A sample of gas at a pressure of 121.59 kPa, a volume of 31 L, and a temperature of 360 K contains how many moles of gas? A. 0.71 mol B. 0.96 mol C. 1.3 mol D. 1.8 mol

Answer:

a. P atoms introduce additional mobile negative charges

Explanation:

Silicon atoms have a valency of 4, this means that there are 4 electrons in their outermost orbital. Thus, in silicon crystals, each Si atom is connected to 4 different Si atoms.

When we replace a few of these Si atoms with P atoms, which have a valency of 5, 4 out of these 5 outermost electrons will be bonded with the surrounding Si atoms. The fifth electron would not be bonded, meaning it would be free to move, acting as a mobile negative charge carrier (See the picture below).

Just to be clear, regarding option c), it's true that Ge atoms have more electrons that Si atoms, however that it's not a reason for an increase in conductivity.

When a small fraction of silicon (Si) atoms in a crystal are replaced with a different element, such as phosphorus (P) or germanium (Ge), the conductivity of the crystal increases. This increase in conductivity is due to P atoms introducing additional mobile negative charges, which allow electrons to move freely and contribute to the conductivity. This is known as n-type doping.

When a small fraction of silicon (Si) atoms are replaced with those of a different element, such as phosphorus (P) or germanium (Ge), the conductivity of the Si crystals increases. The correct reason for this increase in conductivity is that P atoms introduce additional mobile negative charges. These extra electrons from the P atoms are able to move freely within the crystal lattice, contributing to the conductivity of the material. This type of doping with impurities is called n-type doping, where the primary carriers of charge are negative electrons.

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Hydrogen was the most common gas in the Earth's early atmosphere. The introduction of oxygen happened much later with the process of photosynthesis. Despite being present later, oxygen was not a primary component of the initial atmosphere.

The gas that was most likely present in Earth's earliest atmosphere was hydrogen. This is based on scientific research, which suggests that the early atmosphere of the Earth was predominantly composed of light gases like hydrogen and helium. Over time, the Earth's atmosphere evolved and changed due to various geological and biological processes.

For instance, the process of photosynthesis introduced oxygen into the atmosphere much later in Earth's history. It's also important to note that despite being present, oxygen was not one of the primary components of Earth's early atmosphere.

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Answer:

26.18 g

Explanation:

The molar volume of oxygen at 1 atm and 0ºC is 22.4 L/mol.

If you want to react 7 L it means you will use

7/22.4 = 0.3125 moles of oxygen

the balanced equation is 3 Fe + 2 O2 = Fe3O4

which means that 2 moles of oxygen reacts with 3 moles of iron

If you have 0.3125 moles of oxygen, 0.468 moles of iron will be needed.

Iron molecular weight is 55.84 and than, 0.3125 moles corresponds to a mass of iron equal to 26.18

To determine the mass of iron required to react with 7.0 L of oxygen, we need to use the balanced equation and the molar ratios between iron and oxygen.

To determine the mass of iron required to react with 7.0 L of oxygen, we need to use the balanced equation and the molar ratios between iron and oxygen. The balanced equation for the reaction is:

4Fe + 3O2 → 2Fe2O3

From the equation, we can see that it takes 4 moles of iron to react with 3 moles of oxygen to produce 2 moles of Fe2O3. First, we need to convert the volume of oxygen to moles by using the ideal gas law. At 0 degrees Celsius and 1 atm, the volume can be calculated as follows:

V = nRT/P

V = (7.0 L)(0.0821 L/mol·K)(273 K) / (1 atm)

V = 17.109 moles

Next, we need to use the molar ratios to determine the moles of iron required. Since the ratio is 4 moles of iron to 3 moles of oxygen, we can set up a proportion:

4 moles Fe / 3 moles O2 = x moles Fe / 17.109 moles O2

Solving for x:

x = (4 moles Fe)(17.109 moles O2) / 3 moles O2

x = 22.812 moles Fe

Lastly, we can convert the moles of iron into grams using the molar mass of iron:

(22.812 moles)(55.85 g/mol) = 1276.06 grams

Therefore, approximately 1276.06 grams of iron is required to react with 7.0 L of oxygen to produce Fe3O4.

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When a chemical reaction take up energy then this type of reaction is called endothermic reaction. Therefore,  heat is being transferred from the surroundings to the system.

In chemistry there are various type of reaction out of which the two main types are the exothermic reaction and endothermic reaction.

Exothermic reaction is the one in which energy releases in form of heat respiration reaction is an example of exothermic reaction. In respiration food that we eat are broken down in glucose with release of energy.

Endothermic reaction is the one in which energy is taken out during the reaction. Photosynthesis is an example of endothermic reaction where sunlight energy is taken by the plants to make food.

H[tex]_2[/tex]+ Br[tex]_2[/tex]→ 2 HBr  ΔH = 72.80 kJ.

72.80 kJ of heat should be transferred when 38.2g of liquid brownie reacts with excess hydrogen gas to form hydrogen bromine.  Heat is being transferred from the surroundings to the system.

Therefore,  heat is being transferred from the surroundings to the system.

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The question contains a typo and seems to refer to a chemical reaction between hydrogen gas and bromine to form hydrogen bromide. Without specific data on the enthalpy change, one cannot conclusively say how much heat is transferred or definitively whether the reaction is exothermic or endothermic.

The original question seems to have a typo referring to 'liquid brownie' reacting with hydrogen gas, which is non-sensical in a chemistry context. Assuming the question pertains to the reaction of hydrogen gas (H₂) with bromine (Br₂) to form hydrogen bromide (HBr), as per the given example:

H₂(g) + Br₂(g) → 2 HBr(g)

To determine how much heat is transferred during this reaction and whether the reaction is exothermic or endothermic, we need additional data such as the enthalpy change (ΔH) for the reaction, which is not provided in the question. In general, if ΔH is negative, the reaction is exothermic, meaning heat is transferred from the system to the surroundings. Conversely, if ΔH is positive, the reaction is endothermic, and heat is transferred from the surroundings to the system.

Given the typical nature of reactions between halogens and hydrogen, one could infer that the reaction is likely exothermic, but without specific data on enthalpy change, this is an educated guess.

Answer:

11.6 mL

Explanation:

First we need to calculate the number of moles of Zn2+ present in the solution:

[tex]n=V*C\\n_{Zn^{2+}}=0.65*0.012=7.8x10^{-3}moles[/tex]

As the charge of ion zinc is 2+ and the charge of hydroxide is 1-, we need double moles of NaOH:

[tex]n_{NaOH}=0.0156moles[/tex]

As we have the concentration and the moles, we can calculate the volume:

[tex]V=\frac{n}{C} \\\\V=0.01164L=11.6mL[/tex]

Answer:

43.668 kg

Explanation:

First we set the equation:

[tex]Al_{2}O_{3} + 6NaOH + 12HF \longrightarrow 2Na_{3}AlF_{6}+9H_{2}O[/tex]

Now, we need to now the kmoles for each reactant:

[tex]M_{Al_{2}O_{3}}=101.96kg/kmol\\M_{NaOH}=40kg/kmol\\M_{HF}=20.01kg/kmol\\n_{Al_{2}O_{3}}=0.104kmol\\n_{NaOH}=1.285kmol\\n_{HF}=2.57kmol[/tex]

With this, we can see that the limit reactant is the aluminum oxide, so, with the equation for the reaction we know that 1 kmol of aluminum oxide, produces 2 kmol of cryolite, so we set a rule of three and see that 0.208 kmoles of cryolite are produced, the we proceed to calculate the mass:

[tex]M_{Na_{3}AlF_{6}}=209.94kg/kmol\\n_{Na_{3}AlF_{6}}=0.208kmol\\m_{Na_{3}AlF_{6}}=43.668kg[/tex]

Answer:

1. evaporation of ethanol

2. condensation of ethanol

Explanation:

The first and second process imply a change of phase, which is essentially a  physical change because the matter change from liquid to gas and from gas to liquid respectively and it can ve reversed.

The options 3 to 6 are related to a chemical reaction because the outcomes of the process are different compounds than the ones we had at the beginning. The main reaction that implies process 3 to 6 is represented by:

C6H12O6 +  impurities → 2 C2H5OH + 2 CO+ C

For option 5. burning of natural gas , we have:

Natural gas(Mainly a mix of CH4 and C2H10) +O2→COn+H2O

Therefore, justo 1 and 2 are a physical change

The evaporation and condensation of ethanol are physical changes as these do not alter the substance's chemical composition. The formation of carbon deposit, carbon dioxide gas, the burning of natural gas, and ethanol formation all involve chemical reactions and are thus considered chemical changes.

In the experiment described, both physical and chemical changes occur. Physical changes involve a change in physical state without altering the substance's chemical composition. Therefore, the evaporation of ethanol (#1), and the condensation of ethanol (#2) are physical changes. The substance remains ethanol in both cases; it merely changes from liquid to gas or from gas to liquid.

Chemical changes, on the other hand, involve a chemical reaction where a new substance is formed. Therefore, the formation of a carbon deposit inside the flask from decomposition of glucose (#3), the formation of carbon dioxide gas from glucose (#4), the burning of natural gas (#5), and the formation of ethanol from glucose by yeast (#6) are examples of chemical changes.

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Answer: The value of n =2.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given molecules}}{\text {Avogadro's number}}=\frac{8.06\times 10^{20}}{6.023\times 10^{23}}=0.0013moles[/tex]

0.0013 moles of [tex]XeF_n[/tex] weigh = 0.227 grams

Thus 1 mole of [tex]XeF_n[/tex]  will weigh = [tex]\frac{0.227}{0.0013}\times 1=174.62[/tex]  grams

Molar mass of [tex]XeF_n[/tex] = [tex]1(131.3)+n(19)=174.62[/tex]

[tex]n=2[/tex]

Thus the value of n =2

The xenon fluoride compound is likely XeF2. This was deduced by relating the given mass and number of molecules of the compound to its molar mass and Avogadro's number. The calculated molar mass fits best with that of XeF2.

The subject of your question lies in the area of analytical chemistry, specifically in calculations involving the concept of the mole. Here we want to find the molecular formula for the xenon fluoride, XeFn using the data provided. We know the Avogadro's number which states that one mole of any substance contains 6.02 x 1023 molecules.

In this question, we're given that 8.06 x 1020 molecules of XeFn weigh 0.227 g. We can calculate the molar mass of this specific compound. To elaborate, if one mole weighs 'M' grams and contains 6.02 x 1023 molecules, then 0.227 g would have (0.227/M) moles or (0.227/M) x 6.02 x 1023 molecules. If we set this equal to 8.06 x 1020 and solve for 'M', the molar mass of the compound, we can calculate a value of approximately 169 g/mol.

Given that the molar mass of xenon (Xe) is about 131 g/mol and that of fluorine (F) is approximately 19 g/mol, we can reason that the molar mass of the compound XeFn is closest to that of XeF2, since 131 + 2(19) = 169 g/mol. Therefore, the xenon fluoride compound being referred to in the problem is XeF2.

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Answer:

How many grams of CCL4 were formed? 116.9 g

How many grams of Cl2 reacted with the CH4? 243.8 g

Explanation:

First we need to know the molar mass for every element or compound in the reaction:

[tex]M_{CH_{4}}=16 g/mol\\M_{CH_{3}Cl}=50.49g/mol\\M_{CH_{2}Cl_{2}}=84.93g/mol\\M_{CHCl_{3}}=119.38g/mol\\M_{CCl_{4}}=153.82g/mol[/tex]

Now we proceed to calculate the amount of moles produced, per product:

[tex]n_{CH_{4}}=2.4\\n_{CH_{3}Cl}=0.18\\n_{CH_{2}Cl_{2}}=0.55\\n_{CHCl_{3}}=0.91\\n_{CCl_{4}}=n_{CH_{4}}-(n_{CH_{3}Cl}+n_{CH_{2}Cl_{2}}+n_{CHCl_{3}})\\n_{CCl_{4}}=0.76mol\\m_{CCl_{4}}=n_{CCl_{4}}*M_{CCl_{4}}\\m_{CCl_{4}}=116.9g[/tex]

To calculate the mass of chlorine we just need to make a mass balance:

[tex]m_{CH_{4}}+m_{Cl_{2}}=m_{CH_{3}Cl}+m_{CH_{2}Cl_{2}}+m_{CHCl_{3}}+m_{CCl_{4}}\\m_{Cl_{2}}=m_{CH_{3}Cl}+m_{CH_{2}Cl_{2}}+m_{CHCl_{3}}+m_{CCl_{4}}-m_{CH_{4}}\\m_{Cl_{2}}=243.8g[/tex]

Final answer:

To find the grams of CCl4 formed, we need to calculate the grams of Cl2 reacted with CH4 and then convert it to grams of CCl4 using the stoichiometry of the balanced equation. 2.4 mol of Cl2 reacted with 38.4 g of CH4. Therefore, 369.17 grams of CCl4 were formed.

Explanation:

To find the grams of CCl4 formed, we need to calculate the grams of Cl2 reacted with CH4 and then convert it to grams of CCl4 using the stoichiometry of the balanced equation. We start by calculating the molar mass of CH4, which is 16.04 g/mol. Since all the CH4 reacted, we can use its mass to calculate the moles of CH4. 38.4 g CH4 / 16.04 g/mol = 2.4 mol CH4.

Next, we use the balanced equation to determine the moles of Cl2 that reacted with CH4. From the equation, we know that 1 mole of CH4 reacts with 4 moles of Cl2. So, 2.4 mol CH4 x (4 mol Cl2 / 1 mol CH4) = 9.6 mol Cl2.

Now, we can convert the moles of Cl2 to grams using its molar mass of 70.90 g/mol. 9.6 mol Cl2 x 70.90 g/mol = 681.84 g Cl2 reacted. This is the answer to the second part of the question.

Finally, we use the stoichiometry of the balanced equation to calculate the grams of CCl4 formed. From the equation, we know that 1 mole of CH4 reacts to produce 1 mole of CCl4. So, the moles of CCl4 formed is equal to the moles of CH4 reacted, which is 2.4 mol.

Finally, we convert the moles of CCl4 to grams using its molar mass of 153.82 g/mol. 2.4 mol CCl4 x 153.82 g/mol = 369.17 g CCl4. Therefore, 369.17 grams of CCl4 were formed.

Answer:

b) Solvent B was not a good mobile phase because it could not dissolve the ink.

Explanation:

The base for chromatography is that the mixture to analyze is soluble in the solvent, if this happen, there will be separation. The only thing that may affect the separation, is that the mixture is not soluble in the solvent.

Answer: Option B

Explanation:

The paper chromatography can be defined as the technique in which there are two phase. One is stationary phase that lies in the cellulose fiber of the paper.

The mobile phase is the solvent that moves on the stationary phase. This is the usually solvent.

The interaction between the mobile phase and stationary phase must be good so as to carry the stationary phase along with the mobile phase.

If the stationary phase is not carried along with the mobile phase then the interaction between the stationary phase and mobile phase is not strong.

Answer:

Because they are normally dissolved in water

Explanation:

The phospholipids can act as molecules that are carry inside or outside the cells, to transport it they are dissolved in another hydrophilic media, in this way the hydrophilic part can be in the outside part of the group of moleucles stick together like in cell membranes that are conformed by a double lipidic layer, in which the hydrophobic part it in the inside part. The cells normally are surrounded by different water solutions as the blood or another solutions.

Answer: The mass of bromine reacted is 160.6 grams.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]        .....(1)

Given mass of aluminium = 18.1 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of aluminium}=\frac{18.1g}{27g/mol}=0.670mol[/tex]

The chemical equation for the reaction of aluminium and bromide follows:

[tex]2Al+3Br_2\rightarrow 2AlBr_3[/tex]

By Stoichiometry of the reaction:

2 moles of aluminium reacts with 3 moles of bromine gas

So, 0.670 moles of aluminium will react with = [tex]\frac{3}{2}\times 0.670=1.005mol[/tex] of bromine gas.

Now, calculating the mass of bromine gas, we use equation 1:

Moles of bromine gas = 1.005 moles

Molar mass of bromine gas = 159.81 g/mol

Putting values in equation 1, we get:

[tex]1.005mol=\frac{\text{Mass of bromine}}{159.81g/mol}\\\\\text{Mass of bromine}=(1.005mol\times 159.81g/mol)=160.6g[/tex]

Hence, the mass of bromine reacted is 160.6 grams.

Final answer:

To find out how many grams of bromine react with 18.1g of aluminum, we first determine the mass ratio from a previous reaction and then use that to calculate the mass of bromine. The proportional relationship indicates that approximately 159.8 grams of bromine would react with 18.1 grams of aluminum.

Explanation:

The student asks how many grams of bromine react with 18.1g of aluminum. The chemical reaction is similar in behavior to the reaction of aluminum with chlorine where aluminum bromide is formed. Given a previous reaction scale of 27g aluminum yielding 266.7g aluminum bromide, we use stoichiometry to find the grams of bromine reacting with 18.1g aluminum.

First, determine the molar mass of aluminum (Al) and aluminum bromide (AlBr3). Aluminum has a molar mass of approximately 27g/mol and aluminum bromide has a molar mass of roughly 267g/mol based on the atomic weights of aluminum (approximately 27) and bromine (approximately 80 per bromine atom, with three bromine atoms per molecule). Find the mole ratio of Al to AlBr3 from the balanced chemical equation, which is 2:2 (or 1:1 for simplicity). From the initial reaction scale, you can calculate the ratio of aluminum to bromine involved in the reaction.

Next, set up a proportional relationship:

To solve for x (the mass of aluminum bromide produced from 18.1g of aluminum), cross-multiply and divide:

27g Al * x g AlBr3 = 18.1g Al * 266.7g AlBr3

x = (18.1g Al * 266.7g AlBr3) / 27g Al

x ≈ 177.7g AlBr3

Finally, determine the mass of bromine involved. Since 27g of Al produces 266.7g of AlBr3, the mass of bromine can be calculated by subtracting the mass of aluminum from the total mass of the product:

266.7g AlBr3 - 27g Al = 239.7g Br2

Now, calculate the mass of bromine that would react with 18.1g of Al using the same proportion:

Cross-multiply and solve for y:

y = (18.1g Al * 239.7g Br2) / 27g Al

y ≈ 159.8g Br2

Therefore, approximately 159.8 grams of bromine would react with 18.1 grams of aluminum.

Answer:

The length (quantitative measurement) of the pencil is 5.5

Explanation:

Quantitative measurement are type of measurements which result in numbers    

with or without appropriate unit .

To obtain the length of the pencil , the pencil starts from 0 of the scale and ends at 5.5 , hence the value is 5.5.

Quantitative measurement in the picture is the LENGTH of the pencil shown in the picture.

The value of quantitative measurement obtained from the picture is 5.5 , therefore length of the pencil is 5.5 .

Answer:

The answer to your question is: the last option ( 2 mole Fe : 3 mol Cl2)

Explanation:

The mole ratio in a chemical reaction is expressed by the integers before each element, for example if we have

                                                           2H2SO4 + 3NaOH,

The mole ratio between, H2SO4 and NaOH is 2H2SO4 : 3 NaOH

In your question, the question is the ratio between Fe and Cl2, so we look at their coefficients and then:

                    2 mole Fe : 3 mol Cl2

Answer:

1st: Light excites an electrom from photosystem II

2nd: Light excites an electrom from photosystem I

3rd: Electrons pass through an electron chain, which generates a H+ gradient used to make ATP

4th: Electron reduce NADP+ to NADPH

Explanation:

While the light simultaneously excites both photosystems, it must first occur in photosystem II and then be able to transfer the e-energized to photosystem I

1st Step:

Within the photosystems we find different photosynthetic pigments, that is, capable of absorbing light. These pigments are classified according to the maximum absorption wavelengths.

When the light hits the photosystems they absorb it and the delocalized -e (electrons) are energized or "excited".

Then these energized ones are transferred to molecules within the membrane that houses the pigments.

The e- that takes the photosystem I are provided by the photosystem II

2nd and 3rd Step:

The -e energized from photosystem II are transferred to a transport chain of -e within the membrane containing the pigments. As these -e circulate, they lose energy that is used to translocate H + (protons).

The accumulation of H + within the membrane generates an electrochemical gradient.

H + return to the stroma through the enzyme ATP synthase. This operation is called chemosmosis.

This enzyme uses H + to catalyze the synthesis of ATP (ADP + Pi), a process called phosphorylation.

4th Step:  

The e-energized of photosystem I are used to reduce NADP + and generate NADPH that are used in conjunction with ATP to generate "light independent reactions"

Those lost from photosystem I are replaced by e-de-energized from photosystem II, while those lost from photosystem II are replaced by e-released from water by photolysis.

Water is divided by the energy of light into H + (used in chemosmosis) and oxygen (released as a byproduct)

The correct order of the steps for the light reactions in photosynthesis is: Light excites an electron from photosystem II, electrons pass through an electron transport chain, excitation of an electron in photosystem I, and finally, the reduction of NADP+ to NADPH i.e. 2, 4, 1, 3.

The correct order of the steps for the light reactions in photosynthesis is:

This process starts with the absorption of light in photosystem II, which excites electrons. These electrons pass through an electron transport chain that uses the energy from the electrons to pump H+ ions and establish a gradient. This H+ gradient is used to synthesize ATP. Finally, light is absorbed by photosystem I which leads to the reduction of NADP+ to NADPH.

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Final answer:

The limiting reagent in this reaction is CuSO4.

Explanation:

The limiting reagent in this reaction is CuSO4.

To determine the limiting reagent, we need to compare the number of moles of each reactant with the stoichiometric ratio of the balanced equation. First, calculate the number of moles of Na2S and CuSO4 using their molar masses. Then, compare the moles of each reactant to the stoichiometric ratio in the balanced equation. The reactant that produces fewer moles of the product is the limiting reagent.

In this case, calculate the moles of Na2S and CuSO4. The stoichiometric ratio is 1:1 for Na2S and CuSO4. Since the moles of CuSO4 is smaller than the moles of Na2S, CuSO4 is the limiting reagent.

Answer:

a) 1.43 g/L

b) 1.27 g/L

Explanation:

Oxygen is an ideal gas, so, using the ideal gas equation:

PV = nRT  

Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature (always in Kelvin!).

n = mass (m)/molar mass (MM), so:

[tex]PV = \frac{m}{MM}RT[/tex]

PVMM = mRT

[tex]PMM = \frac{m}{V} RT[/tex]

m/V is the density (d), so:

d = [tex]\frac{PMM}{RT}[/tex]

R = 0.082 atm.L/(mol.K) and MM of O2 = 2x 16 = 32 g/mol

a) for STP, P = 1 atm and T = 0ºC = 273 K

d = [tex]\frac{1x32}{0.082x273}[/tex]

d = 1.43 g/L

b) P = 1 atm and T = 35ºC + 273 = 308 K

d = [tex]\frac{1x32}{0.082x308}[/tex]

d = 1.27 g/L

As oxygen is an ideal gas, the ideal gas equation is as follows:

[tex]PV= nRT[/tex]

P = pressure, V = volume, n = number of moles, R = gas constant, and T = temperature. n= m/MM

Thus, answers for calculated densities are 1.43g/L and 1,27g/L.

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Answer:

0.133 M

Explanation:

The volume of the solution is given, so in order to find concentration, the number of moles must be found, since C = n/V.

The balanced reaction equation is:

HI + KOH ⇒ H₂O + KI

Thus, the moles of KOH added to neutralize all of the HI will be equal to the moles of HI that must have been present.

The amount of KOH that was added is calculated as follows.

n = CV = (0.145 mol/L)(45.7 mL) = 6.6265 mmol KOH = 6.6265 mmol HI

Since HI and KOH are related in a 1:1 molar ratio, the same amount of HI must have been present.

Finally, the concentration of HI is calculated:

C = n/V = (6.6265 mmol) / (50.0 mL) = 0.133 mol/L = 0.133 M

The final molarity of the hydroiodic acid solution is 0.133 M.

To determine the molarity of the hydroiodic acid (HI) solution, we can use the relationship between the moles of HI and KOH used in the titration process. The balanced equation for the neutralization reaction is:

HI(aq) + KOH(aq) → KI(aq) + H₂O(l)

The reaction shows a 1:1 molar ratio between HI and KOH. Given the volume and molarity of the KOH solution, we can calculate the moles of KOH used:

Moles of KOH = Molarity of KOH × Volume of KOH (in liters) = 0.145 M × 0.0457 L = 0.0066265 mol

Since the ratio is 1:1, the moles of HI will be the same as the moles of KOH, which is 0.0066265 mol. Now we can find the molarity of the HI solution using the volume of HI solution:

Molarity of HI = Moles of HI / Volume of HI (in liters) = 0.0066265 mol / 0.0500 L = 0.13253 M

Therefore, the molarity of the hydroiodic acid solution is 0.133 M (rounded to three significant figures).

Answer:

Based on the conditions given as vapor pressure is 256 torr,the methanol-water interactions are stronger than the methanol-methanol and water-water interactions is the correct option.

The solution of methanol and water is non-ideal as the actual vapor pressure is lower than the expected value from Raoult’s law, indicating that the solute-solvent interactions are stronger than those within the pure components.

A solution of methanol and water with a mole fraction of water of 0.312 and a total vapor pressure of 21 torr at 39.9 degrees C is not displaying ideal behavior. In an ideal solution, Raoult's law predicts that the total vapor pressure is the sum of the partial pressures of the components, each calculated as the product of the pure component's vapor pressure and its mole fraction in the solution.

In this case, if we assume the solution is ideal, the expected vapor pressure would be higher because methanol and water have vapor pressures of 256 torr and 55.3 torr, respectively. The given total vapor pressure (21 torr) is significantly less than the expected value calculated using Raoult’s law, indicating that the solution exhibits non-ideal behavior.

This observation implies that the intermolecular interactions between methanol and water are stronger than those in the pure substances (methanol with methanol and water with water). Thus, the interactions in the mixture reduce the tendency to escape to the vapor phase, resulting in a lower total vapor pressure than predicted for an ideal solution.

Answer:

b

Explanation:

Answer:

The answer to your question is p O2 = 17.2 kPa, letter b

Explanation:

Remember that in a mixture of gases, total pressure is equal to the sum of the pressure of individual gases.

Then

          Pt = p He + p CO2 + p O2

Data

Pt = 101.3 kPa

pHe = 84kPa

pCO2 = 0.1 kPa

p O2 = ?

       Substitution

       101.3 = 84 + 0.1 + pO2

       pO2 = 101.3 - 84 - 0.1

        pO2 = 17.2 kPa

Answer:

mouth, stomach

Explanation:

The carbohydrates start the digestion in the mouth, with the enzymes dissolved in the saliva, and during the rest of the digestive system it continues until the intestine.

Most of the digestion of lipids is in the intestine, with the lipid enzymes that come from the bile. But it starts in the mouth with the absorption of the small ones.

The chemical digestion of the proteins begins in the stomach because the peptide enzyme it turns active with the acidic environment of the stomach.

Answer:

c) kg

Explanation:

Kilograms stands alone. It has to be hooked up to another unit for it to be a derived unit.

I am joyous to assist you anytime.

Answer:

The answer to your question is c) kg

Explanation:

Derived units is when to units are combined to get a new one, For example, combining length and length gives lenght² and then we measured area, this is a derived unit.

a) cm³ this answer is wrong because we are combining length x length x length so we get the units of volume, this is a derived unit.

b) g/mL this answer is wrong because we are combining mass and volume so we can measure density, this units are derived.

c)kg this answer is correct because it measures mass and only that, then it's a fundamental unit.

The names have been correctly sorted as follows;

Molecule 1 - 3-chloro-4-iodo-2-methyl hexane

Molecule 2 - 4-chloro-3-iodo-2-methyl hexane

Molecule 4 - 3-chloro-2-iodo-4-methyl hexane

The IUPAC naming system for organic compounds

To name organic compounds, identify the longest continuous chain of carbon atoms, the "parent chain". Look up the root name of this chain based on its length, and identify each substituent and its position on the chain with a number.

Combine the root name, substituent prefixes with their positions, and functional group suffixes to form the complete name. Lastly, use commas and hyphens to separate and connect different parts of the name.

Answer:

Concentration of A at equilibrium = 1 - 0.5 = 0.5 M

Explanation:

[tex]A \leftrightharpoons B + C[/tex]

Equilibrium constant = 0.5

Initial concentration of A = 1 M

             [tex]A \leftrightharpoons B + C[/tex]

Initial      1            0     0

At equi.  1-x         x      x

Equilibrium constant = [tex]\frac{[B][C]}{[A]}[/tex]

[tex]0.5 = \frac{x \times x}{1-x} \\0.5(1-x) = x^2\\0.5 -0.5x = x^2\\x^2+0.5x - 0.5 = 0[/tex]

on solving,

x = 0.5 M

Concentration of A at equilibrium = 1 - 0.5 = 0.5 M

Final answer:

Use the equilibrium constant and an ICE table to set up the equation Keq = [B][C]/[A] and solve for x to determine the equilibrium concentration of A after adding more of the substrate to the mixture.

Explanation:

To calculate the concentration of A when the reaction reaches equilibrium after adding 0.50 moles of A, we use an ICE table for the equilibrium process and the equilibrium constant given as Keq = 0.5. Initially, we have 1.0 M of A without B. Upon addition, we have 1.50 M of A. If we let x be the change in concentration of A as it reacts to form B and C, at equilibrium, [A] = 1.50 - x, [B] = x, and [C] = x because the mole ratio of A:B:C is 1:1:1. Setting up the equilibrium expression as Keq = [B][C]/[A], and substituting the terms with the expressions in terms of x, we get 0.5 = x²/(1.50 - x). We must solve this quadratic equation to find the value of x and, consequently, the equilibrium concentration of A.

Equilibrium constant = ([B][C])/([A])

0.5 = (x * x)/(1-x) \n0.5(1-x) = x^2\n0.5 -0.5x = x^2\nx^2+0.5x - 0.5 = 0

on solving,

x = 0.5 M

Concentration of A at equilibrium = 1 - 0.5 = 0.5 M

Answer:

The answer to your question is: 85.458 amu

Explanation:

data

Rubidium-85   A = 84.9118 amu   abundance = 72.15%  

Rubidium - 87  A = 86.9092 amu abundance = 27.85%

Atomic weight = ?

Atomic weight = 84.9118(0.7215) + 86.9092(0.2785)

Atomic weight = 61.2538 + 24.2042

Atomic weight = 85.458 amu

Rubidium has two naturally occurring isotopes, rubidium -85 ( atomic mass = 84.9118 amu; abundance = 72.15%) and rubidium -87 (atomic mass = 86.9092; abundance = 27.85%). The atomic weight of rubidium is 85.46 amu.

To calculate the average atomic mass of element use this formula

Average Atomic Mass = f₁M₁ + f₂M₂

where,

f = Fraction of natural abundance of isotope

M = Mass number of isotope

Isotope Rubidium -85 (Atomic mass = 84.9118 amu and Abundance = 72.15)

Abundance = [tex]\frac{72.15}{100}[/tex]

                    = 0.7215

Isotope Rubidium- 87 (Atomic mass = 86.9092 amu and Abundance = 27.85%)

Abundance = [tex]\frac{27.85}{100}[/tex]

                    = 0.2785

Now put the value in above formula, we get:

Average Atomic Mass = f₁M₁ + f₂M₂

                                     = (84.9118 × 0.7215) + (86.9092 × 0.2785)

                                     = 61.26 + 24.20

                                     = 85.46 amu

Thus from the above conclusion we can say that Rubidium has two naturally occurring isotopes, rubidium -85 ( atomic mass = 84.9118 amu; abundance = 72.15%) and rubidium -87 (atomic mass = 86.9092; abundance = 27.85%). The atomic weight of rubidium is 85.46 amu.

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Cheddar cheese, iced tea with no ice, chicken noodle soup, vanilla ice cream, antifreeze, rocky road ice cream, and trail mix are examples of homogeneous and heterogeneous mixtures.

Homogeneous mixtures:

Heterogeneous mixtures:

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Homogeneous mixtures are uniform throughout; examples include iced tea with no ice and antifreeze. Heterogeneous mixtures aren't uniform and individual components can be seen; examples include cheddar cheese, chicken noodle soup, bread pudding, vanilla ice cream, rocky road ice cream, and trail mix.

The question pertains to the classification of various items as either homogeneous or heterogeneous mixtures. A homogeneous mixture is a combination of substances with a composition that is uniform throughout. In such a mixture, you cannot see the individual components, and there's no variation from point to point. On the other hand, a heterogeneous mixture is a collection of substances whose composition can vary from point to point and where the individual components can be seen.

From the provided list, the examples of homogeneous mixtures are Iced tea with no ice and Antifreeze because their compositions are the same throughout. Meanwhile, Cheddar Cheese, Chicken noodle soup, Bread pudding, Vanilla ice cream, Rocky road ice cream, and Trail mix could be considered examples of heterogeneous mixtures as their compositions are not uniform and the different components can be seen distinctly.

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The ion with a 2+ charge is the one with 10 protons, 9 neutrons, and 8 electrons because there are two more protons, which are positively charged, than electrons, which are negatively charged thus giving it a positive charge.

The correct answer is A: an ion with 10 protons, 9 neutrons, and 8 electrons. The charge of an ion is determined by the difference in the number of protons (positive charges) and electrons (negative charges). Note that the numbers of protons and electrons match exactly in a neutral atom. Therefore, an ion with 10 protons and 8 electrons would have a charge of 2+, since there are two more protons than electrons, giving it an overall positive charge.

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An ion with a charge of 2+ is an ion with 10 protons, 9 neutrons, and 8 electrons (option A).

To determine which ion has a charge of 2+, we need to understand that an ion with a 2+ charge has lost two electrons compared to its neutral state. Let's examine the options:

The correct answer is A, as it is the only ion with a 2+ charge.

Answer:

The answer is: Law of multiple proportions        

Explanation:

The law of multiple proportions is a law of chemical combination given by Dalton in 1803.

According to this law, if more than one chemical compound is formed by combining two elements, then the mass of an element that combines with the fixed mass of other element is represented in the form of small whole number ratio.

Therefore, is an illustration of the law of the law of multiple proportions.

Methane and ethane are both made up of carbon and hydrogen. In methane, there are 12.0 g of carbon for every 4.00 g of hydrogen, a ratio of 3:1 by mass. In ethane, there are 24.0 g of carbon for every 6.00 g of hydrogen, a ratio of 4:1 by mass. This is an illustration of the law of multiple proportions.

The law of multiple proportions states that when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other element are in a ratio of small whole numbers.

 In the given example, methane [tex]CH_3[/tex] and ethane [tex]C_2H_6[/tex] are compounds formed by carbon (C) and hydrogen (H). The mass ratios of carbon to hydrogen in these compounds are as follows:

For methane :

- Mass of carbon (C) = 12.0 g

- Mass of hydrogen (H) = 4.00 g

- Mass ratio of C to H = 12.0 g / 4.00 g = 3:1

 For ethane

- Mass of carbon (C) = 24.0 g

- Mass of hydrogen (H) = 6.00 g

- Mass ratio of C to H = 24.0 g / 6.00 g = 4:1

The ratios of the masses of carbon that combine with a fixed mass of hydrogen (12.0 g of carbon for methane and 24.0 g of carbon for ethane) are in a simple whole number ratio of 1:2 (or 3:6 if we consider the hydrogen masses as well). This exemplifies the law of multiple proportions, as the mass of carbon combining with a fixed mass of hydrogen increases in a small whole number ratio from one compound to the other.

Final answer:

To exceed the FDA's assumed sodium intake limit of 2,300 mg per day for an adult, one would have to consume approximately 44.23 liters of softened water with a sodium concentration of 5.2×10⁻²% by mass.

Explanation:

The question pertains to the consumption of softened water and its sodium concentration relative to the FDA recommendations for sodium intake. The FDA's recommended limit is not specified in the question, so for this explanation, we will assume a commonly referenced guideline limit of 2,300 mg sodium per day for an adult. The concentration of sodium in the softened water is given as 5.2×10⁻²% by mass. Given this and the density of water (1.0 g/mL), we can calculate the volume of softened water consumed that would exceed the FDA's recommended sodium intake limit.

If the softened water contains 5.2×10⁻²% sodium by mass, this means there are 0.052 grams of sodium in every 100 grams of the softened water. Since 1 mL of water weighs 1 gram, this also means there are 0.052 grams of sodium per 100 mL of water.

To find the volume (V) of water that contains 2,300 mg (2.3 grams) of sodium, we set up the following equation:

V (in mL) × 0.052 g/100 mL = 2.3 g.

Solving for V, we get:

V = (2.3 g × 100 mL) / 0.052 g

V = 44,230.77 mL.

Therefore, one would have to consume approximately 44.23 liters of this softened water to exceed the FDA recommendation for sodium intake.

Answer:

1)

Explanation:

Fe+2 + 2e -> Fe

The element that is reduced gains electrons. In this case Fe is gaining electrons

Answer: The half reaction that describes the reduction reaction is [tex]Fe^{2+}(aq.)+2e^-\rightarrow Fe(s)[/tex]

Explanation:

Oxidation reaction is defined as the reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

[tex]X\rightarrow X^{n+}+ne^-[/tex]

Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

[tex]X^{n+}+ne^-\rightarrow X[/tex]

For the given chemical reaction:

[tex]Mg(s)+Fe^{2+}(aq.)\rightarrow Mg^{2+}(aq.)+Fe(s)[/tex]

The half reactions follows:

Oxidation half reaction:  [tex]Mg(s)\rightarrow Mg^{2+}(aq.)+2e^-[/tex]

Reduction half reaction:  [tex]Fe^{2+}(aq.)+2e^-\rightarrow Fe(s)[/tex]

Hence, the half reaction that describes the reduction reaction is [tex]Fe^{2+}(aq.)+2e^-\rightarrow Fe(s)[/tex]