1. Alginate is
a. Reversible Hydrocolloid impression material
b. Irreversible Hydrocolloid impression material
c. Polysulfideimpression material
d. Polyvinyl siloxane impression material

Answer:

Explanation:

Increase of boiling point, decrease of freezing point, and decrease of vapor pressure are all colligative properties.

The colligative properties are the physical properties of the solutions that depends on the number of particles of solute dissolved and not on the nature of the particles.

Regarding the vapor pressure of a solution, since the solute particles occupy part of the surface of the solution, less solvent particles are in the surface and so the vapor pressure decreases. The higher the number of solute particles, the greater the decrease on the vapor pressure.

As for the boiling boiling point, take into account that the boiling points is the temperature at which the vapor pressure of the substance reachs the atmospheric pressrue. Since, the presence of solute particles diminishes the vapor pressure of the solution, the solutions will need to absorb more energy to reach the atmospheric pressure, meaning that the boiling point will be higher. The higher the number of solute particles, the greater the increase on the boiling point.

As for the freezing point, the solutions freeze at lower temperatures when a solute is added. The higher the number of particles of solute, the lower the freezing point of the solution.

Then, you need to find which out of three 1.5 m (molal) solutions of glucose, sodium sulphate, and ammonium phosphate will produce the greater number of particles.

Glucose is a molecular compound (not ionic), so each molecule of glucose will yield one particle.

Sodium sulphate is Na₂SO₄; a ionic compound that ionizes into 3 ions: 2 Na⁺ and 1 SO₄²⁻.

Ammonium phospahte is (NH₄)₃ PO₄; a ionic compound that ionizes into 4 particles: 3 NH₄⁺ and 1 PO₄³⁻.

Therefore, since every unit of ammonium phosphate yields to more particles than every unit of glucose or sodium sulphate, the 1.5 m solution of ammonium phosphate will have the most impact on the colligative properties.

Answer:

the answer is D

Explanation:

:0)

Answer:

The answer to your questions are

a) 190.96 gr

b) 0.387 moles

c) 2.332 x 10²³ molecules

d) 10 atoms

Explanation:

a)              192.217 u = (192.963)(0.627) + 0.373x

            0.373x = 192.217 - 120.99

             0.373x = 71.23

                   x = 190.96 g

b) MW C5H10O5 = (12 x 5) + ((10 x 1) + (16 x 5) = 150 g

                            = 60 + 10 + 80

                           = 150 g

                150 g ---------------- 1 mol

                  58 g  -------------    x

x = 58 x 1/150 = 0.387 moles

                   1 mol ---------------- 6 .023 x 10²³ molecules

                0.387 mol -----------   x

 x = 0.387 x 6 .023 x 10²³/1 = 2.331 x 10²³ molecules

The answer is option B "era." A era can be up to several millions of years. It isn't period because period can be days to years, it isn't years because it's only one year, wouldn't be option D either because a epoch is up to three million years.

Hope this helps.

Answer:

Explanation:

So, the formula for the compound should be:

[tex]XCl_{6}[/tex]

Now we assume that we have 1 mol of substance, so we can make calculations to know the molar mass of element X, as follows:

[tex]M_{Cl}=35.45g/mol\\[/tex]

So we have that 6 moles weight 212.7g, and we can make a rule of three to know the weight of compound X:

[tex]212.7g\rightarrow 86.9\%\\x\rightarrow 13.1\%\\\\x=\frac{212.7g*13.1}{86.9} =32.06g[/tex]

As we used 1 mol, we know that the molar mass is 32.06g/mol

So the element has a molar mass of 32.06 g/mol and an oxidation state of +6, with this information, we can assure that the element X is sulfur, so the compound is [tex]SCl_{6}[/tex]

Final answer:

To identify element X in a compound with chlorine, where X constitutes 13.10% of the compound's mass, and the atom ratio of X to Cl is 1:6, we calculate based on moles and percent composition, leading to the identification of phosphorus (P) as element X.

Explanation:

To find element X in a compound composed of element X and chlorine, where the compound is 13.10% X by mass, and each molecule of the compound contains six times as many chlorine atoms as X atoms, we can use the concept of percent composition and atomic weights. Assuming a 100 g sample, we can calculate that there are 13.10 g of element X and 86.90 g of chlorine. Given that the molar mass of chlorine (Cl) is approximately 35.45 g/mol, we can estimate the moles of chlorine and use the given ratio to find the moles and eventually, the molar mass of element X. If we assume that there is 1 mole of X and 6 moles of Cl in the molecule, we can calculate the molar mass of X using its percent composition and the total molar mass of chlorine present.

Step-by-step:

Answer:

C

Explanation:

Answer:

A. 97.8 mm  

Explanation:

There are 10 divisions between 9 cm (90 mm) and the end of the ruler (10 cm or 100 mm).

Each division equals 1 mm.

Each 5 mm mark has a longer tick, and the dashed red line is between the 97 mm and 98 mm marks.

You would normally estimate to the nearest tenth of a division. An estimate of 0.8 mm is reasonable.

The length of the object is 97.0 mm + 0.8 mm = 97.8 mm.

B is wrong. You can't possibly estimate to the nearest hundredth of a division.  

C is wrong. The dashed red line slightly before the 98 mm mark.

D is wrong. If the dashed red line were exactly on the 98 mm mark, you would record the measurement as 98.0 mm. This indicates that you measured the object to zero tenths on either side of the mark.

The caloric content of peanut butter, given the data from the bomb calorimetry experiment, is approximately 5.8 Cal/g.

The subject of the question is determining the caloric content of peanut butter in Cal/g using bomb calorimetry. You can do this by using the information provided on the temperature rise and the calorimeter's heat capacity. The heat absorbed by the calorimeter is calculated by multiplying the heat capacity of the calorimeter by the change in temperature.

In this case, the heat absorbed is 130 kJ/degrees C * (25.2 degrees C - 22.4 degrees C) = 364 kJ. To convert this value to Calories, we need to remember that 1 Calorie equals 4.184 kJ. Therefore, the caloric energy produced by the combustion of peanut butter is approximately 87 Cal.

Since the mass of the peanut butter is 15g, the caloric content of peanut butter is 87 Cal / 15 g gives approximately 5.8 Cal/g.

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The food caloric content of peanut butter can be calculated using bomb calorimetry. The calorimeter used in this experiment has a heat capacity of 130 kJ/°C, and the temperature change in the calorimeter is from 22.4 °C to 25.2 °C. The caloric content of peanut butter is determined to be 6.743 Cal/g.

The food caloric content of peanut butter can be calculated using the concept of bomb calorimetry. In this case, the mass of the peanut butter is 15g and it is combusted in a calorimeter with a heat capacity of 130 kJ/°C. The change in temperature of the calorimeter is from 22.4 °C to 25.2 °C.

To calculate the food caloric content, we can use the formula:

Caloric content = (Heat capacity of calorimeter × Change in temperature) / Mass of peanut butter

Substituting the given values:

Caloric content = (130 kJ/°C × (25.2 °C - 22.4 °C)) / 15g

Now, we can convert kJ to Cal using the conversion factor 1 kJ = 0.239 kcal:

Caloric content = (130 × 0.239 × (25.2 - 22.4)) / 15g

Simplifying the equation gives:

Caloric content = 6.743 Cal/g

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Answer:

Explanation:

Given parameters:

Mass of CS₂ = 0.15g

Volume of O₂ = 285mL

Condition of reaction = at standard temperature and pressure.

Solution:

 Balanced reaction equation is shown below:

                CS₂ + 3O₂ → CO₂ + 2SO₂

For gases at standard temperature and pressure(STP);

     number of moles = [tex]\frac{volume occupied}{22.4}[/tex]

To find the final mixture of the gases at STP, we must identify the limiting reactant here first. The limiting reactant determines the extent of the reaction since it is the one used up.

    Number of moles of CS₂ = [tex]\frac{mass}{molar mass}[/tex]

     molar mass of CS₂ = 12 + (2 x 32) = 76g/mol

    Number of moles of CS₂ = [tex]\frac{0.15}{76}[/tex] = 0.00197mole

Number of moles of O₂ =  [tex]\frac{volume occupied}{22.4}[/tex]

   converting mL to dm³;

     1000mL = 1dm³

       285mL = 0.285dm³

Number of moles of O₂ =  [tex]\frac{0.285}{22.4}[/tex] = 0.01272mole

From the reaction equation:

       1 mole of CS₂ reacted with 3 moles of O₂;

  0.00197 mole of CS₂ will require 3 x 0.00197; 0.00591mole to react with;

We see that we have more than enough O₂. CS₂ is the limiting reagent.

Number of moles of excess O₂ =  0.01272 - 0.00591 = 0.00681moles

Now, we use the information of the limiting reagent to determine the product,

  1 mole of CS₂ will produce 1 mole of CO₂;

   0.00197 mole of CS₂  will produce 0.00197 mole of CO₂

 1 mole of CS₂ will produce 2 mole of SO₂

  0.00197mole of CS₂ will produce 2 x 0.00197; 0.00394mole of SO₂

The moles of the final mixture will = number of moles of excess O₂ + number of moles of CO₂ + number of moles of SO₂ =      0.00681+0.00197+0.00394

  = 0.01272moles

Volume of the final mixture = number of moles  x 22.4 = 0.01272 x 22.4

 Volume of final mixture = 0.2849dm³ or 284.9mL

Answer:

A) Rapid chemical reaction where gas is released.

Explanation:

Edge 2020 GOT IT RIGHT!

The process of an antacid tablet reacting in water is a rapid chemical reaction characterized by the release of gas. This occurs when components like calcium carbonate in the antacid neutralize stomach acid, producing calcium chloride, water, and carbon dioxide gas.

When an antacid tablet is placed in water, the best description of the process is a rapid chemical reaction where gas is released. Antacid tablets typically contain Arrhenius bases like Al(OH)3, Mg(OH)2, CaCO3, and NaHCO3 that react with and neutralize stomach acid. For example, calcium carbonate reacts with hydrochloric acid in the stomach to produce calcium chloride, water, and carbon dioxide gas, leading to a reduction in acid and sometimes a belch from the released CO2.

CaCO3(s) + 2HCl(aq) = CaCl2(aq) + H2O(l) + CO2(g)

This reaction is an example of a neutralization reaction, which is a type of acid-base reaction, and in the case of antacids, it typically happens rapidly upon the dissolution and reaction of the antacid in the stomach's hydrochloric acid.

Explanation:

the process where the acids and bases neutralize each other to form salt and water is known as neutrilazation

Answer:

Na+ flows into the cell and then K+ flows out of the cell

Explanation:

When the cell is at rest potential, Na+ is more highly concentrated on the outside of the cell, and K+, on the inside of the cell. During an action potential, the plasma membrane depolarizes, Na+ channels are opened and Na+ starts flowing into the cell, in favor of its electrochemical gradient. When this happens, K+ will also be opened and K+ will start flowing outside the cell, causing the membrane potential to go back to its rest value.

Answer:

It is an example of a precipitation reaction. Soap contains the conjugate bases of fatty acids, and these anions (carboxylates of fatty acids) tend to form insoluble salts with magnesium and calcium cations.

Explanation:

Usual soaps are made by a reaction called saponification, in which triglycerides (which are the main components of animal and vegetal fats and oils) react with a strong base (like KOH or NaOH), breaking the triglyceride into three fatty acid conjugate bases (carboxylates) and a molecule of glycerol.

These fatty acid conjugate bases are formed by a long saturated carbon chain and an ionised tip (i attached the structure of a generic fatty acid carboxylate). These anions form very insoluble salts with Calcium of Magnesium cations, and thus a precipitate is formed, therefore the answer is B.

Final answer:

The formation of a ring in a bathtub due to the reaction of hard water with soap is an example of a precipitation reaction.

Explanation:

The formation of a ring in a bathtub due to the reaction of hard water with the solute anions in soap is an example of a precipitation reaction. Hard water, which contains Mg2+ and Ca2+ ions, reacts with the anions in soap to form insoluble salts. These salts precipitate and create a ring in the bathtub.

Answer:

5.122 × 10⁻¹³

Explanation:

Sphere volume is calculated with the formula:

[tex]V = \frac{4}{3}[/tex] π [tex]r^{3}[/tex]

V - volume

r - radius

radius = diameter / 2

radius of atom= 1 / 2 = 0.5 Å

radius of nucleus = (8 × 10⁻⁵) / 2 = 4 × 10⁻⁵ Å

now we may calculate the volume for the atom and the nucleus

atom volume = (4/3) × 3.14 × (0.5)³ = 0.523 ų

nucleus volume = (4/3) × 3.14 × (4 × 10⁻⁵)³ = 2.679 × 10⁻¹³ ų

now the fraction of volume of the atom that is taken up by the nucleus:

fraction = nucleus volume / atom volume

fraction = 2.679 × 10⁻¹³ / 0.523 = 5.122 × 10⁻¹³

Answer:

5400 cans

Explanation:

First we convert the total weight, 1 ton, to grams:

[tex]1ton=1x10^{6}g[/tex]

Now we need to know the mass of aluminum:

[tex]m_{Al}=\frac{10^{6}*8.1}{100} =81000g[/tex]

Now we make the relation between the mass of aluminum in 1 ton of the earth's crust and the mass of aluminum per can:

[tex]n=\frac{81000g}{15g/can} =5400cans[/tex]

5,400 cans could be made from 1 ton of the Earth’s crust of aluminum.

We can calculate the number of cans by dividing the total mass of aluminum in earth's crust to the mass of aluminum in one can.

We know that relation between tons and gram is as follow:

1 ton = 10⁶ grams

Given % mass of aluminum = 8.1%

Now, the mass of aluminum in 1 ton can be calculated as:

mass = 8.1 × 10⁶ / 100 = 81,000 grams

Given mass of aluminum in one can = 15 grams

Therefore number of cans = 81,000 / 15 = 5,400

Hence, 5,400 cans can be made.

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Explanation:

An enzyme is defined as a substance which acts as a catalyst and it is produced by a living organism to bring about a specific biochemical reaction.

An enzyme actually helps in increasing the rate of a reaction by combining at a specific site and like a catalyst an enzyme does not get consumed in a chemical reaction.

An enzyme binds with the substrate and results in the formation of a stable enzyme-substrate complex. Hence, [tex]\Delta G[/tex] will be lower when compared to both uncatalyzed and loosely bound reaction.

An enzyme also helps in lowering the activation energy so that more number of molecules can participate in the reaction. Hence, it leads to increase in the rate of reaction.

Thus, we can conclude that qualities of a chemical reaction which are affected by enzymes are as follows.

The quality of a chemical reaction that is affected by enzymes would be the energy required to start the reaction.

The energy required to start a reaction is known as the activation energy. It is a barrier that the reactants must cross before they can become products.

Enzymes do not interfere with the speed of reaction directly. Instead, they lower the energy barrier, the activation energy, that the reactants must cross and as such, products are arrived at quicker.

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Answer: Two answers apply to physical change:

1-Condensation of Ethanol

2-Evaporation of Ethanol

Explanation:

Physical and Chemical changes are two very different types of change. However, both processes are possible because of the same thing: the investment of energy.

Physical change requires a lower investment of energy and this is the reason why these kind of changes can be reverted and the molecular structure of the matter does not change.

For example: I can easily change the state of water with fire or a fridge, and go from ice to liquid to vapor and back as many times I want. This is possible because I'm not giving enough energy to break or modify bonding forces inside the molecule of water, but enough energy to modify the interaction between two different molecules of water.

This is why Ethanol can be evaporated and then condensated into liquid again, representing this physical changes.

Chemical change, on the other hand, requires a higher investment of energy and because of this is that the bonds between atoms inside a molecule can be affected (bonds between atoms in the molecule are stronger). Therefore, when the molecules themselves are modified, the matter is transformed from a compound into a new one. This is why glucose becomes carbon dioxide, ethanol or carbon. These three compounds appear when the bonds between Carbons in the glucose are broken.

There is 35.14 kg of carbon dioxide is produced upon the complete combustion of 18.9 L of propane.

Given that;

The density of the liquid propane in the tank is 0.621 g/mL.

Now write the balanced equation for the combustion of propane (C₃H₈):

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

According to the balanced equation, one mole of propane (C₃H₈) produces three moles of carbon dioxide (CO₂) upon complete combustion.

Given:

Volume of propane = 18.9 L

Density of liquid propane = 0.621 g/mL

First, let's determine the mass of propane (C₃H₈) using its density:

Mass of propane = Volume of propane × Density of liquid propane

Moles of propane = (18.9 L × 1000 mL/L × 0.621 g/mL) / 44.10 g/mol

Moles of propane ≈ 266.14 mol

Moles of carbon dioxide = Moles of propane (3 moles CO₂ / 1 mole C₃H₈)

Moles of carbon dioxide ≈ 266.14 mol (3 mol CO₂ / 1 mol C₃H₈)

Moles of carbon dioxide ≈ 798.42 mol

Now, let's calculate the mass of carbon dioxide:

Mass of carbon dioxide = Moles of carbon dioxide × Molar mass of carbon dioxide

The molar mass of carbon dioxide = 44.01 g/mol

Mass of carbon dioxide ≈ 798.42 mol × 44.01 g/mol

Mass of carbon dioxide ≈ 35,138.5 g

Since the mass is in kilograms, convert grams to kilograms:

Mass of carbon dioxide ≈ 35,138.5 g / 1000

Mass of carbon dioxide ≈ 35.14 kg

Therefore, approximately 35.14 kg of carbon dioxide is produced upon the complete combustion of 18.9 L of propane.

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The balanced equation for combustion of propane is used to determine the stoichiometric relationship between propane and carbon dioxide. Using the given density, the volume of propane is converted to moles and then to grams of CO2. Finally, grams of CO2 are converted to kg to answer the question.

The subject of this question is the combustion reaction of propane. The balanced equation for the combustion of propane (C3H8) is that propane reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O), in a 3:1 ratio of CO2 to C3H8.

Therefore, the formula is: C3H8 + 5O2 -> 3CO2 + 4H2O

To determine the amount of carbon dioxide produced, we need to start by converting the volume of liquid propane to moles using the given density and the molar mass of propane (44.1 g/mol). Then, use the stoichiometric relationship from the balanced equation to convert from moles of propane to moles of carbon dioxide. Finally, convert the moles of CO2 to grams (and then to kg) using the molar mass of CO2 (44.01 g/mol).

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Answer:

the answer is the option "a"

Explanation:

in petroleum polycyclic aromatic hydrocarbons are found. These organic compounds are composed of aromatic rings. PAH (Polycyclic Aromatic Hydrocarbons), are formed during the incomplete combustion of any type of organic matter. In general, exposure will be to a mixture of PAHs.

Answer:

She added the ingredients in the wrong order, this will make the ingredients to mix in a different way and that can lead to different errors, formation of clumps is one of them.

Answer:

Explanation:

The acid present in the lemon will react with the protein from the milk so it will clabben it forming this little clumps. It could be also the Corn Starch that will not dissolve correctly if add all in a go, due to the polarity.

Answer:

The answer to your question is: letter D. 1.33 L

Explanation:

Data

V1 = 50 ml

C1 = 19.3

To solve this problem use the formula  C₁V₁ = C₂V₂

                            C2 = C1V1 / V2

C = concentration

V = volume

a) 1.15 L

                     C2 = (19.3)(50) / 1150

                      C2 = 0.84 M

b) No right answer

c) V2= 0.80 L

                     C2 = (19.3)(50) / 800

                     C2 = 1.2 M

d) V2 = 1.33 L

                        C2 = (19.3)(50) / 1330

                       C2 = 0.72 M

e) V2 = 350 ml

                         C2 = (19.3)(50) / 350

                         C2 = 2.75 M  

Answer:

The answer to your question is:

Explanation:

2.45 g of Phosphorus

MW P = 31 g

MW P2O5 = 2(31) + 5(16) = 142 g

From the balance reaction

               4 P      ⇒ 2 P2O5

Then     4(31) g P    ⇒  2 (142)  g P2O5

       124g  of P       ⇒  284 g   of P2O5              Rule of three

            2.45g P      ⇒      x

x = 2.45 x 284/124 = 695.8/124 = 5.61 g of P2O5

Answer:

5.619  grams of diphosphorus pentoxide are formed when 2.45 g of phosphorus reacts with excess oxygen.

Explanation:

[tex]4P(s)+5O-2(g)\rightarrow 2P_2O_5(s)[/tex]

Mass of  phosphorus = 2.45 g

Moles of phosphorous = [tex]\frac{2.45 g}{31 g/mol}=0.7903 mol[/tex]

According to reaction 4 moles of phosphorus gives 2 moles of diphosphorus pentoxide.

Then 0.7903  moles of phosphorus will give:

[tex]\frac{2}{4}\times 0.7903 mol=0.03957 mol[/tex] of diphosphorus pentoxide

Mass of 0.03957 moles of  diphosphorus pentoxide :

[tex]0.03957 mol\times 142 = 5.619 g[/tex]

5.619  grams of diphosphorus pentoxide are formed when 2.45 g of phosphorus reacts with excess oxygen.

Answer:

The answer to your question is: 63.5 u

Explanation:

Data                  Percent                         Molecular mass

63 Cu               69.2%                             62.9296 u

65Cu                 30.8%                             64.9278 u

Average atomic mass = ?

Formula

Average atomic mass = (0.692 x 62.9296) + (0.308 x 64.9278)

                                     = 43.55 + 19.99

                                      = 63.5 u

The average atomic mass of copper is calculated by multiplying the atomic mass of each isotope (63 Cu and 65 Cu) by its natural abundance, then adding those values together. The result is approximately 63.57 u.

The average atomic mass of a chemical element like copper is calculated by multiplying the atomic mass of each isotope by its natural abundance (as a decimal), then adding those values together. In this case, copper has two common isotopes: 63Cu and 65Cu.

Here's how you'd calculate it:

So, the average atomic mass of copper is approximately 63.57 u.

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Answer:

D

Explanation:

Ketone is R-CO-R' Which is D

A is R-CHO

B is R-COO-R'

C is R-COOH

Answer : The correct option is, (D)

Explanation :

A ketone is an organic compound which contain a functional group with (-CO) i.e R-CO-R'

R-CO-R' functional group consists of a carbonyl center that means a carbon which is double-bonded to the oxygen. The carbon of this carbonyl group bonded with a carbon chain, (R) to one end and a carbon chain, (R') to another end.

For example : [tex]H_3C-CO-CH_3[/tex]

In the given options we conclude that, option D is a ketone.

Hence, the correct option is, (D)

Answer:

molar composition for liquid

xb= 0.24

xt=0.76

molar composition for vapor

yb=0.51

yt=0.49

Explanation:

For an ideal solution we can use the Raoult law.

Raoult law: in an ideal liquid solution, the vapor pressure for every component in the solution (partial pressure) is equal to the vapor pressure of every pure component multiple by its molar fraction.  

For toluene and benzene would be:

[tex]P_{B}=x_{B}*P_{B}^{o}[/tex]

[tex]P_{T}=x_{T}*P_{T}^{o}[/tex]

Where:

[tex]P_{B}[/tex] is partial pressure for benzene in the liquid  

[tex]x_{B}[/tex] is benzene molar fraction in the liquid  

[tex] P_{B}^{o}[/tex] vapor pressure for pure benzene.  

The total pressure in the solution is:

[tex]P= P_{T}+ P_{B}[/tex]

And  

[tex]1=x_{B}+x_{T}[/tex]

Working on the equation for total pressure we have:

[tex]P=x_{B}*P_{B}^{o} + x_{T}*P_{T}^{o}[/tex]

Since [tex]x_{T}=1-x_{B}[/tex]

[tex]P=x_{B}*P_{B}^{o} + (1-x_{B})*P_{T}^{o}[/tex]

We know P and both vapor pressures so we can clear [tex]x_{B}[/tex] from the equation.

[tex] x_{B}=\frac{P- P_{T}^{o}}{ P_{B}^{o} - P_{T}^{o}}[/tex]

[tex] x_{B}=\frac{35- 22}{75-22} = 0.24[/tex]

So  

[tex]x_{T}=1-0.24 = 0.76[/tex]

To get the mole fraction for the vapor we know that in the equilibrium:

[tex]P_{B}=y_{B}*P[/tex]

[tex]y_{T}=1-y_{B}[/tex]

So  

[tex]y_{B} =\frac{P_{B}}{P}=\frac{ x_{B}*P_{B}^{o}}{P}[/tex]

[tex] y_{B}=\frac{0.24*75}{35}=0.51[/tex]

[tex]y_{T}=1-0.51=0.49[/tex]

Something that we can see in these compositions is that the liquid is richer in the less volatile compound (toluene) and the vapor in the more volatile compound (benzene). If we take away this vapor from the solution, the solution is going to reach a new state of equilibrium, where more vapor will be produced. This vapor will have a higher molar fraction of the more volatile compound. If we do this a lot of times, we can get a vapor that is almost pure in the more volatile compound. This is principle used in the fractional distillation.

1) Answer:

moles of oxygen (n) = 1.86 moles

Explanation:

according to  Boyle's law the formula to solve this problem is:

PV=nRt

when P is the pressure which equal 1.25 atm

and V is the volume which equal 37.5 L

n is the number of moles which we need to calculate it

R is constant which equal 0.082

t is the temperature in kelvin

By substitution:

1.25*37.5 = n * 0.082 * 307

So n = 1.86 moles

2) Answer:

the volume of oxygen gas = 34 L

Explanation:

at standard temperature and pressure (STP) 1 mole of gas will equal = 22.4L

So when we have 1.5 moles of oxygen at standard temperature and pressure (STP) so we will estimate it like that

1.5 moles *22.4 L/ 1 mole = approximately 34 L

3) Answer:

The volume of H2 = 2.29 L

Explanation:

according to the Balanced equation we can see that the molar ratio between Zn and H2 1 : 1

so to know the number of moles of H2 we will get it for Zn first :

number of moles Zn = mass of Zn / molar mass Zn

                             = 5.98 / 65.39 =0.0914 moles

so number of moles H2 = 0.09 moles

by  substitution in the following formula:

PV = nRT

0.978 * V = 0.09 * 0.082 * 298

so The volume of H2 = 2.29 L

4) Answer:

Volume of O2 = 1.4 L

Explanation:

first we have to balance  the equation:

2Na2O2 +2CO2 → 2Na2CO3 + O2

2 mole CO2 give 1 mole of O2 so the molar ratio is 2:1

at STP 1 mole of gas will equal = 22.4 L

             ??? moles of CO2 = 2.8

n CO2 = 0.125 moles so n O2 = 0.125 /2 = 0.0625 moles

so when 1 mole of as = 22.4

              0.0625 moles O2 = ???

Volume of O2 =0.0625 moles * 22.4 L/ 1 mole

                        = 1.4 L

5) Answer:

the initial quantity of sodium metal used = 17.2 gram

Explanation:

at STP 1 mole of gas will equal = 22.4 L

so  moles of H2 equal ?? when 8.40 liters of H2 gas were produced

so moles of H2  = 8.4/22.4 =0.375 moles

and according to the balanced equation the molar ratio between H2 ans Na is 1 : 2

so number of moles for Na = 0.375 *2 = 0.75 moles

to get the initial quantity of sodium metal (mass Na) = number of moles * molar mass

mass Na = 0.75 moles * 23 gm/mole= 17.25

6) Answer:

False

Explanation:

because STP means standard temperature and pressure. and the Standard temperature must be 273 K and the standard pressure must be 1 atm but in the question the temperature is 298 K not 273 K so , It is not a standard temperature

7) Answer:

22.4 liters

Explanation:

1 mole of gas will equal = 22.4 L

because at the Standard temperature must be 273 K and the standard pressure must be 1 atm

so V = nRT/P

       =1 mole * 0.082 * 273 K / 1 aTm

        = 22.4 liters

Final answer:

The questions involve applications of the ideal gas law and concepts of molar volume at standard temperature and pressure (STP). Answers were calculated using ideal gas law formulas, including conversions of conditions to find moles of gases and volumes at STP.

Explanation:

To answer these questions, let's apply the ideal gas law PV = nRT, where P is pressure in atmospheres (atm), V is volume in liters (L), n is the number of moles of gas, R is the ideal gas constant (0.0821 L atm/mol K), and T is temperature in Kelvin (K). Additionally, knowledge of the molar volume of a gas at standard temperature and pressure (STP, defined as 0°C or 273.15 K and 1 atm pressure), which is 22.4 L/mol, is vital for some of the questions.

Question 1

Using the ideal gas law, rearranged to n = PV/RT, with V = 37.5 L, P = 1.25 atm, and T = 307 K, we can calculate the moles of oxygen produced. This calculation yields approximately 1.86 mol O2.

Question 2

At STP, 1 mole of any gas occupies 22.4 L. Therefore, 1.5 moles of oxygen gas would occupy 34 L (1.5 * 22.4 L/mol).

Question 3

First, convert the mass of Zn to moles, then apply the stoichiometry of the reaction to determine moles of H2 produced. Using the ideal gas law under the given conditions, the volume of H2 can be calculated, resulting in approximately 2.29 L H2.

Question 4

O2 produced from 2.80 liters of CO2 at STP would be half, due to the mole ratio of CO2 to O2 in the equation. Therefore, the correct answer is 1.40 liters of O2.

Question 5

Knowing the volume of H2 gas produced and using its molar volume at STP, calculate the moles of H2, and from there, use stoichiometry to determine the mass of Na metal initially used, yielding 10.0 grams of sodium metal.

Question 6

This statement is false; while at STP (273.15 K, 1 atm), one mole of a gas occupies 22.4 L, the conditions specified in the question (298 K, 1 atm) do not define STP.

Question 7

At STP, the volume of one mole of gas is 22.4 liters.

Answer:

The answer to yourquestion is: a) H₂CNN

Explanation:

Remember that H has one valence electron

                          C has four valence electrons

                          N has five valence electrons

I recommend you draw the Lewis structure to decide which isthe best arrengement. The Lewis structure is in the attachment.

The most plausible atomic arrangement for cyanamide (CH 2 N 2 ) is likely H2NCN, as the formal charges of the atoms in this configuration is closest to their natural state. The formal charge is calculated using an equation that considers valence and bonding electrons. In H2NCN, both nitrogen atoms have a charge of -1 while the carbon remains neutral.

To determine the most plausible atomic arrangement for cyanamide (CH 2 N 2 ), we need to look at their formal charges. For the three arrangements given, H2CNN, H2NCN, and HNNCH, we can calculate the formal charges by using the formula: Formal Charge = [# of valence electrons on atom in free state] - [non-bonding electrons + 1/2(bonding electrons)].

Considering cyanamide, nitrogen typically has a charge of -3, carbon is typically neutral, and hydrogen is typically +1. Therefore, looking at the three arrangements, H2NCN seems to have the most plausible arrangement based on the formal charges. In H2NCN, both nitrogen atoms have a charge of -1 while the carbon remains neutral, which is closest to the natural state of these elements.

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Answer:

63.5 amu

Explanation:

Answer via Educere/ Founder's Education

The relative atomic mass (r.a.m.) of copper (Cu) is 63.5 atomic mass units (amu).

In the element copper (Cu), the Relative Atomic Mass, often abbreviated as r.a.m, is a value obtained by comparing a given atom's mass to the atomic mass of the isotope carbon-12. Therefore, the r.a.m. of copper (Cu) is 63.5 amu (atomic mass units). This value, depending on the nature of the atoms and molecules involved, may vary slightly. Nevertheless, in most general context, you can use 63.5 amu as the r.a.m. of copper.

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Answer:

Th answer to your question is:

a)  3.5 x10⁻¹⁰ meters; 0.35 nm

b) 6857142.86 atoms

c) Volume = 2.06 x 10⁻²³ cm³

Explanation:

a) data

Uranium atoms = 3.5A°

meters

           1 A° ----------------  1 x 10 ⁻¹⁰ m

         3.5A° ---------------  x

 x = 3.5(1 x10⁻¹⁰)/ 1 = 3.5 x10⁻¹⁰ meters

          1 A° ------------------ 0.1 nm

        3.5 A° ---------------- 0.35 nm

b) 2.4 mm

Divide 2,40 mm / uranium diameter

But, first convert 3,5A° to mm   = 3.5 x 10⁻⁷ mm

# of uranium atoms = 2.4 / 3.5 x 10⁻⁷ = 6857142.86

c) volume in cubic cm

Convert 3.5A° to cm  = 3.5 x 10⁻⁸

Volume = 4/3 πr³ = (4/3) (3.14)(1.7 x10⁻⁸)³

Volume = 2.06 x 10⁻²³ cm³

a) The radius of a uranium atom is approximately 1.75 Å, which is equivalent to 17.5 nm. b) To span a distance of 2.40 mm, approximately 6.86 billion uranium atoms would be needed. c) The volume of a single uranium atom, assuming it is a sphere, is (4/3) * π * (1.75 [tex]Å)^3.[/tex]

a.) To find the radius of a uranium atom, we can divide the given diameter by 2. The formula to convert the radius from angstroms (Å) to meters (m) is:

(radius in m) = (diameter in m) / 2

Given that the diameter is 3.50 Å, we can calculate the radius:

(radius in m) = (3.50 Å / 2) = 1.75 Å

Next, to convert the radius from angstroms to nanometers (nm), we can use the conversion factor:

(radius in nm) = (radius in Å) * 10

Therefore, the radius of a uranium atom is 1.75 Å (in meters) and 17.5 nm (in nanometers).

b.) To find how many uranium atoms would be needed to span 2.40 mm, we can divide the distance by the diameter of a uranium atom:

Number of uranium atoms = (2.40 mm) / (diameter of a uranium atom)

To calculate this, we need to convert the diameter from angstroms to millimeters using the conversion factor:

1 mm = 10000000 Å

Therefore, the number of uranium atoms needed to span 2.40 mm is:

Number of uranium atoms = (2.40 mm) / (3.50 Å * (1 mm / 10000000 Å))

c.) To find the volume of a single uranium atom, assuming it is a sphere, we can use the formula for the volume of a sphere:

Volume = (4/3) * π * (radius)^3

Using the radius we calculated in part a, we can substitute the value into the formula and solve for the volume:

Volume = (4/3) * π * (1.75[tex]Å)^3[/tex]

Since the volume is given in cubic centimeters (cm³), we need to convert the unit of length from angstroms to centimeters:

1 Å = 1E-8 cm

Therefore, the volume of a single uranium atom is:

Volume = (4/3) * π * (1.75 Å * (1E-8 cm [tex]/ 1 Å))^3[/tex]

Once we calculate this volume, we will have the answer in cubic centimeters (cm³).