Draw the structure of the neutral product formed in the reaction of dimethyl malonate and methyl vinyl ketone.

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.

Answer:

1. Nuts

2. Canned meats and seafood

3. Dried grains

4. Dark chocolate

5. Protein powders

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)

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Explanation:

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 % .

Answer:

I think you can make 26, hope this helped.

Explanation:

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

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

Learn more about crystalline solids here ;

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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:

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.

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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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)

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:

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:

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).

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 .

Answer:

irreversible.

I hope this will help you

Answer:

sosksjsjjs

Explanation:

even i know how to type şïllily

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.

Molar mass of Acetone

Now

Moles of Acetone:-

Amount of heat:-

Answer:

I do not speak Spanish.

Explanation:

D.22

is my answer than welcome

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

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:

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

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.

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:

[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!

Explanation:

the carbon would be sp3 hybridized, and it doesn't matter which carbon, since either of them have a full octet