A student weighed 0.550 g of lithium chloride, LiCl, to use in a reaction. How many moles is this? a. 5.11 b. 42.39 c. 23.31 d. 77.08 e. 0.0130

Answer:

B. Colloid

Explanation:

Colloid is a mixture that is made up of very small homogeneous particles, thus can not be separated into fractions. Its particles do not settle as other solution, but are evenly dispersed in the medium. Some examples are: liquid milk, jelly, rubber etc

Since colloid is a solution, thus it is made up of two parts; its particles and dispersing medium.

Answer:

HCl.

Explanation:

Of all the type of bond, Ionic bonds are the strongest. Among all the compounds in the options only HCl has ionic bond, rest have covalent bonds. So, the strongest intermolecular forces is in HCl.  Ionic bond are formed with the transfer of electrons, one atom is donor and the other is acceptor.

Answer:

Cu(NO3)2(aq)+Pb(s) ⇌ Pb(NO3)2(aq)+Cu(s)

Explanation:

If we look at the both reactions closely, we will quickly discover that the reaction CuSO4(aq)+Pb(s) ⇌ PbSO4(s)+Cu(s) involves PbSO4.

The compound PbSO4 is insoluble in water and sinks to the bottom of the reaction vessel. When this occurs, the concentration of Pb^2+ becomes low. This will bring about a low voltage in the cell.

On the other hand, Pb(NO3)2 is soluble in water hence the cell voltage in this case is higher than the former.

In the case when we look at both reactions so here we discovered that the reaction of

CuSO4(aq)+Pb(s) ⇌ PbSO4(s)+Cu(s) includes PbSO4.

The compound PbSO4 should be insoluble in water and sinks to the bottom of the reaction vessel. At the time When this occurs, the concentration of Pb^2+ becomes low. This will bring about a low voltage in the cell.

On the other hand, Pb(NO3)2 is soluble in water so the cell voltage in this case is more than the former.

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Answer:

Hey!I think I can help....

Simple Distillation cannot be used to separate a mixture of two or more miscible liquids Fractional Distillation is used to separate such a mixture into its component parts or fractions. The fractions distil over in ascending order of their boiling points,starting with the fraction with the lowest boiling points. For efficient fractional distillation,the difference in the boiling points between successive fractions must be more than 10°C.In the laboratory,the apparatus for fractional distillation is same as the apparatus for simple distillation, except that a fractioning column is introduced between the distillation flask and the condenser.The fractionating column is packed with glass beads,and does the actual separation. The upper part of the column is at a lower temperature than the lower part .Only the vapour with the same temperature as the upper part of the fractionating column passes on the condenser.(This is the fraction with the lowest boiling point)Vapours with higher boiling points condense as they enter the upper part of the fractionating column and the flow back into the distillation flask.This process is repeated until the fraction with the lowest boiling point is distilled over completely. Then,the temperature in the upper part of yhe column rises until the boiling point of the next fraction is reached. This fraction then starts distilling over.The fractional distillation process continues until Al the component fractions in the mixture distil over.Crude oil or petroleum contains many important natural products such as petrol,kerosene,diesel, heavey oil,natural gases and bitumen.Each of these products can be recovered in its pure form by fractional distillation. The fractional distillation plant is made up of two maim parts----boiler,where the crude oil is kept boiling, and the fractionating tower,where yhe actual separation takes place.The fractions with the lower boiling points will emerge near the top part of the tower while those with higher boiling points will come out near the lower part of the tower..... Thank you for the question, I hope it helps you...Thank you.

Answer:

B. a beaker used in an experiment

Explanation:

Answer:

Mass, m = 1.51 grams

Explanation:

It is given that,

The circumference of Aluminium cylinder, C = 13 mm = 1.3 cm

Length of the cylinder, h = 4.2 cm

We know that the density of the Aluminium is 2.7 g/cm³

Circumference, C = 2πr

[tex]r=\dfrac{C}{2\pi}\\\\r=\dfrac{1.3}{2\pi}\\\\r=0.206\ cm[/tex]

Density is equal to mass per unit volume.

[tex]d=\dfrac{m}{V}[/tex]

m is mass of the cylinder

V is the volume of the cylinder

[tex]V=\pi r^2h\\\\V=\dfrac{22}{7}\times0.206^2\times 4.2\\\\V=0.5601\ cm^3[/tex]

So,

[tex]m=d\times V\\\\m=2.7\times 0.5601\\\\m=1.51\ g[/tex]

So, the mass of the cylinder is 1.51 grams.

Answer:

37.1 mL

Explanation:

To accurately read the measurement using a cylinder, we read the measurement at the meniscus i.e measurement under the curve as shown in the diagram above.

The meniscus i.e the measurement under the curve as shown above is 37.1 mL.

Therefore, the volume of the liquid as obtained from the cylinder is 37.1 mL

Answer:

37.1 ml

Explanation:

a graduated cylinder's measurement is taken at the bottom of the meniscus

Answer:

[tex]v_{rms}=866.32m/s[/tex]

Explanation:

Hello,

In this case, since the rms speed of the molecules is computed by:

[tex]v_{rms}=\sqrt{\frac{3RT}{M} }[/tex]

Whereas the absolute temperature is computed from the internal energy (by using the Cp of helium (3.1156 J/g*K) as shown below:

[tex]U=nCvT\\\\T=\frac{U}{nCv}=\frac{15kJ*\frac{1000J}{1kJ} }{10mol*\frac{4.00g}{1mol} *3.1156\frac{J}{g*K} } \\\\T=120.36K[/tex]

Thereby, the rms speed results:

[tex]v_{rms}=\sqrt{\frac{3*8.314\frac{kg*m^2}{s^2*mol*K}*120.36K}{4.00\frac{g}{mol}*\frac{1kg}{1000} } } \\\\v_{rms}=866.32m/s[/tex]

Regards.

Answer:

Please find the attachment to this question:

Explanation:

By eliminating the agent produced in the reaction, it will give a structure of the alkene. please find the given attachment.

Answer:

The correct answer is 0.058

Step-by-step explanation:

The mole fraction of KI (X) in the solution is the number of moles of KI divided into the total number of moles of the solution (nt):

X= moles KI/nt

First we have to calculate the molecular weights of the components of the solution: KI and water (H₂O):

Molecular weight (Mw) of KI:

Mw(KI)= 39 g/mol + 127 g/mol = 166 g/mol

Molecular weight of H₂O:

Mw(H₂O)= (1 g/mol x 2) + 16 g/mol = 18 g/mol

Then, we calculate the number of moles there is of each component in the solution by dividing the mass into Mw:

Moles of KI= mass KI/Mw(KI) = 3.4 g/166 g/mol = 0.020 mol KI

Moles of H₂O= mass H₂O/Mw(H₂O)= 5.8 g/18 g/mol = 0.322 mol H₂O

Thus, the total moles of solution will be:

nt= moles KI + moles H₂O= 0.020 mol + 0.322 mol = 0.342 mol

Finally, we divide the moles of KI into the total moles of solution (nt) to obtain the mole fraction (X):

X= moles KI/nt = 0.020 mol/0.342 mol = 0.058

Answer:

Nonmetallic elements form ions by gaining valence electrons to complete their outer shell.

The more valence electrons an element has in its outer shell, the easier it is to complete.

The fewer electron shells an element has, the easier it is to fill its outermost shell.

Reactivity in nonmetals increases as you go from left to right in a group, and

decreases as you go from top to bottom.

Explanation:

Nonmetallic elements form ions by gaining valence electrons to complete

their outer shell.

The more valence electrons an element has in its outer shell, the easier it is to complete.

The fewer electron shells an element has, the easier it is to fill its outermost shell.

Reactivity in nonmetals increases as you go from left to right in a group, and decreases as you go from top to bottom.

When there are more valence electrons then it makes it easy for the element

to complete its outermost shell as against if it was less as it would be harder

due to a bigger number of electrons needed.

Non metals are elements which accepts valence electrons to complete its

outermost shell and becomes negatively charged when this happens.

When an element has fewer electron shells then filling the outermost shell

will be easier as the electron shells require a lesser number of electrons and

increases as the number of shells increases.

This is because the tendency to accept electrons increases from left to right

and decreases down the group.

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Answer:

All non-zero digits are significant

Explanation:

Answer:

[tex]V_2=19.23L[/tex]

Explanation:

Hello,

In this case, by using the Avogadro's law which allows us to understand the volume-moles behavior as a directly proportional relationship:

[tex]\frac{V_2}{n_2} =\frac{V_1}{n_1}[/tex]

We can compute the volume of 34.3 g of argon by representing it in mole as shown below:

[tex]n_1=1 mol\\\\n_2=34.3g*\frac{1mol}{39.95g} =0.859mol[/tex]

Thus, we find:

[tex]V_2=\frac{V_1*n_2}{n_1}=\frac{22.4L*0.859mol}{1mol} \\\\V_2=19.23L[/tex]

Best regards.

Answer:

Molar solubility of magnesium fluoride (MgF2) = 8.26 × 10⁻¹⁰ M

Explanation:

Given:

Ksp = 5.16 × 10⁻¹¹

NaF = 0.250

Find:

Molar solubility of magnesium fluoride (MgF2)

Computation:

Molar solubility of magnesium fluoride (MgF2) = Ksp / NaF²

Molar solubility of magnesium fluoride (MgF2) = 5.16 × 10⁻¹¹ / (0.250)²

Molar solubility of magnesium fluoride (MgF2) = 8.26 × 10⁻¹⁰ M

The molar solubility of magnesium fluoride (MgF2) is 8.26 × 10⁻¹⁰ M.

Since

Ksp = 5.16 × 10⁻¹¹

NaF = 0.250

Now we know that

Molar solubility of magnesium fluoride (MgF2) = Ksp / NaF²

= 5.16 × 10⁻¹¹ / (0.250)²

= 8.26 × 10⁻¹⁰ M

hence, The molar solubility of magnesium fluoride (MgF2) is 8.26 × 10⁻¹⁰ M.

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Answer:

5.99 moles of [tex]UF_6[/tex]

Explanation:

In this case, we can start with the decomposition of [tex]UF_6[/tex], so:

[tex]UF_6~->~U~+~3F_2[/tex] (Reaction 1)

The [tex]F_2[/tex] can react with carbon to produce [tex]CF_4[/tex]:

[tex]CF_4~+~2F_2~->~CF_4[/tex] (Reaction 2)

If we 8.99 mol of [tex]CF_4[/tex], we can calculate the moles of [tex]F_2[/tex] that we need. In reaction 2 we have a molar ratio of 1:2 (2 moles of [tex]F_2[/tex] will produce 1 mol of [tex]CF_4[/tex]):

[tex]8.99~mol~CF_4\frac{2~mol~F_2}{1~mol~CF_4}~=~17.98~mol~F_2[/tex]

With this value and using the molar ratio in reaction 1 (3 moles of [tex]F_2[/tex] are producing by each mol of [tex]UF_6[/tex]), so:

[tex]17.98~mol~F_2\frac{1~mol~UF_6}{3~mol~F_2}~=~5.99~mol~UF_6[/tex]

So, we will need 5.99 moles of [tex]UF_6[/tex] to produce 8.99 mol of [tex]CF_4[/tex].

I hope it helps!

This means a release of free energy from the system corresponds to a negative change in free energy, but to a positive change for the surroundings.

Answer:

option a. Temperature is equal to 273 k. Is the correct one

A boils when its vapour pressure is equal to, or greater than, the external pressure pushing on it. Therefore, option (b) is correct.

A boiling point of a liquid can be defined as the temperature at which the vapour pressure becomes equal to the atmospheric pressure of the liquid. The liquid is converted into a vapour at this temperature.

With the addition of heat that the liquid changes into vapour without any increase in temperature. The vapour pressure increases until it is equal to the pressure of the gas surrounding it as a liquid is heated. Bubbles of vaporized liquid are formed within the liquid and rise to the surface.

At the boiling temperature, the vapour inside the bubble will exert enough pressure to prevent the bubble from collapsing. The molecules of a liquid will able to overcome the forces of attraction between molecules to create vapour.

Therefore, at boiling point, the vapour pressure is equal to, or greater than, the external pressure pushing on it.

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Answer:

The answer to your question is given below.

Explanation:

When there is an adequate supply of oxygen available to react with the fuel of a bunsen burner, the flame produce is a non luminous bunsen flame.

This flame i.e non luminous flame produced is much hotter than the luminous and three zones can be seen in the flame. This zones are

1. The zone of unburnt gas.

2. The luminous zone.

3. The outermost non luminous zone.

When an adequate supply of oxygen is available to react with the fuel of a Bunsen burner, there is complete combustion of the fuel and a non-luminous flame is produced.

Complete combustion occurs where there is plentiful supply of oxygen. During the reaction, more heat energy is produced as the flame is non-luminous and less soot is produced.

A non-luminous flame is a flame which produces less light but more heat energy. It is usually blue in color and consists of three zones. These zones are:

1. The zone of unburnt gas.

2. The luminous zone.

3. The outermost non luminous zone.

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Answer:

2CH3OH + 3O2 --> 2CO2 + 4H2O

This is called Combustion of Methanol

Explanation:

Count the numbers of atoms on each sides and try to get the same amount of quantity on each sides, so you'll eventually get "balanced".

or

Watch this video: "How to Balance CH3OH + O2 = CO2+ H2O (Combustion of Methanol)"

Good luck with future work! Keep studying chemistry!

- D

Answer:

Hey there!

10000 gallons would be 37.85 cubic meters.

Let me know if this helps :)

Answer: 37.8501 m³

Explanation:

 10000.0 gal × 1 m³/264.2 gal

=37.8501 m³ (significant digits)

Allene (1,2-propadiene) has point group D2d, itself is achiral because it has two planes of symmetry. ... An allene with substituents on one terminal carbon atom are unlike and substituent on other terminal carbon atoms are same, allene will be achiral. It will have one symmetry plane.

Hope this helped :)

Answer:

He studied the way in which the elements combined during chemical reactions

Explanation:

Which method did Dalton use to develop his atomic theory?

D.) He studied the way in which elements combined during chemical reactions.

Answer: It’s D

Explanation:

Answer:

The correct answer is 84.81 °C.

Explanation:

Based on the given information, the volume or mass of coffee is 250 ml, and the mass of metal spoon is 80 grams. The specific heat capacity of metal spoon is 0.80 J/g °C, and the specific heat capacity of coffee is 4.18 J/g °C. In the given case, heat is lost from the coffee and is gained by the spoon.  

For finding the final temperature of the coffee, the formula to be used is Q = msΔT, here m is the mass, s is the specific heat, and ΔT is the change in temperature.  

Heat lost from the coffee = Heat gained by the spoon

250 × 4.18 (90 - T2) = 80 × 0.8 (T2 - 0)

1045 (90 - T2) = 64 (T2 -0)

94050 - 1045 T2 = 64 T2

94050 = 1045 T2 + 64 T2

94050 = 1109 T2

T2 = 84.84 °C

Answer:

[tex]Q=108.55kJ[/tex]

Explanation:

Hello,

In this case, we can compute the required heat by using the following equation:

[tex]Q=mCp(T_2-T_1)[/tex]

Thus, by considering the given mass, specific heat and both initial and final temperature, we obtain:

[tex]Q=500g*1.67\frac{J}{g\°C} *(150.0\°C-20.0\°C)*\frac{1kJ}{1000J} \\\\Q=108.55kJ[/tex]

Regards.

Answer:

pH = 12.15

Explanation:

To determine the pH of the HCl and KOH mixture, we need to know that the reaction is  a neutralization type.

HCl  +  KOH  →  H₂O  +  KCl

We need to determine the moles of each compound

M = mmol / V (mL) → 30 mL . 0.10 M = 3 mmoles of HCl

M = mmol / V (mL) → 40 mL . 0.10 M = 4 mmoles of KOH

The base is in excess, so the HCl will completely react and we would produce the same mmoles of KCl

HCl  +  KOH  →  H₂O  +  KCl

3 m       4 m                       -

             1 m                      3 m

As the KCl is a neutral salt, it does not have any effect on the pH, so the pH will be affected, by the strong base.

1 mmol of KOH has 1 mmol of OH⁻, so the [OH⁻] will be 1 mmol / Tot volume

[OH⁻] 1 mmol / 70 mL = 0.014285 M

- log [OH⁻] = 1.85 → pH = 14 - pOH → 14 - 1.85 = 12.15

The pH of the solution is 12.15.

The following equation should be

HCl  +  KOH  →  H₂O  +  KCl

Here we have to measured the moles of each compound i.e.

M = mmol / V (mL) → 30 mL . 0.10 M = 3 mmoles of HCl

M = mmol / V (mL) → 40 mL . 0.10 M = 4 mmoles of KOH

So, it be like

3 m       4 m                       -

           1 m                      3 m

Now the pH should be like

= [OH⁻] 1 mmol / 70 mL

= 0.014285 M

Now

= - log [OH⁻] = 1.85

pH = 14 - pOH

= 14 - 1.85

= 12.15

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Answer:

pH will be increased.

Explanation:

When you mixed sodium acetate and acetic acid,  pH will be maintained and will have slight changes. This constitutes what is known as a buffer solution.

But if you analize the reactions:

First of all, you have a weak acid, acetic:

CH₃COOH   +  H₂O  ⇄  CH₃COO⁻  +   H₃O⁺      Ka

And the salt can be dissociated as:

NaCH₃COO →  CH₃COO⁻  +   Na⁺

As you are adding acetate, this is the conjugate strong base from the weak acid and as every base, it brings hydroxides to medium, so the pH will be increased. Protons from the acid will be neutralized, thouhgh.

CH₃COO⁻  +   H₂O  ⇄  CH₃COOH + OH⁻     Kb

Answer:

[tex]1.53~moles~CH_2Cl_2[/tex].

Explanation:

We can start with the reaction, if we know the formula for each compound:

-) Methane: [tex]CH_4[/tex]

-) Carbon tetrachloride: [tex]CCl_4[/tex]

-) Dichloromethane: [tex]CH_2Cl_2[/tex]

With this in mind, we can write the reaction:

[tex]CH_4~+~CCl_4~->~CH_2Cl_2[/tex]

Now, we can balance the reaction:

[tex]CH_4~+~CCl_4~->~2CH_2Cl_2[/tex]

After this, we have 2 carbon atoms on each side, 4 hydrogen atom on each side, and 4 chlorine atoms on each side.

If we want to know how many moles of [tex]CH_2Cl_2[/tex] would be produced with .766 moles of [tex]CH_4[/tex], we have to check the balanced reaction and use the molar ratio. In this case, the molar ratio is 1 mol [tex]CH_4[/tex] will produce 2 moles of [tex]CH_2Cl_2[/tex] (1:2). So:

[tex]0.766~moles~CH_4\frac{2~moles~CH_2Cl_2}{1~mol~CH_4}=1.53~moles~CH_2Cl_2[/tex]

We wil have [tex]1.53~moles~CH_2Cl_2[/tex].

I hope it helps!

Answer:

d. sodium t-butoxide + bromomethane

Explanation:

Sodium t-butoxide is a strong base with the chemical formula  (CH3)3CONa and it is a non-nucleophilic base while bromomethane with the chemical formula CH3Br.

They both together gives 2-ethoxypropane (C5H12O).

(CH3)3CONa + CH3Br => C5H12O + NaBr

Hence, the correct option is "d".

Answer:

Thinking...

o-O

(ﾉ◕ヮ◕)ﾉ*:･ﾟ✧

Answer:

[tex][Cu]^{remaining}=0.357 M[/tex]

Explanation:

Hello,

In this case, we may use the following equation in order to compute the moles of copper that are processed, considering it goes from Cu⁰ to Cu²⁺, so two electrons are transferred:

[tex]n_{Cu}=\frac{2.68C/s*282s}{96500C/mol*2} =3.9x10^{-3}mol[/tex]

After that, we compute the initial moles of copper in the solution, considering that the concentration of copper (II) equals the concentration of copper:

[tex]n_{Cu,0}=0.366mol/L*0.425L=0.156mol[/tex]

In such a way, we can subtract the process moles to the initial moles to compute the remaining moles of copper:

[tex]n_{Cu}^{remaining}=0.156mol-0.0039mol=0.152mol[/tex]

Finally, the concentration is:

[tex][Cu]^{remaining}=0.152mol/0.425L[/tex]

[tex][Cu]^{remaining}=0.357 M[/tex]

Regards.