Suppose you are using distillation to separate cyclohexane and toluene. The boiling point of cyclohexane is Choose... oC and the boiling point of toluene is Choose... oC. Therefore, the liquid collected first should be Choose... .

Answers

Answer 1

Answer:

81°C

111°C

cyclohexane

Explanation:

Distillation is a process of separating two liquids based on differences in Bolling point. For two substances having different boiling points, they are collected as they are converted into vapour, condensed and move down the condenser one after the other.

Since the boiling point of cyclohexane is less than that of toluene, cyclohexane is collected first before toluene.


Related Questions

When 1 mole of CO(g) reacts with H2O(l) to form CO2(g) and H2(g) according to the following equation, 2.80 kJ of energy are absorbed. CO(g) + H2O(l)CO2(g) + H2(g) Is this reaction endothermic or exothermic? _________ What is the value of q? kJ

Answers

Answer: Endothermic, 2.80 kJ

Explanation

Since this reaction absorbs heat, it is endothermic.

The energy absorbed per mole CO is 2.80 kJ and this reaction is already balanced. q= 2.80 kJ

Hope this helps:)

Complete the balanced dissociation equation for the compound below in aqueous solution. If the compound does not dissociate, write NR after the reaction arrow.
HI (aq) -->

Answers

Answer:

[tex]{ \bf{HI _{(aq)} \: → \: H {}^{ + } _{(aq)} \: + \: \: I {}^{ - } _{(aq)} }}[/tex]

Propane gas reacts with oxygen according to this balanced equation: C subscript 3 H subscript 8 space (g )space plus space 5 space O subscript 2 space (g )space rightwards arrow 3 space C O subscript 2 space (g )space plus space 4 space H subscript 2 O space (g )How many liters of carbon dioxide are produced at STP when 44 g of C3H8 completely reacts with oxygen

Answers

Explanation:

The balanced chemical equation of the reaction is:

[tex]C_3H_8(g)+ 5O_2 (g)->3CO_2(g)+4H_2O(g)[/tex]

From the balanced chemical equation,

1 mole of propane forms ------ 3 mol. of [tex]CO_2[/tex] gas.

The molar mass of propane is 44.1 g/mol.

One mole of any gas at STP occupies --- 22.4 L.

Hence, 44 g of propane forms (3x22.4 L=) 67.2 L of CO2 gas at STP.

Answer:

Thus, 67.2 L of CO2 is formed at STP.

A metal (C = 0.2158 cal/g· °C) is removed from a hot (350. °F) oven in which it had achieved thermal equilibrium. The metal is placed into 200. mL acetic acid. The temperature of the acid increases to 90.3 °C from 24.3 °C. What is the mass of the metal? (dacetic acid = 1.04 g/cm3; Cs, acetic acid = 2.055 J/g·°C) Group of answer choices 120. g 362 g 1452 g 347 g 281 g

Answers

Answer:

362g

Explanation:

heat lost by metal= heat gained by acetic acid

tfs are the same so you cando delta T

convert Cal/gc to J/gc

thectgod ig follow

What is the maximum mass of PH3 that can be formed when 62.0g of phosphorus reacts with
4.00g of hydrogen?

P4(g)+ 6H2(g) → 4PH3(g)

Answers

Answer: The mass of [tex]PH_3[/tex] produced is 45.22 g

Explanation:

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

For [tex]P_4[/tex]:

Given mass of [tex]P_4[/tex] = 62.0 g

Molar mass of [tex]P_4[/tex] = 124 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of }P_4=\frac{62.0g}{124g/mol}=0.516mol[/tex]

For [tex]H_2[/tex]:

Given mass of [tex]H_2[/tex] = 4.00 g

Molar mass of [tex]H_2[/tex] = 2 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of }H_2=\frac{4.0g}{2g/mol}=2mol[/tex]

The chemical equation follows:

[tex]P_4(g)+6H_2(g)\rightarrow 4PH_3(g)[/tex]

By stoichiometry of the reaction:

If 6 moles of hydrogen gas reacts with 1 mole of [tex]P_4[/tex]

So, 2 moles of hydrogen gas will react with = [tex]\frac{1}{6}\times 2=0.333mol[/tex] of [tex]P_4[/tex]

As the given amount of [tex]P_4[/tex] is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, hydrogen gas is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 6 moles of [tex]H_2[/tex] produces 4 mole of [tex]PH_3[/tex]

So, 2 moles of [tex]H_2[/tex] will produce = [tex]\frac{4}{6}\times 2=1.33mol[/tex] of [tex]PH_3[/tex]

We know, molar mass of [tex]PH_3[/tex] = 34 g/mol

Putting values in equation 1, we get:

[tex]\text{Mass of }PH_3=(1.33mol\times 34g/mol)=45.22g[/tex]

Hence, the mass of [tex]PH_3[/tex] produced is 45.22 g

what is valency of an atom?​

Answers

The number of replaceable electrons in an atom is called its valency.

Examples

Monovalent - HydrogenDivalent - Oxygen

Valency = 8 - Number of electron in last shell [When number of electrons in last shell > 4]

Valency = Number of electron in last shell [When number of electrons in last shell < 4]

Thanks !

☺️☺️☺️☺️☺️☺️☺️

Answer:

the combining capacity if an atom is know as valency.

the property of an element that determines the number of other atimd with an aton if the element can combine.

How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?

Answers

Answer:

2.0202 grams

Explanation:

1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.

so ?g glucose = 144.3 mL soln

Now apply the conversion factor, and you have:

?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).

so you have (144.3x1.4/100) g glucose= 2.0202 grams

Which is an example of using an open-ended question to uncover a problem? O a) "Do you have a problem you'd like addressed today?" b) "Is there a problem? C) "What seems to be the problem?" O d) "Can I help you?"

sublimation is a change from the solid phase to the____phase

Answers

Answer:

solid to gaseous or gaseous to solid

Explanation:

Sublimation is the transition of a substance directly from the solid to the gas state, without passing through the liquid state. sublimation is most often used to describe the process of snow and ice changing into water vapor in the air without first melting into water.

Can someone teach me step by step how finding the oxidation number in this problem:

Fe in Fe(CIO2)3​

Answers

Answer:

+3

Explanation:

u see sum of oxidation number in all situations have to be 0

ClO2 =-1

so Fe is +3

pls help name any of these compounds​

Answers

Answer:

D. Propanol

Explanation:

C3H7OH the presence of alcohol functional group makes it propanol

Which subshells are found in each of the following shells
electron subshell - M shell

Answers

Answer:

3

Explanation:

The electron shells are labelled as K,L,M,N,O,P, and Q or 1,2,3,4,5,6, and 7.

As we go from innermost shell outwards, this number denotes the number of subshell in the shell. Electrons in outer shells have higher average energy and travel farther from the nucleus than those in inner shells.

Hence, M shell contains s,p and d subshells.

Forcus on the yellow highlighted texts, your help is appreciated.
[tex]{ \sf{ \red{no \: pranks}}}[/tex]

Answers

Answer:

Transition temperature is the temperature at which a substance changes from one state to another.

Allotropy is the existence of an element in many forms.

Identify what reagents you would use to achieve each transformation: Conversion of 2-methyl-2-butene into a secondary alkyl halide. Br2, ROOR Br2, H2O HBr, ROOR HBr Conversion of 2-methyl-2-butene into a tertiary alkyl halide. Br2, H2O HBr Br2, ROOR HBr, ROOR

Answers

Answer:

Conversion of 2-methyl-2-butene into a secondary alkyl halide - ROOR, HBr

Conversion of 2-methyl-2-butene into a tertiary alkyl halide - HBr

Explanation:

The addition of HBr to 2-methyl-2-butene occurs in accordance to Markovnikov rule in the absence of peroxide.

According to Markovnikov rule; ''the negative part of the addendum is attached to the carbon atom bearing the least number of hydrogen atoms.'' Following the Markovnikov rule, the tertiary alkyl halide is obtained.

In the presence of peroxide, this rule is not followed and the reaction proceeds in an anti-Markovnikov way to yield a secondary alkyl halide.

Help me in this question!!!​

Answers

Answer:

d. End product is that product with a ketone and carboxylic acid.

Explanation:

[tex]{ \sf{NaBH_{4} : }}[/tex]

Sodium borohydride is a reducing agent, it reduces the ketone to a primary alcohol.

[tex]{ \sf{H _{2} O \: and \: H {}^{ + } }}[/tex]

Then acidified water is an oxidising mixture which reverses the reduction reaction.

Explanation:

Option D is your answer

Hope it helps

What minimum mass of HCl in grams would you need to dissolve a 2.2 g iron bar on a
padlock?

Answers

I think your answer should be 2.8 g

2.8 g is the minimum mass of HCl in grams that would you need to dissolve a 2.2 g iron bar on a padlock.

What is dissolution?

When a solute is dissolved in a solvent, a solution is created. Dissolution is the process through which solutes, or dissolved parts, combine to form a solution inside a solvent. In this procedure, the gas, liquid, or solid dissolves inside the original solvent and forms a solution.

In some polymer applications, dissolution is also an issue since it results in swelling, a loss of strength and stiffness, and a change in volume. Whether a chemical process is man-made or natural, dissolution is crucial. Catalysts are tested using dissolution. 2.8 g is the minimum mass of HCl in grams that would you need to dissolve a 2.2 g iron bar on a padlock.

Therefore, 2.8 g is the minimum mass of HCl in grams that would you need to dissolve a 2.2 g iron bar on a padlock.

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What are the effects of global warming?​

Answers

the effects are: temperature rises, water shortages, and increased fire threats

Calculate the mass of water produced when 7.49 g of butane reacts with excess oxygen.

Answers

Answer:

[tex]m_{H_2O}=12.9gH_2O[/tex]

Explanation:

Hey there!

In this case, according to the given information, it turns out possible for us to solve this problem by firstly writing out the reaction whereby butane is combusted in the presence of excess oxygen:

[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]

Thus, we can evidence a 2:10 mole ratio of butane to water, and thus, the stoichiometric setup to calculate the mass of produced water is:

[tex]m_{H_2O}=7.49gC_4H_{10}*\frac{1molC_4H_{10}}{52.12gC_4H_{10}} *\frac{10molH_2O}{2molC_4H_{10}}*\frac{18.02gH_2O}{1molH_2O}\\\\m_{H_2O}=12.9gH_2O[/tex]

Regards!

What is alkaline and what is acidic pH

Answers

Answer:

An alkaline is a substance that dissolves in water to produce hydroxyl ions (OH-)

Explanation:

The pH range of an alkaline is from 8–14.

Acidic pH ranges from 0–6.9.

The molecular ion is not visible in the mass spectrum of 2-chloro-2- methylpropane. At what m/z value would the molecular ion be if it were visible? What evidence is there in the mass spectrum that suggests that the peak at m/z= 77 contains a chlorine atom?

Answers

Answer:  hello the complete question is attached below

Visibility of molecular ion = m/z value of 77

Explanation:

For The molecular ion to be visible, it has to be at an m/z value of 77 and this is because molecular ions will have an m/z ratio =  molecular mass of given molecule in most cases but not always in all cases.

And the visibility is possible after the removal of CH₃ ion.

ii) Evidence in the mass spectrum that suggests peak at m/z = 77

attached below

You have been contracted to determine how different salts affect the pH of water. Which of the solids in the following set should you test to investigate for the effects of cations on pH?

a. AlBr3
b. Rb2SO3
c. MgCl2
d. RbBrO
e. CH3NH3Br

Answers

Answer:

Hence the solids that should test to investigate the effects of cations on pH is

[tex]AlBr_{3}[/tex] (Cation is Al 3+)  

[tex]MgCl_{2}[/tex]  ( Cation is Mg 2+)  

[tex]CH_{3} NH_{3} Br[/tex] ( Cation is NH2+).

Explanation:

The solids in the following should you test to investigate the effects of cations on pH.  

[tex]AlBr_{3}[/tex] contains (Cation is Al 3+)  

[tex]MgCl_{2}[/tex] contains ( Cation is Mg 2+)  

[tex]CH_{3} NH_{3} Br[/tex] contains( Cation is NH2+ )

The atoms or the molecules containing the positive charge that gets attracted to the cathode are called cations. The compounds a. [tex]\rm AlBr_{3}[/tex], c. [tex]\rm MgCl_{2}[/tex] and e. [tex]\rm CH_{3}NH_{3}Br[/tex] should be investigated.

What are cations and pH?

Cations are the positive charge containing molecules and atoms that have more protons in their nucleus than the number of electrons in their shells. They are formed when they lose one or more electrons to another atom.

The addition or release of the electrons of the cations and anions affects the pH system as absorption of the cation decreases the pH and absorption of the anions increases the pH.

Hence, [tex]\rm Al^{3+}[/tex], [tex]\rm Mg^{2+}[/tex] and [tex]\rm NH^{2+}[/tex] are the cation that should be investigated. The addition of the cations will reduce the pH of the reaction.

Therefore, absorption of the cation reduces the pH.

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If 7 mol of copper reacts with 4 mol of oxygen, what amount of copper (II) oxide is produced? What amount of the excess reactant remains?

Answers

Answer:

7 mol CuO

0.5 mol O₂

Explanation:

Step 1: Write the balanced equation

2 Cu + O₂ ⇒ 2 CuO

Step 2: Identify the limiting reactant

The theoretical molar ratio (TMR) of Cu to O₂ is 2:1.

The experimental molar ratio (EMR) of Cu to O₂ is 7:4 = 1.75:1.

Since TMR > EMR, Cu is the limiting reactant

Step 3: Calculate the amount of CuO produced

7 mol Cu × 2 mol CuO/2 mol Cu = 7 mol CuO

Step 4: Calculate the excess of O₂ that remains

The amount of O₂ that reacts is:

7 mol Cu × 1 mol O₂/2 mol Cu = 3.5 mol O₂

The excess of O₂ that remains is:

4 mol - 3.5 mol = 0.5 mol

. Which of the following statement is not related to a chemical reaction ? A. New substances are formed B. Atoms of the elements transform into atoms of other elements C. The properties of the new substances will be different D. There will be bond breaking and bond forming​

Answers

Answer:

the answer should be B because elements do not tranform into other elements in a chemical reaction

am I right please?

potassium and chlorine react to form potassium chloride. a.it is a redox reaction,explain why. b.see if u can write a balanced equation for it.​

Answers

Answer:

K+ClKCl

Explanation:

because the reaction is between metal Potassium and Non-metal Chlorine

Answer:

Explanation:

a) It is a redox reaction because KCl is an ionic compounds with K having a + charge and Cl having a - charge. Originally, both have an oxidation state of 0 and not K has 1+ and Cl has 1-. Therefore, one species was oxidized and one was reduced which is indicative of a redox reactions.

b)

2K + Cl2 => 2KCl

Given the following balanced equation:
3Cu(s) + 8HNO3(aq) = 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)
Determine the mass of copper (II) nitrate that would be formed from the complete reaction
of 35.5g of copper with an excess of nitric acid.

Answers

Answer: The mass of copper (II) nitrate produced is 105.04 g.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of copper = 35.5 g

Molar mass of copper = 63.5 g/mol

Plugging values in equation 1:

[tex]\text{Moles of copper}=\frac{35.5g}{63.5g/mol}=0.560 mol[/tex]

The given chemical equation follows:

[tex]3Cu(s)+8HNO_3(aq)\rightarrow 3Cu(NO_3)_2(aq)+2NO(g)+4H_2O(l)[/tex]

By the stoichiometry of the reaction:

If 3 moles of copper produces 3 moles of copper (II) nitrate

So, 0.560 moles of copper will produce = [tex]\frac{3}{3}\times 0.560=0.560mol[/tex] of copper (II) nitrate

Molar mass of copper (II) nitrate = 187.56 g/mol

Plugging values in equation 1:

[tex]\text{Mass of copper (II) nitrate}=(0.560mol\times 187.56g/mol)=105.04g[/tex]

Hence, the mass of copper (II) nitrate produced is 105.04 g.

Buffer solutions that maintain certain levels of pH or acidity are widely used in biochemical experiments. One common buffer system uses sodium dihydrogenphosphate and sodium monohydrogenphosphate. What are the formulas of these two compounds

Answers

Answer:

Sodium dihydrogenphosphate = NaH₂PO₄

Sodium monohydrogenphosphate = Na₂HPO₄

Explanation:

A buffer solution is a solution is a solution that resists changes to its oH when a little quantity of strong acid or strong base is added to it.

They are solutions of weak acids or weak bases and their salts known as conjugate base or conjugate acids respectively for the weak acids and weak bases.

For example, a solution of the weak acid ethanoic acid and its salt or conjugate base, sodium ethanoate serves as a buffer solution.

In biochemical experiments, where the pH of the reaction medium is kept as constant and as close as possible to that of the internal environment, buffer solutions are widely used. One of the commonly used buffers is the phosphate buffer. The phosphate buffer consists of the acid salts sodium dihydrogenphosphate and sodium monohydrogenphosphate. Sodium dihydrogenphosphate serves as the weak acid while sodium monohydrogenphosphate serves as the conjugate base.

The formulas of these two compounds are given below:

Sodium dihydrogenphosphate = NaH₂PO₄

Sodium monohydrogenphosphate = Na₂HPO₄

What mass of NaNO3 must be dissolved to make 838mL of a 1.25 M solution

Answers

Answer:

89.04 g of NaNO₃.

Explanation:

We'll begin by converting 838 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

838 mL = 838 mL × 1 L / 1000 mL

838 mL = 0.838 L

Next, we shall determine the number of mole of NaNO₃ in the solution. This can be obtained as follow:

Volume = 0.838 L

Molarity = 1.25 M

Mole of NaNO₃ =?

Mole = Molarity × volume

Mole of NaNO₃ = 1.25 × 0.838

Mole of NaNO₃ = 1.0475 mole

Finally, we shall determine the mass of NaNO₃ needed to prepare the solution. This can be obtained as follow:

Mole of NaNO₃ = 1.0475 mole

Molar mass of NaNO₃ = 23 + 14 + (16×3)

= 23 + 14 + 48

= 85 g/mol

Mass of NaNO₃ =?

Mass = mole × molar mass

Mass of NaNO₃ = 1.0475 × 85

Mass of NaNO₃ = 89.04 g

Therefore, 89.04 g of NaNO₃ is needed to prepare the solution.

Too many objects inside a laboratory fume hood can disrupt the airflow and possibly compromise you safety. Which of the following are considered best practices in the use of a laboratory fume hood?

a. Open the sash as much as possible
b. Work at least 25 cm inside the hood
c. Use fast, quick movements to limit your exposure
d. Place objects to one side—work on other side
e. Use a raised along the back of the hood

Answers

Best practices for fume hoods: work 25 cm inside, organize items to one side, use raised ledge; avoid open sash and quick movements.

Laboratory fume hoods must be used safely. Workers should operate at least 25 cm within the hood to preserve ventilation and avoid dangerous chemicals. Place things on one side of the hood to preserve ventilation and prevent clogging.

A raised ledge on the rear of the hood prevents things from falling in and impeding airflow. Avoid fully opening the sash to maintain ventilation and containment. Fast, rapid motions can interrupt airflow, so prevent them. These practises guarantee the fume hood contains harmful compounds, making the lab safer. Therefore, option (B), (D) and (E) are correct.

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The density of a gas cannot be measured.
True
False

Answers

Answer:

False

Explanation:

Answer:

False

Explanation:

Elimination of the pharmaceutical IV antibiotic gentamicin follows first-order kinetics. If the half-life of gentamicin is 1.5 hours for an adolescent. What fraction of the original reactant concentration will remain after 8 hours if the original concentration was 8.4 x 10-5 M.

Answers

Explanation:

The given data is:

The half-life of gentamicin is 1.5 hrs.

The reaction follows first-order kinetics.

The initial concentration of the reactants is 8.4 x 10-5 M.

The concentration of reactant after 8 hrs can be calculated as shown below:

The formula of the half-life of the first-order reaction is:

[tex]k=\frac{0.693}{t_1_/_2}[/tex]

Where k = rate constant

t1/2=half-life

So, the rate constant k value is:

[tex]k=\frac{0.693}{1.5 hrs}[/tex]

The expression for the rate constant is :

[tex]k=\frac{2.303}{t} log \frac{initial concentration}{concentration after time "t"}[/tex]

Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:

[tex]\frac{0.693}{1.5 hrs} =\frac{2.303}{8 hrs} x log \frac{8.4x10^-^5}{y} \\ log \frac{8.4x10^-^5}{y} =1.604\\\frac{8.4x10^-^5}{y}=10^1^.^6^0^4\\\frac{8.4x10^-^5}{y}=40.18\\y=\frac{8.4x10^-^5}{40.18} \\=>y=2.09x10^-^6[/tex]

Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.

In an analysis of interhalogen reactivity, 0.350 mol ICl was placed in a 5.00 L flask and allowed to decompose at a high temperature.
2 ICl(g) I2(g) + Cl2(g)
Calculate the equilibrium concentrations of I2, Cl2, and ICl. (Kc = 0.110 at this temperature.)
I2 M
Cl2 M
ICl M

Answers

Answer:

[ICl] = 0.0420 M

[I₂]  = [Cl₂] = 0.0140 M

Explanation:

Step 1: Calculate the initial concentration of ICl

[ICl] = 0.350 mol / 5.00 L = 0.0700 M

Step 2: Make an ICE chart

        2 ICl(g) ⇄ I₂(g) + Cl₂(g)

I        0.0700     0         0

C        -2x          +x        +x

E    0.0700-2x      x          x

The concentration equilibrium constant (Kc) is:

Kc = 0.110 = [I₂] [Cl₂] / [ICl]² = x² / (0.0700-2x)² = (x/0.0700-2x)²

0.332 = x/0.0700-2x

x = 0.0140

The concentrations at equilbrium are:

[ICl] = 0.0700-2x = 0.0700-0.0280 = 0.0420 M

[I₂]  = [Cl₂] = x = 0.0140 M

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