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Different boiling point.

Because in distillation the substance with lower boiling point is evaporated, leaving the other material(s).

Distillation relies on different boiling points to separate out parts of a solution.

Answer:

7.640 kg

Explanation:

Step 1: Write the balanced complete combustion equation for ethanol

C₂H₆O + 3 O₂ ⇒ 2 CO₂ + 3 H₂O

Step 2: Calculate the moles corresponding to 4 kg (4000 g) of C₂H₆O

The molar mass of C₂H₆O is 46.07 g/mol.

4000 g × 1 mol/46.07 g = 86.82 mol

Step 3: Calculate the moles of CO₂ released

86.82 mol C₂H₆O × 2 mol CO₂/1 mol C₂H₆O = 173.6 mol CO₂

Step 4: Calculate the mass corresponding to 173.6 moles of CO₂

The molar mass of CO₂ is 44.01 g/mol.

173.6 mol × 44.01 g/mol = 7640 g = 7.640 kg

Answer:

a) Covalent bonds

b) Covalent and ionic bonds

c) Covalent bonds

d) Covalent bonds

Explanation:

Metals and non-metals form ionic bonds (electrons are transferred), whereas nonmetals and nonmetals form covalent bonds.

Identify whether the compound contains ionic bonds, covalent bonds, or both.

a) CBr₄. C and Br are nonmetals. Thus, they form covalent bonds.

b) copper(II) sulfate. Sulfate contains S and O (nonmetals), which are bonded through covalent bonds. Sulfate is bonded to copper (metal) through an ionic bond.

c) N₂O₃. N and O are nonmetals. Thus, they form covalent bonds.

d) phosphorous trichloride. P and Cl are nonmetals. Thus, they form covalent bonds.

Given is the ratio of conjugate base and conjugate acid of phosphoric acid. pH of a substance is the concentration of the hydrogen ions in its solution and higher this concentration lower is the value of pH.

pKa value is a measure of the strength of acid, it is the negative log of acid dissociation constant Ka.

Answer:

To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN

Explanation:

Dilution is the reduction of the concentration of a chemical in a solution and consists simply of adding more solvent.

In a dilution the amount of solute does not vary. But as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

In a solution it is fulfilled:

Ci* Vi = Cf* Vf

where:

In this case:

Replacing:

1.40 M* 20 mL= 0.088 M* Vf

Solving:

[tex]Vf=\frac{1.40 M* 20 mL}{0.088 M}[/tex]

Vf= 318.18 mL

To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN

Answer: The balanced half-equations for silver + oxygen= silver oxide are:

Oxidation-half reaction: [tex]Ag \rightarrow 2Ag^{+} + 2e^{-}[/tex]

Reduction-half reaction: [tex]O_{2} + 2e^{-} \rightarrow 2O^{-}[/tex]

Explanation:

The word equation is as follows.

silver + oxygen = silver oxide

In terms of chemical formulas this equation can be written as follows.

[tex]Ag + O_{2} \rightarrow Ag_{2}O[/tex]

The removal on electron(s) from an atom, ion or molecule in a chemical reaction is called oxidation.

The gain of electron(s) by an atom, ion or molecule in a chemical reaction is called reduction.

Hence, half-reaction equations for the given reaction is as follows.

Oxidation-half reaction: [tex]Ag \rightarrow 2Ag^{+} + 2e^{-}[/tex]

Reduction-half reaction: [tex]O_{2} + 2e^{-} \rightarrow 2O^{-}[/tex]

As the number of atoms participating in the reaction are equal. Hence, the half-equations are balanced.

Thus, we can conclude that the balanced half-equations for silver + oxygen = silver oxide are:

Oxidation-half reaction: [tex]Ag \rightarrow 2Ag^{+} + 2e^{-}[/tex]

Reduction-half reaction: [tex]O_{2} + 2e^{-} \rightarrow 2O^{-}[/tex]

Answer:

Compound X has a molar mass of 316.25 g*mol^-1 and the following composition:

element & mass %

phosphorus & 39.18%

sulfur & 60.82%

Write the molecular formula of X.

Explanation:

The given molecule of phosphorus and sulfur has molar mass --- 316.25 g.

Empirical formula calculation:                      

element:              phosphorus                       sulfur

co9mposition:      39.185%                            60.82%

divide with

atomic mass:          39.185/31.0 g/mol           60.82/32.0g/mol

                              =1.26mol                           1.90mol

smallest mole ratio:   1.26mol/1.26mol =1      1.90mol/1.26 mol =1.50

multiply with 2:          2                                         3

Hence, the empirical formula is:

P2S3.

Mass of empirical formula is:

158.0g/mol

Given, molecule has molar mass --- 316.25 g/mol

Hence, the ratio is:

316.25g/mol/158.0 =2

Hence, the molecular formula of the compound is :

2 x (P2S3)

=[tex]P_4S_6[/tex]

Answer:

ΔH = -2446J

Explanation:

Based on the reaction:

2 Na(s) + Cl2(g) → 2NaCl

We can find the enthalpy of this reaction using Hess's law:

The enthalpy of a reaction is equal to the sum of the enthalpy of products times their reaction quotient subtracting the enthalpy of reactants times their reaction quotient. For the reaction of the problem:

ΔH = 2ΔH(NaCl) - [2ΔH(Na) + ΔHCl2)]

ΔH(NaCl) = 670J

ΔH(Na) = 235cal * (4.184J/1cal) = 983J

ΔHCl2 = 435cal * (4.184J/1cal) = 1820J

ΔH = 2*670J - [2*983J + 1820J]

ΔH = 1340J - [3786J]

Answer:

the heat content of a system at constant pressure

Explanation:

Answer:

Draw the skeletal structure for: (E)-hept-5-en-2-one

Explanation:

The root word hept indicates that the given compound has seven carons in its longest chain.

-en- primary suffix indicates that the compound has one double bond in it.

2-one indicates that the compound has -C=O bond in the second carbon.

The prefix (E) indicates that the highest priority groups are on the opposite direction of the double bond.

The structure of the given molecule is:

Answer:

2.29 M

Explanation:

Equation of the reaction;

Ca(OH)2(aq) + 2HClO4(aq) → 2H2O(l) + Ca(ClO4)2(aq)

Concentration of acid CA = 1.49 M

Concentration of base CB= ????

Volume of acid VA= 43.0 ml

Volume of base VB= 14.0 ml

Number of moles of acid NA = 2 moles

Number of moles of base NB = 1 mole

CAVA/CBVB = NA/NB

CAVANB =CBVBNA

CB= CAVANB/VBNA

CB= 1.49 × 43.0 × 1/14.0 × 2

CB= 2.29 M

Answer:

Butanoic acid

Explanation:

The IUPAC name of CH3CH2CH2COOH is:

The IUPAC name for a given compound is Butanoic acid.

Answer:

The ratio of carbon and hydrogen atoms = 3.5 : 8

= 7 : 16

Then,the empirical formula is C7H16

Answer:

-46.67 J.

Explanation:

We are given;

Initial Pressure = 15atm = 15 × 10^(3) J

Final pressure = 1atm = 1 × 10^(3) J

Temperature = 300k

The pressures were converted to Joules.

Formula for the entropy change is;

∆S_system = ∆S_surrounding = -(dQ)/T

-(dQ)/T = (-(15 × 10^(3)) - (1 × 10^(3))/300)

= -46.67 J.

Answer:

NH4Cl > Li2SO4 > CoCl3

Explanation:

Let us recall that the freezing point depression depends on the molality of the solution and the number of particles present.

Let us also recall that freezing point depression is a colligative property. It depends on the number of particles present in solution.

Usually, the more the number of particles present, the lower the freezing point. Hence, NH4Cl which has only two particles will have the highest freezing point while CoCl3 which has four particles will have the lowest freezing point.

Answer:

19000 torr

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 0.150 L

Initial pressure (P₁) = 1520 torr

Final volume (V₂) = 0.012 L

Temperature = constant

Final pressure (P₂) =?

The final pressure of the gas can be obtained by using the Boyle's law equation as illustrated below:

P₁V₁ = P₂V₂

1520 × 0.150 = P₂ × 0.012

228 = P₂ × 0.012

Divide both side by 0.012

P₂ = 228 / 0.012

P₂ = 19000 torr

Thus, the final pressure of the gas is 19000 torr

Answer:

[Na⁺] = 1.99 M

Explanation:

Na₂S is a ionic salt that can be dissociated.

Dissociation equation is:

Na₂S  →  2Na⁺  +  S⁻²

1 mol of sodium sulfide can give 2 moles of sodium cation.

We convert moles of salt: 78 g . 1mol / 78.06 g = 0.999 moles

As ratio is 1:2, after dissociation we have (0.999 . 2) = 1.998 moles of Na⁺

Molarity is a type of concentration.

It indicates moles of solute in 1 L of solution and in this case, we have 1 L as final voulme.

Moles of Na⁺ are 1.998 moles. Then molarity (mol/L) is:

M =1.99 mol/L  

Answer:

CH4< CH4I< NaCl

Explanation

NaCl has the boiling point of 1,413°C ( 2,575°F )

CH3I has a boiling point of 42°C ( 107°F )

CH4 has the boiling point of -161.6°C ( -258.9°F )

Answer:

a. eletrophile

b. able to impose regioselectivity and stereo selectivity.

c. faster

Explanation:

Necleophile reaction is chemical reaction in which electron rich chemical specie replaces functional group with another electron deficient molecule. Lewis acid catalyst is organic chemical reaction which lewis acid act as electron pair acceptor. Nucleophile reaction proceeds about 25 times more faster than benzene.

Answer:

1.20x10⁻²⁰μL

Explanation:

1cm³ is equal to 1milliliter. As we must know, 1milliliter = 1000 microliters, 1000μL. To convert the 1.20x10⁻²³mL we need to use the conversion factor: 1mL = 1000μL.

The volume of tantalum in μL is:

1.20x10⁻²³mL * (1000μL /1L) = 1.20x10⁻²⁰μL

Answer:

6+9=15

mass number =15

mass number = no. protons + no. of neutrons

Answer:

S + 2CO = SO2 + 2C

First, look for the amount of substance of sulfur:

n(S) = m / M

n(S) = 14.8 g/32 g / mol = 0.4625 mol

n(CO) = m (CO) / M (CO)

M(CO) = 12 + 16 = 28 g/mol

n(CO) = 19.9 g/28 g/mol = 0.71 mol

S in excess, so for calculating we take CO:

n(SO2) = n(CO)/2 = 0.71 mol/2 = 0.355 mol

m(SO2) = M(SO2)*n(SO2)

M(SO2) = 32 + 16*2 = 64 g/mol

m(SO2) = 64 g/mol * 0.355 mol = 22.74 g

Answer:

63.05% of MgCO3.3H2O by mass

Explanation:

of MgCO3.3H2O in the mixture?

The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:

Mass water:

3.883g - 2.927g = 0.956g water

Moles water -18.01g/mol-

0.956g water * (1mol/18.01g) = 0.05308 moles H2O.

Moles MgCO3.3H2O:

0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =

0.01769 moles MgCO3.3H2O

Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-

0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O

Mass percent:

2.448g MgCO3.3H2O / 3.883g Mixture * 100 =

Answer:

Ba(OH)₂ + 2 HBr ⇒ BaBr₂ + 2 H₂O

Explanation:

We have the products of a reaction and we have to predict the reactants. Since the products are binary salt and water, this must be a neutralization reaction. In neutralizations, acids react with bases. The acid that gives place to Br⁻ is HBr, while the base the gives place to Ba²⁺ is Ba(OH)₂. The balanced chemical equation is:

Ba(OH)₂ + 2 HBr ⇒ BaBr₂ + 2 H₂O

Answer:

Substrate

Explanation:

In biochemical sciences, a substrate is a substance that is acted upon by an enzyme to yield a product. Enzymes are known for catalyzing biochemical reactions. The substances that are usually worked with during this catalytic process are termed as SUBSTRATES.

Substrates, which are usually changed during the process, binds to the active site on the enzyme and form an enzyme-substrate complex.

According to this question, pyruvate is converted to lactate in a reaction that is catalyzed by the enzyme lactate dehydrogenase. This means that the compound on which the enzyme works, pyruvate, is called the SUBSTRATE.

Answer:

1.02 Hz

Explanation:

frequency= (1/t) = (1/0.98) = 1.02 hz

Answer:

Electrons

Explanation:

Electrons are produced when light with enough energy is directed onto the cathode where electrons are ejected as beam towards the screen.

Answer:

0.465

Explanation:

To find the volume of a substance, divide the mass by the density.

M/D = V

10.0 / 21.5 = 0.4651163

Then round to 3 significant figures: and the density is 0.465

Answer:

A sample of salt has 1.74 moles of sodium chloride. how many formula units of the ionic compound are in the sample?

Explanation:

Given, 1.74 moles of NaCl.

Since one mole of NaCl consists of --- [tex]6.023 * 10^2^3[/tex] formula units.

Then, 1.74mol of NaCl contains how many formula units of NaCl?

[tex]1.74 mol x \frac{6.023x10^2^3}{1 mol} \\=10.5x10^2^3[/tex]formula units.

Hence, the given sample has 10.5x10^23 formula units.