Which of the following would be useful for converting g/mol to g/L?
A. Mass percent
B. Avogadro's number
C. Molarity
D. Molar mass

Answer:

a lower temperature to liquefy

Explanation:

Answer:

Thrust.

Explanation:

hope this helps you :)

Answer:

thrust

Explanation:

Answer:

Metallic bonding

Explanation:

Ionic bonding involves the transfer of electrons from a highly electropositive metal to a highly electronegative nonmetal.

The metallic bond is somewhat similar to the ionic bond since there are also charged positive metal ions. The only difference is that there isn't any electronegative element that accepts the electrons.

In a metallic bond, the positively charged metal ions are bound together by a sea of mobile electrons. The attractive force between the metal ions and the mobile electrons hold the metallic crystal lattice together.

Answer:

Mg²⁺

Explanation:

Electronic configuration = 1s22s22p6 (or [Ne] )

Charge = -2

This means the element has two extra electrons. So total electrons = 12.

The lement is Magnesium and the ion is Mg²⁺

Answer:

[tex]Pb(CO_{3})_{2} \\Pb(NO_{3})_{4} \\FeCO_{3}\\Fe(NO_{3})_{2}[/tex]

Explanation:

[tex]Pb^{4+}(CO_{3}^{2-})_{2} --->Pb(CO_{3})_{2} \\Pb^{4+} (NO_{3}^{-})_{4} --->Pb(NO_{3})_{4} \\Fe^{2+} CO_{3}^{2-} --->FeCO_{3}\\Fe^{2+} (NO_{3}^{-})_{2}--->Fe(NO_{3})_{2}[/tex]

Answer:

d) El agua pesada contiene mas elementos

Explanation:

La diferencia fundamental entre el agua pesada y el agua ligera es que la primera tiene una proporción mayor de deuterio que la segunda. El deuterio es un ión del hidrógeno que tiene un peso atómico mayor que el hidrógeno común y corriente. Por ende, la opción D ofrece la mejor aproximación.

Answer:

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Explanation:ki

Answer:

A, B and C are compounds

Explanation:

First of all, I need to establish that when carbon is burnt in excess oxygen, carbon dioxide is obtained as shown by this equation; C(s) + O2(g) ----> CO2(g).

Looking at the presentation in the question, A was said to be heated strongly and it decomposed to B and C. Only a compound can decompose when heated. Elements can not decompose on heating. Secondly, compounds usually decompose to give the same compounds that combined to form them. Compounds hardly decompose into their constituent elements.

Again from the information provided, the compound A is a white solid. This is likely to be CaCO3. It decomposes to give another white solid. This may be CaO and the gas was identified as CO2.

Hence;

CaCO3(s)--------> CaO(s) + CO2(g)

Answer:

[tex]=C_3H_4O_3[/tex]

Explanation:

When percentage composition is given, and asked for the empirical formula, it is simplest to  assume 100 g of material. Thus,

Mass C = 40.92 g.  Moles C = 40.92 g x 1 mole/12 g = 3.41 moles C

Mass H = 4.58 g.  Moles H = 4.58 g x 1 mole/1.0 g = 4.58 moles H

Mass O = 54.50 g.  Moles O = 54.50 g x 1 mole/16 g = 3.41 moles O

Now, we want to get the moles into whole numbers, so we begin by dividing all by the smallest, i.e. divide all values by 3.41.

Moles C = 3.41/3.41 = 1

Moles H = 4.58/3.41 = 1.34

Moles O = 3.41/3.41 = 1

Now, in order to get 1.34 to be a whole number we multiply it (and all others) by 3

Moles C = 1x3 = 3

Moles H = 1.34x3 = 4

Moles O = 1x3 = 3

Empirical Formula [tex]=C_3H_4O_3[/tex]

Answer:

Kinetic energy = 1/2mv²

where m is the mass

v = velocity

m = 5.31 × 10^-23 kg

v = 1.00 × 10^2 m/s

Kinetic energy = 1/2 × 5.31 × 10^-23 × ( 1.00 × 10^2)²

= 2.655 × 10^-19 Joules

Hope this helps

Answer:

33.3% of Sn in the sample

Explanation:

The addition of SO₄⁻ ions produce the selective precipitation of Pb²⁺ to produce PbSO₄.

Moles of PbSO₄ (molar mass 303.26g/mol) in 2.93g are:

2.93g ₓ (1mol / 303.26) = 9.66x10⁻³ moles PbSO₄ = Moles Pb²⁺.

As molar mass of Pb is 207.2g/mol, mass in 9.66x10⁻³ moles of Pb²⁺ is:

9.66x10⁻³ moles of Pb²⁺ ₓ (207.2g / mol) = 2.00g of Pb²⁺

As mass of the sample is 3.00g, mass of Sn²⁺ is 3.00g - 2.00g = 1.00g

And the percentage of Sn in the sample is:

1.00g / 3.00g ₓ 100 =

Combining quantum theory with wave motion of atomic particles is: Wave Mechanics

Answer:

61.54%

Explanation:

Hello,

To calculate the percent yield of a product, we express it as ratio between the actual yield to the theoretical yield multiplied by 100.

Percent yield = (actual yield / theoretical yield) × 100

Actual yield = 400mg

Theoretical yield = 650mg

Percent yield = (400 / 650) × 100

Percent yield = 0.6154 × 100

Percent yield = 61.54%

Percent yield of guaifenesin in the drug is 61.54%

Answer:

[tex]x_{acetone}=7.970x10^{-3}[/tex]

Explanation:

Hello,

In this case, for the given molarity, we can assume a volume of 1 L of solution, to obtain the following moles of acetone:

[tex]n=0.422mol/L*1L=0.422mol[/tex]

Then, with the density of solution, we can compute the mass of the solution for the selected 1-L volume basis:

[tex]m_{solution}=1L*\frac{1000mL}{1L}*\frac{0.970g}{1mL}=970g[/tex]

After that, we compute the mass of water in the solution, considering the mass of acetone (molar mass = 58.08 g/mol):

[tex]m_{H_2O}=970g-0.422molAcetone*\frac{58.08g\ Acetone}{1mol\ Acetone} =945.49gH_2O[/tex]

Next, the moles of water:

[tex]n_{H_2O}=945.49g*\frac{1molH_2O}{18gH_2O} =52.53molH_2O[/tex]

Finally, the mole fraction:

[tex]x_{acetone}=\frac{n_{acetone}}{n_{acetone}+n_{H_2O}}=\frac{0.422mol}{0.422mol+52.53mol}\\ \\x_{acetone}=7.970x10^{-3}[/tex]

Regards.

Answer:

3 RbOH + H₃PO₄ → Rb₃PO₄ + 3 H₂O

Explanation:

Let's consider the double-replacement reaction between rubidium hydroxide and phosphoric acid to form rubidium phosphate and water. The cation rubidium replaces the cation hydrogen and the anion hydroxyl replaces the anion phosphate. The balanced chemical reaction is:

3 RbOH + H₃PO₄ → Rb₃PO₄ + 3 H₂O

Answer:

Explanation:

Volume of silver cube = 2.42³ = 14.17 cm³

mass of silver cube = volume x density

= 14.17 x 10.49 = 148.64 gm

Volume of gold cube = 2.75³ = 20.8  cm³

mass of gold cube =  20.8 x 19.3 = 401.44 gm

specific heat of silver and gold are .24 and .129 J /g°C

mass of 112 mL water = 112 g

Heat absorbed = heat lost = mass x specific heat x temperature fall or rise

Heat lost by metals

= 148.64 x .24 x ( 85.4 -T) + 401.44 x .129 x ( 85.4 - T )

= (35.67 + 51.78 ) x ( 85.4 - T )

87.45 x ( 85.4 - T )

= 7468.23 - 87.45 T

Heat gained by water

= 112 x 1 x ( T - 20.5 )

= 112 T - 2296

Heat lost = heat gained

7468.23 - 87.45 T = 112 T - 2296

199.45 T = 9764.23

T = 48.95° C

Answer: B. 430 K

Explanation:

According to Gibb's equation:

[tex]\Delta G=\Delta H-T\Delta S[/tex]

[tex]\Delta G[/tex] = Gibbs free energy  

[tex]\Delta H[/tex] = enthalpy change  = +62.4 kJ/mol

[tex]\Delta S[/tex] = entropy change  = +0.145 kJ/molK

T = temperature in Kelvin

[tex]\Delta G[/tex]  = +ve, reaction is non spontaneous

[tex]\Delta G[/tex]  = -ve, reaction is spontaneous

[tex]\Delta G[/tex]  = 0, reaction is in equilibrium

[tex]\Delta H-T\Delta S=0[/tex] for reaction to be spontaneous

[tex]T=\frac{\Delta H}{\Delta S}[/tex]

[tex]T=\frac{62.4kJ/mol}{0.145kJ/molK}=430K[/tex]

Thus the Reaction is spontaneous when temperature is 430 K.

Answer:

430 K

Explanation:

i just took the test on a pex :)

Answer:

[tex]V_2 = 4.87 * 10^3[/tex]

Explanation:

This question is an illustration of ideal Gas Law;

The given parameters are as follows;

Initial Temperature = 25C

Initial Volume = 4.5 * 10³L

Required

Calculate the volume when temperature is 50C

NB: Pressure remains constant;

Ideal Gas Law states that;

[tex]PV = nRT[/tex]

The question states that the pressure is constant; this implies that the constant in the above formula are P, R and n

Divide both sides by PT

[tex]\frac{PV}{PT} = \frac{nRT}{PT}[/tex]

[tex]\frac{V}{T} = \frac{nR}{P}[/tex]

Represent [tex]\frac{nR}{P}[/tex] with k

[tex]\frac{V}{T} = k[/tex]

[tex]k = \frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]

At this point, we can solve for the required parameter using the following;

[tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]

Where V1 and V2 represent the initial & final volume and T1 and T2 represent the initial and final temperature;

From the given parameters;

V1 = 4.5 * 10³L

T1 = 25C

T2 = 50C

Convert temperatures to degree kelvin

V1 = 4.5 * 10³L

T1 = 25 +273 = 298K

T2 = 50 + 273 = 323K

Substitute values for V1, T1 and T2 in [tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]

[tex]\frac{4.5 * 10^3}{298} = \frac{V_2}{323}[/tex]

Multiply both sides by 323

[tex]323 * \frac{4.5 * 10^3}{298} = \frac{V_2}{323} * 323[/tex]

[tex]323 * \frac{4.5 * 10^3}{298} = V_2[/tex]

[tex]V_2 = 323 * \frac{4.5 * 10^3}{298}[/tex]

[tex]V_2 = \frac{323 * 4.5 * 10^3}{298}[/tex]

[tex]V_2 = \frac{1453.5 * 10^3}{298}[/tex]

[tex]V_2 = 4.87 * 10^3[/tex]

Hence, the final volume at 50C is [tex]V_2 = 4.87 * 10^3[/tex]

The number of moles of carbon atoms in 0.500 mol of ethane (C₂H₆) is equal to one mole.

A mole can be defined as a scientific unit that is utilized to calculate the quantities such as atoms, molecules, ions, or other particular particles. The mass of one mole of a given chemical element is atomic mass and that of 1 mole of a chemical compound is molar mass.

The number of entities found in one mole is equal to 6.023 × 10 ²³ which is known as Avogadro’s constant.

Given, the number of moles of C₂H₆ = 0.500 mole

One molecule of ethane has carbons = 2

One mole of ethane has moles of carbons = 2 moles

0.500 mol of ethane has moles of carbon atoms = 0.500×2 = 1 mol

Therefore, one mole of carbon atoms is present in 0.500 mol of ethane C₂H₆.

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Answer: [H2SO4] = 0.5M;

              [KOH] = 1M

Explanation: Molarity is the solution concentration defined by:

molarity = [tex]\frac{mol}{L}[/tex] or M

To determine the concentration of the mixture, find how many mols of each compound there are in the mixture:

50 mL = 0.05L

H2SO4

1 mol/L * 0.05L = 0.05mol

KOH

2mol/L * 0.05L = 0.1 mol

The mixture has a total volume of:

V = 50 + 50 = 100 mL = 0.1 L

The concentration of the resullting solutes:

[H2SO4] = [tex]\frac{0.05}{0.1}[/tex] = 0.5 M

[KOH] = [tex]\frac{0.1}{0.1}[/tex] = 1 M

Concentration of H2SO4 is 0.5M while for KOH is 1M.

Answer:

Path A-B-D involves a catalyst and is slower than A-C-D

Explanation:

The diagram above illustrates both the catalyzed path and the uncatalyzed path of a chemical reaction.

The catalysed path is the path expressed with broken lines and the uncatalyzed path is the path expressed with thick small line as shown in the diagram above.

The catalyzed path has a higher activation energy than the uncatalyzed path.

Therefore, the catalyzed path will be slower that the uncatalyzed path because, the catalyzed path will require a higher energy to overcome the activation energy in order for the reaction to proceed to product.

On the other hand, the uncatalyzed path has a lower activation energy and a lesser amount of energy is needed to overcome it in order for the reaction to proceed to product.

Answer:

0.623 M

Explanation:

Step 1: Given data

Step 2: Calculate the percent by volume (%m/v)

We will use the following expression.

[tex]\%m/v = \%m/m \times \rho = 7.00\% \times 1.071g/mL = 7.50g\%mL[/tex]

Step 3: Calculate the molarity

7.50 g of magnesium sulfate are dissolved in 100 mL of the solution. The molarity is:

[tex]M = \frac{7.50g}{120.37g/mol \times 0.100L } = 0.623 M[/tex]

Answer:

C. Grinding the solute down into tiny particles.

Explanation:

The dissolution of a solute has something to do with particle size. The size of solute particles usually determines how quickly a solute dissolves in a solvent. When large solute particles are introduced into the solvent, the large solute particles do not easily interact with solvent particles hence preventing easy dissolution in the solvent.

However, when the solute is ground into tiny particles, smaller solute particles interact more effectively with solvent particles hence dissolution is faster.

Therefore, tiny solute particles will dissolve faster in a solvent than a lump of solute. Summarily, small particle size enhances dissolution of a solute in the appropriate solvent.

Answer: stirring the solution vigorously

Grinding the solute down into tiny particles

gently heating the solution

Explanation:

A dissolution will proceed more readily when heated . Breaking up the solute as much as possible will aid in overcoming the solute-solute interaction, as will stirring the solution

Answer:

1.30atm

Explanation:

P1/T1 = P2/T2

1.07/393 = P2/478

Answer: the first one is correct

Explanation:

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Answer:

C. Potential energy

Explanation:

Kinetic energy and gravitational potential energy are both forms of potential energy. Potential energy is stored energy, when an object is not in motion it has stored energy. When an object is an motion it has kinetic energy. An object posses gravitational potential energy when it is above or below the zero height.

Answer:

E) Al³⁺

Explanation:

A reaction involving a gain of electrons is known as a reduction reaction because the oxidation number of the species gaining the electron is reduced.

In the given question, the oxidation number (charge) of particle accepting two electrons will decrease by 2. From the given options;

A. K is a neutral atom with oxidation number of 0. If is accepts two electrons, its oxidation number becomes -2.

K + 2e⁻  ----> K⁻²

B) Fe²⁺ has a charge of +2. If it accepts two electrons, its charge comes 0.

Fe⁺ + 2e⁻  ----> Fe

C) O²⁻ has a charge of -2. if it accepts two electrons, it will have a charge of -4.

O²⁻ + 2e⁻  ---->  O⁴⁻

D) Ne has a charge of zero. If it accepts two electrons, its charge becomes -2.

Ne + 2e⁻   ---->   Ne²⁻

E) Al³⁺ has a charge of +3. If it gains two electrons, its charge becomes +1.

Al³⁺ + 2e⁻   ---->   Al⁺

Answer:

21.88mL is the volume of base required for the titration.

Explanation:

For an acid-base titration trying to find the concentration of an acid, you must add a known quantity of the acid and titrate it with an standarized base.

If you know the moles of base you add to the acid solution, these moles are equal to moles of acid.

In the buret of the titration, initial volume is 1.94mL and final volume is 23.82mL. The volume you are adding is the difference between initial and final volume, that is:

23.82mL - 1.94mL

Answer:

3-bromobenzoic acid

Explanation:

In this case, we have to remember that the [tex]Br_2/FeBr_3[/tex]  is a reaction in which we add Br into the molecule an electrophilic aromatic substitution. Additionally, we have a carboxylic acid in the benzene. This carboxylic acid is an ortho director because is a deactivating group (it removes electrons from the benzene ring). With this in mind, a "Br" atom would be added in an ortho position respect to the COOH group and we will obtain 3-bromobenzoic acid.

See figure 1.

I hope it helps!

To create 3-bromobenzoic acid, a "Br" atom would be placed at an orthogonal position to the COOH group according to electrophilic aromatic substitution.

Electrophilic aromatic substitution is a type of organic reaction in which an atom or group in an aromatic ring is substituted with an electrophile. It is a fundamental reaction in aromatic chemistry that happens due to the aromatic system's high electron density.

It is an electrophilic aromatic substitution process in which Br is incorporated into the molecule. In addition, the benzene contains a carboxylic acid. Because it removes electrons from the benzene ring, this carboxylic acid functions as an ortho director. To create 3-bromobenzoic acid, a "Br" atom would be placed at an orthogonal position to the COOH group. The product is seen in the photographs below.

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Answer:

"The active site of the enzyme has a complementary rigid structure" belongs to the key and lock system

"The conformation of the enzyme changes when it binds to the substrate so that the active site conforms to the substrate." belongs to the induced fit system.

"The substrate binds to the enzyme at the active site, forming an enzyme-substrate complex" belongs to both, that is, the key and lock system and the induced fit system.

"The substrate binds to the enzyme through non-covalent interactions" can belong to both enzyme systems.

Explanation:

Enzymatic key and lock systems bear this name because the enzyme at its site of union with the substrate has an ideal shape so that its fit is perfect, similar to a headbreaker, so once they are joined they are not It can bind another substrate to the enzyme, since they are generally associated with strong chemical bonds.

The shape of the enzyme's active site is a negative of what the shape of the substrate would be.

On the other hand, in the mechanism or enzyme system of induced adjustment, the enzyme has an active site that is where it binds with the substrate and another site where another chemical component binds, which when this chemical component binds this enzyme changes its morphology and becomes "active" to bond with your substrate.

This happens a lot in the inactive enzymes that are usually activated in digestive processes since the fact that these enzymes are constantly active would be dangerous, therefore the body takes the induced enzyme system as a control mechanism, where a molecule or chemical compound induces change morphological of an enzyme by means of the allosteric union so that it joins its substrate and catalyzes or analyzes it, depending on the enzymatic character of the enzyme.

Answer:

677.7 mmHg

Explanation:

The first empirical study on the behaviour of a mixture of gases was carried out by John Dalton. He established the effects of mixing gases at different pressures in the same vessel.

Dalton's law states that,the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the individual gases present in the mixture of gases. When a gas is collected over water, the gas also contains some water vapour. The partial pressure of the gas will now be given as; total pressure of gas mixture - saturated vapour pressure of water (SVP) at that temperature.

Given that;

Total pressure of gas mixture = 692.2 mmHg

SVP of water at 17°C = 14.5 mmHg

Therefore, partial pressure of oxygen = 692.2-14.5

Partial pressure of oxygen = 677.7 mmHg

Answer:

Explanation:

a. change of colour:

A chemical reaction rearranges the constituent atoms of the reactants to create different substances as products. The products have different molecular structures than the reactants. Different atoms and molecules radiate different colours of light. Hence, there usually is a change in colour during a chemical reaction.

Eg: copper reactions with the elements

b. Evolution of gas:

A gas evolution reaction is a chemical reaction in which one of the end products is a gas such as oxygen or carbon dioxide.

Eg: ammonium hydroxide breaks down to water and ammonia gas.

c. Change of smell :

Production of an Odor Some chemical changes produce new smells.  ... The formation of gas bubbles is another indicator that a chemical change may have occured.

Eg: The chemical change that occurs when an egg is rotting produces the smell of sulfur.

d. Change of state:

A chemical reaction is a process in which one or more substances, also called reactants, are converted to one or more different substances, known as products.

Eg: candle wax (solid) melts initially to produce molten wax (liquid)

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