Ammonium carbamate (NH2COONH4) is a salt of carbamic acid that is found in the blood and urine of mammals. At 202°C, Kc = 0.008 for the following equilibrium: If 6.82 g of NH2COONH4 is put into a 0.5-L evacuated container, what is the total pressure (in atm) at equilibrium? Enter a number to 2 decimal places.

This problem is about the Photoelectric Effect in quantum physics, calculating the maximum kinetic energy of emitted electrons and the maximum number of freed electrons due to light exposure.

The questions are related to the Photoelectric Effect, a fundamental concept in quantum physics.

a.) The relationship between the frequency of light and the maximum kinetic energy of emitted electrons is given by the formula: E = hf – W, where E is the maximum kinetic energy, h is Planck’s constant, f is the frequency of light, and W is the work function of the material. The frequency can be found from the speed of light divided by the given wavelength (439 nm).

b.) The maximum number of electrons freed by light depends on the energy of the photons and the material's work function. You can find this by dividing the total energy of the burst of light (1.00μJ) by the energy required to remove one electron.

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Answer:

35.6 g of W, is the theoretical yield

Explanation:

This is the reaction

WO₃  +  3H₂  →   3H₂O  +  W

Let's determine the limiting reactant:

Mass / molar mass = moles

45 g / 231.84 g/mol = 0.194 moles

1.50 g / 2 g/mol = 0.75 moles

Ratio is 1:3. 1 mol of tungsten(VI) oxide needs 3 moles of hydrogen to react.

Let's make rules of three:

1 mol of tungsten(VI) oxide needs 3 moles of H₂

Then 0.194 moles of tungsten(VI) oxide would need (0.194  .3) /1 = 0.582 moles (I have 0.75 moles of H₂, so the H₂ is my excess.. Then, the limiting is the tungsten(VI) oxide)

3 moles of H₂ need 1 mol of WO₃ to react

0.75 moles of H₂ would need (0.75 . 1)/3 = 0.25 moles

It's ok. I do not have enough WO₃.

Finally, the ratio is 1:1 (WO₃ - W), so 0.194 moles of WO₃ will produce the same amount of W.

Let's convert the moles to mass (molar mass  . mol)

0.194 mol . 183.84 g/mol = 35.6 g

Answer: Mass of one mole of viruses in grams is [tex]54\times 10^{8}[/tex]  

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

Given : One virus has mass of = [tex]9.0\times 10^{-12}mg=9.0\times 10^{-15}g[/tex]     [tex]1mg=10^{-3}g[/tex]

One mole of virus [tex]6.023\times 10^{23}[/tex] has mass of = [tex]\frac{9.0\times 10^{-15}}{1}\times 6.023\times 10^{23}=54\times 10^{8}g[/tex]  

Thus mass of one mole of viruses in grams is [tex]54\times 10^{8}[/tex]  

To find the mass of one mole of viruses in grams, convert the given mass of a virus from mg to grams and then multiply it by Avogadro's number, 6.022 × 10^23 particles/mol.

The mass of one mole of viruses can be calculated by converting the given mass of a virus to grams and then multiplying it by Avogadro's number, which represents the number of particles in one mole. Avogadro's number is approximately 6.022 × 10^23 particles per mole.

First, we convert the given mass of a virus from mg to grams:

9.010 × 10^-12 mg = 9.010 × 10^-15 g

Next, we multiply the mass of one virus by Avogadro's number:

9.010 × 10^-15 g × 6.022 × 10^23 particles/mol =

5.42 × 10^9 g

Therefore, the mass of one mole of viruses is approximately 5.42 × 10^9 grams.

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Answer: 1.14

Explanation:

[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]

To calculate the molarity of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HCl[/tex]

are the n-factor, molarity and volume of base which is NaOH.

We are given:

[tex]n_1=1\\M_1=?\\V_1=10.0mL\\n_2=1\\M_2=0.1M\\V_2=7.2mL[/tex]

Putting values in above equation, we get:

[tex]1\times M_1\times 10.0=1\times 0.1\times 7.2\\\\M_1=0.072M[/tex]

To calculate pH of gastric juice:

molarity of [tex]H^+[/tex] = 0.072

[tex]pH=-log[H^+][/tex]

[tex]pH=-log(0.072)=1.14[/tex]

Thus the pH of the gastric juice is 1.14

The pH of the gastric juice is 1.14

We'll begin by calculating the molarity of the HCl needed for the reaction.

HCl + NaOH —> NaCl + H₂O

From the balanced equation above,

The mole ratio of the acid, HCl (nA) = 1

The mole ratio of the base, NaOH (nB) = 1

Molarity of base, NaOH (Mb) = 0.1 M

Volume of base, NaOH (Vb) = 7.2 mL

Volume of acid, HCl (Va) = 10 mL

MaVa / MbVb = nA/nB

(Ma × 10) / (0.1 × 7.2) = 1

(Ma × 10) / 0.72 = 1

Cross multiply

Ma × 10 = 0.72

Divide both side by 10

Ma = 0.72 / 10

HCl (aq) —> H⁺(aq) + Cl¯(aq)

From the balanced equation above,

1 mole of HCl contains 1 mole of H⁺.

Therefore,

0.072 M HCl will also contain 0.072 M H⁺

Hydrogen ion concentration, [H⁺] = 0.072 M

pH = –Log [H⁺]

pH = –Log 0.072

Thus, the pH of the gastric juice is 1.14

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Answer:

Explanation:

to begin, three steps are highlighted below which are used to write the Lewis structures of the given compounds, viz;

the image below gives a step by step explanation as to answering this question.

i hope this was helpful, cheers.

Drawing Lewis structures entails representing the arrangement of electrons in a molecule, observing the octet rule, and assigning formal charges. The formal charge is the hypothetical charge an atom would possess if electrons in bonds are evenly distributed. The negative formal charges are preferably located on the most electronegative atoms in the molecule or ion when there are multiple possible structures.

Drawing Lewis structures and assigning formal charges requires an understanding of the octet rule and the nature of the molecules. Lewis structures depict the arrangement of electrons in a molecule, particularly illustrating the bonding between atoms and the lone pairs of electrons that may exist. The octet rule suggests that atoms are stable when their outermost (valence) shell is full, typically with eight electrons.

The formal charge on an atom in a molecule is the hypothetical charge the atom would have if we could redistribute the electrons in the bonds evenly between the atoms. We calculate formal charge as follows: Formal Charge = [# of valence electrons on atom] – [non-bonded electrons + number of bonds]. Lewis structures are most reliable when adjacent formal charges are zero or of the opposite sign, and if there are several possible structures for a molecule or ion, the one with the negative formal charges on the more electronegative atoms is often the most accurate.

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An appropriate alkyne starting material A could be 2-butyne (C4H6). An intermediate product B could be but-2-en-1-yne (C4H4). Both the starting material and the intermediate product have the same number of carbon atoms.

An appropriate alkyne starting material A could be 2-butyne (C4H6).

An intermediate product B could be but-2-en-1-yne (C4H4).

Both the starting material and the intermediate product have the same number of carbon atoms.

Answer:

(a) Atomic Mass = 35.24

(b) 2.95x10²¹ Cl₂ molecules

Explanation:

In the attached picture you may read your question, written in a different format.

(a) We calculate the atomic weight of the enriched Cl using the abundances and masses of the isotopes:

(b) We use the previously calculated atomic mass and use it as molar mass (35.24 g/mol), alongside Avogadro's number:

Answer:

Option I. E2 and SN1

Explanation:

First, let's discard the options.

Option II cannot be because sodium ethoxide, although is a good nucleophyle, it's also a strong base, so it can take place a acid base reaction, and ethoxide act as base to substract an electrophyle from the iodohexane, therefore, it can go through a mechanism of elimination.

Option III cannot be either because the above explanation. Also a reaction in basic conditions can actually go through bimolecular reactions, so it has to be E2 only. E1 is in acidic conditions mostly and involves a carbocation, which in basic medium cannot be.

Because of the above explanation, option IV cannot be either.

Technically option 1 cannot be either because a reaction if it's bimolecular, then it has to be Sn2 and E2 only.

but it's the only option that has sense above all.

The mechanism is as follow:

The branch of science which deals with chemicals and bonds is called chemistry.

The correct option is A

The other option is wrong because of the following reason:-

Hence, the correct option is 1 that is E2 +SN1

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Answer:

500 years

Explanation:

The original carbon-14 was 100%, and after 100 years, it decays 1.202%. So after 100 years it goes to 98.798%, after more 100 years (200 years), it will be 97.596% (98.798 - 1.202), thus, after n 100 years "package", the percenatge will be:

100% - actual% = n*1.202

n = (100% - actual%)/1.202

n = (100% - 94%)/1.202

n = 4.992

So, the number of years is n*100 = 4.992*100 = 499.2 ≅ 500 years.

Answer:

5.426 x 10⁹ g

5.55 mol

Explanation:

This type of problems involve the use of proportions , and the use of conversion of units to solve them.

In the first part we have to determine the mass in grams of a mole of virus, so we have to convert the mass to grams and  multiply by avogadro´s number.

9.010x 10⁻¹² mg x  ( 1 g / 1000 mg ) = 9.010 x 10⁻¹⁵ g

mass of 1 mol viruses:

9.010 x 10⁻¹⁵ g/ virus x ( 6.022 x 10²³ virus/mol ) = 5.426 x 10⁹ g /mol

(Note we rounded to 4 significant figures since 9.010 has 4 significant figures.)

For the second part convert the mass of the oil tanker to grams, and make use of the previous result to determine the # of moles of viruses which have the same mass.

mass oil tanker = 3.01 x 10⁷ Kg x ( 1000 g /Kg ) = 3.01 x 10¹⁰ g

3.01 x 10¹⁰ g x ( 1 mol virus / 5.426 x 10⁹ g ) = 5.55 mol

( Note here we rounded to three significant figures since in the multiplication we have 3.01 with three significant figures. )

The result is amazing and it is due to the very small mass of the virus. Imagine only 5.55 mol of virus in the same mass as that of an oil tanker !!!

Final answer:

The mass of one mole of viruses is 5.42 grams, calculated by converting the mass of a single virus to grams and then multiplying by Avogadro's number. To equate to the mass of an oil tanker, there would be 5.54 × 10^9 moles of viruses.

Explanation:

To calculate the mass of one mole of viruses, we use the given mass of a single virus and Avogadro's number. Since we have the mass of one virus as 9.0 × 10-12 mg, we first convert this mass to grams by dividing by 1,000,000 (since there are 1,000,000 micrograms in a gram), giving us 9.0 × 10-18 g. Then, we multiply this mass by Avogadro's number (6.022 × 1023 particles/mole) to get the mass of one mole of viruses: 5.42 g.

To find out how many moles of viruses have a mass equal to that of the oil tanker, we first convert the mass of the oil tanker to grams (3.0 × 107 kg is equal to 3.0 × 1010 g because there are 1,000 kg in a tonne and 1,000 g in a kg). Now, we divide this mass by the mass of one mole of viruses (5.42 g/mole), giving us: 5.54 × 109 moles.

Answer:

The initial velocity as a percentage of Vmax is 1100%

Explanation:

From Michaelis Menten equation,

Vo = Vmax[S]/Km+[S]

Vo/Vmax = [S]/Km+[S]

Expressing the initial velocity as a percentage of Vmax is

Vmax/Vo = Km+[S]/[S] × 100

[S] = Km/10

Km = 10[S]

Vmax/Vo = 10[S]+[S]/[S] × 100 = 11[S]/[S] ×100 = 11×100 = 1100%

The initial velocity of an enzyme, as a percentage of Vmax, is 10% when the substrate concentration is equal to Km/10.

The initial velocity of an enzyme, as a percentage of Vmax, can be calculated using the Michaelis-Menten equation. Let's assume that the substrate concentration [S] is equal to Km/10. The Michaelis constant (Km) represents the substrate concentration at which the velocity is half of Vmax.

Therefore, if [S] = Km/10, the velocity would be 10% of Vmax. This means that the initial reaction rate is only 10% of the maximum rate that can be achieved when all enzyme active sites are saturated.

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Answer : The correct option is, (C) 10-mL volumeric pipet.

Explanation :

Graduated cylinder : It is a measuring cylinder that is used to measure the volume of a liquid. It has a narrow cylindrical shape. The marked line drawn on the graduated cylinder shows the amount of liquid that has been measured.

Pipet : It is a type of laboratory equipment that is used to measure the volume of a liquid. It is small glass tube and the marked line drawn on the pipet. It is used to accurately measure and transfer of volume of liquid from one container to another.

Volumetric flask : It is a type of laboratory tool that is also used for measuring the volume of liquid. It is used to make up a solution to a known volume. It measure volumes much more precisely than beakers.

Beaker : It is a type of laboratory equipment that has cylindrical shape and it is used for the mixing, stirring, and heating of chemicals.

As per question, we conclude that the pipet is most precise than other devices because in pipet the marking lines are more accurate. Thus, it can be used to measure volume to precision.

Hence, the correct option is, (C) 10-mL volumeric pipet.

Answer:

B. only one copy of the field in memory

Explanation:

A static method is sort of a description of a class but is not part of the objects that it generates. Crucial: A program may perform a static method without constructing an object first! All other functions (those not static) only occur when they're member of an object. Thus it is necessary to build an object before they could be executed.

Therefore, when an static field is declared static, there will be:

B. only one copy of the field in memory

Answer:

hydrogen Flouride, hydrogen Tennesside

Explanation:

The strongest bond among hydrogen halides is found in hydrogen fluoride, while the longest bond is found in hydrogen iodide. This is due to the decreasing bond strength and increasing bond length with increasing size of the halide ion.

Among the hydrogen halides, the strongest bond is found in hydrogen fluoride (HF) and the longest bond is found in hydrogen iodide (HI). The bond strength in hydrogen halides typically decreases as the size of the halide ion increases, with fluorine being the smallest and therefore, the strongest. Conversely, the bond length increases with the size of the halide ion, hence why iodine, the largest halide, forms the longest bond with hydrogen.

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Answer:

[Cl₂] in equilibrium is 1.26 M

Explanation:

This is the equilibrium:

PCl₃(g) + Cl₂(g) ⇋ PCl₅(g)

Kc = 91

So let's analyse, all the process:

                PCl₃(g)        +        Cl₂(g)     ⇋        PCl₅(g)

Initially     0.24 M                 1.50M                 0.12 M

React           x                           x                         x

Some amount of compound has reacted during the process.

In equilibrium we have

              0.24 - x                  1.50 - x                  0.12 + x

As initially we have moles of product, in equilibrium we have to sum them.

Let's make the expression for Kc

Kc = [PCl₅] / [Cl₂] . [PCl₃]

91 = (0.12 + x) / (0.24 - x) ( 1.50 - x)

91 = (0.12 + x) / (0.36 - 0.24x - 1.5x + x²)          

91 (0.36 - 0.24x - 1.5x + x²) = (0.12 + x)

32.76 - 158.34x + 91x² = 0.12 +x

32.64 - 159.34x + 91x² = 0

This a quadratic function:

a = 91; b= -159.34; c = 32.64

(-b +- √(b² - 4ac)) / 2a

Solution 1 = 1.5

Solution 2 = 0.23 (This is our value)

So [Cl₂] in equilibrium is 1.50 - 0.23 = 1.26 M

Answer:

a) 25 vials

b) 3 vials

c) 3 bags

Explanation:

When a concentrated solution is diluted to form another solution, the new concentration can be calculated by:

C1V1 = C2V2

Where C is the concentration, V is the volume, 1 represents the initial solution, and 2 the final solution. The multiplication CV is constant because it represents the amount of matter in the solution which will not be changed.

a) Let's identify the volume (V1) needed when the concentrated solution has C1 = 0.5%, and the final solution has C2 = 0.125% and V2 = 100 mL

0.5*V1 = 0.125*100

0.5V1 = 12.5

V1 = 25 mL

Thus, this is the volume for 1 bag, for 30 bags, V = 30*25 = 750 mL. The vials needed is the volume divided by the volume of one vial:

750/30 = 25 vials.

b) Doing the same thing, now with C1 = 50 mg/mL, C2 = 1 mg/mL, and V2 = 100 mL:

50*V1 = 1*100

V1 = 2 mL

The volume for 30 bags is then 30*2 = 60 mL. The number of vials is:

60/20 = 3 vials.

c) In this case, the concentration of sodium chloride is the same in both solutions. Thus, the volume of it is the total volume of the bag (100 mL) less the volume of the other substances:

100 - 25 - 2 = 73 mL

The volume for 30 bags is 30*73 = 2190 mL. Thus, if the concentrated bag has 1 L = 1000 mL, the bags needed are:

2190/1000 = 2.19

Thus, 3 bags are needed (the 3rd bag will not be totally used!).

This is an incomplete question, here is a complete question.

The heat of combustion of bituminous coal is 2.50 × 10² J/g. What quantity of the coal is required to produce the energy to convert 106.9 pounds of ice at 0.00 °C to steam at 100 °C?

Specific heat (ice) = 2.10 J/g°C

Specific heat (water) = 4.18 J/g°C

Heat of fusion = 333 J/g

Heat of vaporization = 2258 J/g

A) 5.84 kg

B) 0.646 kg

C) 0.811 kg

D) 4.38 kg

E) 1.46 kg

Answer : The correct option is, (A) 5.84 kg

Explanation :

The process involved in this problem are :

[tex](1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\\\(3):H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)[/tex]

The expression used will be:

[tex]Q=[m\times \Delta H_{fusion}]+[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{vap}][/tex]

where,

[tex]Q[/tex] = heat required for the reaction = ?

m = mass of ice = 106.9 lb = 48489.024 g      (1 lb = 453.592 g)

[tex]c_{p,l}[/tex] = specific heat of liquid water = [tex]4.18J/g^oC[/tex]

[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = [tex]333J/g[/tex]

[tex]\Delta H_{vap}[/tex] = enthalpy change for vaporization = [tex]2258J/g[/tex]

Now put all the given values in the above expression, we get:

[tex]Q=145903473.2J[/tex]

Now we have to calculate the quantity of the coal required.

[tex]m=\frac{Q}{\Delta H}[/tex]

[tex]m=\frac{145903473.2J}{2.50\times 10^4J/g}[/tex]

[tex]m=5836.138929g=5.84kg[/tex]      (1 g = 0.001 kg)

Thus, the quantity of the coal required is, 5.84 kg

Alumina, corundum, and bauxite contain significant amounts of aluminum oxide. Alumina is a form of aluminum oxide, corundum is a crystalline form that usually contains traces of other elements, and bauxite is the principal ore of aluminum.

The substances which contain significant amounts of aluminum oxide are alumina, corundum, and bauxite.

Alumina, by its nature, is a form of aluminum oxide. Corundum is a crystalline form of aluminum oxide and typically contains traces of iron, titanium, vanadium, and chromium. Bauxite is the principal ore of aluminum and it primarily consists of one or more aluminum hydroxide minerals, which are often structurally composed of aluminum oxide.

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Answer:

102 L

Explanation:

Data of the solution

The moles of HCl in the solution are:

1.25 L × 3.20 mol/L = 4.00 mol

The gas must contain 4.00 moles of HCl

Data of the gas

745 torr × (1 atm/760 torr) = 0.980 atm

We can the volume (V) of HCl gas using the ideal gas equation.

P × V = n × R × T

V = n × R × T/P

V = 4.00 mol × (0.08206 atm.L/mol.K) × 303.2 K/ 0.980 atm

V = 102 L

Answer:

The neutral organic product of the reaction is the 2-methoxy-2-methylpropane, shown in the attached figure.

Explanation:

By reacting an alkene (2-methylprop-1-ene) with an alcohol (methanol) in the presence of an acid it forms an ester (2-methoxy-2-methylpropane) through Markovnikov’s rule. First, a protonation of the alkene occurs where a tertiary carbocation is formed, then the nucleophile (methanol) attacks the previously formed carbocation, and finally ocurr the deprotonation of the hydrogen bound to the oxygen, thus forming the ester, in our case the neutral organic product 2-methoxy-2-methylpropane.

Without information about the reactants and reaction conditions, it is impossible to predict the neutral organic product of a reaction.

The question refers to a reaction and asks for the prediction of the neutral organic product. However, the reaction and reactants are not provided, so it is impossible to provide a specific answer. In organic chemistry, reactions can involve different reactants and conditions, leading to various possible products.

To predict the neutral organic product of a reaction, it is necessary to know the reactants and the specific reaction conditions.

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Answer:

2-methoxybutane

Explanation:

This reaction is an example of Nucleophilic substitution reaction. Also, the reaction of (S)-2-bromobutane with sodium methoxide in acetone, is bimolecular nucleophilic substitution (SN2). The reaction equation is given below.

(S)-2-bromobutane + sodium methoxide (in acetone) → 2-methoxybutane

Answer:

1.60 g

Explanation:

A chemist determines by measurements that 0.0500 moles of oxygen gas, that is, O₂, participate in a chemical reaction. The molar mass of oxygen is 32.00 g/mol. We can find the mass corresponding to 0.0500 moles using the following expression.

m = n × M

where

m is the mass

n are the moles

M is the molar mass

m = n × M

m = 0.0500 mol × 32.00 g/mol

m = 1.60 g

Answer:

Total 20 mL of solution is needed in which 2.5 m L will be of Albuterol solution and 17.5 mL will be of normal saline solution.

Explanation:

Amount of Albuterol in 1 vial = 2.5 mg/0.5 mL

Volume of dose in which 12.5 mg of Albuterol is present be x.

So,

[tex]\frac{2.5 mg}{0.5 mL}=\frac{12.5 mg}{x}[/tex]

x = 2.5 mL

Volume of  Albuterol solution is 2.5 mL.

If the output flow of your continuous nebulizer is 10 mL per hour.Then in 2 hours total volume of solution delivered = T

T = 10 × 2 mL = 20 mL

Volume of normal saline solution needed = y

T = x + y

y = T - x = 20 mL - 2.5 mL = 17.5 mL

Total 20 mL of solution is needed in which 2.5 mL will be of Albuterol solution and 17.5 mL will be of normal saline solution.

Answer:

Moles of [tex]ScCl_3[/tex] = 6 moles

Explanation:

The reaction of [tex]Sc[/tex] and [tex]Cl_2[/tex] to make [tex]ScCl_3[/tex] is:

[tex]2Sc+3Cl_2[/tex]⇒[tex]2ScCl_3[/tex]

The above reaction shows that 2 moles of Sc  can react with 3 moles of [tex]Cl_2[/tex] to form [tex]ScCl_3.[/tex]

Mole Ratio= 2:3

For 10 moles of Sc we need:

Moles of [tex]Cl_2[/tex] = [tex]Moles of Sc *\frac{3 moles of Cl_2}{2 Moles of Sc}[/tex]

Moles of [tex]Cl_2[/tex] = [tex]10 *\frac{3 moles of Cl_2}{2 Moles of Sc}[/tex]

Moles of [tex]Cl_2[/tex] =15 moles

So 15 moles of [tex]Cl_2[/tex] are required to react with 10 moles of [tex]Sc[/tex] but we have 9 moles of [tex]Cl_2[/tex] , it means [tex]Cl_2[/tex] is limiting reactant.

[tex]Moles of ScCl_3=Given\ Moles\ of\ Cl_2 *\frac{2\ Moles\ o\ fScCl_3}{3\ Moles\ of\ Cl_2}[/tex]

[tex]Moles\ of\ ScCl_3=9 *\frac{2\ Moles\ of\ ScCl_3}{3\ Moles\ of\ Cl_2}[/tex]

Moles of ScCl_3= 6 moles

Answer:

6.00 moles of ScCl3 will be produced.

Explanation:

Step 1: Data given

Moles of Sc = 10.00 moles

Moles of Cl2 = 9.00 moles

Molar mass of Sc = 44.96 g/mol

Molar mass of Cl2 = 70.9 g/mol

Step 2: The balanced equation

2Sc + 3Cl2 → 2ScCl3

Step 3: Calculate the limiting reactant

For 2 moles Sc we need 3 moles Cl2 to produce 2 moles ScCl3

Cl2 is the limiting reactant. It will completely be consumed (9.00 moles)

Sc is the limiting reactant. There will react 2/3 * 9.00 = 6.00 moles

There will remain 10.00 - 6.00 =4.00 moles Sc

Step 4: Calculate moles ScCl3

For 3 moles Cl2 we'll have 2 moles ScCl3

For 9.00 moles we'll have 6.00 moles of ScCl3

6.00 moles of ScCl3 will be produced.

Answer: Use of Etherificarion followed by fractional distinction.

Explanation:

It is done by reacting a mixture of tetrahydrofurfuryl alcohol and up to about one equivalent of at least one low work function element, Reacting the said mixture with a halide; fractionally distilling said reacted mixture to yield the first distillate; reacting said first distillate with an excess amount of at least one low work function element; and, fractionally distilling said reacted first distillate to obtain the purified ether wherein said at least one low work function element is an elemental metal or a metal hydride which has a φ of less than about 3.0 eV.

Answer: c. oxygen-17

Explanation:

The isotopic representation of an atom is: [tex]_Z^A\textrm{X}[/tex]

where,

Z = Atomic number of the atom

A = Mass number of the atom

X = Symbol of the atom

In a nuclear reaction, the total mass and total atomic number remains the same.

For the given nuclear reaction:

[tex]^{14}_{7}\textrm{N}+^4_2\textrm{He}\rightarrow ^A_Z\textrm{X}+^{1}_{1}\textrm{H}[/tex]

To calculate A:

Total mass on reactant side = total mass on product side

14 + 4= A + 1

A = 17

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

7+ 2 = Z + 1

Z = 8

The isotopic symbol of element is [tex]_{17}^{8}\textrm{O}[/tex]

Thus a proton is produced along with oxygen-17.

A proton is produced along with:

c. oxygen-17

[tex]^AX_Z[/tex]

where,

Z = Atomic number of the atom

A = Mass number of the atom

X = Symbol of the atom

In a nuclear reaction, the total mass and total atomic number remains the same.

For the given nuclear reaction:

[tex]^{14}N_7+^4He_2----- > ^AX_Z+^1H_1[/tex]

To calculate A:

Total mass on reactant side = total mass on product side

14 + 4= A + 1

A = 17

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

7+ 2 = Z + 1

Z = 8

The isotopic symbol of element is: [tex]^8O_{17}[/tex]

Thus, A proton is produced along with oxygen-17

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Answer:

The empirical formula is C3H4O

Explanation:

Step 1: Data given

Mass of the compound = 1.000 grams

The compound contains:

- Carbon

- hydrogen

- oxygen

The combustion of this compound gives:

2.360 grams of CO2

0.640 grams of H2O

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 2.360 grams / 44.01 g/mol

Moles CO2 = 0.05362 moles

In CO2 we have 1 mol

This means for 1 mol CO2 we have 1 mol C

For 0.05362 moles CO2 we have 0.05362 moles C

We have 0.05362 moles of C in the compound  

Step 3: Calculate mass of C

Mass C = moles C * molar mass C

Mass C = 0.05362 moles * 12.0 g/mol

Mass C = 0.643 grams  

Step 4: Calculate moles of H2O  

Moles H2O = 0.640 grams / 18.02 g/mol

Moles H2O =  0.0355 moles H2O

For 1 mol H2O we have 2 moles of H

For 0.0355 moles H2O we have 2*0.0355 =  0.071 moles H  

Step 5: Calculate mass of H

Mass H = moles H * molar mass H

Mass H = 0.071 moles * 1.01 g/mol

Mass H = 0.072 grams

Step 6: Calculate mass of O

Mass of O = Mass of compound - mass of C - mass of H

Mass of O = 1.000 g - 0.643 - 0.072 = 0.285 grams

Step 7: Calculate moles of O

Moles O = 0.285 grams / 16.0 g/mol

Moles O = 0.0178 moles

Step 8: Calculate mol ratio

We divide by the smallest amount of moles

C:  0.05362 / 0.0178 = 3

H: 0.071 / 0.0178 = 4

O: 0.0178/0.0178 = 1

The empirical formula is C3H4O

The study of chemicals and bonds is called chemistry.

The correct answer is C3H4O

All the data is given in the question, these data is as follows:-

The compound contains:

The combustion of this compound gives:

The formula to calculate the moles is as follows:-

[tex]Moles CO2 = \frac{mass}{molar mass}[/tex]

[tex]Moles \ CO2 = \frac{2.360} {44.01} Moles\ CO2 = 0.05362 moles [/tex]

In CO2 we have 1 mole, Which means for 1 mole CO2 we have 1 mole Carbon. For 0.05362 moles CO2, we have 0.05362 moles Carbon, We have 0.05362 moles of C in the compound  

lets calculate the mass of C

Mass C = moles C * molar mass C

Mass C = [tex]0.05362 moles * 12.0 g/mol[/tex]

Mass C = 0.643 grams  

The moles of H2O is as follows

Moles H2O = [tex]\frac{0.640}{18.02} [/tex]

Moles H2O = 0.0355 moles H2O

For 1 mole H2O, we have 2 moles of H. For 0.0355 moles H2O we have 2*0.0355 =  0.071 moles H  

let's Calculate the mass of H

Mass H = moles H * molar mass H

Mass H = 0.071 moles * 1.01 g/mol

Mass H = 0.072 grams

let's Calculate mass of O

Mass of O = Mass of compound - mass of C - mass of H

Mass of O = 1.000 g - 0.643 - 0.072 = 0.285 grams

let's Calculate moles of O

Moles O = 0.285 grams / 16.0 g/mol

Moles O = 0.0178 moles

The mole ratio is as follows:-

We divide by the smallest amount of moles

C:  0.05362 / 0.0178 = 3

H: 0.071 / 0.0178 = 4

O: 0.0178/0.0178 = 1

Hence, The empirical formula is C3H4O

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Option A states that 0.1 moles of HCl remain unreacted. This proves that option A is incorrect. The volume of hydrogen gas produced is 0.22 L. Thus option D is correct.

From the given reaction, 1 mole of Fe and 2 moles of HCl reacts to form 1 mole ferric chloride and 1-mole hydrogen gas.

The number of moles of HCl in 30 ml 1 M solution are:

Moles = molarity [tex]\times[/tex] volume (L)

Moles of HCl = 1 [tex]\times[/tex] 0.03

Moles of HCl = 0.030 moles.

The moles of 0.56 grams Fe powder are :

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

Moles of Fe = [tex]\rm \dfrac{0.56}{56}[/tex] moles

Moles of Fe = 0.01 moles

For the reaction of 0.01 moles of Fe, moles of HCl required = 0.01 [tex]\times[/tex] 2

Moles of HCL reacted = 0.02 moles

Total moles of HCL = 0.03 moles

Moles of HCl unreacted = 0.03 - 0.02

Moles of HCl unreacted = 0.01 moles

Option A states that 0.1 moles of HCl remain unreacted. This proves that option A is incorrect.

0.01 moles of Fe form, 0.01 mole of Hydrogen gas.

From the ideal gas equation:

PV = nRT

1 [tex]\times[/tex] Volume = 0.01 [tex]\times[/tex] 0.0821 [tex]\times[/tex] 273

Volume of Hydrogen gas = 0.22 L.

The volume of hydrogen gas produced is 0.22 L. Thus option D is correct.

For more information about the chemical reaction, refer to the link:

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Option D is correct since exactly 0.22 L of H₂ is produced, which matches the stoichiometric calculations for the reaction.

To determine why A is incorrect, we need to perform stoichiometric calculations based on the reaction: Fe(s) + 2HCl(aq) --> FeCl₂(aq) + H₂(g). First, calculate the moles of Fe and HCl:

The balanced equation shows that 1 mole of Fe reacts with 2 moles of HCl. Therefore, 0.0100 mol of Fe will react with 0.0200 mol of HCl, leaving an excess of 0.0100 mol of HCl (0.0300 mol - 0.0200 mol). However, this contradicts option A which states 0.100 mol HCl remains unreacted.

Now, for option D: 0.0100 mol of Fe will produce 0.0100 mol of H₂. Using the ideal gas law at standard conditions (273 K and 1.0 atm), the volume of H₂ produced is:

This confirms that D is correct.

The complete question is:

Fe(s) + 2HCl(aq) --> FeCl₂(aq) + H₂(g)

When a student adds 30.0 mL of 1.00 M HCl to 0.56 g of powdered Fe, a reaction occurs according to the equation above. When the reaction is complete at 273 K and 1.0 atm, which of the following is true?

A) HCl is in excess, and 0.100 mol of HCl remains unreacted.

D) 0.22 L of H₂ has been produced.

The correct question is:

To the reaction mixture having reaction as 25 Fe(s) + 2 HCI(aq) -> FeCl₂(aq) + H₂(g), When a student adds 30.0 mL of 1.00 M HCI to 0.56 g of powdered Fe, a reaction occurs according to the equation above. When the reaction is complete at 273 K and 1.0 atm, which of the following is true?

(A) HCI is in excess, and 0.100 mol of HCI remains unreacted.

(B) HCI is in excess, and 0.020 mol of HCI remains unreacted.

(C) 0.015 mol of FeCl₂ has been produced.

(D) 0.22 L of H₂ has been produced.

The question is incomplete, here is the complete question:

Calculate the volume in liters of a 0.13 M potassium dichromate solution that contains 200. g of potassium dichromate . Round your answer to 2 significant digits.

Answer: The volume of solution is 5.2 L

Explanation:

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]

We are given:

Molarity of solution = 0.13 M

Given mass of potassium dichromate = 200. g

Molar mass of potassium dichromate = 294.15 g/mol

Putting values in above equation, we get:

[tex]0.13M=\frac{200}{294.15\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{200}{294.15\times 0.13}=5.23L[/tex]

Hence, the volume of solution is 5.2 L

The volume of a potassium dichromate solution containing 1.8 moles cannot be calculated without the molarity of the solution. Once molarity is known, use V = n / M and convert to liters, considering significant figures for accuracy.

To calculate the volume of a potassium dichromate solution that contains 1.8 moles of potassium dichromate, we must first determine the molarity (M) of the solution, which is not provided in the question. Assuming we have this information, the formula to find the volume (V) when the number of moles (n) and molarity (M) are known is:V = n / M

Without a given molarity, we cannot proceed with this calculation. If molarity is provided, we'd convert it to liters. Since the molarity and actual number of grams of potassium dichromate were not provided in your question, we cannot complete this calculation precisely. However, the reference provided for preparing a different solution with KBrO3 and the details about significant figures indicate that calculations should be made using suitable glassware for precision and rounding off to the correct number of significant digits.

Explanation:

Most reagent forms are going to absorb water from the air; they're called "hygroscopic".  Water presence can have a drastic impact on the experiment being performed  For fact, it increases the reagent's molecular weight, meaning that anything involving a very specific molarity (the amount of molecules in the final solution) will not function properly.

Heating will help to eliminate water, although some chemicals don't react well to heat, so it shouldn't be used for all.  A dessicated environment is simply a means to  "dry."  That allows the reagent with little water in the air to attach with.