law.
P₁ × V₁ / T₁ = P₂ × V₂ / T₂
V₂ = P₁ × V₁ × T₂/ T₁ × P₂
V₂ = 745 Torr × 1.41 × 10⁴ L × 225 K/ 294 K × 63.1 Torr
V₂ = 1.27 × 10⁵ L
CAN HF USED TO CLEAVE ETHERS EXPLAIN
Answer:
no
Explanation:
Fluoride is not nucleophilic (having the tendency to donate electrons) enough to allow for the use of HF to cleave ethers in protic media(protic solvents are polar liquid compounds that have dissociable hydrogen atoms). The rate of reaction is comparably low, so that heating of the reaction mixture is required.
Write the functional isomers of C2H6O?
Answer:
See explanation
Explanation:
Isomers are molecules which have the same molecular formula but different structural formulas. Sometimes, isomers may even contain different functional groups.
The formula C2H6O may refer to an ether or an alcohol. The compound could be CH3CH2OH(ethanol) or CH3OCH3(methoxymethane).
Hence, the functional isomers of the formula C2H6O are ethanol and methoxymethane.
How to solve this problem step by step
Answer:
[tex]V_2= 736mL[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by using the combined gas law:
[tex]\frac{P_2V_2}{T_2}= \frac{P_1V_1}{T_1}[/tex]
Thus, we solve for the final volume by solving for V2 as follows:
[tex]V_2= \frac{P_1V_1T_2}{T_1P_2}[/tex]
Now, we plug in the variables to obtain the result in milliliters and making sure we have both temperatures in Kelvins:
[tex]V_2= \frac{1.20atm*735mL*279K}{(112+273)K*660/760atm}\\\\V_2= \frac{1.20atm*735mL*279K}{(112+273)K*660/760atm}=736mL[/tex]
Regards!
How much of a 24-gram sample of Radium-226 will remain unchanged at the end of three half-life periods?
Answer:
The right answer is "3 g".
Explanation:
Given:
Initial mass substance,
[tex]M_0=24 \ g[/tex]
By using the relation between half lives and amount of substances will be:
⇒ [tex]M=\frac{M_0}{2^n}[/tex]
[tex]=\frac{24}{2^3}[/tex]
[tex]=3 \ g[/tex]
Thus, the above is the correct answer.
Draw the structure of the neutral product formed in the reaction of dimethyl malonate and methyl vinyl ketone.
Answer:
Explanation:
The reaction between dimethyl malonate which is an active methylene group with an (∝, β-unsaturated carbonyl compound) i.e methyl vinyl ketone is known as a Micheal Addition reaction. The reaction mechanism starts with the base attack on the β-carbon to remove the acidic ∝-hydrogens and form a carbanion. The carbanion formed(enolate ion) attacks the methyl vinyl ketone(i.e. a nucleophilic attack at the β-carbon) to give a Micheal addition product, this is followed by the protonation to give the neutral product.
Consider the following equation for the combustion of acetone (C3H6O), the main ingredient in nail polish remover.
C3H6O(l) + 4O2(g) → 3CO2(g) + 3H2O(g) ΔHrxn = −1790kJ
If a bottle of nail polish remover contains 143 g of acetone, how much heat would be released by its complete combustion? Express your answer to three significant figures.
Molar mass of Acetone
C3H6O3(12)+6+1658g/molNow
1 mol releases -1790KJ heat .Moles of Acetone:-
143/58=2.5molAmount of heat:-
2.5(-1790)=-4475kJ9. How can you separate sugar from a sugar solution contained in a glass without taste? Explain
Answer:
See explanation
Explanation:
Sugar is a polar crystalline substance. The sugar crystal is capable of dissolving in water since it is polar.
When sugar dissolves in water, a sugar solution is formed. If I want to separate the sugar from the water in the solution, I have to boil the solution to a very high temperature.
When I do that, the water in the sugar solution is driven off and the pure sugar crystal is left behind.
How many atoms are in .45 moles of P4010
Answer:
5×6.02×1023
Explanation:
there are 5×6.02×1023 molecules of p4010 in 5mole. there are four P atoms in a single molecule of p4010
C. A sample may contain any or all of the following ions Hg2 2, Ba 2, and Al 3. 1) No precipitate forms when an aqueous solution of NaCl was added to the sample solution. 2) No precipitate forms when an aqueous solution of Na2SO4 was added to the sample. 3) A precipitate forms when the sample solution was made basic with NaOH. Which ion or ions were present. Write the net ionic equation(s) for the the reaction (s).
Answer:
Al^3+
Explanation:
Solubility rules tell us what substances are soluble in water. Since NaCl was added and no precipitate was observed, the mercury II ion is absent.
Addition of Na2SO4 does not form a precipitate meaning that Ba^2+ is absent.
If a precipitate is formed when NaOH is added, the the ionic reaction is as follows;
Al^3+(aq) + 3 OH^-(aq) ------> Al(OH)3(s)
How many grams of sodium nitrate (NaNO3) are needed to
prepare 100 grams of a 15.0 % by mass sodium nitrate
solution?
Answer:
15.0 g
Explanation:
15.0% =0.150
100.0 g × 0.150= 15.0g
Sodium nitrate is "an inorganic compound with the formula of NaNO₃.
What is an inorganic compound?Inorganic compound is "a chemical compound that lacks carbon–hydrogen bonds".
15% = 0.15
100.0 g × 0.15= 15g
Hence, 15g of Sodium nitrate are needed to prepare 100 gms of a 15% by mass sodium nitrate.
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Si enfriamos mercurio de 100C. Calcular la cantidad de calor que se debe restar sabiendo que la masa de mercurio es de 1800gr
Answer:
I do not speak Spanish.
Explanation:
Which of the following are examples of physical properties of ethanol? Select all that apply.
The boiling point is 78.37°C
It is a clear, colorless liquid
It is flammable
It is a liquid at room temperature
2.50 L of 0.700 M phosphoric acid reacts with 5.25 moles of sodium hydroxide. How many moles of hydrogen ions will completely neutralize the moles of hydroxide ions present in this amount of sodium hydroxide? a) 0.583 b) 1.75 c) 3.00 d) 15.75 e) 5.25
Answer:
5.25 moles of protons. Option e
Explanation:
Reaction between phosphoric acid and sodium hydroxide is neutralization.
We can also say, we have an acid base equilibrium right here:
H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
Initially we have 5.25 moles of base.
We have data from the acid, to state its moles:
M = mol/L, so mol = M . L
mol = 1.75 moles of acid
If we think in the acid we know:
H₃PO₄ → 3H⁺ + PO₄⁻³
We know that 1 mol of acid can give 3 moles of protons (hydrogen ions)
If we have 1.75 moles of acid, we may have
(1.75 . 3) /1 = 5.25 moles of protons
These moles will be neutralized by the 5.25 moles of base
H₃O⁺ + OH⁻ ⇄ 2H₂O Kw
In a titration of a weak acid and a strong base, we have a basic pH
PLEASE HELP ASAP MOLES TO MOLECULES
Answer:
4.77mol is the correct answer
During the postabsorptive state, metabolism adjusts to a catabolic state.
a. True
b. False
Answer:
The postabsorptive state (also called the fasting state) occurs when the food is already digested and absorbed, and it usually occurs overnight, when you sleep (if you skip meals for some days, you will enter in this state).
The catabolic state is the metabolic breakdown of molecules into simpler ones, releasing energy (heat) and utilizable resources.
Now, when you are in a postabsorptive state, the glucose levels start to drop, then the body starts to depend on the glycogen stores, which are catabolized into glucose, this is defined as the start of the postabsorptive state.
So yes, as the postabsorptive states, catabolic processes start to happen, so the statement is true.
Balance the following chemical equation.
CCl4 -> ___ C+ ___ Cl2
Answer:
Explanation:
CCl4 => C + 2Cl2
Solutions of Cu2+ turn blue litmus red because of the equilibrium: Cu(H2O)62+(aq) + H2O(l) ↔ Cu(H2O)5(OH)+(aq) + H3O+(aq) for which Ka = 1.0 x 10-8. Calculate the pH of 0.10 M Cu(NO3)2(aq).
Answer: The pH of 0.10 M [tex]Cu(NO_{3})_{2}(aq)[/tex] is 4.49.
Explanation:
Given: Initial concentration of [tex]Cu(H_{2}O)^{2+}_{6}[/tex] = 0.10 M
[tex]K_{a} = 1.0 \times 10^{-8}[/tex]
Let us assume that amount of [tex]Cu(H_{2}O)^{2+}_{6}[/tex] dissociates is x. So, ICE table for dissociation of [tex]Cu(H_{2}O)^{2+}_{6}[/tex] is as follows.
[tex]Cu(H_{2}O)^{2+}_{6} \rightleftharpoons [Cu(H_{2}O)_{5}(OH)]^{+} + H_{3}O^{+}[/tex]
Initial: 0.10 M 0 0
Change: -x +x +x
Equilibrium: (0.10 - x) M x x
As the value of [tex]K_{a}[/tex] is very small. So, it is assumed that the compound will dissociate very less. Hence, x << 0.10 M.
And, (0.10 - x) will be approximately equal to 0.10 M.
The expression for [tex]K_{a}[/tex] value is as follows.
[tex]K_{a} = \frac{[Cu(H_{2}O)^{2+}_{6}][H_{3}O^{+}]}{[Cu(H_{2}O)^{2+}_{6}]}\\1.0 \times 10^{-8} = \frac{x \times x}{0.10}\\x = 3.2 \times 10^{-5}[/tex]
Hence, [tex][H_{3}O^{+}] = 3.2 \times 10^{-5}[/tex]
Formula to calculate pH is as follows.
[tex]pH = -log [H^{+}][/tex]
Substitute the values into above formula as follows.
[tex]pH = -log [H^{+}]\\= - log (3.2 \times 10^{-5})\\= 4.49[/tex]
Thus, we can conclude that the pH of 0.10 M [tex]Cu(NO_{3})_{2}(aq)[/tex] is 4.49.
DATA SHEET p 45. TRIAL 1 TRIAL 2 1. Mass of the ground pretzel 1.00 gram 1.03 g 2. Initial volume of the AgNO3 solution 0.00 mL 9.10 mL 3. Final volume of the AgNO3 solution 9.10 mL 17.25 mL 4. Volume of AgNO3 solution used 9.10 mL 8.15 mL Line 3 – Line 2 5. Volume of AgNO3 solution in liters _____ L _____ L 6. Molarity of AgNO3 solution 0.01 M 0.01 M (given) 7. Number of moles of AgNO3 ______ mol _____ mol (Line 5 × Line 6) 8. Number of mol of NaCl present in pretzel ______ mol _____ mol (Line 7) number of mol NaCl = number of mol AgNO3 9. Mass of NaCl present in the titrated sample ______ gram _____ gram (Line 8) × 58.5 g/mol
Answer:
1. 1.00 gm
2. 50 ml
3. 38.93 ml
4. 11.07 ml
5. 0.01107 L
6. 0.010 moles / L
7. 0.0001107 moles
8. 0.0001107 moles
9. 0.00647042 grams
Explanation:
Silver nitrate can react with various compounds to form different products. The weight of products may be different from the original solution introduced due to combustion reaction, as heat energy is released during the chemical process.
An endothermic reaction will start when the required
energy is received from the environment or solution.
AH
activation
thermal
kinetic
Answer:
A: ΔH
Explanation:
Endothermic reactions are this that occur as a result of absorption of heat energy from the surroundings by the reactants to form new products.
Thus, we can say it is one with an increase in enthalpy (ΔH) of the system.
Thus, option A is correct.
When butane reacts with Br2 in the presence of Cl2, both brominated and chlorinated products are obtained. Under such conditions, the usual selectivity of bromination is not observed. In other words, the ratio of 2-bromobutane to 1-bromobutane is very similar to the ratio of 2-chlorobutane to 1-chlorobutane. Can you offer and explanation as to why we do not observe the normal selectivity expected for bromination
Answer:
Bromine radical formation is carried out in the presence of Br₂ and Cl₂ causing the normal selectivity not to be observed ( this causes the difference in activation energy to be reduced )
Explanation:
Why the normal selectivity expected for bromination is not observed
On the basis of selectivity and applying the Arrhenius equation the greater the difference between the activation energies the more the selectivity.
as seen in the formation of primary and secondary radicals in the Bromine radical formation. this difference is caused mainly by the propagation step ( exothermic ) . But the main reason why the the usual selectivity of bromination is not observed is because it Bromine radical formation is carried out in the presence of Br₂ and Cl₂ ( this causes the difference in activation energy to be reduced )
Which statement is true with respect to standard reduction potentials?
SRP values that are greater than zero always represent a reduction reaction.
SRP values that are less than zero always represent a reduction reaction.
Half-reactions with SRP values greater than zero are spontaneous.
Half-reactions with SRP values greater than zero are nonspontaneous.
Answer:
C). Half-reactions with SRP values greater than zero are spontaneous.
Explanation:
SRPs or Standard Reduction Potentials are characterized as the ability of a probable distinction among the anode and cathode of a usual/standard cell. It aims to examine the capacity of chemicals to reduce themselves.
The third statement asserts a true claim regarding the SRPs(Standard Reduction Potentials) that the 'half-reactions which take place with the SRP possesses the values higher than zero and they are unconstrained.' The other statements are incorrect as they either show the estimation of SRPs more than 0 or display them as being restricted. Thus, option C is the correct answer.
0.50 g of hydrogen chloride (HCl) is dissolved in water to make 4.0 L of solution. What is the pH of the resulting hydrochloric acid solution
Explanation:
Given the mass of HCl is ---- 0.50 g
The volume of solution is --- 4.0 L
To determine the pH of the resulting solution, follow the below-shown procedure:
1. Calculate the number of moles of HCl given by using the formula:
[tex]number of moles of a substance=\frac{given mass of the substance}{its molecular mass}[/tex]
2. Calculate the molarity of HCl.
3. Calculate pH of the solution using the formula:
[tex]pH=-log[H^+][/tex]
Since HCl is a strong acid, it undergoes complete ionization when dissolved in water.
[tex]HCl(aq)->H^+(aq)+Cl^-(aq)[/tex]
Thus, [tex][HCl]=[H^+][/tex]
Calculation:
1. Number of moles of HCl given:
[tex]number of moles of a substance=\frac{given mass of the substance}{its molecular mass}\\=0.50g/36.5g/mol\\=0.0137mol[/tex]
2. Concentration of HCl:
[tex]Molarity of HCl=\frac{number of moles of HCl}{its molar mass}\\=\frac{0.0137 mol}{4.0 L} \\= 0.003425 M[/tex]
3. pH of the solution:
[tex]pH=-log[H^+]\\=-log(0.003425)\\=2.47[/tex]
Hence, pH of the given solution is 2.47.
Identify the oxidation half-reaction for this reaction:
Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)
A. Fe2+ + 2e → Fe(s)
O B. H2(g) → 2H+ + 2e
O C. Fe(s) → Fe2+ + 2e
O D. 2H+ + 2e → H2(9)
Answer:
Fe(s)->Fe2+2e-
Explanation:
A.p.e.x
The oxidation half-reaction for the given reaction is Fe(s) → Fe²⁺ + 2e⁻ Hence, Option (C) is correct
What is Oxidation reaction ?Oxidation reaction is a chemical reaction which can be described as follows ;
Addition of oxygen Removal of hydrogen Loss of ElectronAddition of electronegative atomRemoval of Electropositive elementIn the given reaction ;
Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g)
Fe at RHS got converted to Fe²⁺ state at LHS which shows the gain of electron by Fe with in the reaction.
Therefore,
The oxidation half-reaction for the given reaction is Fe(s) → Fe²⁺ + 2e⁻ Hence, Option (C) is correct
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5. A beam of photons with a minimum energy of 222 kJ/mol can eject electrons from a potassium surface. Estimate the range of wavelengths of light that can be used to cause this phenomenon. Show your calculations with units of measure (dimensional analysis) and briefly explain your reasoning.
Answer: The range of wavelengths of light that can be used to cause given phenomenon is [tex]8.953 \times 10^{21} m[/tex].
Explanation:
Given: 222 kJ/mol (1 kJ = 1000 J) = 222000 J
Formula used is as follows.
[tex]E = \frac{hc}{\lambda}[/tex]
where,
E = energy
h = Planck's constant = [tex]6.625 \times 10^{-25} Js[/tex]
c = speed of light = [tex]3 \times 10^{8} m/s[/tex]
Substitute the values into above formula as follows.
[tex]E = \frac{hc}{\lambda}\\222000 J = \frac{6.625 \times 10^{-34}Js \times 3 \times 10^{8} m/s}{\lambda}\\\lambda = 8.953 \times 10^{21} m[/tex]
Thus, we can conclude that the range of wavelengths of light that can be used to cause given phenomenon is [tex]8.953 \times 10^{21} m[/tex].
