Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq) , as described by the chemical equation

MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

How much MnO2(s) should be added to excess HCl(aq) to obtain 185 mL Cl2(g) at 25 °C and 745 Torr ?

mass of MnO2:

Answers

Answer 1

Answer:

0.605 g

Explanation:

MnO₂(s) + 4HCl(aq) ⟶ MnCl₂(aq) + 2H₂O(l) + Cl₂(g)

First we calculate how many Cl₂ moles need to be produced, using the PV=nRT formula:

P = 745 Torr ⇒ 745 / 760 = 0.980 atmV = 185 mL ⇒ 185 / 1000 = 0.185 Ln = ?R = 0.082 atm·L·mol⁻¹·K⁻¹T = 25 °C ⇒ 25 + 273.16 = 298.16 K

Inputting the data:

0.980 atm * 0.185 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 Kn = 0.00696 mol

Then we convert 0.00696 moles of Cl₂ to MnO₂ moles:

0.00696 mol Cl₂ * [tex]\frac{1molMnO_2}{1molCl_2}[/tex] = 0.00696 mol MnO₂

Finally we convert 0.00696 moles of MnO₂ to grams, using its molar mass:

0.00696 mol MnO₂ * 86.94 g/mol = 0.605 g

Related Questions

Suppose that you add 24.3 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 3.14 oC compared to pure benzene. What is the molar mass (in g/mol) of the unknown compound

Answers

Solution :

We know that :

[tex]$\Delta T_f = k_f.m$[/tex]  and   [tex]$m=\frac{w_2}{m_2 \times w_1}$[/tex]

Then, [tex]$\Delta T_f = k_f.\frac{w_2}{m_2.w_1}$[/tex]   ..................(1)

Where,

[tex]w_1[/tex] = amount of solvent (in kg)

[tex]w_2[/tex] = amount of solute (in kg)

[tex]m_2[/tex] = molar mass of solute (g/mole)

[tex]m[/tex] = molality of solution (mole/kg)

Given :

[tex]\Delta T_f[/tex] = [tex]3.14\ ^\circ C[/tex],   [tex]k_f= 5.12\ ^\circ C/m[/tex]

                              [tex]=5.12 \ ^\circ C/mole/kg[/tex]

                              [tex]=5.12 \ ^\circ C \ kg/mole[/tex]

[tex]w_1[/tex] = 0.250 kg,  [tex]w_2[/tex] = 24.3 g

Then putting this values in the equation is (1),

[tex]$3.14 = \frac{5.12 \times 24.3}{m_2 \times 0.250}$[/tex]

[tex]$m_2 = \frac{5.12 \times 24.3}{3.14 \times 0.250}$[/tex]

[tex]m_2= 158.49[/tex]  g/mole

So, the molar mass of the unknown compound is 158.49 g/mole.

A mixture of argon and neon gases at a total pressure of 874 mm Hg contains argon at a partial pressure of 662 mm Hg. If the gas
mixture contains 12.0 grams of argon, how many grams of neon are present?

Answers

Answer:

6.684g

Explanation:

Here, we can use the mole ratio of the gases to calculate.

We know that the mole ratio of the gases equate to their number of moles.

Firstly, we calculate the number of moles of the oxygen gas. The number of moles of the oxygen gas is equal to the mass of the oxygen gas divided by the molar mass of the oxygen gas. The molar mass of the oxygen gas is 32g/mol

Thus, the number of moles produced is 5.98/32 = 0.186875

Where do we move from here?

We know that if we place the partial pressure of oxygen over the total pressure, this would be equal to the number of moles of oxygen divided by the total number of moles. Now let’s do this.

449/851 = 0.186875/n

n =(0.186875 * 851)/449

n = 0.3542

Now we do the same for argon to get the number of moles of argon.

Firstly, we use dalton’s partial pressure law to get the partial pressure of argon. In the simplest form, the partial pressure of argon is the total pressure minus the partial pressure of oxygen.

P = 851 - 449 = 402 mmHg

We now use the mole ratio relation.

402/851 = n/0.3542

n = (402 * 0.3542) / 851

n = 0.1673

Since we now know the number of moles of argon, we can use this multiplied by the atomic mass of argon to get the mass.

The atomic mass of argon is 39.948 amu

The mass is thus 39.948 * 0.1673 = 6.684g

Consider the reaction: NaNO3(s) + H2SO4(l) NaHSO4(s) + HNO3(g) ΔH° = 21.2 kJ

How much heat must absorbed by the reaction system to convert 100g of NaNO3 into NaHSO4(s)?

Answers

Answer:

endet nach selam nw

4gh7

Please help me! I am a bit stuck on this.

Answers

Option 2 is the correct answer

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. The metal and water come to the same temperature at 24.6 °C. How much heat did the metal give up to the water?

Answers

Answer:

1087.84 J

Explanation:

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

Mass of metal (Mₘ) = 70 g

Temperature of metal (Tₘ) = 80 °C

Mass of water (Mᵥᵥ) = 100 g

Temperature of water (Tᵥᵥ) = 22 °C

Equilibrium temperature (Tₑ) = 24.6 °C

Heat lost by metal (Qₘ) =?

NOTE: Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC

Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)

Qₘ = Qᵥᵥ

Thus, we shall determine the heat gained by water. This can be obtained as follow:

Qᵥᵥ = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

Qᵥᵥ = 100 × 4.184 (24.6 – 22)

Qᵥᵥ = 418.4 × 2.6

Qᵥᵥ = 1087.84 J

Thus, the heat gained by water is 1087.84 J.

Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)

Qₘ = Qᵥᵥ

Qᵥᵥ = 1087.84 J

Qₘ = 1087.84 J

Therefore, the heat lost by the metal is 1087.84 J

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.

What is a calorimeter?

A calorimeter is an object used for calorimetry, or the process of measuring the heat of chemical reactions or physical changes as well as heat capacity.

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. The final temperature of the system is 24.6 °C.

Let's use the following expression to calculate the heat absorbed by the water.

Qw = c × m × ΔT

Qw = (4.184 J/g.°C) × 100 g × (24.6 °C - 22.0 °C) = 1.08 kJ

where,

Qw is the heat absorbed by the water.c is the specific heat capacity of water.m is the mass of water.ΔT is the change in the temperature for water.

According to the law of conservation of energy, the sum of the heat absorbed by the water and the heat released by the metal (Qm) is zero.

Qw + Qm = 0

Qm = -Qw = -10.8 kJ

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.

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A solution is prepared by dissolving 6.60 g of an nonelectrolyte in water to make 550 mL of solution. The osmotic pressure of the solution is 1.84 atm at 25 °C. The molecular weight of the nonelectrolyte is ________ g/mol.

Answers

Answer:

160 g/mol

Explanation:

Step 1: Calculate the molarity of the solution

We will use the following expression.

π = M × R × T

where,

π: osmotic pressure of a nonelectrolyteM: molarityR: ideal gas constantT: absolute temperature (25 °C = 298 K)

M = π / R × T

M = 1.84 atm / (0.0821 atm.L/mol.K) × 298 K = 0.0752 mol/L

Step 2: Calculate the moles of solute in 550 mL (0.550 L)

0.550 L × 0.0752 mol/L = 0.0413 mol

Step 3: Calculate the molecular weight of the nonelectrolyte

0.0413 moles weigh 6.60 g.

6.60 g/0.0413 mol = 160 g/mol

Carbon dioxide gas is collected at 27.0 oC in an evacuated flask with a measured volume of 30.0L. When all the gas has been collected, the pressure in the flask is measured to be 0.480atm. Calculate the mass and number of moles of carbon dioxide gas that were collected.

Answers

Answer:

[tex]M_{CO_2}= 25.7g[/tex]

Explanation:

From the question we are told that:

Temperature [tex]T=27.0[/tex]

Volume [tex]V=30L[/tex]

Pressure [tex]P=0.480atm[/tex]

Generally the equation for Ideal gas is mathematically given by

PV=nRT

Therefore

[tex]n=\frac{0.480 x 30}{0.08205 x 300}[/tex]

[tex]n=0.59moles[/tex]

Generally Mass of CO2 is given as

[tex]M_{CO_2}= 0.59 * 44 g/mol[/tex]

[tex]M_{CO_2}= 25.7g[/tex]

Write a balanced half-reaction for the oxidation of manganese ion (mn2 ) to solid manganese dioxide (mno2) in acidic aqueous solution. Be sure to add physical state symbols where appropriate.

Answers

Answer:

Mn2+(aq) + 2H2O(l)  ⇒ MnO2(s) + 4H+(aq) + 2e-

Explanation:

Step 1: Data given

The oxidation number of manganese ion (Mn2+ ) is +2

The oxidation number of manganese dioxide  is +(MnO2)4

This means the oxidation number from Mn will go from +2 to +4, since it's increased, this is an oxidation reaction

Mn2+(aq)  ⇒ MnO2(s)

We have to balance both sides. Mn is already the same. But on the right side we have O atoms. T obalance both sides we have to add O atoms to the left side. This by adding 2x H2O

Mn2+(aq) + 2H2O(l)  ⇒ MnO2(s)

Now the amount of O atoms is balanced, but we have H- atoms at the left side. To balance we have to add 4 H atoms to the right side

Mn2+(aq) + 2H2O(l)  ⇒ MnO2(s) + 4H+(aq)

Now the amount of atoms is balanced at both sides. We also have to check if the charge on both sides is the same.

Since the left side has a charge of +2, and right has a charge of +4, we have to add 2 electrons to balance this.

Mn2+(aq) + 2H2O(l)  ⇒ MnO2(s) + 4H+(aq) + 2e-

how can we convert plastic garbage energy into electric energy​

Answers

Answer:

Unfortunately, we don`t know how to convert plastic material into electricity yet. I suppose an idea is for someone to invent a machine similar to biomass, where dead plants created energy, but here it`s plastic. The only issue is that it could release deadly chemicals.

Sorry if this isn`t much help, but there isn`t really an answer.   :/

Answer:

Plastics are among the most valuable waste materials – although with the way people discard them, you probably wouldn’t know it. It’s possible to convert all plastics directly into useful forms of energy and chemicals for industry, using a process called “cold plasma pyrolysis”.

Hope this helps you ❤️

MaRk mE aS braiNliest ❤️

A base which can be used to relieve indigestion

Answers

Explanation:

Antacids are medications used to manage the symptoms of indigestion and heartburn. Antacids contain active ingredients that are bases. These allow antacids to neutralize any stomach acid which could be causing digestive discomfort.

QUESTION 11
Identify the reaction type.
KOH + HNO3 -> H2O + KNO3
O combustion
O decomposition
O combination
O single displacement
O double displacement

Answers

O single displacement
Is correct answer
The answer is a single displacement. (C)

Determine the kinds of intermolecular forces that are present in each of the following. Part A Xe Check all that apply. dispersion forces dipole-dipole forces hydrogen bonding Request Answer Part B N2 Check all that apply. dispersion forces dipole-dipole forces hydrogen bonding Request Answer Part C CO Check all that apply. dispersion forces dipole-dipole forces hydrogen bonding Request Answer Part D HF Check all that apply. dispersion forces dipole-dipole forces hydrogen bonding

Answers

Answer:

Part A

dispersion forces

Part B

dispersion forces

Part C

dispersion forces

dipole-dipole forces

Part D

dispersion forces

dipole-dipole forces

hydrogen bonding

Explanation:

Dispersion forces occur in all molecules. They result from momentary shifts in the electron cloud of molecules which induces a dipole in another molecule. This induced dipole eventually spreads throughout the molecule.

For Xe which is a noble gas and N2 which is a diatomic molecule, dispersion forces is the only kind of intermolecular force present in the molecule.

CO is a polar molecule hence in addition to dispersion forces, dipole-dipole forces also exist in the molecule.

HF is a polar molecule hence it possesses dipole-dipole forces in addition to dispersion forces. In this molecule, hydrogen is bonded to a highly electronegative atom (fluorine). Hence, hydrogen bonding is a dominant intermolecular interaction in the molecule.

How many moles of (CH3)3NH+ are in 6.0 g of (CH3)3NH+?

Answers

Answer:

0.1 mol

Explanation:

6/(15*3+15)

0.1 mol moles of (CH3)3NH+ are in 6.0 g of (CH3)3NH+

What is mole?

The mole, symbol mol, exists as the SI base unit of the amount of substance. The quantity amount of substance exists as a measure of how many elementary entities of a provided substance exist in an object or sample.A mole corresponds to the mass of a substance that includes 6.023 x 1023 particles of the substance. The mole exists the SI unit for the amount of a substance. Its symbol stands mol.

The compound trimethylamine, (CH3 )3N, exists as a  weak base when dissolved in water.

A mole exist expressed as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole exists as a convenient unit to utilize because of the great number of atoms, molecules, or others in any substance.

To find the amount of the substance (CH3)3NH+ to calculate its molar mass:

M((CH3)3NH+) = (12+3)*3 + 14+1 = 60 g/mol

n((CH3)3NH+) = m/M

m((CH3)3NH+) = 6g

Thus,

n((CH3)3NH+) = 6g/60 g/mol = 0.1 mol

Hence,

n((CH3)3NH+) = 0.1 mol

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How is magma formed?

Answers

Answer:

“Magma” is exclusively found and formed beneath the earth’s surface. Once magma is on or above the surface of the earth it is referred to as “lava.” Magma is typically formed by extreme temperature melting solid rock within the earth. Pressure and rock composition can also affect magma formation. High pressure can help magma be “squeezed” from partially molten rock. Likewise, as rocks are usually composed of different minerals with different melting points, magma formation from rocks is usually only partial and uneven.

Explanation:

Magma can also be created when hot, liquid rock intrudes into Earth's cold crust. As the liquid rock solidifies, it loses its heat to the surrounding crust. Much like hot fudge being poured over cold ice cream, this transfer of heat is able to melt the surrounding rock (the “ice cream”) into magma. Hope this helps

Different vinegars can be 5-20% acetic acid solutions and have been used for medicinal purposes for thousands of years. If a person takes 2.0 tablespoons of vinegar a day and the Molarity of the vinegar is .84 M, then how many grams of acetic acid (HC2H3O2) will be consumed? 1 Tablespoon is 15 mL.

.013 g
.026 g
.76 g
1.5 g

Answers

Answer:

1.5g

Explanation:

Remember that Molarity = (#moles of solute)/(#liters of solution)

This problem informs us that the Molarity of the vinegar is 0.84 and that the solution is 15mL.

First let's get your SI units to the correct ones.

15mL (1L/1000mL) = 0.015L

Molarity = (#moles of solute)/(#liters of solution) ~

(Molarity)(#liters of solution) = #moles of solute

(0.84M)(.015L) = 0.0126moles of acetic acid per tablespoon

2 tablespoons a day = 0.0126moles*2 =  0.0252 moles of acetic acid.

Now that we have the # of moles of acetic acid we need to get our answer into grams. The molecular weight of HC2H3O2 is 60g/mole.

0.0252mole HC2H3O2 (60g HC2H3O2/1mole HC2H3O2) = 1.512g ~ 1.5g HC2H3O2.

15. In the image given below, magnesium metal is coiled as a thin ribbon. What property of metal is exhibited by it? A Ductility B Lustrous C Sonorous D Malleability​

Answers

Answer: The property of magnesium that is exhibited by it is DUCTILITY. The correct option is A.

Explanation:

Magnesium is a member of the alkaline earth metals. It occurs in nature, only in the combined state, as Epsom salt, dolomite and in many trioxosilicates( IV) including talc and asbestos. They have the following physical properties:

--> Appearance: they are silvery-white solids

--> Relative density: It has a relative density of 1.74

--> DUCTILITY: it's very ductile in nature

--> melting point: it has a melting point of 660°C.

--> Conductivity: They are good conductor of heat and electricity.

Furthermore, DUCTILITY is the physical property of a metal associated with the ability to be hammered thin or stretched into wire without breaking. A metal such as magnesium can therefore be coiled as a thin ribbon without fracturing due to its ductile physical properties.

Activation energy is:
A. The energy needed to begin breaking the bonds of reactants.
B. None of these.
C. The maximum amount of energy reactants can hold.
D. The energy needed to begin breaking the bonds of products.

Answers

Your answer is most definitely letter a

Activation energy is the energy needed to begin breaking the bonds of reactants. Hence, option A is correct.

What is activation energy?

Activation energy is defined as the minimum amount of energy necessary to initiate a chemical reaction.

Hence, activation energy is the energy needed to begin breaking the bonds of reactants.

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6. In a particular atom, an electron moves from n = 3 to the ground state (n = 1), emitting a photon with frequency 5.2 x 1015 Hz as it does so. What is the difference in energy between n = 3 and n = 1 in this atom? g

Answers

Answer: The question wants you to determine the energy that the incoming photon must have in order to allow the electron that absorbs it to jump from  

n

i

=

2

to  

n

f

=

6

.

A good starting point here will be to calculate the energy of the photon emitted when the electron falls from  

n

i

=

6

to  

n

f

=

2

by using the Rydberg equation.

1

λ

=

R

(

1

n

2

f

1

n

2

i

)

Here

λ

si the wavelength of the emittted photon

R

is the Rydberg constant, equal to  

1.097

10

7

 

m

1

Plug in your values to find

1

λ

=

1.097

10

7

.

m

1

(

1

2

2

1

6

2

)

1

λ

=

2.4378

10

6

.

m

1

This means that you have

λ

=

4.10

10

7

.

m

So, you know that when an electron falls from  

n

i

=

6

to  

n

f

=

2

, a photon of wavelength  

410 nm

is emitted. This implies that in order for the electron to jump from  

n

i

=

2

to  

n

f

=

6

, it must absorb a photon of the same wavelength.

To find the energy of this photon, you can use the Planck - Einstein relation, which looks like this

E

=

h

c

λ

Here

E

is the energy of the photon

h

is Planck's constant, equal to  

6.626

10

34

.

J s

c

is the speed of light in a vacuum, usually given as  

3

10

8

.

m s

1

As you can see, this equation shows you that the energy of the photon is inversely proportional to its wavelength, which, of course, implies that it is directly proportional to its frequency.

Plug in the wavelength of the photon in meters to find its energy

E

=

6.626

10

34

.

J

s

3

10

8

m

s

1

4.10

10

7

m

E

=

4.85

10

19

.

J

−−−−−−−−−−−−−−−−−  

I'll leave the answer rounded to three sig figs.

So, you can say that in a hydrogen atom, an electron located on  

n

i

=

2

that absorbs a photon of energy  

4.85

10

19

 

J

can make the jump to  

n

f

=

6

.

Explanation:

a) Define typical polyfunctional acid ?

b) Show the equations of dissociation mechanism of phosphoric acid as an example.

c) Write the equation for calculating the [H3O*].​

Answers

a) A polyfunctional acid is an acid that has more than one functional group.

b) The equations of dissociation of phosphoric acid are:    

H₃PO₄ + H₂O ⇄ H₂PO₄⁻ + H₃O⁺   H₂PO₄⁻ + H₂O ⇄ HPO₄²⁻ + H₃O⁺  HPO₄²⁻ + H₂O ⇄ PO₄³⁻ + H₃O⁺  

c) The equation for calculating the concentration of H₃O⁺ is [tex] [H_{3}O^{+}] = (\frac{K_{1}K_{2}K_{3}[H_{3}PO_{4}]}{[PO_{4}^{-3}]})^{1/3} [/tex]

       

a) A polyfunctional acid can be defined as an acid that has more than one functional group. Phosphoric acid (H₃PO₄) is an example of polyfunctional acid since it is composed of three hydroxyl groups joined to a phosphorus atom, which is also joined to an oxygen atom by a double bound. In that structure, the three hydrogen atoms of the hydroxyl groups give the acidic behavior to this compound.                  

b) Phosphoric acid has three equations of dissociation:  

H₃PO₄ + H₂O ⇄ H₂PO₄⁻ + H₃O⁺    (1)H₂PO₄⁻ + H₂O ⇄ HPO₄²⁻ + H₃O⁺   (2)HPO₄²⁻ + H₂O ⇄ PO₄³⁻ + H₃O⁺   (3)  

The dissociation constants for the three above equations are:

[tex] K_{1} = \frac{[H_{2}PO_{4}^{-}][H_{3}O^{+}]}{[H_{3}PO_{4}]} [/tex]   (4)

[tex] K_{2} = \frac{[HPO_{4}^{2-}][H_{3}O^{+}]}{[H_{2}PO_{4}^{-}]} [/tex]    (5)

[tex] K_{3} = \frac{[PO_{4}^{3-}][H_{3}O^{+}]}{[HPO_{4}^{2-}]} [/tex]    (6)

c) We can calculate the concentration of H₃O⁺ for each equilibrium with the equations (4), (5), and (6).    

The general reaction of dissociation of phosphoric acid is given by the sum of equations (1), (2), and (3):

H₃PO₄ + 3H₂O ⇄ PO₄³⁻ + 3H₃O⁺   (7)  

The concentration of H₃O⁺ for the total dissociation reaction (eq 7) can be found as follows:  

[tex] K_{t} = \frac{[PO_{4}^{-3}][H_{3}O^{+}]^{3}}{[H_{3}PO_{4}]} [/tex]   (8)

Where:

[tex] K_{t} = K_{1}*K_{2}*K_{3} [/tex]

Hence, by knowing the dissociation constants K₁, K₂ and K₃, and the concentrations of PO₄³⁻ and H₃PO₄, the [H₃O⁺] is:

[tex][H_{3}O^{+}] = (\frac{K_{1}K_{2}K_{3}[H_{3}PO_{4}]}{[PO_{4}^{-3}]})^{1/3}[/tex]

         

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There are three isotopes of carbon. They have mass number of 12, 13 and 14. The average atomic mass of carbon is 12.0107 amu. What does this say about the relative abundances of the three isotopes?​

Answers

Answer:

lots more of the carbon 12 than the others

havent calculated it percentage-wise but you can see its very close to 12 meaning it is of far greater abundance that carbon 13 and 14

Explanation:

A solution is made by dissolving 5.84 grams of NaCl in enough distilled water to give a final volume of 1.00 L. What is the molarity of the solution
Group of answer choices

0.0250 M

0.400 M

0.100 M

1.00 M

Answers

Answer:

Explanation:

1. A solution is made by dissolving 5.84g of NaCl is enough distilled water to a give a final volume of 1.00L. What is the molarity of the solution? a. 0.100 M b. 1.00 M c. 0.0250 M d. 0.400 M 2. A 0.9% NaCl (w/w) solution in water is a. is made by mixing 0.9 moles of NaCl in a 100 moles of water b. made and has the same final volume as 0.9% solution in ethyl alcohol c. a solution that boils at or above 100°C d. All the above (don't choose this one) 3. In an exergonic process, the system a. gains energy b. loses energy c. either gains or loses energy d. no energy change at all

Answer:

[tex]\boxed {\boxed {\sf 0.100 \ M }}[/tex]

Explanation:

Molarity is a measure of concentration in moles per liter.

[tex]molarity = \frac{moles \ of \ solute}{liters \ of \ solution}}[/tex]

The solution has 5.84 grams of sodium chloride or NaCl and a volume of 1.00 liters.

1. Moles of Solute

We are given the mass of solute in grams, so we must convert to moles. This requires the molar mass, or the mass of 1 mole of a substance. These values are found on the Periodic Table as the atomic masses, but the units are grams per mole, not atomic mass units.

We have the compound sodium chloride, so look up the molar masses of the individual elements: sodium and chlorine.

Na: 22.9897693 g/mol Cl: 35.45 g/mol

The chemical formula (NaCl) contains no subscripts, so there is 1 mole of each element in 1 mole of the compound. Add the 2 molar masses to find the compound's molar mass.

NaCl: 22.9897693 + 35.45 = 58.4397693 g/mol

There are 58.4397693 grams of sodium chloride in 1 mole. We will use dimensional analysis and create a ratio using this information.

[tex]\frac {58.4397693 \ g\ \ NaCl} {1 \ mol \ NaCl}[/tex]

We are converting 5.84 grams to moles, so we multiply by that value.

[tex]5.84 \ g \ NaCl *\frac {58.4397693 \ g\ NaCl} {1 \ mol \ NaCl}[/tex]

Flip the ratio. It remains equivalent and the units of grams of sodium chloride cancel.

[tex]5.84 \ g \ NaCl *\frac {1 \ mol \ NaCl}{58.4397693 \ g\ NaCl}[/tex]

[tex]5.84 *\frac {1 \ mol \ NaCl}{58.4397693 }[/tex]

[tex]0.09993194823 \ mol \ NaCl[/tex]

2. Molarity

We can use the number of moles we just calculated to find the molarity. Remember there is 1 liter of solution.

[tex]molarity= \frac{moles \ of \ solute}{liters \ of \ solution}[/tex]

[tex]molarity= \frac{ 0.09993194823 \ mol \ NaCl}{1 \ L}[/tex]

[tex]molarity= 0.09993194823 \ mol \ NaCl/L[/tex]

3. Units and Significant Figures

The original measurements of mass and volume have 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 9 in the ten-thousandths place tells us to round the 9 to a 0, but then we must also the next 9 to a 0, and the 0 to a 1.

[tex]molarity \approx 0.100 \ mol \ NaCl/L[/tex]

1 mole per liter is 1 molar or M. We can convert the units.

[tex]molarity \approx 0.100 \ M \ NaCl[/tex]

The molarity of the solution is 0.100 M.

Determine whether the statement about identifying a halide is true: Regardless of any concentration of ammonium solution, the precipitates in the reaction solution of my unknown halide after 0.1M AgNO3 remain because my unknown halide solution contains Br. Select one: True False

Answers

Answer:

False

Explanation:

The statement ; Regardless of any concentration of ammonium solution the precipitate of unknown halide after 0.1M AgNO3 will remain is FALSE

This is Because the remaining concentration of AgNO3 is dependent on the solubility of Ag⁺

compared to an atom of C-14, an atom of C-12 has a lesser

atomic number

number of protons

number of electrons

number of neutrons

Answers

Answer:

mass number

Explanation:

because the mass

number is the number of protons plus the number of neutron and the number of proton in an elements is always the same , therefore and atom of C-14 has greater mass number

A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 6.90kg of water at 34.7 degrees C . During the reaction 57.1kJ of heat flows out of the bath and into the flask.
Calculate the new temperature of the water bath. You can assume the specific heat capacity of water under these conditions is 4.18J.g^(-1).K^(-1) . Round your answer to significant digits.

Answers

Answer:

[tex]T_2= 36.7 \textdegree C[/tex]

Explanation:

Mass of Water [tex]m_w=6.90kg[/tex]

Temperature [tex]T=34.7 degrees[/tex]

Heat Flow [tex]H=57.1kJ[/tex]

Specific heat capacity of water [tex]\mu= 4.18J.g^(-1).K^(-1)[/tex]

Generally the equation for Final Temperature is mathematically given by

[tex]M*\mu *T_1 + Q = M*\mu *T_2[/tex]

[tex]T_2=\frac{M*\mu *T_1 + Q }{M*\mu}[/tex]

Therefore

[tex]T_2=\frac{6.90*4.18*34.7 + 57.1}{6.90*4.18}[/tex]

[tex]T_2= 36.7 \textdegree C[/tex]

The half life of radium-226 is 1600 years. If you have 200 grams of radium today how many grams would be present in 8000 years?

Answers

Answer:

Half life is the time taken by a radio active isotope to reduce by half of its original amount. Radium-226 has a half life of 1602 years meaning that it would take 1602 years for a mass of radium to reduce by half.

Number of half lives in 9612 years = 9612/1602 = 6 half lives

New mass = Original mass x (1/2)n where n is the number of half lives.

Therefore, New mass= 500 x (1/2)∧6

                                 = 500 x 0.015625

                                 = 7.8125 g

Hence the mass of radium after 9612 years will be 7.8125 grams.

Explanation:

Answer:

[tex]\boxed {\boxed {\sf 6.25 \ grams}}[/tex]

Explanation:

We are asked to find the mass of a sample of radium-226 after half-life decay. We will use the following formula:

[tex]A= A_o *\frac{1}{2}^{\frac{t}{h}}[/tex]

In this formula, [tex]A_o[/tex] is the initial amount, t is the time, and h is the half-life.

For this problem, the initial amount is 200 grams of radium-226, the time is 8,000 years, and the half-life is 1,600 years.

[tex]\bullet \ A_o= 200 \ g \\\\bullet \ t= 8,000 \ \\\bullet \ h= 1,600[/tex]

Substitute the values into the formula.

[tex]A= 200 \ g * \frac{1}{2} ^{\frac{8.000}{1,600}[/tex]

Solve the fraction in the exponent.

[tex]A= 200 \ g * \frac{1}{2}^{5}[/tex]

Solve the exponent.

[tex]A= 200 \ g *0.03125[/tex]

[tex]A= 6.25 \ g[/tex]

In addition, we can solve this another way. First, we find the number of half-lives by dividing the total time by the half-life.

8,000/1,600= 5 half-lives

Every half-life, 1/2 of the mass decays. Divide the initial mass in half, then that result in half, and so on 5 times.

1.  200 g/2= 100 g2. 100 g / 2 = 50 g3. 50 g / 2 = 25 g 4. 25 g / 2 = 12.5 g5. 12.5 g / 6.25 g

After 8,000 years, 6.25 grams of radium-226 remains.

A hot pot of water is set on the counter to cool. After a few minutes it has lost 495 J of heat energy. How much heat energy has the surrounding air gained?

_____unit_____

Answers

Answer:

495 J

Explanation:

When the hot pot was set on the counter to cool, heat energy was lost from the pot. Note that according to the first law of thermodynamics, heat is neither created nor destroyed.

This implies that, the heat energy lost from the pot must be gained by the surrounding air. Therefore, if 495 J of energy is lost from the pot, then 495 J of energy is gained by the surrounding air.

The turbines in a hydroelectric plant are fed by water falling from a 50 m height. Assuming 91% efficiency for conversion of potential to electrical endrgy, and 8% loss of the resulting power in transmission, what is the mass flow rate of water required to power a 200 W light bulb? ​

Answers

From  the information given;

the height of the water stream = 50 mthe efficiency of conversion from potential energy to electrical energy is  91%loss of power transmission = 8%

To determine the mass flow rate, let's start by understanding some concepts and parameters.

The power is known to be the energy per unit of time. Mathematically, it can be written as:

[tex]\mathbf{Power = \dfrac{Energy}{Time}}[/tex]

[tex]\mathbf{P =\dfrac{E_p}{time}}[/tex]

[tex]\mathbf{P =\dfrac{m\times g\times z}{time}}[/tex]

where;

[tex]\mathbf{E_p}[/tex] is the potential energy of the streamm = mass flow rateg = acceleration under gravityz = height

Thus;

[tex]\mathbf{E_p}[/tex] = m × 9.81 m/s² × 50 m

[tex]\mathbf{E_p}[/tex] = m × 490.5 (m²/s²)

Recall that:

The power P = 200 W, and;the conversion of the P.E = 91% = 0.91

[tex]\mathbf{E_p}[/tex] = 0.91 × 490.5m (m²/s²)

[tex]\mathbf{E_p}[/tex] = 446.355m (m²/s²)

Since the resulting power transmission is said to be 8%

Then;

the loss in the power transmission (P) = 100% - 8% ×  446.355m (m²/s²)

the loss in the power transmission (P) = 92%  ×  446.355m (m²/s²)

the loss in the power transmission (P) = 0.92  ×  446.355m (m²/s²)

the loss in the power transmission (P) =  410.65m (m²/s²)

Finally;

P = 410.65m (m²/s²)

[tex]\mathbf{P = 410.65 \times m (\dfrac{m^2}{s^2})}[/tex]

replacing the values, we have:

[tex]\mathbf{200 = 410.65 \times m (\dfrac{m^2}{s^2})}[/tex]

[tex]\mathbf{m = \dfrac{200 watt}{410.65\times (\dfrac{m^2}{s^2})}}[/tex]

[tex]\mathbf{m = \dfrac{200 \dfrac{J}{s}}{410.65\times (\dfrac{m^2}{s^2})}}[/tex]

since 1 J/s = 1 kgm²/s²)

Then:

[tex]\mathbf{m = \dfrac{200 \dfrac{\dfrac{kg\times m^2}{s^2}}{s}}{410.65\times (\dfrac{m^2}{s^2})}}[/tex]

[tex]\mathbf{m = \dfrac{200 \ {kg}}{410.65 \ s}}[/tex]

mass flow rate of the water (m) = 0.487 kg/s

Therefore, we can conclude that the mass flow rate of the water required to power a 200 W bulb light is 0.487 kg/s

Learn more about the hydroelectric plant here:

https://brainly.com/question/2635539?referrer=searchResults


Dung dịch nào sau đây chỉ chứa các ion (bỏ qua sự điện li của nước, các chất điện li mạnh phân li hoàn toàn)?
A. HBr, Na2S, Mg(OH)2, Na2CO3.
B. H2SO4, NaOH, NaCl, HF.
C. HNO3, H2SO4, KOH, K2SiO3.
D. Ca(OH)2, KOH, CH3COOH, NaCl.

Answers

Answer:

Dung dịch nào sau đây chỉ chứa các ion (bỏ qua sự điện li của nước, các chất điện li mạnh phân li hoàn toàn)?

A. HBr, Na2S, Mg(OH)2, Na2CO3.

B. H2SO4, NaOH, NaCl, HF.

C. HNO3, H2SO4, KOH, K2SiO3.

D. Ca(OH)2, KOH, CH3COOH, NaCl.

molecular weight of K2SO3

Answers

Explanation:

the molecular weight of K2SO3 is 158. 2598 m/s.

Oxygen is composed of three isotopes: oxygen-16, oxygen-17 and oxygen-18 and has an average atomic mass of 15.9982 amu. Oxygen-17 has a mass of 16.988 amu and makes up 0.032% of oxygen. Oxygen-16 has a mass of 15.972 amu and oxygen-18 has a mass of 17.970 amu. What is the percent abundance of oxygen-18?

Answers

Answer:

The percent abundance of oxygen-18 is 1.9066%.

Explanation:

The average atomic mass of oxygen is given by:

[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]

Where:

m: is the atomic mass

%: is the percent abundance

Since the sum of the percent abundance of oxygen isotopes must be equal to 1, we have:  

[tex] 1 = \%_{16} + \%_{17} + \%_{18} [/tex]

[tex] 1 = x + 3.2 \cdot 10^{-4} + \%_{18} [/tex]

[tex] \%_{18} = 1 - x - 3.2 \cdot 10^{-4} [/tex]

Hence, the percent abundance of O-18 is:  

[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]  

[tex]15.9982 = 15.972*x + 16.988*3.2 \cdot 10^{-4} + 17.970*(1 - 3.2 \cdot 10^{-4} - x)[/tex]

[tex] x = 0.980614 \times 100 = 98.0614 \% [/tex]                                                              

Hence, the percent abundance of oxygen-18 is:

[tex]\%_{18} = (1 - 3.2 \cdot 10^{-4} - 0.980614) \times 100 = 1.9066 \%[/tex]                      

Therefore, the percent abundance of oxygen-18 is 1.9066%.

I hope it helps you!                                                      

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