Select the number of valence electrons for hydrogen.

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!

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.

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 )

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.

Answer:

Explanation:

CCl4 => C + 2Cl2

Answer:

1.27 × 10⁵ L

Explanation:

Step 1: Given data

Step 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

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.

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

Oxidation reaction is a chemical reaction which can be described as follows ;

In 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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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)

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

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.

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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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]

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.

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!

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.

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.

Answer:

4.77mol is the correct answer

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.

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.

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.

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].

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]

Answer:

I do not speak Spanish.

Explanation:

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.

Molar mass of Acetone

Now

Moles of Acetone:-

Amount of heat:-

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

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.

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.

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.