A gas effuses 4.0 times faster than oxygen (o2). what is the molecular mass of the gas? 1.0 g/mol 1.0 g/mol 2.0 g/mol 2.0 g/mol 4.0 g/mol 4.0 g/mol 7.67 g/mol

Answers

Answer 1

Answer:

32(molecular mass has no unit )

Explanation:

(16)(o2)

16×2

=32


Related Questions

A container holds 265 mL of chlorine gas (Cl2). If the gas sample is at standard temperature and pressure (STP), what is its mass in grams?

Answers

The mass, in grams, of the chlorine gas will be 0.84 grams

Mass of gases at STP

At standard temperature and pressure, 22.4 L of gas has 1 mole of the gas in it.

This time, what we have is 265 mL of the gas at STP. 265 ml is equivalent to: 265/1000 = 0.265 L

If.

22.4 L = 1 mole

Then,

0.265 L = 0.265 x 1/22.4 = 0.01183 moles

This means that 0.01183 moles of chlorine is present in the gas at STP.

Mass of 0.01183 moles of chlorine = moles x molar mass.

The molar mass of chlorine gas is 71

Mass of the chlorine gas = 0.01183 x 71 = 0.84 grams

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Draw the major product formed in the reaction involving an acid anhydride dissolved in alcohol.

Answers

Ester is the major product formed in the reaction involving an acid anhydride dissolved in alcohol.

What is the chemistry of acid anhydrides?Acid chlorides and anhydrides both go through a number of similar reactions when subjected to nucleophilic action. Anhydrides react with water to generate carboxylic acids, with alcohols to form esters, and with amines to form amides, though they do so more slowly than acid chlorides. By using hydride reduction, anhydrides can also be converted to primary alcohols. As many anhydrides are created by coupling two carboxylic acids, one equivalent of the carboxylic acid is lost in anyhydride reactions as a leaving group. As a result, reactions are often only carried out using affordable, widely accessible anhydrides, such as acetic or benzoic. The advantage of anydrides over acid chlorides is that they are simpler to work with.

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In general ethers have a much lower boiling point than their isomeric alcohols. why?

Answers

Ethers have a much lower boiling point than their isomeric alcohols because ether have not hydrogen bond that why ethers have lower boiling point .

Ethers will not have a hydrogen atom attached to the oxygen atom as alcohols do. Because ether molecules can not form intermolecular hydrogen bonds, their boiling points are much shorter than those of alcohol with such a comparable mass.

Owing to the combination of intermolecular H-bonding in alcohols though not in ethers, alcohols possess higher boiling temperatures than isomeric ethers.

Therefore, ethers have a much lower boiling point than their isomeric alcohols because ether have not hydrogen bond that why ethers have lower boiling point .

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Determine the molecular formula of a compound whose molecular mass is 60.00 g/mol and has an empirical formula of ch4n. ch4n c2h5n2 c2h8n2 c3h12n3

Answers

Molecular formula of a compound whose molecular mass is 60.00 g/mol and has an empirical formula of C₂H₈N₂.

We know the empirical formula and thus the molar mass of the empirical formula, we simply need to find out how many of these fit into the molar mass of the molecular formula.

In this problem, we have an empirical formula of CH₄N

so the molar mass is 12 + 4 + 14 = 30 g/mol.

Molecular formula mass/Empirical formula mass=60 g/mol/30 g/mol=2

The molecular formula is TWICE that of the empirical formula.

Molecular formula = 2XCH₄N = C₂H₈N₂

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Answer:

C2H8N2

Explanation:

edge 2023

A chemist reacted 0. 2 moles sodium benzoate with 0. 25 moles of hydrochloric acid. if she generated 22 g benzoic acid, what was her percent yield? (mw of benzoic acid = 122. 12 g mol-1)

Answers

A chemist reacted 0. 2 moles sodium benzoate with 0. 25 moles of hydrochloric acid. if she generated 22 g benzoic acid, the percent yield is 90%.

Reactant + HCl ------ Product

0.2 mol 0.25 22g

mol

Number of moles of reactant = 0.2mol

Molar mass of product = 122.12g

Number of moles of product = given mass/ molar mass

= 22g/ 122.12 g

= 0.180mol

Percent yield = (number of moles of product/ number of moles of reactant) × 100

= (0.180/0.2) × 100

= 90%

Thus we concluded that the percent yield of the given solution is 90%.

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Draw the alkane formed when 4,5,5‑trimethyl‑1‑hexyne or 4,5,5‑trimethylhex‑1‑yne is treated with two equivalents of hbr.

Answers

Answer:

2,2 - dibromo - 4,5,5 - trimethylhexane

Explanation:

When 2 equivalents of HBr are added to 4,5,5‑trimethyl‑1‑hexyne, the alkane (at the top right of the image) is formed. The bromine branches are added to the second carbon because the intermediate carbocations are the most stable on the secondary carbon. Below I have drawn the mechanism.

A current of 5. 68 a is passed through a Fe(NO3)2 solution. How long, in hours, would this current have to be applied to plate out 7. 20 g of iron?

Answers

There are  1.2 hr would this current have to be applied to plate out 7. 20 g of iron .

Calculation ,

Given ; Current ( I ) = 5. 68 A

In [tex]Fe(NO_{3} )_{2}[/tex] , the valancy of Fe is +2 .

2 moles of [tex]e^{-}[/tex] are required for the decomposition of 1 mole of Fe .

7. 20 g  of Fe in moles  = 7. 20 g /55.845 g/mol =0.12 mole

x moles of  [tex]e^{-}[/tex] are required for the decomposition of 0.128 mole of Fe .

moles of   [tex]e^{-}[/tex] are required = 0.256 moles

Charge on 1 mole  of   [tex]e^{-}[/tex] = 96500 C

Charge on 0.256  mole  of   [tex]e^{-}[/tex] = 24704 C

Current ( I )= Q/t  

t =Q / I = 24704 C/5. 68 A = 4349 sec = 1.2 hr

Therefore , there are  1.2 hr would this current have to be applied to plate out 7. 20 g of iron .

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What point defects are possible for al2o3 as an impurity in mgo? how many al3 ions must be added to form each of these defects?

Answers

Point defects are possible for al2o3 as an impurity in mgo. then 2 Al3+ must be added to form 1 Mg2+  must be added to form each of these defects

Point defects are zero-dimensional lattice defects, meaning they lack lattice structure in any dimension. Impurity atoms in a pure metal, vacancies, and self-interstitials are examples of common point defects.

Point defects are zero-dimensional lattice defects, which means they lack any kind of lattice structure. Point defects include, but are not limited to, atoms of impurity in a pure metal, vacancies, and self-interstitials.

The term "point defect" refers to the flaw or imperfection that results from the placement of a point-like constituent particle, such as an atom, an ion, or a molecule, in a crystalline structure.

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If the sodium oxide (Na2O) in soda-lime glass was replaced by potassium oxide (K2O), this would result in . . .

Answers

it would result in a explosion

Calculate the ph of a 0.377 m solution of ethylenediamine ( h2nch2ch2nh2 ). the pa values for the acidic form of ethylenediamine ( h 3nch2ch2nh 3 ) are 6.848 ( pa1 ) and 9.928 ( pa2 ).

Answers

The final answer is 11.71

The concentration of a given solution (H2NCH2CH2NH2)  is 0.314 M

pa1 = 6.848

a1 = 1.42 X [tex]10^{-7}[/tex]

Pa2 = 9.928

a2 = 1.18 X [tex]10^{-10}[/tex]

The equation is given as;

NH2CH2CH2NH2 + H2O ---> NH2CH2CH2NH3+ + OH-

Initial 0.314 0 0

Change -x +x +x

Equilibrium 0.314 -x x x

b1 = Kw / Ka2 = 10^-14 / 1.18 X 10^-10 = 8.47 X 10^-5 = [NH2CH2CH2NH3+][OH-] / [NH2CH2CH2NH2]

8.47 X 10^-5 = x^2 / 0.314 -x

8.47 X 10^-5 X 0.314 - 8.47 X 10^-5x = x^2

x^2 - 2.66 X 10^-5 + 8.47 X 10^-5x = 0

on solving

x = 0.0051 M = [NH2CH2CH2NH3+] = [OH-]

[NH2CH2CH2NH2] = 0.314 - 0.0051 = 0.3089 M

For the second dissociation,

NH2CH2CH2NH3+ + H2O ---> +NH3CH2CH2NH3+ + OH-

Initial 0.0051 0 0.0051

Change -x +x +x

Equilibrium 0.0051 -x x 0.0051 +x

b2 = Kw / Ka1 = 10^-14 / 1.42 X 10^-7 = 7.04 X 10^-8 = [+NH3CH2CH2NH3+] [OH-] / [+NH3CH2CH2NH2]

7.04 X 10^-8 = x (0.0051+x) / (0.0051-x)

We may ignore x in the denominator as b2 is very low

7.04 X 10^-8 = x (0.0051+x) / (0.0051)

3.59 X 10^-10 = 0.0051x + x^2

x = 7.039 X 10^-8 M

[OH-] = 0.0051 + x = 0.0051 approx

[+NH3CH2CH2NH3+] = x = 7.039 X 10^-8 M

NH2CH2CH2NH3+ = 0.0051 +x = 0.0051 M approx

pOH = --log [OH-] = -log 0.0051 = 2.29

pH = 14 - pOH = 11.71

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What is the ph of a 0. 10 m solution of NaCN at 25°C?(ka= 4. 9 x 10^-10for hcn at 25°C. )

Answers

The pH of a  a 0. 10 m solution of NaCN at 25°C is 11.15

Calculation

The reaction in the solution is given below

                   [tex]CN^{-} + H_{2} O[/tex] → HCN + [tex]OH^{-}[/tex]

initial                    0.1

change           ( -x)                (+x)

equilibrium     (  0.1 - x )         x

Kb = [HCN] [[tex]CN^{-}[/tex] ]/[[tex]CN^{-}[/tex] ]

Kb × Ka = Kw = 1.0 × [tex]10^{-14}[/tex]

Kb =  1.0 × [tex]10^{-14}[/tex] / 4.9  × [tex]10^{-10}[/tex] =  [HCN] [[tex]CN^{-}[/tex] ]/[[tex]CN^{-}[/tex] ] = [tex]x^{2}[/tex]/ (  0.1 - x )

Kb =  2.04× [tex]10^{-5}[/tex] = [tex]x^{2}[/tex]/ (  0.1 - x )

Since , [NaCN] /Kb > 100 , we can simplify the above equation to

=  2.04× [tex]10^{-5}[/tex] = [tex]x^{2}[/tex]/ (  0.1 )

x = 1.43 × [tex]10^{-3}[/tex] M = [HCN] = [ [tex]OH^{-}[/tex]]

Then pOH = 2.84

pH + pOH = 14

pH = 14- pOH = 14 - 2.84 = 11.15

Therefore , the pH is 11.15

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What is the theoretical yield (mass of Al2O3 formed) for this reaction?
(answer is not 204)

Answers

Answer:

102 g Al₂O₃

Explanation:

The theoretical yield is the mass of product calculated via the molar masses and balanced chemical equation.

The limiting reagent is the reactant that is completely reacted before the other reactant(s) are used up. Since Al produces the smaller amount of product, it appears that Al is the limiting reactant. You can only make as much product as the limiting reactant allows. As such, the theoretical yield is 102 grams Al₂O₃.

help me with this question please like right now

Answers

Using the gas laws, the volume of nitrogen produced is 6.6 L.

What is the product of a reaction?

We know that the product of a reaction is obtained from the balanced reaction equation. The reactants combine together to yield the products hence we have; N2 + 3H2 ----> 2NH3

We know that 1 mole of hydrogen occupies 22.4 L

x moles of hydrogen occupies 25500 L

x = 1 mole * 25500 L/ 22.4 L

x = 1138.4 moles

If 3 mole of hydrogen reacts with 1 mole of nitrogen

1138.4 moles  reacts with 1138.4 moles * 1 mole/ 3 moles = 379.5 moles

Mass of nitrogen = 379.5 moles * 28 g/mol = 10626 g

Now 3 moles of hydrogen produces 44.8 L of NH3

1138.4 moles  produces 1138.4 moles * 44.8 L/ 3 moles = 17000 L

Now;

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2 = 1 * 17000 * 848/5.5 * 273

V2 =9601 L

2) 2NH4NNO3(s) ---> 4H2O(g) + 2N2(g) + O2(g)

Number of moles = 12g/ 80 g/mol = 0.15 moles

Now;

2 moles of NH4NNO3 produced 2 moles of nitrogen hence 0.15 moles of N2 was produced.

1 mole of N2 occupies 22.4 L

0.15 moles of N2 occupies 0.15 moles  *  22.4 L/1 mole = 3.36 L

P1 = 760 torr

V1 = 3.36 L

T1 = 273 K

P2 = 745 torr

T2 = 527 degrees or 800 K

V2 = ?

V2 =  P1V1T2/P2T1

V2 =  760 *  3.36 * 527/745 * 273

V2 = 6.6 L

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Which best explains why the trend in noble gas boiling points increases down the group?.

Answers

The atoms become larger, the intermolecular forces between the atoms become stronger and more energy is needed to overcome these forces. Hence, the boiling point of noble gases increases down the group.

What are noble gases?

Group 18 of the modern periodic table consists of noble gases. Helium, neon, argon, krypton, xenon, and radon are noble gases. These gases are monoatomic and chemically inert under normal conditions and because of this they are also called inert gases. These gases are present in very small quantities in the atmosphere, and so they are also called rare gases.

Trends in Physical Properties of Noble Gases:

Atomic Radii (Atomic size): As we move down the group from helium to radon the atomic size of Noble Gasses keeps on increasing. This is because when going down the group, the number of occupied shells with valence electrons increases.Boiling and Melting Points: As we move down the group, the boiling and melting point of the Noble Gasses increases due to the following reason:When moving down in the group, the atomic radius increases which results in the formation of a strong Van Der Waals force of attraction between the atoms.As to overcome the interatomic force of attraction which becomes stronger, it also requires more energy during melting and boiling.Density: All elements of group 18 have low densities. When going down the group, density increases as the atomic mass keeps on increasing.Ionization Energy: In the periodic table, As we move down the group the first ionization energy keeps on decreasing.Ionization Potential: When going down the group, the atomic radii increases which increases the attractive force and ultimately results in the increase of polarity and decrease in the ionization potential. This is because the larger atom of the group in the valence electrons are held together less tightly by the atom as they are situated far away from the nucleus.

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number of oxygen atoms in 16 g of So2??

Answers

Now , we know that one mole of SO² Contains 2 Moles ( two ) of atoms of oxygen . Hence , 16 Grams of SO² Contains 0.5 Mole(s) of Oxygen

0.5

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Consider the reaction below.

H O single bonded to C double bonded above to O and to the right to the bottom left of a benzene ring. The ring is bonded from the upper right to a C double bonded above to O and to the right to O H. H O bonded to C H Subscript 2 Baseline C H Subscript 2 Baseline bonded to O H. Arrow pointing with bond H Subscript 2 Baseline O beside it pointing down to H O bonded to C double bonded above to O and to the right to the bottom left of a benzene ring. The ring is bonded from the upper right to a C double bonded above to O and to the right to O bonded to C H Subscript 2 Baseline C H Subscript 2 Baseline bonded to O H.
This reaction eventually forms this product.

A bond from left to O single bonded to C double bonded above to O and to the right to the bottom left of a benzene ring. The ring is bonded from the upper right to a C double bonded above to O and to the right to O bonded to C H Subscript 2 Baseline C H Subscript 2 Baseline bonded to O bonded to C double bonded above to O and right to the upper left corner of a benzene ring. The ring is bonded from the lower left to C double bonded above to O and single bonded right to O bonded to CH Subscript 2 Baseline C H Subscript 2 Baseline bonded right to O bonded to C double bonded above to O and right to the lower left of a benzene ring, which is bonded from upper right to C double bonded above to O and single bonded right to O bonded to C H Subscript 2 Baseline C H Subscript 2 Baseline bonded to O.
Which type of reaction is represented by these diagrams?

elimination
substitution
addition polymerization
condensation polymerization

Answers

The type of reaction that is represented by these diagrams (attached) are given as:  "condensation polymerization" (Option D)

What is condensation polymerization?

Any type of polymer in polymer science that undergoes a condensation reaction during the polymerization process is referred to as a condensation polymer (i.e. a little molecules, methanol or water is derived as a metabolite).

The provided reaction is an example of condensation polymerization because it involves the combination of two monomers, which produces a big polymer and water as a byproduct.

Condensation polymers make significant contributions to the packaging, insulation, and textile sectors as far as real-world application of same is concerned.

Also examples of times that are condensation polymers and that have silicon rather than carbon as part of their molecular or structural make up are;

Silicone Oils and Rubbers.

In conclusion, it is to be noted that The type of reaction that is represented by these diagrams (attached) are given as:  "condensation polymerization"

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Answer:

condensation polymerization

Explanation:

Calculate the ph after 10. 0 ml of 0. 40 m naoh is added to 20. 0 ml of 0. 50 m hc.

Answers

The pH is 13.31.

The smaller the pH value the more acidic, the larger the pH value the more basic.

The range is 0 to 14, with 7 representing neutrality. Acidity is indicated by pH values below 7, whereas baseness is shown by pH values above 7.

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per liters of a solution.

Molarity of NaOH = 0.40 M

Volume of NaOH = 10.0 mL

Molarity of HCL = 0.50 M

Volume of HCL = 20.0 ml

[tex][OH^-] = (M_2V_2 -M_1V_1)/(V_1+V_2)\\[/tex]

         [tex]= (0.50*20.0 - 0.40*10.0) / (20+10)[/tex]

         [tex]= (10 -4)/ 30\\[/tex]

         [tex]= 6/30[/tex]

         [tex]=0.2[/tex]

[tex]pOH = - log (OH ^-)[/tex]

[tex]pOH = - log (0.2)\\[/tex]

[tex]pOH = - ( - 0.698)[/tex]

[tex]pOH = 0.69[/tex]

[tex]pH + pOH = 14[/tex]

[tex]pH = 14 -0.69[/tex]

[tex]pH = 13.31[/tex]

Therefore, the pH after 10.0 mL of 0.40 M NaOH is 13.31.

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Calculate the ph of a solution formed by mixing 250. 0 ml of 0. 15 m nh4cl with 200. 0 ml of 0. 12 m nh3. the kb for nh3 is 1. 8 × 10-5.
i. 9. 45
ii. 4. 74
iii. 9. 06
iv. 04. 55
v. 9. 26

Answers

The potential of the hydrogen in water gives the pH of the acidity and basicity of the solution. The pH of the solution is 9.06. Thus, option iii is correct.

What is pH?

The pH of a substance is given by the subtraction of the pOH from 14 which is the range of the pH scale.

The dissociation reaction for ammonia is given as,

NH₃ (aq) + H₂O (l)  ⇄  NH₄⁺ (aq) + OH⁻ (aq)

Here, the concentration of ammonia is [NH₃] - x, ammonium ion is [NH₄⁺] + x, and hydroxide ion is x.

The molar concentration of ammonia is,

M = (0.12 M × 0.2 L) ÷ 0.45 L = 0.053 M

The molar concentration of ammonium ion is,

M = (0.15 M × 0.25 L) ÷ 0.45 L = 0.083 M

From the base dissociation constant and previous concentration from the reaction, the value of x or hydroxide ion is calculated as,

Kb = [NH₄⁺][OH⁻] ÷ [NH₃]

1.8 × 10⁻⁵ (0.053 - x) - (0.083 + x) × x = 0

x = [OH⁻] = 1.15 x 10⁻⁵

The pH from the hydroxide ion is calculated as,

pOH = - log [OH⁻]

= - log (1.15 x 10⁻⁵)

= 4.94

Further,

pH = 14 - pOH

= 14 - 4.94

= 9.06

Therefore, option iii. 9.06 is the pH of the solution.

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What mass of silver can be plated onto an object in 33. 5 minutes at 8. 70 a of current? ag (aq) e- → ag(s) 19. 6 g 9. 78 g 0. 326 g 3. 07 g 0. 102 g

Answers

Mass of silver can be plated onto an object in 33. 5 minutes at 8. 70 a of current is 19.55 g

Given, I=8.70A, t=33.5min=33.5*60=2010 sec, m of Ag = 19600g M=107.87 g/mol

From  faradays law we can write q=I*t= 8.70*2010=17487 C.

moles of Ag =17487/ 19600=0.1812 moles

Ag deposited = 0.1812*107.87 = 19.55 g

Silver

Having the atomic number 47 and the Latin word argentum, which means "bright" or "white," silver is a chemical element. Its symbol is Ag. The highest electrical, thermal, and reflectivity of any metal are displayed by this transition metal, which is soft, white, and lustrous. The metal can be found in the Earth's crust in three different forms: as an alloy with other metals like gold and in minerals like argentite and chlorargyrite. It can also be found in its pure, free elemental form, also known as "native silver." During the refining of copper, gold, lead, and zinc, the majority of silver is created as a byproduct. As a valuable metal, silver has long been treasured.

What mass of silver can be plated onto an object in 33. 5 minutes at 8. 70 a of current? ag (aq) e- → ag(s) 19. 6 g 9. 78 g 0. 326 g 3. 07 g 0. 102 g

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Find the percent ionization of a 0. 337 m hf solution. The ka for hf is 3. 5 × 10^-4.

Answers

Percent ionization = 3.17%

We utilize the supplied acid equilibrium constant (Ka) to calculate the percentage of the acid that is ionized. It is the ratio of the acid and dissociated ion equilibrium concentrations. The HF acid would dissociate in the manner described below:

HF = H+ + F-

The following is how the acid equilibrium constant might be written:

Ka = [H+][F-] / [HF] = 3.5 x 10-4

We utilize the ICE table to determine the equilibrium concentrations.

        HF             H+              F-

I      0.337           0                 0

C      -x              +x               +x

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

E    0.337-x        x                   x

3.5 x 10-4 = [H+][F-] / [HF]

3.5 x 10-4 = [x][x] / [0.337-x]

Solving for x,

x = 0.01069 = [H+] = [F-]

percent ionization = 0.01069 / 0.337 x 100  = 3.17%

percent ionization = 3.17%

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The molar volume of copper (63.5 g.mol^–1 ) at 25°C is 7.09 cm^3.mol^–1. What is the density of copper at 25°C in g.cm^3?

Answers

Considering the definition of molar volume and density,the density of copper at 25°C is 8.96 [tex]\frac{g}{cm^{3} }[/tex].

Molar volume

Molar volume is an intensive property that indicates how much space one mole of a certain substance or compound occupies. In other words, the molar volume is the space occupied by one mole of a substance.

The expression to calculate the molar volume is:

Vm= V÷ n

where:

V is the occupied volume.n is the molar amount of the substance.

Density

Density is defined as the property that matter, whether solid, liquid or gas, has to compress into a given space.

In other words, density is a quantity that allows us to measure the amount of mass in a certain volume of a substance.

Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

density= mass÷ volume

Relationship between molar volume and density

The relationship between molar volume and density is reflected in the following expression, where the molar volume is calculated by dividing the molar mass (M) by the mass density (ρ):

Vm= M÷ρ

Density of copper

In this case, you know:

Vm= 7.09 [tex]\frac{cm^{3} }{mol}[/tex]M= 63.5 [tex]\frac{g}{mol}[/tex]ρ= ?

Replacing in the relationship between molar volume and density:

7.09 [tex]\frac{cm^{3} }{mol}[/tex]= 63.5 [tex]\frac{g}{mol}[/tex]÷ρ

Solving:

7.09 [tex]\frac{cm^{3} }{mol}[/tex]×ρ= 63.5 [tex]\frac{g}{mol}[/tex]

ρ= 63.5 [tex]\frac{g}{mol}[/tex]÷7.09 [tex]\frac{cm^{3} }{mol}[/tex]

ρ= 8.96 [tex]\frac{g}{cm^{3} }[/tex]

Finally, the density of copper at 25°C is 8.96 [tex]\frac{g}{cm^{3} }[/tex].

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if there are 12 moles of hydrogen how many moles of nitrogen are needed to react completely with the hydrogen? (N2(g)+3H2(g)->2NH3(g))

Answers

Answer:

Explanation:

Take a look at the balanced chemical equation for this synthesis reaction

N

2(g]

+

3

H

2(g]

2

NH

3(g]

Notice that you have a

1

:

3

mole ratio between nitrogen gas and hydrogen gas. This means that, regardless of how many moles of nitrogen gas you have, the reaction will always consume twice as many moles of hydrogen gas.

So, if you have

2

moles of nitrogen taking part in the reaction, you will need

2

moles N

2

3

moles H

2

1

mole N

2

=

6 moles H

2

0. 34 ll of hno3hno3 is titrated to equivalence using 0. 14 ll of 0. 1 mnaohmnaoh. what is the concentration of the hno3hno3 ?

Answers

When 0.34 of HNO₃ is titrated to equivalence using 0.14 l of 0.1 m NaOH then the concentration of HNO₃ is 0.041 M

The reaction of neutralization of HNO₃ with NaOH is

HNO₃ + NaOH → H₂O + NaNo₃

When 1 mole of HNO₃ react with 1 mole of NaOH, based on chemical rection the moles of NaOH at equivalence point are equal to moles of HNO₃ present in solution: -

With the mole and volumes, we can find molarity as follows:

            Moles of NaOH = moles HNO₃

            ⁼ 0.14 L X (0.1 mol NaOH/L) = 0.014 mole NaOH

            =0.014 mol HNO₃

Molarity: -

              [tex]\frac{(Mole of HNO₃)}{(volume of HNO₃)}= \frac{0.014}{0.34}[/tex]

               = 0.041 M

Thus, from above solution we concluded that the concentration of HNO₃ solution is 0.041 M.

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Given the following reaction: 2O3---->3O2. DeltaH of reaction=-286kJ/mol
What is the Delta H reaction of the reaction 3O2----->2O3

Answers

The enthalpy of reaction (ΔH) of the above chemical reaction is equal to -143 kJ/mol.

What is a chemical reaction?

A chemical reaction can be defined as a chemical process that involves the continuous transformation (rearrangement) of the ionic, atomic or molecular structure of a chemical element by breaking down and forming chemical bonds, in order to produce a new chemical compound while new bonds are formed.

What is a balanced equation?

A balanced chemical equation can be defined as a chemical equation wherein the number of atoms on the reactant (left) side is equal to the number of atoms on the product (right) side.

Next, we would write the properly balanced chemical equation for this chemical reaction:

                                    3/2O₂ -----> 2/2O₃

Therefore, the enthalpy of reaction (ΔH) of the above chemical reaction is half the enthalpy of reaction (ΔH) of the initial chemical reaction:

Enthalpy of reaction (ΔH) = ½ × -286 kJ/mol

Enthalpy of reaction (ΔH) = -143 kJ/mol.

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A piece of plastic has a net charge of 2. 00 μc. How many more protons than electrons does this piece of plastic have?

Answers

The piece of plastic has   1.256× 10¹³  protons  more than  that of electrons.

What are protons and electrons?

A proton is a subatomic particle with a defined mass of 1 and a charge of +1 (positive charge). The symbol for the proton is either p or p+.

An electron is a subatomic particle that has a negative electric charge and that actively orbits the nucleus of the atom .

To calculate ,

given ,

magnitude of charge , q=+2μC=+2×10⁻⁶

Since, protons carry positive charge. Then, we know that standard number of proton in 1 Coulomb of charge is,

1C=6.28×10¹⁸ protons

Number of protons in 1 microcoulomb charge is,

=6.28×10¹⁸×10⁻⁶

=6.28×10¹²

Then for the net charge of q, the number of protons is given as,

np= q×6.28×10¹²

np= 2×6.28×10¹²

np=1.256×10¹³ protons

Thus, the piece of plastic has 1.256×10¹³   more protons than electrons.

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What is the kinetic product obtained from the addition of 1 mole of bromine to 1,3-butadiene?

Answers

From the addition of 1 equivalent of HBr to 1,3-butadiene, 3-bromo-1-butene is obtained as a product.

What is moles?

Moles are defined as the smallest base unit that is equivalent to [tex]6.02214 X 10^{23}[/tex] units.

Moles are chemical entities, which are used to measure the mass or molecular mass of the compound, molecule, or atom. The moles are, for example can be explained as equal to the mass of the atom. For example, there is one mole in 12 grams of carboncarbon.

Thus the kinetic product formed is 3-bromo-1-butene.

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How many milliliters of a 9. 0 mh2so4 solution are needed to make 0. 45 l of a 3. 5 m solution?

Answers

There are 1157.14 milliliters of a 9. 0 M [tex]H_{2} SO_4}[/tex] solution are needed to make 0. 45 L of a 3. 5 M solution.

Calculation,

The formula for the dilution of a solution is given as:

[tex]M_{1} V_{1}[/tex] = [tex]M_{2} V_{2}[/tex]

Where M is molarity and V is the volume of the solution in liters ( L ).

Given data,

[tex]M_{1}[/tex] =  9. 0 M

[tex]V_{1}[/tex] = 0. 45 L

[tex]M_{2}[/tex] = 3. 5 M

[tex]V_{2}[/tex] = ?

9. 0 M ×0. 45 =  3. 5 M × [tex]V_{2}[/tex]

[tex]V_{2}[/tex] in Lit = 9. 0 M ×0. 45/ 3. 5 M = 1.15 L

[tex]V_{2}[/tex] in the ml = 1.15 × 1000ml = 1157.14 ml

Volume in milliliters is 1157.14 .

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How many peaks are in the proton spin decoupled 13c nmr spectrum of 1,3,5-trinitrobenzene?

Answers

There are 2  peaks are in the proton spin decoupled 13C NMR spectrum of 1,3,5-trinitrobenzene .

The 13C NMR spectrum give the peak which is directly about the carbon skeleton not just the proton attached to it  . The number of signals tell us how many different carbons or set of equivalent carbons . The splitting of a signal tells us how many hydrogens  are attached to each carbon.

In 1,3,5-trinitrobenzene molecule , there are 2  peaks are in the proton spin decoupled 13C NMR spectrum three carbon give one signal and another three carbon give another one signal .

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Determine the name for aqueous H2CO3. carbonous acid dihydrogen carbonate carbonic acid hydrocarbonic acid hydrocarbide acid

Answers

The name for aqueous [tex]H_{2} CO_{3}[/tex] is carbonic acid.

So, option C is correct one.

Binary hydrogen compounds with non metals may form H+ (proton) and an anion dissolved in water. The acidic solutions are named as if they were molecular acids , using the usual name for the compound itself , replacing hydrogen with hydro- and suffix -ide with ic. The word acid is then use.

The inorganic acid is an acid drive from one or more inorganic compounds. All inorganic acids form hydrogen ions and the conjugate base ions when dissolved in water.

Example: carbonic acid ( [tex]H_{2} CO_{3}[/tex] )

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A metal object is to be gold-plated by an electrolytic procedure using aqueous AuCl3 electrolyte. How much gold may be deposited in 3.0 min by a constant current of 10. A

Answers

Answer:

This is the answer

Explanation:

charges passed = current x time = 10 x 3 x 60

= 1800 C

mole of charge = 1800 / 96500

= .01865 moles

Au+3 contains 3 positive charges

3 mole of charge will deposit 1 mole of Au .01865 moles will deposit .01865 / 3 mole

= 6.2167 x 10-3 moles.

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