what is the best course of action if solid material remains in the flask after the heating step of recrystallization

Answer:

Cooking a pot of soup

Explanation:

id say that because when you freeze ice cream, its already frozen, so no heat is being released. melting ice wouldn't be the answer because, once again, it is already frozen, and no heat is being released.

Answer:

the correct answer is freezing ice cream

Explanation:

i took the test & got this question correct. also, heat energy is released when freezing because there is no heat energy involved.

Answer:

Hope this helps...

Good luck on your assignment...

Answer:

Copper (II) chloride.

Explanation:

Hello,

In this case, considering the described reaction which is also given as:

[tex]2Al + 3CuCl_2 \rightarrow 2AlCl_3 + 3Cu[/tex]

For us to identify the limiting reactant we first compute the available moles of aluminium:

[tex]n_{Al}=512gAl*\frac{1molAl}{27gAl}=19.0molAl[/tex]

Next, we compute the consumed moles of aluminium by the 1147 grams of copper (II) chloride by using their 2:3 molar ratio:

[tex]n_{Al}^{consumed}=1147gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2}*\frac{2molAl}{3molCuCl_2} =5.69molAl[/tex]

Thereby, we can infer aluminium is in excess since less moles are consumed than available whereas the copper (II) chloride is the limiting reactant.

Best regards.

3.97×1023 molecules C2H6          1 mol  C2H6  

------------------------------------------ x ------------------------------------   = 0.66 mol C2H6

                                                    6.022 x 1023 molec. C2H6

Answer:

the mass of HBr that would react is 25.41 g of HBr

Explanation:

attached is the calculations.

Answer:

the density if vinegar will also be needed

Explanation:

Because this is an experiment of volumetric analysis

Answer:

1.22 × 10³ g

Explanation:

Step 1: Write the balanced equation

P₄O₁₀ + 6 H₂O ⇒ 4 H₃PO₄

Step 2: Calculate the moles of H₃PO₄ produced by 3.10 moles of P₄O₁₀

The molar ratio of P₄O₁₀ to H₃PO₄ is 1:4. The moles of H₃PO₄ produced are 4/1 × 3.10 mol = 12.4 mol

Step 3: Calculate the mass corresponding to 12.4 moles of H₃PO₄

The molar mass of H₃PO₄ is 97.99 g/mol.

[tex]12.4 mol \times \frac{97.99g}{mol} = 1.22 \times 10^{3} g[/tex]

Answer:

2ErF3 + 3Mg → 2Er + 3MgF2

Explanation:

Erbium metal is a member of the lanthaniod series. It reacts with halogens directly to yield erbium III halides such as erbium III chloride, Erbium III fluoride etc.

Erbium metal (Er) can be prepared by reacting erbium(III) fluoride with magnesium; the products are erbium metal and magnesium fluoride. This is a normal redox process in which the Erbium metal is reduced while the magnesium is oxidized. The balanced reaction equation of this process is; 2ErF3 + 3Mg → 2Er + 3MgF2

Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

[tex]n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC[/tex]

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

[tex]n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH[/tex]

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

[tex]m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH} =0.1900gO[/tex]

4. Compute the moles of oxygen by using its molar mass:

[tex]n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO[/tex]

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

[tex]C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1[/tex]

6. Search for the closest whole number (in this case multiply by 2):

[tex]C_3H_6O_2[/tex]

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

[tex]M=12*3+1*6+16*2=74g/mol[/tex]

Which is about three times in the molecular formula, for that reason, the actual formula is:

[tex]C_9H_{18}O_6[/tex]

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.

Answer:

[tex]C_{base}=0.302M[/tex]

Explanation:

Hello,

In this case, we can evidence that when calcium hydroxide solution reacts with hydrochloric acid solution, the balanced neutralization reaction turns out:

[tex]2HCl(aq)+Ca(OH)_2\rightarrow CaCl_2(aq)+2H_2O(l)[/tex]

Moreover, the concentration of neutralized calcium hydroxide can be computed by using the 2:1 mole ratio between the base and the acid:

[tex]C_{acid}V_{acid}=2*C_{base}V_{base}\\\\C_{base}=\frac{C_{acid}V_{acid}}{2*V_{base}} =\frac{0.225M*26.85mL}{2*10.0mL}\\ \\C_{base}=0.302M[/tex]

Regards.

Answer:

the answer is monerans

Explanation:

When Carl Woese developed the modern system of classification, he broke the previous kingdom of Monera into the two kingdoms of Bacteria and Archaea.

Some biologists believed it made sense to classify prokaryotes as belonging to their own kingdom, the Monera. That served as the foundation for Richard Whittaker  and Lynn Margulis's five-kingdom proposal, which enhanced the Haeckel plan by include a kingdom of fungus.

Protists, protozoa, monera, fungi, and viruses have long been proposed as belonging to different kingdoms, but traditional evolutionists during the majority of the 20th century had given none of them any thought.

Later, the Monera kingdom was split into Eubacteria and Archaebacteria by Carl Woese .  Moreover, he divided the five kingdoms into three domains: Eukaryotes, Archaea, and Bacteria.

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Your question is incomplete. But your complete question is as follows:

When Carl Woese developed the modern system of classification, he broke the previous kingdom of into the two kingdoms of  _____  into  Bacteria and Archaea.

Answer:

254.23 g/mol

Explanation:

Atomic mass for Osmium tetroxide would be 254.23 g/ml

Answer:254.2276

Explanation:

Answer:

[tex]r_{NH_3,abs} =6.00\frac{M}{s}[/tex]

Explanation:

Hello,

In this case, we can write the law of mass action for the undergoing chemical reaction, based on the rates and the stoichiometric coefficients:

[tex]\frac{1}{-2}r_{NH_3} =\frac{1}{1} r_{N_2}=\frac{1}{3}r_{H_2}[/tex]

In such a way, knowing the rate of formation hydrogen (H₂), we can know the  rate of change of ammonia, that must be negative for consumption:

[tex]r_{NH_3} =\frac{-2}{3}r_{H_2}=\frac{-2}{3}*9.00\frac{M}{s} \\\\r_{NH_3} =-6.00\frac{M}{s}[/tex]

Nevertheless, the absolute magnitude will be positive:

[tex]r_{NH_3,abs} =6.00\frac{M}{s}[/tex]

Best regards.

Considering the definition of bond and the different type of bonds, valence is not one of  the types of bonds.

A chemical bond is defined as the force by which the atoms of a compound are held together. These are electromagnetic forces that give rise to different types of chemical bonds.

In other words, a chemical bond is the force that joins atoms to form chemical compounds and confers stability to the resulting compound.

The covalent bond is the chemical bond between atoms where electrons are shared, forming a molecule. Covalent bonds are established between non-metallic elements, such as hydrogen H, oxygen O and chlorine Cl. These elements have many electrons in their outermost level (valence electrons) and have a tendency to gain electrons to acquire the stability of the electronic structure of noble gas. The shared electron pair is common to the two atoms and holds them together.

An ionic bond is produced between metallic and non-metallic atoms, where electrons are completely transferred from one atom to another. During this process, one atom loses electrons and another one gains them, forming ions.

Metallic bonds are a type of chemical bond that occurs only between atoms of the same metallic element. In this way, metals achieve extremely compact, solid and resistant molecular structures, since the atoms that share their valence electrons.

In summary, valence is not one of  the types of bonds. The types of bonds are covalent, ionic and metallic.

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Answer:

Ba(NO₃)₂(s) → Ba²⁺ + 2NO₃⁻

Explanation:

A strong electrolyte is a salt (A compound that has an anion and a cation and are neutral) that, in water, dissociates completely in its ions.

In Barium nitrate, Ba(NO₃)₂, the cation is Ba²⁺ (Alkaline earth metal), and the anion is the nitrate ion, NO₃⁻.

Thus, when Ba(NO₃)₂ (s) is dissolved in water, its transformation is:

When solid barium nitrate (Ba(NO₃)₂) dissolves in water, it undergoes a dissociation process where the compound breaks apart into its constituent ions.

Dissociation refers to the process in which a compound breaks apart into its constituent ions when dissolved in a solvent, typically water. In this process, the chemical bonds within the compound are disrupted, resulting in the separation of positive and negative ions.

The dissociation occurs due to the interaction between the solute particles and the solvent molecules, leading to the formation of hydrated ions.

The transformation can be represented as follows:

Ba(NO₃)₂(s) → Ba²⁺(aq) + 2NO₃⁻(aq)

In this process, the barium nitrate compound dissociates into barium ions (Ba²⁺) and nitrate ions (NO₃⁻) in the aqueous solution. The resulting ions are free to move and conduct electricity, indicating that barium nitrate is a strong electrolyte when dissolved in water.

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Answer:

1a. 0.89 gcm¯³

1b. Yes.

1c. Tetrahydrofuran.

2. 0.54 g/mL

Explanation:

1. Data obtained from the question include:

Volume = 0.988 L = 988 cm³

Mass = 879 g

1a. Determination of the density

Density = mass /volume

Density = 879/ 988

Density = 0.89 gcm¯³

Therefore, the density of the liquid is 0.89 gcm¯³

1b. From the given data, it is possible to determine the identity of the liquid.

1c. The density of the liquid is 0.89 gcm¯³. Comparing the density of the liquid obtained with those given in the table, the liquid is tetrahydrofuran

2. Data obtained from the question include:

Mass of empty cylinder = 5.25 g

Mass of cylinder and sodium thiosulfate = 75.82 g

Volume = 130.63 mL

Next, we shall determine the mass of sodium thiosulfate. This can be obtain as follow:

Mass of empty cylinder = 5.25 g

Mass of cylinder and sodium thiosulfate = 75.82 g

Mass of sodium thiosulfate =.?

Mass of sodium thiosulfate = Mass of cylinder and sodium thiosulfate – Mass of empty cylinder

Mass of sodium thiosulfate = 75.82 – 5.25

Mass of sodium thiosulfate = 70.57 g

Finally, we shall determine the concentration of the sodium thiosulfate as follow:

Mass = 70.57 g

Volume = 130.63 mL

Concentration =?

Concentration = mass /volume

Concentration = 70.57/130.63

Concentration = 0.54 g/mL

The concentration of the solution is 0.54 g/mL

Answer:

The answer to your question is given below

Explanation:

The balanced equation for the reaction is given below:

CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)

1. Writing an expression for the equilibrium constant, K.

The equilibrium constant, K for a reaction is simply the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.

Thus, we can write the equilibrium constant, K for the reaction as follow:

CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)

K = [CaCO3] [H2]⁴ / [CaO] [CH4] [H2O]²

2. Based on the value of K, more products will be in the equilibrium mixture since the value of K is a positive large number.

Answer:

427°C .

Explanation:

Step 1:

Data obtained from the question. This include the following:

Initial temperature (T1) = 77°C

Initial pressure (P1) = P

Final pressure (P2) = 2P

Final temperature (T2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature.

This is illustrated below:

T(K) = T (°C) + 273

Initial temperature (T1) = 77°C

Initial temperature (T1) = 77°C+ 273 = 350K

Step 3:

Determination of the new temperature. The new temperature can be obtained as follow:

P1/T1 = P2/T2

P/350 = 2P/T2

Cross multiply

P x T2 = 350 x 2P

Divide both side by P

T2 = (350 x 2P ) / P

T2 = 700K

Step 4:

Conversion of Kelvin temperature to celsius temperature.

This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 700K

T(°C) = 700 – 273

T(°C) = 427°C

Therefore, the new temperature of the gas is 427°C

D. The name for the set of independent and dependent variables that will be controlled by the scientist.

The statement, that describes the ‘control’ in an experiment is "the name for the set of independent and dependent variables that will be controlled by the scientist."

A control is an element in an experiment that remains intact or unaffected by other variables. An experiment or observation aiming to minimise the influence of variables other than the independent variable is referred to as a scientific control. It serves as a standard or point of reference against which other test findings are measured.

In a scientific experiment, an independent variable is the variable that is modified or manipulated in order to assess the effects on the dependent variable. In a scientific experiment, the dependent variable is the variable that is being tested and measured. The designation given to the set of independent and dependent variables that the scientist will regulate.

Hence the correct option is D.

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Answer:

H₂(g) + Cu²⁺(aq) → 2H⁺(aq) + Cu(s)

Explanation:

In a redox reaction, one half-reaction is the oxidation (where the atom loss electrons) whereas the other reaction is the reduction (Where the atom is gaining electrons.

In the reactions:

H₂(g) → 2H⁺(aq) + 2e⁻ oxidation

Here, the reaction is written as the oxidation because the hydrogen H₂ is in oxidation state 0 and H⁺ in +1. That means each atom is loosing one electron.

Cu²⁺(aq) + 2e⁻ → Cu(s) reduction

And here, the Cu²⁺ is in +2 oxidation state and after the reaction is in Cu(s) 0 state. Thus, each atom is gaining 2 electrons.

The sum of both reactions is:

H₂(g) + Cu²⁺(aq) + 2e⁻ → 2H⁺(aq) + 2e⁻ + Cu(s)

Subtracting the electrons in both sides of the reaction:

Answer:

[tex]\Delta _cH=-1328.3kJ/mol[/tex]

Explanation:

Helllo,

In this case, for the given chemical reaction in gaseous state:

[tex]CH_3OCH_3+3O_2\rightarrow 2CO_2+3H_2O[/tex]

We comoute the combustion enthalpy as the reaction enthalpy for one mole of fuel (dimethyl ether) considering the formation enthalpy of each given substance and whether they are reactants (subtracting) or products (adding), therefore we write:

[tex]\Delta _cH=2*\Delta _fH_{CO_2}+3*\Delta _fH_{H_2O}-\Delta _fH_{CH_3OCH_3}-3*\Delta _fH_{O_2}[/tex]

Whereas the formation enthalpies for carbon dioxide, water, dimethyl ether and oxygen are -393.5, -241.8, -184.1 and 0 kJ/mol respectively, thereby, the combustion enthalpy turns out:

[tex]\Delta _cH=2(-393.5)+3*(-241.8)-(-184.1)-3(0)\\\\\Delta _cH=-1328.3kJ/mol[/tex]

Notice that enthalpy of formation of oxygen is zero since forming an element has no chemical sense, it just exists as it has been early demonstrated.

Regards.

Answer:

a. boiling

Explanation:

Answer:

here, as we have known the elements of group 3A(13) such as aluminium , boron has three valance electron and in perodic table the elements are kept with similar proterties in same place so, their valance electron is 3.

hope it helps...

The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.

The periodic table is organized into groups (vertical columns), periods (horizontal rows), and families (groups of elements that are similar). Elements in the same group have the same number of valence electrons.

Groups are the columns of the periodic table, and periods are the rows. There are 18 groups, and there are 7 periods plus the lanthanides and actinides.

There are two different numbering systems that are commonly used to designate groups, and you should be familiar with both.

The traditional system used in the United States involves the use of the letters A and B. The first two groups are 1A and 2A, while the last six groups are 3A through 8A. The middle groups use B in their titles.

Therefore, The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.

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Answer:

Edge length of the unit cell is 4.07x10⁻¹⁰m

Explanation:

In a face-centered cubic structure, the edge, a, could be obtained using pythagoras theorem knowing the hypotenuse of the unit cell, b, is equal to 4r:

a² + a² = b² = (4r)²

2a² = 16r²

a = √8 r

That means edge lenght is = √8 r

adius

As radius of Silver is 144pm = 144x10⁻¹²m:

a = √8 r

a = √8 ₓ 144x10⁻¹²m

a = 4.07x10⁻¹⁰m

Answer:

1) R₃CO , H, H₃C, a carbon free radical

2) high temperature, ultraviolet irradiation

Explanation:

1) Homolysis leads to the formation of free radicals (species having a free electron). Thus, answer is :

R₃CO

H

H₃C

a carbon free radical

2) Homolysis require high temperature, ultraviolet irradiation.

Answer:

2. Due to the possession of resonance

Explanation:

In the benzene ring, the electrons that results in the bonds between the carbon atoms are delocalized. That is, they do not belong to a specific carbon atom. It is this unique feature that enables them to have a bond length between single and double bond lengths.

This feature is as a result of resonance.

The correct option is 2.

Answer:

0.109 mol/tablespoon

Explanation:

6.37 g/ 58.5 mol = 0.10888888 mol (0.109 significantly)

Answer:

A: 0.109

Explanation:

Edge 2020

Answer:

The answer is

Explanation:

Density = mass / volume

mass = density × volume

volume = 2cm³

density = 520g/cm³

mass = 2 × 520

= 1040g

Hope this helps you

Answer:

i. n = 5

ii. ΔE = 7.61 × [tex]10^{-46}[/tex] KJ/mole

Explanation:

1. ΔE = (1/λ) = -2.178 × [tex]10^{-18}[/tex]([tex]\frac{1}{n^{2}_{final} }[/tex] - [tex]\frac{1}{n^{2}_{initial} }[/tex])

    (1/434 × [tex]10^{-9}[/tex]) = -2.178 × [tex]10^{-18}[/tex] ([tex]\frac{n^{2}_{initial} - n^{2}_{final} }{n^{2}_{final} n^{2}_{initial} }[/tex])

⇒ 434 × [tex]10^{-9}[/tex] = (1/-2.178 × [tex]10^{-18}[/tex])[tex]\frac{n^{2}_{final} *n^{2}_{initial} }{n^{2}_{initial} - n^{2}_{final} }[/tex]

But, [tex]n_{final}[/tex] = 2

434 × [tex]10^{-9}[/tex] = (1/2.178 × [tex]10^{-18}[/tex])[tex]\frac{2^{2} n^{2}_{initial} }{n^{2}_{initial} - 2^{2} }[/tex]

434 × [tex]10^{-9}[/tex]  × 2.178 × [tex]10^{-18}[/tex] = [tex](\frac{4n^{2}_{initial} }{n^{2}_{initial} - 4 })[/tex]

⇒ [tex]n_{initial}[/tex] = 5

Therefore, the initial energy level where transition occurred is from 5.

2. ΔE = hf

     = (hc) ÷ λ

    = (6.626 × 10−34 × 3.0 × [tex]10^{8}[/tex] ) ÷ (434 × [tex]10^{-9}[/tex])

    = (1.9878 × [tex]10^{-25}[/tex]) ÷ (434 × [tex]10^{-9}[/tex])

    = 4.58 × [tex]10^{-19}[/tex] J

    = 4.58 × [tex]10^{-22}[/tex] KJ

But 1 mole = 6.02×[tex]10^{23}[/tex], then;

energy in KJ/mole = (4.58 × [tex]10^{-22}[/tex] KJ) ÷ (6.02×[tex]10^{23}[/tex])

         = 7.61 × [tex]10^{-46}[/tex] KJ/mole

The initial energy level is 5  and the energy of this transition in units of kJ/mole is 7.57 * 10^-43 kJ/mole

We must first calculate ΔE as follows;

ΔE = hc/λ

h = Plank's constant = 6.6 * 10^-34 Js

c = speed of light = 3 * 10^8 m/s

λ = wavelength = 434 * 10^-9

ΔE =  6.6 * 10^-34 * 3 * 10^8/434 * 10^-9

ΔE = 0.0456 * 10^-17 J

ΔE = [tex]ΔE = -2.178 x10^-18 (\frac{1}{n^2final} - \frac{1}{n^2initial}) \\ΔE = -2.178 x10^-18 (\frac{1}{2^2} - \frac{1}{n^2initial} )\\\\4.56 * 10^-19/2.178 x10^-18 = (\frac{1}{2^2} - \frac{1}{n^2initial})\\0.210 = (\frac{1}{2^2} - \frac{1}{n^2initial})\\\frac{1}{n^2initial} = 0.25 - 0.210\\\frac{1}{n^2final} = 0.04\\n = (\sqrt{(0.04)^-1} \\n = 5[/tex]

Energy of this transition in units of kJ/mole = 4.56 * 10^-19/ 6.02 * 10^23

= 7.57 * 10^-43 kJ/mole

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Answer:

Explanation: M(PCL5)= 31 + 5(35.5)

=208.5g/mol

M(H20)= 18g/mol

n(PCL5) = 75.5÷208.5

= 0.362mol

n(HCl)/n(PCL5)= 5/1

n(HCl)= 5×0.362

=1.81mol of HCl