A gas occupies a volume of 180 mL at 35 °C and 95.9 kPa. What is the volume of the gas at conditions of STP?

Answers

Answer 1

Answer:

the volume of the gas at conditions of STP = 151.04998 ml

Explanation:

Data given:

V1 = 180 ml

T1 = 35°C or 273.15 + 35 = 308.15 K

P1 = 95.9 KPa

V2  =?

We know that at STP

P2 = 1 atm or 101.3 KPa

T2 = 273.15 K

At STP the pressure is 1 atm and the temperature is 273.15 K

applying Gas Law:

[tex]\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}[/tex]

putting the values in the equation of Gas Law:

[tex]V_2=\frac{P_1V_1T_2}{T_1P_2}[/tex]

V_2 =[tex]\frac{95.9\times180\times273.15}{308.15\times101.3}[/tex]

V2 = 151.04998

therefore, V2 = 151.04998 ml

Answer 2

Answer:

151 mL is the correct answer to the given question .

Explanation:

We know that

[tex]PV =n RT[/tex]

Where P =pressure ,V=volume and T=Temperature

Given

P=95.9 kPa.

V=[tex]180 * 10 ^{-3}[/tex]

R=25/3

T=273 + 35 =308k

Putting these value into the equation we get

[tex]95.9\ * 180\ *\ 10^{-3} \ =\ n * \frac{25}{3} * 308[/tex]

n=[tex]6.72 * 10^{-3}[/tex]

Now using the equation

[tex]n= \ \frac{V}{22.4}[/tex]

[tex]6.72 * 10^{-3} =\frac{V}{22.4}\\ V\ =\ 150.6mL[/tex]

This can be written as  151mL


Related Questions

A student takes a measured volume of 3.00 M HCl to prepare a 50.0 mL sample of 1.80 M HCI. What volume of 3.00 M HCI

did the student use to make the sample?

Use M,V;-MV

3.70 mL

16.7 ml

30.0 mL

83.3 mL

Mark this and return

Save and Exit

Next

Submit

Answers

Answer:

30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.

Explanation:

M1 = 3.00 M

M2 = 1.80 M

V2 = 50 .0 mL = 50 /1000 L = 0.05 L

V1 = unknown

In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.

So we use the equation:

                                  M1 V1 = M2 V2

V1 = M2 V2 / M1

V2 = 1.80 * 0.05 / 3.0

V2 = 0.09 /3.0

V2 = 0.03 L or 30 mL

To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.

Answer:

C on edg 2021

Explanation:

i dont like reading

What is the predicted order of first ionization energies from highest to lowest for lithium (Li), sodium (Na), potassium (K), and rubidium (Rb)?
Rb > K > Na > Li
K > Rb > Na > Li
Li > Na > K > Rb
Rb > K > Li > Na

Answers

Answer:

Li>Na>K>Rb

Explanation:

Answer:

c. Li > Na > K > Rb

Explanation:

edge 2021

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Enter your answer in the provided box. A sample of an unknown gas effuses in 14.5 min. An equal volume of H2 in the same apparatus under the same conditions effuses in 2.42 min. What is the molar mass of the unknown gas

Answers

Answer:

Molar mass = 71.76 g/mol

Explanation:

The relationship between molar mass and rate of effusion is given as;

Vh / Vu = √ (Mu / Mh)

The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles.

Rate = volume /  time (Assuming Volume = 1)

Vh = Rate of effusion of Hydrogen = 1 / 2.42

Vu = Rate of effusion of unknown gas = 1 / 14.5

Mh = Molar mass of hydrogen = 2

Mu = Molar mass of unknown gas = x

Substituting into the formular, we have;

(1 / 2.42) / (1 / 14.5) = √ ( x / 2)

5.99 = √ ( x / 2)

35.88 = x / 2

x = 71.76

g The solution you created in this simulation was a 0.300M NH4Cl solution. The lab also stated that, in g/L, this concentration was 16.0 g/L. Show the calculations that prove that to be true.

Answers

Answer:

16.0473 g/L

Explanation:

0.300 M=

0.300 mol/L x 53.491 grams/mol = 16.0473 grams/L

The concentration of the 0.300M NH₄Cl solution in g/L will be equal to 16.04 g/L.

What is the molarity?

The concentration of the solution can be determined if we have the molecular formula of the compound and its molecular weight. We can easily determine the majority of a solution from the moles of solute and the volume of the solution.

The molarity of a solution can be evaluated from the number of moles of a solute per liter of a solution.

The Molarity can be determined from the formula mentioned below:

Molarity (M) = Moles of solute (n)/Solution's volume ( in L)

Given, the molarity of NH₄Cl solution = 0.300 M

We can also write it as 0.300 mol/L

It means 0.300 moles in one liter.

The molar mass of NH₄Cl  = 53.5 g/mol

Then the mass of 0.300 mol of NH₄Cl  = 0.300 ×53.5 = 16.04 g

Therefore, the concentration of NH₄Cl solution is 16.04g/L is equivalent to 0.300 M.

Learn more about molarity, here:

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The decomposition of Ca(OH)2(s) intoCaO(s) and H2O(g) at constant pressure requires the addition of 109 kJ of heat per mole of Ca(OH)2.
A. Write a balanced thermochemical equation for the reaction.
B. Draw an enthalpy diagram for the reaction, showing the activation energy, Ea, and the enthalpy change for the reaction.

Answers

Answer:

A. Ca(OH)2 ----> CaO + H2O; ∆H = +109KJ

B. Check attached document below for the enthalpy diagram

Explanation:

A thermochemical equation is a balanced chemical equation in which enthalpy change is variable. It tells about the nature of the reaction.

Enthalpy change is the difference between the heat content of the products and reactants in a thermochemical equation.

∆H is negative for an exothermic reaction (a reaction where heat is given out), while it is positive for an endothermic reaction ( a reaction in which heat is added).

The activation energy, Ea, is the minimum amount of energy reactant particles must possess in order for a reaction to proceed towards product formation.

As per the decomposition of the Ca(OH) 2 and the CaO the constant pressure needs to be added of the 109KJ.

The balance thermochemical equation for the reaction is expressed as  Ca(OH)2 ----> Ca O + H2O; ∆H = +109KJThe diagram for the reaction and the activation energy is given above.

Learn more about decomposition.

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11. Caproic acid, which is responsible for the foul odor of dirty socks, is composed of C, H, and O atoms. Combustion of a 0.225-g sample of this compound produces 0.512 g CO2 and 0.209 g H2O. (a) What is the empirical formula of caproic acid

Answers

Answer:

C3H6O

Explanation:

Step 1:

Data obtained from the question include the following:

Mass of the compound = 0.225g

Mass of CO2 = 0.512g

Mass of H2O = 0.209g

Step 2:

Determination of the masses of carbon, hydrogen and oxygen present in the compound.

This is illustrated below:

For Carbon, C:

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 0.512 = 0.1396g

For Hydrogen, H:

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 0.209 = 0.0232g

For Oxygen, O:

Mass of O = 0.225 – (0.1396 + 0.0232)

Mass of O = 0.0622g

Step 3:

Determination of the empirical formula for caprioc acid.

This can be obtain as follow:

C = 0.1396g

H = 0.0232g

O = 0.0622g

Divide by their molar mass

C = 0.1396/12 = 0.0116

H = 0.0232/1 = 0.0232

O = 0.0622/16 = 0.0039

Divide by the smallest

C = 0.0116/0.0039 = 3

H = 0.0232/0.0039 = 6

O = 0.0039/0.0039 = 1

Therefore, the empirical formula for caprioc acid is C3H6O

complite the following reactions. NaOH(aq)+FeBr3(aq)→

Answers

Answer:

3NaOH+FeBr3>3NaBr+

Fe(OH)3

Explanation:

After writing the equation it has to be balanced

How is excitation in spectroscopy brought about​

Answers

Answer: the exciation of molecules is brount by absorption of energy  in spectroscpy

Explanation:

What is the reactant(s) in the chemical equation below?
2Al(s) + 2NaOH(aq) + 2H2O()
2NaAlO2(aq) + 3H2(9)
A. 2Al(s) + 2NaOH(aq) + 2H200)
B. 2NaAlo2(aq) + 3H2(g)
C. 2Al(s)
D. 3H2(g)

Answers

Answer:

A

Explanation:

They are all found in the reactants side

Calculate Keq for these reactions and predict if the equilibrium will lie to the right or to the left as written. (You may enter your answer in scientific notation, e.g. 1.0*10^-9. Enter your answer to two significant figures.) Reaction 1: + + pKa = 9 pKa = 38 Keq = Equilibrium position = _______ Reaction 2: + + pKa = 35 pKa = 25 Keq = Equilibrium position = _______

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

For reaction 1

    [tex]K_{eq} = 10^{29}[/tex]

     The equilibrium position is to the right

For reaction 2

       [tex]K_{eq} = 10^{-6.66}[/tex]

        The equilibrium position is to the left

Explanation:

Generally  [tex]pKa[/tex] is mathematically evaluated as  

[tex]pKa = pKa _ \ {left }} - pKa _ \ {right }}[/tex]

And equilibrium position [tex]K_a[/tex] is mathematically evaluated as [tex]K_{eq} = 10^\ {-pK_a}[/tex]

From the question we are told that

For reaction 1

         [tex]pKa_\ {left}} \ = 9[/tex]

        [tex]pKa_\ {right }} \ = 38[/tex]

So

       [tex]pKa = 9-38[/tex]

       [tex]pKa =-29[/tex]

So  [tex]K_{eq} = 10^{-(-29)}[/tex]

      [tex]K_a = 10^{29}[/tex]

This implies that the equilibrium position is to the right

   For reaction 2

       [tex]pKa_\ {left}} \ = 15.9[/tex]

       [tex]pKa_\ {right }} \ = 9.24[/tex]

So

       [tex]pKa = 15.9-9.24[/tex]

       [tex]pKa = 6.66[/tex]

So  [tex]K_{eq} = 10^{-(6.66)}[/tex]

      [tex]K_{eq} = 10^{-6.66}[/tex]

This implies that the equilibrium position is to the left

Real images can be projected onto a screen.
A. True
B. False

Answers

Answer:

True

Explanation:

A real image can be projected or seen on a screen but a virtual image cannot because a real image is formed when light rays coming from an object actually meet at a point after refraction through a lens while a virtual image is formed when light rays coming from an object only appear to meet at a point when produced ...

Hope this helps you, and Good luck!

A 1.44 L buffer solution consists of 0.137 M butanoic acid and 0.275 M sodium butanoate. Calculate the pH of the solution following the addition of 0.069 moles of NaOH . Assume that any contribution of the NaOH to the volume of the solution is negligible. The Ka of butanoic acid is 1.52×10−5 .

Answers

Answer:

The answer is "[tex]P^{H}=5.379[/tex]".

Explanation:

[tex]\ NaOH \ value = \frac{n}{v}\\\\[/tex]

                     [tex]=\frac{0.069\ moles}{0.144L}\\\\=0.04791\ M[/tex]

[tex]\ Ka=1.52 \times 10^{-5}\\\\P^{ka} = -10g \ ka \\\\[/tex]

      [tex]= -10 \times 1.52 \times 10^{-5}\\\\= 4.82\\[/tex]

Equation:

[tex]CH_3CH_2CH2COOH+NaOH\rightarrow CH_3CH_2CH_2COONa +H_2O\\\\[/tex]

[tex]\boxed{\left\begin{array}{ccccc}I &0.137 &0.04791 &0.275 & -- \\ C &-0.04791 &-0.04791 &+0.04791 & -- \\E &0.08909 &0&0.32291 & -- \end{array}\right}[/tex]

[tex]P^{H}= P^{ka}+\log\frac{CH_3CH_2CH_2COONa}{CH_3CH_2CH_2COOH}\\\\[/tex]

      [tex]= 4.82+\log\frac{0.32291}{0.08909}\\\\=5.379[/tex]

The temperature program for a separation starts at a temperature of 50 °C and ramps the temperature up to 270 °C at a rate of 10 °C/minute. Which statement is NOT true for this separation?
A) At 10 °C/minute, a total of 22 minutes is needed to reach 270 oC.
B) Strongly retained solutes will remain at the head of the column while the temperature is low.
C) Weakly retained solutes will separate and elute early in the separation.
D) The vapor pressure of strongly retained solutes will increase as temperature increases.
E) Strongly retained analytes will give broad peaks.

Answers

Answer:

The correct answer to the question is Option E (Strongly retained analytes will give broad peaks).

Explanation:

The other options are true because:

A. Initial temp = 50 °C

   Final temp =  270 °C

Differences in temp = 270 - 50 = 220°C

Rate =  10 °C/minute.

So, at 10 °C/minute,

total of 220°C /10 °C = number of minutes required to reach the final temp.

220/10 = 22 minutes

B. A column has a minimum and maximum use temperature. Solutes that are already retained would remain stationary while temperatures are low. This would only change if there is an increase in temperature. Heat transfers more energy to the liquid which would make the solute interact with the column phase.

C. Weakly retained solutes may contain larger molecules, will separate by absorbing into the solvent early in separation making the mobile phase separates out into its components on the stationary phase.

D. Retained solute's vapor pressure is higher at higher temperatures making it possible for particle to escape more from the solute when the temperature is high than when it is low.

Consider the addition of an electron to the following atoms from the third period. Rank the atoms in order from the most negative to the least negative electron affinity values based on their electron configurations.
Atom or ion Electron configuration
Br 1s22s22p63s23p64s23d104p5
Ge 1s22s22p63s23p64s23d104p2
Kr 1s22s22p63s23p64s23d104p6

Answers

Answer:

The ranks is

Ge: 3d10 4s2 4p2 (6 electrons in the outer shell)

Br: 3d10 4s2 4p5 (7 electrons in the outer shell)

Kr: 3d10 4s2 4p6 (8 electrons in the outer shell)

Explanation:

Electronic configuration reffers to the distribution of electrons of an atom or molecule in atomic or molecular orbitals. It gives us the understanding of the shape and energy of its electrons. The electronic configuration explain the The electron affinity or propensity to attract electrons

It Should be noted that the most stable configuration in an electronic configuration is attributed to when the last shell is full, i.e. when the last shell has 8 electrons.

When an atom is closer to reach the 8 electrons in the outer shell, then it's electron affinity big.

Considering the given three configuration of the elements above, we can see that "Br"needs requires only 1 electron to have 8 electrons in the outer shell, therefore, it is considered to have the biggest electron affinity among them which is reffers to as the LEAST NEGATIVE.

Ge: with the electronic configuration 3d10 4s2 4p2 has 6 electrons in the outer shell which means it still requires 2 electrons to complete 8 electrons in its outer shell, so it can be deducted that it posses an atom that is more negative than Br.

Kr: with the electronic configuration 3d10 4s2 4p6 which is a noble gas has 8 electrons in the outer shell cannot add more electrons to its outer shell because the 8 electrons is complete posses the least electron affinity among the three elements and it is the MOST NEGATIVE

Trans-4-hexen-3-ol can be synthesized starting from acetaldehyde. One of the key reagents is ethyl grignard.
1. Synthesize ethyl grignard from acetaldehyde in the steps below using the reagents provided.
2. Synthesize (trans)-4-hexen-3-ol from acetaldehyde.

Answers

find the given attachment

Classify the following as Arrhenius, Bronsted-Lowry, or Lewis acid-base reactions. A reaction may fit all, two, one, or none of the categories:I. Cu2+ + 4 Cl− CuCl42−II. Al(OH)3 + 3HNO3 Al3+ + 3H2O + 3 NO3−III. N2 + 3 H2 2NH3IV. CN− + H2O HCN + OH

Answers

Answer:

I. Lewis acid-base reaction

II. Arrhenius, Brønsted-Lowry, and Lewis' acid-base reaction

III. Brønsted-Lowry and Lewis'acid-base reaction

IV. Lewis acid-base reaction

Explanation:

According to Arrhenius, an acid is a substance that dissolves in water to produce H+ ions, and a base is a substance that dissolves in water to produce hydroxide (OH−) ions.

In the reaction below, AH is an avid, BOH is a base reacting together to form a salt(A-B+) and water only.

AH + BOH ---> A-B+ + H2O

According to Brønsted-Lowry definition, an acid is any substance that can donate a proton, and a base is any substance that can accept a proton.

In the reaction below, AH is an acid while B is a base, reacting together to form an acid-base conjugate pair.

AH + B <-----> BH+ + A-

According to Lewis' definition, an acid is a species that accepts an electron pair while a base donates an electron pair resulting in a coordinate covalently bonded compound, also known as an adduct. In the reaction below, A+ is an acid, B- is a base, reacting together to form product A-B.

A+ + B- ------> A-B

Considering the above definitions;

I. Cu²+ + 4 Cl− ---> CuCl4²− is a Lewis acid-base reaction because it involves electron sharing only.

II. Al(OH)3 + 3HNO3 ---> Al3+ + 3H2O + 3 NO3− is an Arrhenius, Brønsted-Lowry, and a Lewis acid-base reaction because it involves protons, electrons and hydroxide ions.

III. N2 + 3 H2 ---> 2NH3 is a Lewis acid-base reaction because it involves sharing of electrons only.

IV. CN− + H2O ---> HCN + OH is both a Lewis and Brønsted-Lowry acid-base reaction because both protons and electrons sharing is involved.

In a Bronsted-Lowry acid-base reaction reaction, an acid donates protons which is accepted by the base.

The following are useful definitions of acids and bases;

An Arrhenius acid produces hydrogen ion as its only positive ion in solution while an Arrhenius base produces hydroxide ion as its only negative ion in solution.A Bronsted-Lowry acid donates hydrogen ions while a Bronsted-Lowry base accepts hydrogen ionsA Lewis acid accepts lone pairs of electrons while a Lewis base donates lone pairs of electrons.

Based on these, we can now classify the reactions accordingly;

Cu^2+ + 4Cl−  ------>[CuCl4[^2−   Lewis acid-base reaction Al(OH)3 + 3HNO3 -----> Al^3+ + 3H2O + 3NO3^− Arrhenius acid-base reactionN2 + 3H2  ----> 2NH3 NoneCN− + H2O ------> HCN + OH^- Bronsted-Lowry acid-base reaction

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Calculate the standard entropy of reaction at 298 K for the reaction Hg(liq) + Cl2(g) → HgCl2(s) The standard molar entropies of the species at that temperature are: Sºm (Hg,liq) = 76.02 J / (K mol) ; Sºm (Cl2,g) = 223.07 J / (K mol) ; Sºm (HgCl2,s) = 146.0 J / (K mol)

Answers

Answer:

−153.1 J / (K mol)

Explanation:

Calculate the standard entropy of reaction at 298 K for the reaction Hg(liq) + Cl2(g) → HgCl2(s) The standard molar entropies of the species at that temperature are: Sºm (Hg,liq) = 76.02 J / (K mol) ; Sºm (Cl2,g) = 223.07 J / (K mol) ; Sºm (HgCl2,s) = 146.0 J / (K mol)

Hg(liq) + Cl2(g) → HgCl2(s)

Given that;

The standard molar entropies of the species at that temperature are:

Sºm (Hg,liq) = 76.02 J / (K mol) ;

Sºm (Cl2,g) = 223.07 J / (K mol) ;

Sºm (HgCl2,s) = 146.0 J / (K mol)

The standard molar entropies of reaction = Sºm[products] - Sºm [ reactants]

= 146.0 J / (K mol) – [76.02 J / (K mol) +223.07 J / (K mol) ]

= -153.09 J / (K mol)

= or -153.1 J / (K mol)

Hence the answer is  −153.1 J / (K mol)

Which process is a physical change

Answers

Answer:

a physical change is something that has not been modified chemically and can possibly be changed back to the state it was once before. A physical change keeps all the same atoms and none of them is modified.

Example:

When a block of clay is morphed into a giraffe statue, it can be morphed back to its original state. If someone burnt the block of clay, the atoms would be modified and it would be unable to go back to its previous state.

-----------------------------------------------------------------------------------------------------------------

(If you're referring to a question with these answers)

A. iron rusting

B. milk turning to curd

C. water boiling

D. paper burning

E. hard water staining pipes

-----------------------------------------------------------------------------------------------------------------

Answer:

C. Water Boiling

(If you are referring to a question with these answers I think this is the correct answer if not I do apologize)

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A microwave has a wavelength of 0.028 m. What us this wavelength in scientific notation

Answers

Answer:

2.8 times 10^ -2

Explanation:

scientific notation is supposed to be a number between 1 and 10

the decimal point moves 2 to the right in order to get 2.8 which makes the exponent negative

C3H7-C(=O)-NH2 IUPAC NAME ?

Answers

Answer:

Amide

Explanation:

O=NH2 is the Amide group versus NH2, which is the amine group.

Answer:

Butamide

Explanation:

C3H7-C(=O)-NH2 IUPAC NAME

C4H9NO

     H   H   H

H - C - C - C - C = O

     H   H   H   N - H

                      H

But amide

Amide because R-CO-NH2 ie C(=O)-NH2

But because 4 Cabon

A chemistry student is given of a clear aqueous solution at . He is told an unknown amount of a certain compound is dissolved in the solution. The student allows the solution to cool to . At that point, the student sees that a precipitate has formed. He pours off the remaining liquid solution, throws away the precipitates, and evaporates the water from the remaining liquid solution under vacuum. More precipitate forms. The student washes, dries and weighs the additional precipitate. It weighs 50,0 g.
Using only the information above, can you calculate the solubility of X in water at 16. C. If you said yes, calculate it.

Answers

Answer:

Solubility cannot be calculated.

Explanation:

To calculate the solubility of X it is necessary to know the value of the mass of the solute (X) that can be dissolved in 100 g of water.

[tex]Solubility = \frac{Solute mass}{100 grams of water}[/tex]

Taking into account that we do not know the value of the mass of the solution, therefore the value of the solubility of the compound cannot be determined.

What happens at this point

Answers

Answer:

What are you referring to exactly

Explanation:

Answer:  C

Explanation:

There is a difference in air pressure.  That's what I put and I got a 92.

What are plastic bottles made of?
Polyethylene

halogen

silicon

Alkyl groups

Answers

Answer:

polyethylenes

Explanation:

the plastic bottles used to hold potable water and other drinks are made from polyethylene because, the material is both strong and light.

hope this helped!

Answer: Polyethylenes

Explanation: I got 100% on the test :)

Calculate the Kc for the following reaction if an initial reaction mixture of 0.500 mole of CO and 1.500 mole of H2 in a 5.00 liter container forms an equilibrium mixture containing 0.198 mole of H2O and corresponding amounts of CO, H2, and CH4.

Answers

Answer:

4.41

Explanation:

Step 1: Write the balanced equation

CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)

Step 2: Calculate the respective concentrations

[tex][CO]_i = \frac{0.500mol}{5.00L} = 0.100M[/tex]

[tex][H_2]_i = \frac{1.500mol}{5.00L} = 0.300M[/tex]

[tex][H_2O]_{eq} = \frac{0.198mol}{5.00L} = 0.0396M[/tex]

Step 3: Make an ICE chart

        CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)

I       0.100      0.300        0            0

C         -x           -3x          +x          +x

E    0.100-x    0.300-3x     x            x

Step 4: Find the value of x

Since the concentration at equilibrium of water is 0.0396 M, x = 0.0396

Step 5: Find the concentrations at equilibrium

[CO] = 0.100-x = 0.100-0.0396 = 0.060 M

[H₂] = 0.300-3x = 0.300-3(0.0396) = 0.181 M

[CH₄] = x = 0.0396 M

[H₂O] = x = 0.0396 M

Step 6: Calculate the equilibrium constant (Kc)

[tex]Kc = \frac{[CH_4] \times [H_2O] }{[CO] \times [H_2]^{3} } = \frac{0.0396 \times 0.0396 }{0.060 \times 0.181^{3} } = 4.41[/tex]

Which process is used to make lime (calcium oxide) from limestone (calcium carbonate)?​

Answers

Answer:

Explanation:

Calcium oxide is fromed by the decomopostion of CaCO3 at high temperature.

CaCO3   ------> CaO  +CO2

Hope this helps you

When 8.1 g of an unknown non-electrolyte is dissolved in 50.0 g of carbon tetrachloride, the boiling point increased by 3.67 degrees C. If the Kbp of the solvent is 4.95 K/m, calculate the molar mass of the unknown solute.

Answers

Answer:

218.3 g/mol

Explanation:

Boiling point elevation occurs when a solute is added to a solvent increasing the boiling point of the solution with regard to the pure solvent.

The law is:

ΔT = Kb×m×i

Where ΔT is change in temperature (3.67°C), Kb is the boiling point constant of the solvent (4.95°C/m), m is molality of the solution and i is Van't Hoff factor (1 for a non-electrolyte).

3.67°C = 4.95°C/m×m×i

0.7414m = molality of the solution (Moles solute / kg solvent).

As the mass of the solvent is 50.0g = 0.0500kg:

0.7414m = Moles solute / 0.0500kg

0.0371 = moles of solute

As the mass of the solute is 8.1g, molar mass of the solute (Ratio between mass in g and moles) is:

8.1g / 0.0371mol =

218.3 g/mol

carbon dioxide is a non-polar molecule true or false​

Answers

Answer:

True

Explanation:

Due to the arrangement of the molecule, a carbon dioxide molecule is non-polar.

An unknown compound, B, has the molecular formula C7H12. On catalytic hydrogenation 1 mol of B absorbs 2 mol of hydrogen and yields 2-methylhexane. B has significant IR absorption band at about 3300 and 2200 cm-1. Which compound best represents B?
a. 5-methyl-1,3-hexadiene
b. 5-methyl-1-hexyne
c. 3-methyl-1-hexyne
d. 5-methyl-2-hexyne
e. 2-methyl-1,5-hexadiene

Answers

Answer:

B and D

Explanation:

If we use the info given we have a band a 3300 cm-1 and 2200 cm-1 this indicates that we have an alkyne functional group. Additionally, the hydrogenation of the unknown molecule will consume two moles of hydrogens this fits with the 2 pi bonds in the alkyne functional group. So, we can discard "a" and "e". The product of this hydrogenation is 2-methylhexane therefore we can discard c because the methyl group is placed on carbon 3. Structures b and d can work.

See figure 1

I hope it helps!

The acetate ion is the conjugate base of the weak acid acetic acid. The value of Kb for CH3COO-, is 5.56×10-10. Write the equation for the reaction that goes with this equilibrium constant.

Answers

Answer: The equation for the reaction that goes with this equilibrium constant is [tex]5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]

Explanation:

[tex]CH_3COOH\rightarrow CH_3COO^-+H^+[/tex]

Here [tex]CH_3COOH[/tex] donates a proton and thus behaves as an acid and forms [tex]CH_3COO^-[/tex] which is called as the conjugate base of [tex]CH_3COOH[/tex]

The dissociation constant of acids is given by the term [tex]K_a[/tex] and the dissociation constant of bases is given by the term [tex]K_b[/tex] and is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

[tex]K_a[/tex] for  [tex]CH_3COOH[/tex] :

[tex]K_a=\frac{[CH_3COO^-]\times [H^+]}{[CH_3COOH]}[/tex]

[tex]CH_3COO^-+H^+\rightarrow CH_3COOH[/tex]

[tex]K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]

[tex]5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]

The equation for the reaction that goes with this equilibrium constant is [tex]K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]

Aspirin is usually packaged with
A. acetic anhydride
B. salicylic acid
C. buffering agents ​

Answers

I believe the answer to s B.

Answer:

Aspirin is usually packaged with C. buffering agents.

Explanation:

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