#6 and #7. How many carbon atoms are in a mixture of 7.00 mol c2F2 and 0.400 mol carbon dioxide and also #7

#6 And #7. How Many Carbon Atoms Are In A Mixture Of 7.00 Mol C2F2 And 0.400 Mol Carbon Dioxide And Also

Answers

Answer 1

Answer:

#6  8.67x10²⁴ atoms

#7  

1. Atom

2. Formula unit

3. Molecule

4. Ion

Explanation:

#6 First we calculate how many carbon moles are there in 7.00 moles of C₂F₂, keeping in mind that there are 2 C moles per C₂F₂ mol:

7.00 mol C₂F₂ * 2 = 14.00 mol C

As for carbon dioxide, there are 0.400 C moles in 0.400 moles of CO₂.

We calculate the total number of C moles:

14.00 mol + 0.400 mol = 14.4 mol C

Finally we calculate the number of atoms in 14.4 C moles, using Avogadro's number:

14.4 mol * 6.023x10²³ atoms/mol = 8.67x10²⁴ atoms

#7

1. Radon - Atom (Ra)2. Formula unit (It is a crystalline solid, BaBr₂)3. Molecule (NH₃)4. Ion (It has a formal charge, +2)

Related Questions

Que es la actividad física y en qué mejora

Answers

La actividad física regular puede mejorar su fuerza muscular y aumentar su resistencia. El ejercicio proporciona oxígeno y nutrientes a sus tejidos y ayuda a que su sistema cardiovascular funcione de manera más eficiente. Y cuando la salud de su corazón y pulmones mejoran, tiene más energía para hacer frente a las tareas diarias. Encantado de ayudarle

If 12.3 g of Cu is deposited at the cathode of an electrolytic cell after 5.50 h, what was the current used?​

Answers

Answer:

1.88 A

Explanation:

Let's consider the reduction of copper in an electrolytic cell.

Cu²⁺ + 2 e⁻ ⇒ Cu

We can calculate the charge used to deposit 12.3 g of Cu using the following relations.

The molar mass of Cu is 63.55 g/mol.1 mole of Cu is deposited when 2 moles of electrons circulate.1 mole of electrons has a charge of 96486 C (Faraday's constant).

The charge used is:

[tex]12.3 g \times \frac{1 molCu}{63.55gCu} \times \frac{2molElectron}{1molCu} \times \frac{96486C}{1molElectron} = 3.73 \times 10^{4} C[/tex]

We can convert 5.50 h to seconds using the conversion factor 1 h = 3600 s.

5.50 h × 3600 s/1 h = 1.98 × 10⁴ s

The current used is:

I = q/t = 3.73 × 10⁴ C/1.98 × 10⁴ s = 1.88 A

Write the number of sig. fig. in four numbers given in the sentence below. An (one) octopus has 8 legs. 13 octopi have 104 legs.
Give four answers.
A. Infinity, Infinity, Infinity, Infinity
B. 1, 1, 2, 3
C. Infinity, Infinity, 2, 3
D. No answer text provided.​

Answers

Answer:

1, 1, 2, 3

Explanation:

The numbers 1 and 8 both have 1 sig. fig.

The number 13 has 2 sig. figs.

The number 104 has 3 sig. figs.


A scientific hypothesis is
ANSWER:
predictive.
testable.
explanatory.
all of the above.

Answers

Answer:

All of the above.

Explanation:

For a scientific hypothesis to be considered a hypothesis, it has to be testable. When conducting a lab experiment, it also allows the tester to predict what might occur during and after the experimentation. They are also explanatory. For example, theories are hypotheses that have been verified and can explain why something in nature takes place.

You are asked to prepare a buffer solution with a pH of 3.50. The following solutions, all 0.100 M, are available to you: HCOOH, CH3COOH, H3PO4 , NaCHOO, NaCH3COO, and NaH2PO4.  What would be the best combination to make the required buffer solution? Select one:
a. NaH2PO4 and NaCHOO  
b. H3PO4 and NaH2PO4
c. NaH2PO4 and HCOOH
d. CH3COOH and NaCH3COO e. HCOOH and NaCHOO
can someone helo me with this​

Answers

Answer:

e. HCOOH and NaCHOO

Explanation:

For a buffer solution, both an acid and its conjugate base are required.

With the information above in mind, we can discard options a) and c), as those combinations are not of an acid and its conjugate base.

Now it is a matter of comparing the pKa (found in literature tables) of the acids of the remaining three acids:

H₃PO₄ pKa = 2.12CH₃COOH pKa = 2.8HCOOH pKa = 3.74

The acid with the pKa closest to the desired pH is HCOOH, so the correct answer is e. HCOOH and NaCHOO

Which equation obeys the law of conservation of
mass?

Answers

Answer:2C4H10+2C12+12O2 4CO2+CC14+H20

bio-chemisty of protain​

Answers

Answer:

Protein biochemistry is the study of proteins. Protein biochemistry is a scientific field dedicated to the study of proteins, complex chains of amino acids which make up the building blocks of all living organisms.

Explanation:

I hope that helped

Copy and Pasted!

Answer:

Listen to what guy said on top.

Explanation:

polypeptide structures consisting of one or more long chains of amino acids residue.....

or my answer

True or false: Boron contains 2s22p1 valence electrons, so only one p orbital is needed to form molecular orbitals.

Answers

Answer:

True

Explanation:

The valence orbitals of boron are 2s2 2p1. We have to recall that all the valence orbitals whether full or empty are involved in the formation of molecular orbitals.

The number of molecular orbitals formed is equal to the number of atomic orbitals that are combined.

Since there are two valence orbitals and there is only one p orbital among the valence orbitals, it is true that only one p orbital is needed to form molecular orbitals in boron.

Linoleic acid is a polyunsaturated fatty acid found, in ester form, in many fats and oils. Its doubly allylic hydrogens are particularly susceptible to abstraction by radicals, a process that can lead to the oxidative degradation of the fat or oil.

a. True
b. Flase

Answers

Answer:

True.

Explanation:

The information presented in the question above regarding linoleic acid is true. Linoleic acid is, in fact, found in many oils and fats in the ester form. In addition, linoleic acid is considered a polyunsaturated fatty acid, due to the presence of two unsaturations in its composition. Its chemical formula is CH3-(CH2)4-CH=CH-CH2-CH=CH-(CH2)7COOH and it is an essential fatty acid for the human body, as it is essential in the composition of arachidonic acid that is responsible for building muscle, managing body fat thermogenesis, and regulating core protein synthesis.

Given the following list of densities, which materials would float in a molten vat of lead provided that they do not themselves melt? Densities (g/mL): lead = 11.4, glass = 2.6, gold = 19.3, charcoal = 0.57, platinum = 21.4.
a. gold and platinum
b. glass and charcoal
c. gold, platinum, glass and coal
d. gold and charcoal
e. None of these

Answers

Answer:

b. glass and charcoal

Explanation:

Step 1: Given data

Density of Pb: 11.4 g/mLDensity of Glass: 2.6 g/mLDensity of Au: 19.3 g/mLDensity of charcoal: 0.57 g/mLDensity of platinum: 21.4 g/mL

Step 2: Determine which material will float in molten lead

Density is an intrinsic property of matter. Less dense materials float in more dense materials. The materials whose density is lower than that of lead and will therefore float on it are glass and charcoal.

Suppose you ran this reaction without triethylamine and simply used an excess of reactant 1. At the end of the reaction, your methylene chloride solution would contain mostly reactant 1 and the product. What would you do to remove reactant 1 from the solution

Answers

ummm is that chemistry?

Answer:

is this chem

Explanation:

What are the uses of Sulphuric acid?

Answers

Answer:

The major use of sulfuric acid is in the production of fertilizers, e.g., superphosphate of lime and ammonium sulfate. It is widely used in the manufacture of chemicals, e.g., in making hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs.

The major use of sulfuric acid is in the production of fertilizers, e.g., superphosphate of lime and ammonium sulfate. It is widely used in the manufacture of chemicals, e.g., in making hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs.

En la fermentación del alcohol, la levadura convierte la glucosa en etanol y dióxido de carbono:
C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)
Si reaccionan 5.97 g de glucosa y se recolectan 1.44 L de CO2 gaseoso, a 293 K y 0.984 atm, ¿cuál
es el rendimiento porcentual de la reacción

Answers

Answer:

88.9%

Explanation:

Primero convertimos 5.97 g de glucosa a moles, usando su masa molar:

5.97 g ÷ 180 g/mol = 0.0332 mol

Después calculamos la cantidad máxima de moles de CO₂ que se hubieran podido producir:

0.0332 mol C₆H₁₂O₆ * [tex]\frac{2molCO_2}{1molC_6H_{12}O_6}[/tex] = 0.0664 mol CO₂

Ahora calculamos los moles de CO₂ producidos, usando los datos de recolección dados y la ecuación PV=nRT:

0.984 atm * 1.44 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 Kn = 0.0590 mol

Finalmente calculamos el rendimiento porcentual:

0.0590 mol / 0.0664 mol * 100% = 88.9%

Rank each of the following gases in order of increasing urms assuming equivalent amounts and all gases are at the same temperature and pressure where 1 has the lowest urms and 4 has the highest urms.

a. Gas 1 : H2S
b. Gas: He
c. Gas 3: NF3
d. Gas 4: H2O

Answers

the answer is option c

The Urms refers to the root mean square speed of the gas. The order of increasing Urms for the gases shown in the question; NF3 < H2S < H2O < He.

What is the Urms?

The Urms refers to the root mean square speed of the gas. This is ultimately dependent on the relative molecular mass of the gases when they are maintained at the same temperature.

Now, let us look at the order of increasing Urms for the gases shown in the question; NF3 < H2S < H2O < He.

Learnmore about Urms: https://brainly.com/question/365923

Identify the possible quantitative analysis you can do using only the 28.02 g/mol as a unit factor. Select one or more:

Answers

Answer:

Calculate the moles of N2 molecules in 3.94 grams of nitrogen.

Calculate the grams of N2 in 5.03 x 1020 moles of nitrogen molecules.

Explanation:

Calculate the moles of N2 molecules in 4.73 liters of nitrogen gas. FALSE. You can't make this conversion using only the conversion factor with units of g/mol. To convert liters to moles are necessaries pressure, temperature and volume of the gas to use PV = nRT

Calculate the grams of N2 in 10.58 liters of nitrogen gas. FALSE. As explained, you need, P,V and T to find the moles of the gas. With the moles you can find the mass using the conversion factor of 28.02g/mol

Calculate the moles of N2 molecules in 3.94 grams of nitrogen. TRUE. You can find the moles of N2 as follows:

3.94g N2 * (1mol/28.02g) = 0.14 moles of N2 molecules

Calculate the grams of N2 in 5.03 x 1020 moles of nitrogen molecules. TRUE. The mass in 5.03x10²⁰ moles of nitrogen molecules is:

5.03x10²⁰ moles * (28.02g/mol) = 1.4x10²²g of nitrogen.

Indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium after reactant or product is added.

2CO(g) + O2(g) ⇌ 2CO2

Answers

Answer:

Indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium after reactant or product is added.

[tex]2CO(g) + O2(g) <=> 2CO2[/tex]

Explanation:

When the reactants concentration increases, then the equilibrium will shift towards products and when the concentration of products increases, then equilibrium will shift towards reactants.

So, increases in concentration of carbon monoxide (CO) shifts the equilibrium to favor the formation of carbondioxide.

Similarly increase in concentration of oxygen also favor the formation of product carbon dioxide.

Increase in concentration of CO2 favors the formation of CO and O2.

Decrease in product concentration also favors the formation of product.

Decrease in reactant concentration favors the formation of reactants only.

what is the machine used to check melting point called?​

Answers

Answer:

Melting-point apparatus

write any two things that should be remembered while writing chemical equation​

Answers

Answer:

the product and the reactant must be balanced

if u are required to give the mechanism if the reaction it must be written

Draw structures corresponding to the following IUPAC names:(a) (Z)-2-Ethyl-2-buten-1-ol (b) 3-Cyclohexen-1-ol(c) trans-3-Chlorocycloheptanol (d) 1,4-Pentanediol(e) 2,6-Dimethylphenol (f ) o-(2-Hydroxyethyl)phenol

Answers

Answer:

Draw structures corresponding to the following IUPAC names:(a) (Z)-2-Ethyl-2-buten-1-ol (b) 3-Cyclohexen-1-ol(c) trans-3-Chlorocycloheptanol (d) 1,4-Pentanediol(e) 2,6-Dimethylphenol (f ) o-(2-Hydroxyethyl)phenol

Explanation:

According to IUPAC rules, the name of a compound is:

Prefix+root word+suffix

1) Select the longest carbon chain and it gives the root word.

2) The substituents give the prefix.

3) The functional group gives the secondary suffix and the type of carbon chain gives the primary suffix.

The structure of the given compounds are shown below:

The metal tantalum becomes superconducting at temperatures below 4.483 K. Calculate the temperature at which tantalum becomes superconducting in degrees Celsius.

Answers

Answer:

The correct answer is "-268.667°C".

Explanation:

Given:

Temperature,

= 4.483 K (below)

Now,

The formula of temperature conversion will be:

⇒ [tex]T(^{\circ} C)=T(K)-273.15[/tex]

By putting the values, we get

⇒            [tex]=4.483-273.15[/tex]

⇒            [tex]=-268.667^{\circ} C[/tex]

Thus the above is the correct answer.

If 0.250 L of a 5.90 M HNO₃ solution is diluted to 2.00 L, what is the molarity of the new solution?

Answers

Answer:

0.74 M

Explanation:

From the question given above, the following data were obtained:

Molarity of stock solution (M₁) = 5.90 M

Volume of stock solution (V₁) = 0.250 L

Volume of diluted solution (V₂) = 2 L

Molarity of diluted solution (M₂) =?

The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:

M₁V₁ = M₂V₂

5.90 × 0.250 = M₂ × 2

1.475 = M₂ × 2

Divide both side by 2

M₂ = 1.475 / 2

M₂ = 0.74 M

Thus, the molarity of the diluted solution is 0.74 M

What is the difference between conjugate acid-base pair?

a. a H atom. c. a mole water
b. a H+ ion d. a OH– ion​

Answers

Answer:

b. a H+ ion

Explanation:

The concept of conjugate acid-base pair is related to Bronsted-Lowry acid-base theory and according to this theory, acid is a proton acceptor.

In short,

conjugate base is formed when an acid donates a proton.

conjugate acid is formed when a base accepts a proton.

1.rain pours from the sky
2.leaves of the plant dried
3.fluffy clouds form in the sky
4.bathing suit dries after swim
5.water puddles disappear

A.Evaporation
B.Condensation
C.Precipitation
D.Transpiration
Yan po pag pipilian

Answers

Answer:

1.Precipitation

2.Transpiration

3.Condensation

4.Evaporation

5.Evaporation

3.Condensation

Explanation:

Rain pours from the sky occurs due to the process of precipitation, leaves of the plant dried due to the process of transpiration in which the water is evaporated from the body of plant, fluffy clouds form in the sky occurs in the process of condensation, bathing suit dries after swim is due to evaporation in which water is removed and goes into the atmosphere and water puddles disappear due to the process of evaporation. Evaporation is the removal of water from the any surface whereas transpiration is the removal of water from plant body parts.

Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 10.3 g of octane is mixed with 23. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

Answers

Answer:

9.36 g

Explanation:

The equation of the reaction is;

C8H18(g) + 25/2 O2(g) ----> 8CO2(g) + 9H2O(g)

Number of moles of octane = 10.3g/ 114 g/mol = 0.09 moles

1 mole of octane yields 9 moles of water

0.09 moles of octane yields 0.09 × 9/1 = 0.81 moles of water

Number of moles of oxygen = 23g/32g/mol = 0.72 moles

12.5 moles of oxygen yields 9 moles of water

0.72 moles of oxygen yields 0.72 × 9/12.5 = 0.52 moles of water

Hence oxygen is the limiting reactant;

Maximum mass of water produced = 0.52 moles of water × 18 g/mol = 9.36 g

For the following reaction, 11.6 grams of sulfur are allowed to react with 23.8 grams of carbon monoxide .

sulfur(s) + carbon monoxide(g) sulfur dioxide(g) + carbon(s)

What is the maximum amount of sulfur dioxide that can be formed?

What is the formula for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete?

Answers

Answer:

S + 2CO = SO2 + 2C

First, look for the amount of substance of sulfur:

n(S) = m / M

n(S) = 14.8 g/32 g / mol = 0.4625 mol

n(CO) = m (CO) / M (CO)

M(CO) = 12 + 16 = 28 g/mol

n(CO) = 19.9 g/28 g/mol = 0.71 mol

S in excess, so for calculating we take CO:

n(SO2) = n(CO)/2 = 0.71 mol/2 = 0.355 mol

m(SO2) = M(SO2)*n(SO2)

M(SO2) = 32 + 16*2 = 64 g/mol

m(SO2) = 64 g/mol * 0.355 mol = 22.74 g

the force of attraction between non polar molecules are what (a)electrovalent bond (b)covalent bond (c)Hydrogen bond (d)Van der waals forces​

Answers

Answer:

d. van der waals force

Explanation:

Van der Waals force :

the weakest intermolecular forceand consist of dipole-dipole force and dispersion force.

Identify the options below that are results of adding a catalyst to a chemical system.
The reaction rates are increased.
The reaction quotient is unaffected.
The reaction quotient decreases.
The equilibrium constant is unaffected.

Answers

Answer:

The correct options are a, b and d

Explanation:

A catalyst is a substance that increases the rate of a chemical reaction by reducing the activation energy. Le Catelier's  principle explains how a substance or an "action" can affect a reaction in equilibrium.

The principle states that when a change is made to the conditions of a reacting system at equilibrium, the position of the equilibrium moves to counteract the change made. These changes are change in temperature, pressure, volume and/or concentration. These changes will either cause the equilibrium to shift forward or backward.

However, the presence of a catalyst DOES NOT affect a chemical equilibrium/equilibrium constant nor does it affect the reaction quotient because the same amount of reactants and products are available just as in uncatalyzed reaction except that the reaction proceeds faster (which does not affect equilibrium).

The rate of reaction is given as the time required by the reactant to convert into the product. The addition of catalyst increases the rate of reaction, while the reaction quotient and the equilibrium remain unaffected.

What is a catalyst?

A catalyst is a chemical or compound that adds to the reaction and lowers the activation energy by providing an alternative path to the reaction.

The catalyst takes part in the reaction but did not consume in the chemical reaction.

The equilibrium and the reaction quotient are dependent on the conversion of the reactant to the product. The catalyst is not used in the reaction and thus did not affect the reaction quotient or the equilibrium.

Hence, options A, B, and D are correct for the use of catalysts in the chemical reaction.

Learn more about catalysts, here:

https://brainly.com/question/17052831

PLEASE HELP!!

How does temperature, agitation, and particle size affect solubility?

Answers

Answer:

At higher temperatures, particles move faster and collide more, increasing solubility rates.

Agitation increases solubility rates as well, by bringing fresh solvent into contact with the undissolved solute

The smaller the particle size, the higher (faster) solubility rate. Vice versa, the bigger the particle size, the lower (slower) solubility rate.

Explanation:

A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was , calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.

Answers

Answer:

molar heat of combustion = -5156 *10³ kJ/mol

Explanation:

A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was 10.17 kJ/°C, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.

Step 1: Data given

Mass of naphthalene = 1.435 grams

Initial temperature of water = 20.28 °C

Final temperature of water = 25.95 °C

heat capacity of the bomb plus water was 10.17 kJ/°C

Molar mass naphtalene = 128.2 g/mol

Step 2:

Qcal = Ccal * ΔT

⇒with Qcal =the heat of combustion

⇒with Ccal = heat capacity of the bomb plus water = 10.17 kJ/°C

⇒with ΔT = the difference in temperature = T2 - T1 = 25.95 - 20.28 = 5.67°C

Qcal = 10.17 kJ/°C * 5.67 °C

Qcal = 57.7 kJ

Step 3: Calculate moles

Moles naphthalene = 1.435 grams / 128.2 g/mol

Moles naphthalene = 0.01119 moles

Step 4: Calculate the molar heat of combustion

molar heat of combustion = Qcal/ moles

molar heat of combustion = -57.7 kJ/ 0.01119 moles

molar heat of combustion = -5156 *10³ kJ/mol

4.005 X 74 X 0.007 = 2.10049

Answers

Answer:

2.07459

Explanation:

this is the correct answer.

Other Questions
Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 10.3 g of octane is mixed with 23. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits. why is this written in passive voiceThe baby was left on the front steps of the orphanage Half a league, half a league,Half a league onward,All in the valley of DeathRode the six hundred:'Forward, the Light Brigade!Charge for the guns' he said:Into the valley of DeathRode the six hundred.'Forward, the Light Brigade!'Was there a man dismay'd ?Not tho' the soldier knewSome one had blunder'd:Theirs not to make reply,Theirs not to reason why,Theirs but to do and die,Into the valley of DeathRode the six hundred.Cannon to right of them,Cannon to left of them,Cannon in front of themVolley'd and thunder'd;Storm'd at with shot and shell,Boldly they rode and well,Into the jaws of Death,Into the mouth of HellRode the six hundred.Flash'd all their sabres bare,Flash'd as they turn'd in airSabring the gunners there,Charging an army whileAll the world wonder'd:Plunged in the battery-smokeRight thro' the line they broke;Cossack and RussianReel'd from the sabre-stroke,Shatter'd and sunder'd.Then they rode back, but notNot the six hundred.Cannon to right of them,Cannon to left of them,Cannon behind themVolley'd and thunder'd;Storm'd at with shot and shell,While horse and hero fell,They that had fought so wellCame thro' the jaws of Death,Back from the mouth of Hell,All that was left of them,Left of six hundred.When can their glory fade?O the wild charge they made!All the world wonder'd.Honour the charge they made!Honour the Light Brigade,Noble six hundred! 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