According to the Kinetic Molecular Theory, which container has helium particles with the greatest average kinetic energy?

Answer:

bro what Is this like I dont even kno

Answer:

O2 is the answer I believe

Answer:

the density of the Gas at STP is 0.227 g/L .

Explanation:

This question involves the combined gas law . The equation for the combined gas law

Answer:

a) The system is in equilibrium

b) No other force is acting on the system

Explanation:

a) The forces in a system is balanced until and unless there is equal and opposite force acting on the system. In this image the upward force is equal to the downward force on the book and the table and hence it is in equilibrium

b) An object which is in static mode, produces downward force because of weight and in response to that a normal reaction is produced in upward direction. Apart from these two forces, no other force is acting on this book and table system.

Answer:

Option D. The number of oxygen atom is the same before and after the reaction.

Explanation:

From the question given above, the following were obtained:

Robin's equation:

H₂ + O₂ —> H₂O

Alex's equation

2H₂ + O₂ —> 2H₂O

To know which equation better represents the reaction, we shall determine which of the equation is balanced.

For Robin:

H₂ + O₂ —> H₂O

Element >>> Reactant >>> Product

H >>>>>>>>> 2 >>>>>>>>>> 2

O >>>>>>>>> 2 >>>>>>>>>> 1

Robin's equation is not balanced because the number of atoms of each element in the reactant and product are not equal.

For Alex:

2H₂ + O₂ —> 2H₂O

Element >>> Reactant >>> Product

H >>>>>>>>> 4 >>>>>>>>>> 4

O >>>>>>>>> 2 >>>>>>>>>> 2

Alex' equation is balanced because the number of atoms of each element in the reactant and product are equal.

Thus, option D gives the right answer to the question.

Answer:

See Explanation

Explanation:

At the anode;

Zn(s) -----> Zn^2+(aq) + 2e

At the cathode;

Cu^2+(aq) + 2e ------> Cu(s)

Overall electrochemical reaction;

Zn(s)  + Cu^2+(aq) ------>  Zn^2+(aq) + Cu(s)

E°cell = E°cathode - E°anode

E°cell = 0.34 - (-0.76)

E°cell = 1.1 V

Q = [0.3 M]/[0.5 M]

Q = 0.6

From Nernst equation;

Ecell =  E°cell - 0.0592/n log Q

Ecell = 1.1 - 0.0592/2 log (0.6)

Ecell = 1.1 - 0.0296 log (0.6)

Ecell = 1.11 V

Answer: a) 525 ml

b) 67.2 ml

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

[tex]P\propto \frac{1}{V}[/tex]     (At constant temperature and number of moles)

[tex]P_1V_1=P_2V_2[/tex]  

where,

a) [tex]P_1[/tex] = initial pressure of gas  = 752 mm Hg

[tex]P_2[/tex] = final pressure of gas = 385 mm Hg

[tex]V_1[/tex] = initial volume of gas  = 250 ml

[tex]V_2[/tex] = final volume of gas = ?

[tex]752\times 250=385\times V_2[/tex]  

[tex]V_2=525ml[/tex]

Therefore, the volume at 385 mm Hg is 525 ml.

b) [tex]P_1[/tex] = initial pressure of gas  = 752 mm Hg

[tex]P_2[/tex] = final pressure of gas = 3.68 atm = 2796.8 mm Hg    (760mmHg=1atm)

[tex]V_1[/tex] = initial volume of gas  = 250 ml

[tex]V_2[/tex] = final volume of gas = ?

[tex]752\times 250=2796.8\times V_2[/tex]  

[tex]V_2=67.2ml[/tex]

Therefore, the volume at 3.68 atm is 67.2 ml.

Answer:

16 g

Explanation:

Step 1: Write the balanced equation

2 CO + O₂ → 2 CO₂

Step 2: Calculate the moles of O₂ required to produce 1.0 moles of CO₂

The molar ratio of O₂ to CO₂ is 1:2.

1.0 mol CO₂ × 1 mol O₂/2 mol CO₂ = 0.50 mol O₂

Step 3: Calculate the mass corresponding to 0.50 moles of O₂

The molar mass of O₂ is 32.00 g/mol.

0.50 mol × 32.00 g/mol = 16 g

[tex]M_{A}V_{A}=M_{B}V_{B}\\(5.25)M_{A}=(4.96)(0.9845)\\M_{A}=\frac{(4.96)(0.9845)}{5.25} \approx \boxed{0.93 \text{ M}}[/tex]

En química, el grupo de algunas sustancias compuestas que poseen propiedades químicas semejantes, denominadas propiedades funcionales, recibe el nombre de función química. ... Además están divididas en ácidos, bases, sales y óxidos; y funciones orgánicas que son las relativas a los compuestos orgánicos.

BRAINLIST

Answer:

The volume is 22.66 mL

Explanation:

D = 22.1 g/cm³, M = 523.1 g, V = ? mL

For Volume

Density = Mass ÷ Volume

Volume = Mass ÷ Density

V = M ÷ D

V = 523.1 g ÷ 22.1 g/cm³

V = 23.66 cm³

Now, 1 cm³ = 1 mL

So,

V = 22.66 mL

Thus, The volume is 22.66 mL

-TheUnknownScientist

Answer: D

Explanation:

Answer:

27.8

Explanation:

[tex]\bcancel{\huge\pink{\fbox{\tt{࿐αɴѕωєя࿐}}}}[/tex]

Answer:

coal, cotton, chinese tradictional medicine

Answer: hcl

Explanation:

Answer:

C. II, III, IV, and V.

Explanation:

Acid buffer is  generally formed by the combination of a weak acid as well as the salt of the conjugate base.

Basic buffer is formed the combination of a weak base and also the salt of the conjugate acid.

On dilution the ration of the concentration terms of the salt and weak acid/base does not change. Hence the pH of the buffer solution does not change.

When acid or base is added to buffer, it resists changes in the pH.

Therefore, option (C) is correct.

Answer:

2.18 mol Ar.

Explanation:

Hello there!

In this case, according to the definition of mole as the amount of particles of a given substance, it is possible to introduce the Avogadro's number to assert that 1 mole of any element contains 6.022x10²³ atoms; thus, the moles of Ar in the given amount of atoms turns out to be:

[tex]1.31x10^{24}Ar atoms*\frac{1molAr}{6.022x10^{23}atoms}\\\\=2.18molAr[/tex]

Best regards!

Answer:

12.8

Explanation:

14 = pOH + pH

pH = 14 - pOH

pH = 14 - 1.2

pH = 12.8

Answer:

1. Ca(OH)₂ : Yes

Highest solubility  = pH of 2

2. Cu Br : Yes

Highest solubility  = pH of 4

3. Ca₃(PO₄)₂ : Yes

Highest solubility = pH of 2

Explanation:

From the common ion effect, the solubility of an ionic compound in a solution which already contains one of the ions in that compound will reduce. This is because, when an external stress is placed on a system in equilibrium,  the equilibrium position will shift to remove the effect of that stress. Therefore, when more of the same ions are added to a solution already containing that ion, equilibrium will shift towards removal of the ion, thereby reducing solubility.

1. For Ca(OH)₂:

Ca(OH)₂ is a fairly soluble ionic compound whose dissociation equation is given below:

Ca(OH)₂ (s) ----> Ca²+ (aq) + 2 OH- (aq)

Increase in pH will result in addition of more OH- ions, therefore, its solubility will decrease. On the other hand, a decrease in pH will remove OH- ions, thereby increasing solubility. The pH of highest solubility is 2.

2. For CuBr:

CuBr is a slightly acidic salt as it is salt of a strong acid, HBr and a weak base Cu(OH)₂. Hydrolysis of the salt will result in an acidic medium:

2 CuBr (s) + 2 H₂O (l) ----> 2 HBr (aq) + Cu(OH)₂ (aq).

Thus, an increase in pH will result in an increase in the solubility of CuBr due to the removal of H+ ions. The highest solubility will be at pH of 4.

3) For Ca₃(PO₄)₂

Calcium phosphate is a salt of slightly basic salt as it is a salt of a weak acid, H₃PO₄ and stronger base Ca(OH)₂.

Hydrolysis of the salt will result in a basic medium:

Ca₃(PO₄)₂ (s) + 6 H₂O (l) ---> 3 Ca(OH)₂ (aq) + 2 H₃PO₄ (aq)

Therefore, its solubility increases with decrease in pH due to the removal of OH- ions. The highest solubility will be at pH of 2.

Answer:

Add more H2O

Explanation:

Took the test

If you wanted to produce more hydrochloric acid (HCl), you should add more [tex]H_2O[/tex] to 4HCl + [tex]O_2[/tex] ⇄ 2[tex]H_2O[/tex] + [tex]Cl_2[/tex]. Hence, option A is correct.

Hydrochloric acid [[tex]H^+[/tex]](aq) [tex]Cl^-[/tex](aq) or [tex]H_3O^+ Cl^-[/tex]], also known as muriatic acid, is an aqueous solution of hydrogen chloride.

For instance, adding a strong acid such as HCl to water results in the reaction HCl + [tex]H_2O[/tex] → [tex]H_3O^+[/tex] + [tex]Cl^-[/tex].

In other words, the proton ([tex]H^+[/tex]) from the acid binds to neutral water molecules to form [tex]H_3O^+[/tex] raising the concentration of [tex]H^+[/tex].

Hence, option A is correct.

Learn more about hydrochloric acid here:

https://brainly.com/question/4902679

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Answer:

Decomposition reaction.

Explanation:

2HI —> H₂ + I₂

From the above equation, we can see clearly that HI undergoes a reaction to produce H₂ and I₂ which are the elements that make up HI.

Therefore, the equation illustrated above is a decomposition reaction because HI breaks into H₂ and I₂

NOTE: Decomposition reaction is a reaction in which a compound splits or breakdown into two or more simple elements or compound.

Answer:

No comparison => John Dalton did NOT postulate an atomic structure.

Explanation:

Dalton's postulates are based upon generalizations of observable phenomena. Micro structure was not proposed.  The atomic structure that's accepted in the scientific community evolved from conclusions of ...

Thompson's raisen pudding model => electrons suspended in a positive matrix

Rutherford's shell model => dense positive nucleus surrounded by an electron cloud. (No fine structure postulated)

Bohr Concentric Ring Model => electrons orbiting a positive nucleus in discrete  energy levels much like planets orbiting the sun.

Schrodinger - Dirac Quantum Models => electrons occupying specific orbital energy levels as based upon statistical wave mechanics.

Answer: The molecular formula will be [tex]C_8N_4H_{10}O_2[/tex]

Explanation:

Mass of C= 49.47 g

Mass of N = 28.85 g

Mass of O = 16.48 g

Mass of H = 5.20 g

Step 1 : convert given masses into moles.

Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{49.47g}{12g/mole}=4.12moles[/tex]

Moles of N =[tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{28.85g}{14g/mole}=2.06moles[/tex]

Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{16.48g}{16g/mole}=1.03moles[/tex]

Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.20g}{1g/mole}=5.20moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = [tex]\frac{4.12}{1.03}=4[/tex]

For N = [tex]\frac{2.06}{1.03}=2[/tex]

For O =[tex]\frac{1.03}{1.03}=1[/tex]

For H = [tex]\frac{5.20}{1.03}=5[/tex]

The ratio of C : N: O: H = 4: 2: 1: 5

Hence the empirical formula is [tex]C_4N_2OH_5[/tex]

The empirical weight of [tex]C_4N_2OH_5[/tex] = 4(12)+2(14)+1(16)+5(1)= 97 g.

The molecular weight = 194.19 g/mole

Now we have to calculate the molecular formula.

[tex]n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{194.19}{97}=2[/tex]

The molecular formula will be = [tex]2\times C_4N_2H_5O=C_8N_4H_{10}O_2[/tex]

The question is incomplete, the complete question is;

When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II) sulfate PbSO₄ is reduced to lead at the cathode and oxidized to solid lead(II) oxide PbO at the anode.

Suppose a current of 96.0 A is fed into a car battery for 37.0 seconds. Calculate the mass of lead deposited on the cathode of the battery. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.

Answer:

3.81 g of lead

Explanation:

The equation of the reaction is;

Pb^2+(aq) + 2e ---->Pb(s)

Quantity of charge = 96.0 A * 37.0 seconds = 3552 C

Now we have that 1F = 96500 C so;

207 g of lead is deposited by 2 * 96500 C

x g of lead is deposited by 3552 C

x = 207 *  3552/2 * 96500

x = 735264/193000

x = 3.81 g of lead

5mg in liter is 5 ppm

Explanation: 1 ppm is one part per million.

1 ppm is 1 mg is one part of million from 1 kg = 1000 000 mg

1 litre water is 1. Kg.

Answer:

The answer is "4.31"

Explanation:

aniline millimoles [tex]= 111 \times 0.37 = 41.07[/tex]

added [tex]HNO_3[/tex]  millimoles [tex]= 79.1 \times 0.35 = 27.685[/tex]

[tex]\to 41.07 - 27.685 = 13.385[/tex] millimoles aniline left

[tex]\to 27.685[/tex] millimoles salt formed

total volume[tex]= 111 + 79.1 = 190.1\ mL\\\\[/tex]

[tex]\to [aniline] = \frac{13.385}{190.1} = 0.07 \ M\\\\\to [salt] =\frac{ 27.685}{ 190.1} = 0.146\ M\\\\\to pOH = pKb + \frac{\log [salt]}{ [base]}\\\\\to pOH = 9.37 + \frac{\log [0.146]}{[0.07]}\\\\\to pOH = 9.69\\\\\to pH = 14 - 9.69\\\\\to pH = 4.31\\[/tex]

Answer:

1. Cr₂O₇²−(aq) + 14H+ (aq) + 6e- ---->⟶2 Cr³+(aq) + 7H₂O (l)

2. CrO₄²− (aq)⟶+ 4H+ (aq) + 3e- ---> Cr(OH)₄ −(aq)

3. Bi³+ (aq) + 3H₂O (l) ---> BiO₃− (aq) + 6 H+ (aq) + 2 e-

4. CIO −(aq)⟶+ 2H+ (aq) + 2e- ---> Cl −(aq) + H₂O

Explanation:

The given equations are redox reaction equations expressed as as half reactions.

The first step is to identify whether the half-reaction is oxidation reduction.

Then the number of electrons gained or lost are added on the right side of the equation.

Appropriate H+ ions and water molecules are added where necessary since the reaction takes place in acidic environment

The atoms of elements involved in the reaction are balanced by adding the correct coefficients.

1. (acidic) Cr2O7 2−(aq)⟶Cr 3+(aq)

The half-reaction is reduction as the oxidation number of chromium changes from +6 to +3. Two Cr⁶+ ions accepts 3 electrons each to form Cr³+ ions

Cr₂O₇²−(aq) + 6e- ---->⟶2 Cr³+(aq)

Cr₂O₇²−(aq) + 14H+ (aq) + 6e- ---->⟶2 Cr³+(aq) + 7H₂O (l)

2. (acidic) CrO₄²− (aq)⟶---> Cr(OH)₄ −(aq)

The half-reaction is a reduction. One Cr⁶+ accepts 3 electrons to become Cr³+

CrO₄²− (aq)⟶+ 3e- ---> Cr(OH)₄ −(aq)

CrO₄²− (aq)⟶+ 4H+ (aq) + 3e- ---> Cr(OH)₄ −(aq)

3, (acidic) Bi³+ (aq)⟶---> BiO₃− (aq)

The half-reaction is an oxidation. One Bi³+ ion gives up two electrons to become Bi⁵+

Bi³+ (aq)⟶---> BiO₃− (aq) + 2e-

Bi³+ (aq) + 3H₂O (l) ---> BiO₃− (aq) + 6 H+ (aq) + 2 e-

4. (acidic) CIO −(aq)⟶---> Cl −(aq)

The half-reaction is a reduction. One Cl+ ion accepts two electrons to become Cl- ion.

CIO −(aq) + 2e-⟶---> Cl −(aq)

CIO −(aq)⟶+ 2H+ (aq) + 2e- ---> Cl −(aq) + H₂O

Answer:

1. AgNO₃ (aq) + NaCl (aq) ----> NaNO₃ (aq) + AgCl (s)

2. Li₂SO₄ (aq) + BaCl₂ (aq) ----> 2 LiCl (aq) + BaSO₄ (s)

3. 2 NaOH (aq) + MgCl₂ (aq) ----> 2 NaCl (aq) + Mg(OH)₂ (s)

Explanation:

The reaction involving the mixing of two soluble solutions to produce a precipitate is known as a precipitation reaction.

A precipitation reaction is double-replacement reaction (a reaction that exchanges the cations or the anions of two ionic compounds) in which one product is a solid precipitate.

Precipitation reactions at useful in the identification of various ions present in a solution. In order to predict the reactions that will produce a precipitate, solubility rules as given in the solubility table below can be used.

From the tables, the reactions that will produce a precipitate, as well as their balanced molecular equations are as follows:

1. AgNO₃ (aq) + NaCl (aq) ----> NaNO₃ (aq) + AgCl (s)

2. Li₂SO₄ (aq) + BaCl₂ (aq) ----> 2 LiCl (aq) + BaSO₄ (s)

3. 2 NaOH (aq) + MgCl₂ (aq) ----> 2 NaCl (aq) + Mg(OH)₂ (s)

Answer:

Please find the solution in the attachment file.

Explanation: