If you are given the molarity of a solution, what additional information would you need to find the weight/weight percent (w/w%)?

Answer:

21.2 mL

Explanation:

Step 1: Write the balanced equation.

NaOH + HBr ⇒ NaBr + H₂O

Step 2: Calculate the reacting moles of HBr

25.0 mL of 0.0720 M hydrobromic acid react.

[tex]0.0250 L \times \frac{0.0720mol}{L} = 1.80 \times 10^{-3} mol[/tex]

Step 3: Calculate the reacting moles of NaOH

The molar ratio of NaOH to HBr is 1:1. The reacting moles of NaOH are 1/1 × 1.80 × 10⁻³ mol = 1.80 × 10⁻³ mol.

Step 4: Calculate the required volume of NaOH

[tex]1.80 \times 10^{-3} mol \times\frac{1,000mL}{0.0850mol} = 21.2 mL[/tex]

Answer:

A) linear / linear

Explanation:

In this case, we have a triple bond beetween the atoms (See figure 1). If we have this triple bond we will have an Sp hybridization (in both carbons). We have to remember the relationship between the geometry and the hybridization:

-) Sp3 = Tetrahedral

-) Sp2 = Trigonal

-) Sp = Linear

Due to the hybridization, we will have a linear structure between the atoms. The angle between the atoms is 180º (See figure 2).

So, if we have a hybridization Sp for both carbons, we will have a linear geometry in each carbon. Therefore, the answer is A.

Radiologists are medical doctors that treat injuries using medical imaging (radiology)

Answer:

a person who uses X-rays or other high-energy radiation, especially a doctor specializing in radiology.

Explanation:

Answer:

The Empirical Formular is given as; Ti₆Al₄V

Explanation:

The percent composition of the material is 64.39% titanium, 24.19% aluminum, and 11.42% vanadium.

Elements                        Titanium            Aluminium        Vanadium

Percentage                    64.39                 24.19                   11.42

Divide all through by their molar mass

                                     64.39 / 47.87      24.19 / 27               11.42 / 50.94

                                       =  1.345                = 0.896                 = 0.224

Divide all though  by the smallest number (0.224)

                                     1.345 / 0.224        0.896 / 0.224       0.224 / 0.224

                                     = 6                         = 4                             = 1

The Empirical Formular is given as; Ti₆Al₄V

Using the stepwise procedure for obtaining the empirical formula of a compound, the empirical formula is [tex] T_{6}Al_{4}V[/tex]

Titanium :

Divide by Molar mass : = 64.39/47.87 = 1.345

Aluminum :

Divide by Molar mass : = 24.19/27 = 0.896

Vanadium :

Divide by Molar mass : = 11.42/50.94 = 0.224

Divide by the smallest :

Titanium = 1.345 / 0.224 = 6.00

Aluminum = 0.896 / 0.224 = 4

Vanadium = 0.224 / 0.224 = 1

Hence, the empirical formula is [tex] T_{6}Al_{4}V[/tex]

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Answer:

35%

Explanation:

Percentage yield = actual yield / theoretical yield × 100.

Given:

Actual yield = 63.07g

Theoretical yield = ?

Mole ratio of C7H6O3 to C4H6O3 = 1 : 1

1 mole of C7H6O3 - 138.1g

Which implies that only 1 mole s[tex]\frac{63.07}{180.2} * 100[/tex]hould be used up in the reaction, yielding 180.2 g of C9H8O4. ⇒ Theoretical yield = 180.2g

∴ % Yield = [tex]\frac{63.07}{180.2} * 100[/tex]

= 35% yield.

Let me know if you found this easy to understand.

Answer:

592 K or 319° C

Explanation:

From the statement of Charles law we know that the volume of a given mass of gas is directly proportional to its absolute temperature at constant pressure. Thus;

V1/T1= V2/T2

Initial volume V1 = 1.75 L

Initial temperature T1= 23.0 +273 = 296 K

Final volume V2= 3.50 L

Final temperature T2 = the unknown

T2= V2T1/V1= 3.50 × 296 / 1.75

T2 = 592 K or 319° C

Answer:

The correct answers are:

c. The concentrations of the products decrease

d. The equilibrium constant decreases

Explanation:

Changes in temperature shift the equilibrium. In this problem, the reaction is endothermic, so it absorbs heat so heat is considered as a reactant:

Reactants + heat ⇒ Products

If the temperature is decreased, the heat is decreased, so reactants are removed from the reaction at equilibrium. According to Le Chaterlier's principle, the system will try to compensate the produced change. If reactants are removed, the systems will form reactants and the equilibrium will shift toward the left (formation of more reactants). In consequence, the amount of products will be decreased.

Thus, acorrect option is: c. The concentrations of the products decrease.

Since the equilibrium constant is given by the ratio of concentration of products over concentration of reactants, if the concentration of products decrease, the equilibrium constant also decreases. So, another correct option is: d. The equilibrium constant decreases.

Answer:

2284.2 g.

Explanation:

The following data were obtained from the question:

Percentage yield = 47%

Theoretical yield = 4860 g

Actual yield =?

The percentage yield is simply defined as the ratio of actual yield to the theoretical yield multiplied by 100. Mathematically, it is expressed as:

Percentage yield = Actual yield /Theoretical yield x 100

With the above formula, we can obtain the actual yield as follow:

Percentage yield = Actual yield /Theoretical yield x 100

47% = Actual yield /4860

Cross multiply

Actual yield = 47% × 4860

Actual yield = 47/100 x 4860

Actual yield = 2284.2 g

Therefore, the actual yield is 2284.2 g.

Answer:

16.4 grams of calcium nitrate can be prepared by the reaction of 18.9 grams of nitric acid with 7.4 grams of calcium hydroxide

Explanation:

The balanced reaction is:

Ca(OH)₂ + 2 HNO₃ → Ca(NO₃)₂ + 2 H₂O

First, you must determine the limiting reagent. The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction):

Being:

Then, the molar mass of the compounds participating in the reaction is:

Then, by stoichiometry of the reaction the following amounts of reagents and products participate:

Then apply the following rule of three: if 126 grams of nitric acid reacts with 74 grams of calcium hydroxide, 18.9 grams of nitric acid with how much mass of calcium hydroxide does it react?

[tex]mass of calcium hydroxide=\frac{18.9 grams of nitric acid*74 grams of calcium hydroxide}{126 grams of nitric acid}[/tex]

mass of calcium hydroxide= 11.1 grams

But 11.1 grams of calcium hydroxide are not available, 7.4 grams are available. Since you have less mass than you need to react with 18.9 grams of nitric acid, calcium hydroxide will be the limiting reagent.

Then, it is possible to determine the amount of mass of calcium nitrate produced by another rule of three:  if 164 grams of calcium nitrate are formed by stoichiometry from 74 grams of calcium hydroxide, how much mass of calcium nitrate will form from 7.4 grams of calcium hydroxide?

[tex]mass of calcium nitrate=\frac{7.4 grams of calcium hydroxide*164 grams of calcium nitrate}{74 grams of calcium hydroxide}[/tex]

mass of calcium nitrate= 16.4 grams

16.4 grams of calcium nitrate can be prepared by the reaction of 18.9 grams of nitric acid with 7.4 grams of calcium hydroxide

Answer:

∆H=  438 KJ/mol

Explanation:

First, we have to find the energy bond values for each compound:

-) Cl-Cl = 243 KJ/mol

-) F-F = 159 KJ/mol

-) F-Cl = 193 KJ/mol

If we check the reaction we can calculate the number of bonds:

[tex]Cl_2_(_g_)~+~3F_2_(_g_)~->~2ClF_3_(_g_)[/tex]

In total we will have:

-) Cl-Cl = 1

-) F-F = 3

-) F-Cl = 6

With this in mind. we can calculate the total energy for each bond:

-) Cl-Cl = (1*243 KJ/mol) = 243 KJ/mol

-) F-F = (3*159 KJ/mol) = 477 KJ/mol

-) F-Cl = (6*193 KJ/mol) = 1158 KJ/mol

Now, we can calculate the total energy of the products and the reagents:

Reagents = 243 KJ/mol + 477 KJ/mol = 720 KJ/mol

Products = 1158 KJ/mol

Finally, to calculate the total enthalpy change we have to do a subtraction between products and reagents:

∆H= 1158 KJ/mol-720 KJ/mol = 438 KJ/mol

I hope it helps!

The approximate enthalpy change is:

∆H=  438 KJ/mol

First, we have to find the energy bond values for each compound:

Cl-Cl = 243 KJ/mol

F-F = 159 KJ/mol

F-Cl = 193 KJ/mol

Balanced chemical reaction:

Cl₂ (g) + 3F₂ (g) ⟶ 2ClF₃ (g)

Total number of bond for each:

Cl-Cl = 1

F-F = 3

F-Cl = 6

Total bond energy will be:

Cl-Cl = (1*243 KJ/mol) = 243 KJ/mol

F-F = (3*159 KJ/mol) = 477 KJ/mol

F-Cl = (6*193 KJ/mol) = 1158 KJ/mol

Now, we can calculate the total energy of the products and the reactants:

Reactants = 243 KJ/mol + 477 KJ/mol = 720 KJ/mol

Products = 1158 KJ/mol

Finally, to calculate the total enthalpy change we have to do a subtraction between products and reagents:

∆H= 1158 KJ/mol-720 KJ/mol = 438 KJ/mol

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Answer:

b) rate = kAB.

Explanation:

Hello,

In this case, considering the given statement, we can notice that the rate law of the overall reaction will be determined for the slowest step, that is:

[tex]A+B \rightarrow AB\ \ (slow)[/tex]

In such a way, we can infer that the rate law will contain both the concentration of A and B to the first power both, since their stoichiometric coefficients in the chemical equation are both one:

[tex]rate=k[A][B][/tex]

Thereby, answer is b) rate = kAB, that should be better rate = k[A][B] by expressing the concentrations.

Best regards.

Answer:

A

Explanation:

If we intend to achieve the anti addition of Hex-3-yne to yield (E) Hex-3-ene, the we must use Na/NH3. The first step of the reaction involves the transfer of an electron from sodium to the alkene; this yields a radical anion. Strong electron replusion ensues between the single electron and the lone pair on the carbon. This now forces the both to be found at a trans position to each other and this is the basis of the stereochemistry of the product.

Secondly, the radical anion abstracts a proton from ammonia. Another sodium atom transfers an electron leading to the formation of a vinyl carbanion, the alkyl groups are now trans to each other.

This carbanion now abstracts a proton from ammonia and the final product is formed.

Answer:

V2 = 4.48L

Explanation:

using charles law

V1/T1=V2/T2

3.4/283=V2/373

0.012=V2/373

V2= 0.012 x 373

V2 = 4.48L

Answer:

C. To determine how efficient reactions are.

D. To determine how much reactant they need.

Explanation:

When you are doing a reaction, you are hoping for a percent yield to close of 100%. You make the reaction and determine how many product you obtain. If you know the percent yield of a reaction you can calculate the amount of reactant you need to obtain a determined amount of product.

Having this in mind:

A. To balance the reaction equation.  false. To calculate percent yield you need to balance the reaction before. You don't use percent yield to balance the reaction

B. To determine how much product they will need.  false. You determine how much product you obtain after the reaction. How much product you need is independent of percent yield

C. To determine how efficient reactions are.  true. A way to determine efficience of a reaction is with percent yield. An efficient reaction has a high percent yield.

D. To determine how much reactant they need. true. If you know percent yield of a reaction you can know how many reactant you must add to obtain  the amount of product you want.

Answer:

[tex]\boxed{\sf Option \ E}[/tex]

Explanation:

All the gases at the same temperature and mass have the same average kinetic energy.

If the masses were different, then the different gases will have different velocities. If the temperature was higher then there would be a greater motion, if the temperature was lower, then there would be less motion.

Answer:

option E

Explanation:

Answer:

D

Explanation:

Double Displacement reaction

Both sides are balanced with option D

The balanced form of the chemical equation shown below is [tex]\rm Ca(OH)_2(aq) + Na_2CO_3(aq) \rightarrow CaCO_3(s) + 2NaOH(aq).[/tex] The correct option is D.

A balanced equation is where the reactant and the product have the number of moles of elements. According to the law, the reaction, and the product have the same number of moles after the reaction, so balancing an equation is important.

To balance an equation, it is significant to see the number of moles of reactant and the same number of moles is in the product side. Here the moles of sodium has to be balanced.

Thus, the correct option is D, [tex]\rm Ca(OH)_2(aq) + Na_2CO_3(aq) \rightarrow CaCO_3(s) + 2NaOH(aq).[/tex]

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Answer:

As little as 2 days

Hope this is correct

HAVE A GOOD DAY!

Answer:

Most stable: 2,3-dimethylpent-2-ene > (E)-3,4-dimethylpent-2-ene > (Z)-3,4-dimethylpent-2-ene > 2-methyl-3-methylenepentane : Least stable

Explanation:

Treatment of NaOH with (R)-3-bromo-2,3-dimethylpentane results in the elimination of HBr. Each H atoms present on each [tex]\beta[/tex]-carbon atoms can be eliminated result in the formation of four possible products: (1) 2,3-dimethylpent-2-ene, (2) (E)-3,4-dimethylpent-2-ene, (3) (Z)-3,4-dimethylpent-2-ene and (4) 2-methyl-3-methylenepentane.

The stability of these alkenes depends on the number of hyperconjugative H atoms present with respect to the double bond. In accordance with this, 2,3-dimethylpent-2-ene is the most stable alkene (11-hyperconjugative H atoms). Then, 3,4-dimethylpent-2-ene is the second most stable alkene (7-hyperconjugative H atoms). Among (E)-3,4-dimethylpent-2-ene and (Z)-3,4-dimethylpent-2-ene, (E)-3,4-dimethylpent-2-ene is more stable due to it's less sterically hindered structure. 2-methyl-3-methylenepentane is the least stable alkene (3-hyperconjugative H atoms).

So, decreasing order of stability of alkenes from most stable to least stable:

2,3-dimethylpent-2-ene > (E)-3,4-dimethylpent-2-ene > (Z)-3,4-dimethylpent-2-ene > 2-methyl-3-methylenepentane

Three bromo As just a result of the nucleophilic substitution mechanism, 2,4 dimethylpentane generates a racemic mix containing both the r and s forms of the molecule.

Please find the attached file.

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Answer:

For carbon, the effective nuclear charge is 3.25 and the valence electrons will reside in the orbitals 2s^2 and 2p^2

For silicon, the effective nuclear charge is 4.15 and its valance electrons will reside in the orbitals 3s^2 and 3p^2

Explanation:

Carbon

The effective nuclear charge of carbon is 3.25

To get the orbitals in which it’s valence electron reside, let’s write the electronic configuration

The atomic number of carbon is 6

So the configuration will be;

1s^2 2s^2 2p^2

So the valence electrons will reside in the orbitals 2s^2 and 2p^2

For silicon;

It’s effective nuclear charge is +4.15

The electronic configuration of silicon with atomic number 14 is;

1s^2 2s^2 2p^6 3s^2 3p^2

So the valence electrons will reside in the orbitals 3s^2 and 3p^2

Answer:

question is not clear please send clear question

Answer:

An addiction could occur, maybe an overdose?, this could lead to death and maybe you would do unreasonable things which could get you fined or arrested.

Explanation:

Answer:

Hope this helped,

Kavitha

Answer:

r = k [X]²

Explanation:

X + Y → Z

Generally, the rate of reaction depends on the concentration of reactants. However, the question stated that the rate depends only on reactant X.

The plot of 1/X versus time giving a straight line signifies that this is a second order reaction.

For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of k.

From this, our rate law is r = k [X]²

Answer:

Explanation:

The given formula is empirical formula

Let the molecular formula of first be

[tex]( CCl )_n[/tex]

molecular weight = n x ( 12 + 35.5 )

= 47.5 n

Given molecular weight = 189.83 so

47.5 n = 189.83

n = 3.99 or 4 approx

Molecular formula =

[tex]( CCl )_4[/tex]

= C₄ Cl₄

Let the molecular formula of second compound  be

[tex]( C_3H_2N)_n[/tex]

molecular weight = n x ( 3 x 12 +2+14 )

= 52 n

Given molecular weight = 156.23  so

52 n = 156.23

n = 3.0044 or 3 approx

Molecular formula =

[tex]( C_3H_2N )_3[/tex]

= C₉H₆ N₃

Answer:

The balanced equation is: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)

Explanation:

Zn(s) is a simple substance (its oxidation number is zero) and it is oxidized to Zn²⁺. It loses two electrons, so the half reaction is the following:

Zn(s) → Zn²⁺(aq) + 2 e-   (oxidation reaction)

Hydrogen ion (H⁺) is reduced to hydrogen gas (H₂). The oxidation number is decreased from +1 to 0 (because H₂ is a simple substance). H⁺ gains 1 electron per H atom, so the half reaction is the following:

2H⁺(aq) + 2 e- → H₂(g) (reduction reaction)

We obtain the overall reaction from the addition of the two half reactions. We write the reduction reaction first and then the oxidation reaction, as follows:

2H⁺(aq) + 2 e- → H₂(g)

+

Zn(s) → Zn²⁺(aq) + 2 e-

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

Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)

The two electrons at both sides of the equation (2 e-) are canceled. The overall reaction is in acidic solution due to the presence of H⁺ ions. The net charge at both sides is the same : +2, so the mass and the charge are balanced.

Answer:

Vapor pressure of water = 23.14torr

Explanation:

When you made a solution, vapor pressure decreases following Raoult's law:

[tex]P_{solution} = X_{solvent} P_{solvent}[/tex]

Where P is vapor pressure and X mole fraction

As vapor pressure of water is 23.77torr we must find the mole fraction of water knowing the solution is 1.500m glucose (That is 1.500 moles of glucose per kg of water = 1000g of water).

1000g of H₂O are, in moles (Molar mass: 18.02g/mol):

1000g H₂O ₓ (1mole / 18.02g) = 55.5 moles of H₂O.

As we know now the solution contains 55.5 moles of water and 1.5 moles of glucose. Thus, mole fraction of water (Solvent) is:

[tex]X_{H_2O} = \frac{55.5molesH_2O}{55.5molesH_2O + 1.5 molesGlucose} = 0.9737[/tex]

Replacing in Raoult's law, pressure of water above the solution is:

[tex]P_{solution} = X_{solvent} P_{solvent}[/tex]

[tex]P_{solution} = 0.9737*23.77torr[/tex]

Answer:

THE HEAT OF COMBUSTION IS 4995.69 kJ/mol OF OCTANE.

Explanation:

Heat capacity = 6.18 kJ/C

Temperature change = 41.5 C - 22.0 C = 19.5 C

Heat required to raise the temperature by 19.5 °C is:

Heat = heat capacity * temperature change

Heat = 6.18 kJ/ C * 19.5 C

heat = 120.51 kJ of heat

120.51 kJ of heat is required to raise the temperature of 2.75 g sample of  a liquid octane.

Molar mass of octane = ( 12* 8 + 1 * 18) = 114 g/mol

So therefore, the heat of the reaction per mole of octane will be:

120.51 kJ of heat is required for 2.75 g of octane

x J of heat will be required for 114 g of octane

x J = 120.51kJ * 114 / 2.75

x = 4995.69 kJ of heat per mole.

In conclusion, the heat of the combustion reaction in kJ / mole of octane is 4995.69 kJ/mol

Answer:

53.1% of hydrogen reacts

Explanation:

The mixture of 2 atoms of H with 1 atom of O produce 1 molecule of H₂.

The mass of hydrogen in 2.32g of H₂O could be obtained using molar mass of H₂O (18.01g/mol) and molar mass of hydrogen (1.01g/mol) as follows:

Moles H₂O: 2.32g H₂O × (1mole / 18.01g) = 0.1288 moles of water

1 mole of H₂O contains 2 moles of H, moles of hydrogen in 0.1288 moles of water are:

0.1288 moles H₂O × (2 moles H / 1 mole H₂O) = 0.2576 moles of H

In mass:

0.2576 moles H × (1.01g/ mol H) = 0.260g H you have in the formed water

As before reaction you had 0.485g of H and just 0.260g reacted, mass percent is:

(Mass that reacts / Mass added) × 100

(0.260g / 0.485g) × 100 =

Answer:

a) the percentage yield will exceed 100%

b) the excess reactant is filtered along with the barium sulphate precipitate. It is possible to recover the excess reactant by carefully washing the precipitate with water.

Explanation:

In the precipitation of barium sulphate, the ions in the reactants exchange partners in the product leading to an insoluble product.

In every reaction, there is a limiting reactant whose amount determines the amount of product that can be obtained. The reactant in excess remains in the system even after the reaction is completed and may be recovered alongside the product which leads to a percentage yield above 100%.

If the excess reactant is soluble in water, it can be recovered from the precipitate if needed by washing the precipitate with water.