10.0 mL of concentrated HNO3 (16.0 M ) is added to 100.0 mL of water.
How many moles of HNO3 dissolve in the 100.0 mL of water?

Answers

Answer 1

10.0 mL of concentrated HNO[tex]_3[/tex] (16.0 M ) is added to 100.0 mL of water. 160moles is the number of moles of  HNO[tex]_3[/tex] dissolve in the 100.0 mL of water.

In chemistry, a mole, usually spelt mol, is a common scientific measurement unit for significant amounts of very small objects like atoms, molecules, and other predetermined particles. The mole designates 6.02214076 1023 units, which is a very large number. For the Worldwide System of Units (SI).

The mole is defined as this number as of May 20, 2019, according a general conference upon Weights and Measures. The total number of atoms discovered through experimentation to be present in 12 grammes of carbon-12 was originally used to define the mole.

Molarity₁×Volume₁=Molarity₂×Volume₂

16.0 ×10.0=Molarity₂× 100.0

160/100.0=Molarity₂

1.6M=Molarity₂

number of moles =1.6×100

                          = 160moles

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Related Questions

Completely describe the electrolytic cell corresponding to the following equation. (Hint: you may need to combine 2 half reactions from Table 17-1 to make one of the half reactions for this cell)
Cr2O7^2– + I^– → Cr^3+ + IO3^–
With work please

Answers

The first half-reaction is the oxidation of Cr2O7^2– to Cr^3+ and the second half-reaction is the reduction of I^– to IO3^–. When combined, the overall reaction is Cr2O7^2– + I^– → Cr^3+ + IO3^–.

The electrolytic cell consists of two electrodes, one anode and one cathode, both of which are immersed in an electrolyte solution. At the anode, the Cr2O7^2– ions are oxidized to Cr^3+ ions, releasing electrons into the external circuit.

At the cathode, the I^– ions are reduced to IO3^– ions, and the electrons from the external circuit are used to drive the reaction. The electrolyte solution must contain both Cr2O7^2– and I^– ions in order to facilitate the transfer of electrons between the electrodes.

The overall reaction is driven by the potential difference between the anode and the cathode, which is created by the flow of electrons through the external circuit.

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16. Balance the given redox reaction by ion electron method (basic medium) 1₂ + OH™ >I+103™ + H₂O​

Answers

The balanced redox reaction is:

6I2 + 42OH- → 6IO3- + 18H2O

How do you balance a redox reaction?

We have to note that we can only say that a redox reaction have been balanced when the number of the electrons that were lost in the oxidation half equation is equal to the number of electrons that have been gained in the reduction half equation.

The balanced half reaction equations are;

I2 -→ IO3- + 6e (oxidation half-reaction)

6OH- + 6e- → 2H2O  (reduction half-reaction)

Thus adding the equations would give the balanced reaction equation.

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For each of the following, convert the word equation into a formula equation AND balance it using the tally method!

*The answers have been provided for you, please upload work showing your use of the tally method to earn credit!*

aluminum + nickel(II) sulfate --> aluminum sulfate + nickel
2Al + 3Ni(SO4) --> Al2(SO4)3 + 3Ni

ammonium phosphate + magnesium nitrite --> magnesium phosphate +ammonium nitrite
2(NH4)3(PO4) + 3Mg(NO2)2 --> Mg3(PO4)2 + 6(NH4)(NO2)

lead + nitrogen --> lead(IV) nitride
3Pb + 2N2 --> Pb3N4

Answers

Answer:

For aluminum + nickel(II) sulfate --> aluminum sulfate + nickel:

Al + NiSO4 --> Al2(SO4)3 + Ni

Using the tally method:

Al: 2

Ni: 1

S: 3

O: 12

The balanced equation is:

2Al + 3Ni(SO4) --> Al2(SO4)3 + 3Ni

For ammonium phosphate + magnesium nitrite --> magnesium phosphate + ammonium nitrite:

(NH4)3PO4 + Mg(NO2)2 --> Mg3(PO4)2 + 6NH4NO2

Using the tally method:

N: 6

H: 28

P: 2

O: 16

Mg: 3

The balanced equation is:

2(NH4)3(PO4) + 3Mg(NO2)2 --> Mg3(PO4)2 + 6(NH4)(NO2)

For lead + nitrogen --> lead(IV) nitride:

3Pb + 2N2 --> Pb3N4

Using the tally method:

Pb: 3

N: 8

The balanced equation is:

3Pb + 2N2 --> Pb3N4

Explanation:

If 0.499 g of NaOH (MM = 40.00 g/mol) is dissolved in 150.00 mL of water, what is the theoretical molarity of NaOH? (do not forget about SF)

Answers

Molarity is an important method which is used to calculate the concentration of a solution. The molarity of 0.499 g of NaOH dissolved in 150.00 mL of water is  0.082 M.

Molarity of a solution is defined as the number of moles of the solute present per litre of the solution. The unit of molarity is mol L⁻¹ and it is represented as 'M'.

Molarity = Number of moles of solute / Volume of solution in L

150.00 mL = 0.15 L

Number of moles:

n = 0.499 / 40.00 = 0.0124 moles

M = 0.0124 / 0.15

M = 0.082

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If a sulfuric acid (H₂SO₄, MM = 98.09 g/mol) solution is completely neutralized with 61.37 mL of the NaOH solution above (from question 2) what was the mass in grams of sulfuric acid in the solution. Remember, you need a balanced chemical reaction for stoichiometry. (do not forget about SF)

Answers

Taking into account the reaction stoichiometry, 3.00713 grams of H₂SO₄ is completely neutralized with 61.37 mL of the NaOH.

Reaction stoichiometry

In first place, the balanced reaction is:

H₂SO₄ + 2 NaOH  → Na₂SO₄ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

H₂SO₄: 1 moleNaOH: 2 moles  Na₂SO₄: 1 moleH₂O: 2 moles

The molar mass of the compounds is:

H₂SO₄: 98 g/moleNaOH: 40 g/moleNa₂SO₄: 142 g/moleH₂O: 18 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

H₂SO₄: 1 mole× 98 g/mole= 98 gramsNaOH: 2 moles× 40 g/mole= 80 gramsNa₂SO₄: 1 mole× 142 g/mole= 142 gramsH₂O: 2 moles× 18 g/mole= 36 grams

Mass of H₂SO₄ required

A sulfuric acid solution is completely neutralized with 61.37 mL (0.06137 L) of the NaOH solution. Assuming the concentration of the NaOH solution is 1.0 M, the moles of NaOH that react is calculated as:

Moles of NaOH reacting = concentration× volume in liters

Moles of NaOH reacting = 1 M× 0.06137 L

Moles of NaOH reacting = 0.06137 moles

Now, the following rule of three can be applied: If by reaction stoichiometry 2 moles of NaOH react with 98 grams of H₂SO₄, 0.06137 moles of NaOH react with how much mass of H₂SO₄?

mass of H₂SO₄= (0.06137 moles of NaOH× 98 grams of H₂SO₄)÷2 moles of NaOH

mass of H₂SO₄= 3.00713 grams

Finally, 3.00713 grams of H₂SO₄ is required.

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2.50 g of As2O3 are titrated with 38.5 mL of KMnO4 to reach the end point.
5As2O3(s)+4MnO−4(aq)+9H2O(l)+12H+(aq)⟶10H3AsO4(aq)+4Mn2+(aq)
Calculate the concentration of the KMnO4 solution.

Answers

50 g of As[tex]_2[/tex]O[tex]_3[/tex] are titrated with 38.5 mL of KMnO[tex]_4[/tex] to reach the end point. 0.26M is the concentration of the KMnO[tex]_4[/tex]  solution.

Concentration in chemistry refers to the quantity of a material in a certain area. The ratio of the solute within a solution to the solvent or whole solution is another way to define concentration. In order to express concentration, mass in unit volume is typically used.

The solute concentration can, however, alternatively be stated in moles or volumetric units. Concentration may be expressed as per unit mass rather than volume.

5As[tex]_2[/tex]O[tex]_3[/tex](s)+4MnO[tex]_4[/tex]⁻(aq)+9H[tex]_2[/tex]O(l)+12H⁺(aq)⟶10H[tex]_3[/tex]AsO[tex]_4[/tex](aq)+4Mn[tex]_2[/tex]⁺(aq)

the stoichiometry ratio between As[tex]_2[/tex]O[tex]_3[/tex] and MnO[tex]_4[/tex]⁻ is 5:4

0.0126 moles of  As[tex]_2[/tex]O[tex]_3[/tex] will react with 4/5×0.0126 moles = 0.01008moles

0.01008moles of MnO[tex]_4[/tex]⁻ is present in 38mL

concentration of KMnO[tex]_4[/tex]= moles×volume

                                        = 0.010/38×1000

                                        =0.26M

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determine whether each molecule will have hydrogen bonding. defend your answer with evidence from your notes.

Answers

A - Has no hydrogen bond since hydrogen is not bonded to an electronegative element

B - Has a hydrogen bond due to the N- H bond

C - Has a hydrogen bond due to the O - H bond

D - Has a hydrogen bond due to the H -I bond

What is the hydrogen bond?

In a hydrogen bond, a weak electrostatic connection is created when the positively charged hydrogen atom is drawn to the negatively charged electronegative atom.

This contact is stronger than van der Waals forces but not as strong as covalent or ionic bonds or other intermolecular forces.

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If you have 155 mL solution of a 0.762 M FeCIs solution, how many grams of FeCIs are contained in this sample?


What is the mass in grams of KBr in 0.400 L of a 0.350 M solution?

Answers

1. The mass (in grams) of FeCl₂ contained in the sample is 14.96 grams

2. The mass (in grams) of KBr present in the solution is 16.66 grams

How do i determine the mass present?

1. The mass of FeCl₂ contained in the sample can be obtained as shown below:

First, we shall obtain the mole

Volume = 155 mL = 155 / 1000 = 0.155 LMolarity = 0.762 MMole of FeCl₂ =?

Mole = molarity × volume

Mole of FeCl₂ = 0.762 × 0.155

Mole of FeCl₂ = 0.118 mole

Finally, we shall determine the mass of FeCl₂ present in the sample. Details below:

Mole of FeCl₂ = 0.118 moleMolar mass of FeCl₂ = 126.75 g/molMass of FeCl₂ = ?

Mass = Mole × molar mass

Mass of FeCl₂ = 0.118 × 126.75

Mass of FeCl₂ = 14.96 grams

2. The mass of KBr present in the solution can be obtained as shown below:

First, we shall obtain the mole

Volume = 0.4 LMolarity = 0.350 MMole of KBr =?

Mole = molarity × volume

Mole of KBr = 0.350 × 0.4

Mole of KBr = 0.14 mole

Finally, we shall determine the mass of KBr in the solution. Details below:

Mole of KBr = 0.14 moleMolar mass of KBr = 119 g/molMass of KBr = ?

Mass = Mole × molar mass

Mass of KBr = 0.14 × 119

Mass of KBr = 16.66 grams

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A. What volume of base was needed to reach the equivalence point. B. What is the pH at the equivalent point?

Answers

From the titration curve that have been shown in the image, the equivalence point is 50 mL

What is the equivalence point on a titration curve?

At the equivalence point on a titration curve, the amount of titrant added is chemically equivalent to the amount of analyte in the sample being evaluated. As a result of the reaction between the titrant and analyte at this point, the entire analyte has been neutralized by the titrant.

You can locate the equivalence point by plotting the pH or any relevant aspect of the sample under examination as a function of the volume of titrant used.

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Hydrogen sulfide will burn \in three different ways, depending upon the amount of oxygen present. In one reaction, sulfur dioxide and water are produced. In the second reaction, water, sulfur dioxide, and sulfur are produced. The third reaction produces water and sulfur. What is maximum amount of sulfur dioxide that can be produced with 50.0g of hydrogen

Answers

The term mole concept is used here to determine the mass of sulfur dioxide.The mass of sulfur dioxide obtained from 50.0 g of hydrogen sulphide is 93.93 g.

One mole of a substance is defined as that quantity of it which contains as many entities as there are atoms exactly in 12 g of carbon - 12. The formula used to calculate the number of moles is:

Number of moles = Given mass / Molar mass

Here balanced reaction is:

2H₂S + 3O₂ → 2SO₂ + 2H₂O

n (H₂S) = n(SO₂)

n = 50.0 / 34.1 = 1.466

Molar mass of sulfur dioxide = 64.067 g/mol

Mass = n × M

m = 1.466 × 64.067 = 93.93 g

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Calculating equilibrium concentrations when the net reaction proceeds forward
Consider mixture B, which will cause the net reaction to proceed forward.
Concentration (M)initial:change:equilibrium:[XY]0.500−x0.500−xnet→⇌[X]0.100+x0.100+x+[Y]0.100+x0.100+x
The change in concentration, x, is negative for the reactants because they are consumed and positive for the products because they are produced.

Based on a Kc value of 0.140 and the given data table, what are the equilibrium concentrations of XY, X, and Y, respectively?

Answers

We must utilize the reaction's equilibrium equation to determine the equilibrium concentrations:

Kc = [X][Y]/[XY]

replacing the specified values:

0.140 = (0.100+x)(0.100+x)/(0.500-x)

Expanding and condensing:

0.14 = (0.01 + 0.2x + x^2)/(0.5 - x)

0.07 - 0.14x = x^2 + 0.2x + 0.01

Changing the order and applying the quadratic formula to find x

x^2 + 0.34x - 0.06 = 0

x = 0.193 or -0.533

X cannot be negative because it denotes a change in concentration. Consequently, x = 0.193 M.

It is now possible to determine the equilibrium concentrations:

[XY] = 0.500 - x = 0.307 M

[X] = [Y] = 0.100 + x = 0.293 M

Therefore, the equilibrium concentrations of XY, X, and Y are 0.307 M, 0.293 M, and 0.293 M, respectively.

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2. Label the different layers of the atmosphere and the separating boundaries between each layer.​

Answers

The layers of the atmosphere with the boundaries that are between them are shown in the image attached here.

Layers of the atmosphere

The troposphere, which rises up to an altitude of roughly 10-15 km, is the layer that is closest to the Earth's surface. The majority of the Earth's air mass is contained here, which is also where weather happens.

The tropopause marks the transition from the troposphere to the stratosphere, the next layer. It designates the altitude at which the temperature stops dropping.

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 If you mixed three plastic cup's contents containing 0,05 L of 1M Kool-Aid, 0.05 L of 2.5M Kool-Aid, and 0.05 L of 0.5M Kool-Aid, what would the molarity be of the resulting solution?

Answers

M = 1.183

Molarity is moles/Liter, so setting up some equations gives you

1) 1M = Xmol/0.05L

2) 2.5M = Xmol/0.05L

3) 0.5M = Xmol/0.05L

Solve for each X

1) X = 0.05mol

2) X = 0.125mol

3) X = 0.0025mol

Now add all the moles and Liters (separately)

XM = (0.05+0.125+0.0025)mol/(0.05+0.05+0.05)L

XM = 0.1775mol/0.15L

X = 1.183M

The products of a gamma decay reaction
are shown below. What was the parent
isotope of this nuclear reaction?
22 0
Ne + Y
10 0

Answers

The product of the gamma decay reaction is 22/10 Ne

What is gamma decay ?

Gamma radiation, commonly referred to as gamma decay, is a type of radioactive decay in which an unstable atomic nucleus generates gamma rays, a form of extremely high-energy electromagnetic radiation.

Gamma rays are neutral and have no mass, in contrast to alpha and beta particles, which have mass and charge.

The result of the gamma decay would not alter the parent nuclei thus the product would have the same mass and charge as the parent nucleus.

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Can anyone help me with this??

Answers

In the region labelled as H, weather fronts are most likely to be located. A weather front comprises a line dividing air masses with various differences in their attributes,

A weather front comprises a line dividing air masses with various differences in their attributes, including air density, winds, temperatures, or humidity. These variations frequently result in disturbed or unstable weather at the boundary.

For instance, although warm fronts usually arrive accompanied by stratiform precipitation with fog, cold fronts can bring belts of thunderstorms with cumulonimbus precipitation and be preceded by squall lines. Dry lines, which are softer humidity gradients, can cause severe weather in the summer. In the region labelled as H, weather fronts are most likely to be located.

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i need a little assistance with understanding this

Answers

Q_w = 307.3 J

C_m = .233 J/g•K

%error = 39.95%

For simplicity's sake, I'm relabeling

Q_w as Q1

m_w as m1

C_w as c1

T_eq as T2

T_w as T1

Q_m as Q2

m_m as m2

C_m as c2

experimental value as exp

actual value as actual

Question 1

Q1 = m1•c1•(T2-T1)

Identify what you know

m1 = 124g

c1 = 4.13 J/g • K

T2 = Final temperature = 22.3°C

T1 = Initial temperature = 21.7°C

Convert Celsius to Kelvin (C+273.15=K)

T2 = 295.45 K

T1 = 294.85 K

Plug in

Q1 = 124g•(4.13 J/g•K)•(295.45K - 294.85K)

Solve

Q1 ≈ 307.3 J

Question 2

-Q1 = Q2 = m2•(c2)•(T2-100)

Ignore Q2 for a second, and you're left with

-Q1 = m2•(c2)•(T2-100)

which is the same thing.

Identify what you know

Q1 = 307.3 J

m2 = 17g

T2 = 22.3°C

Plug in

-(307.3J) = 17g • c2 • (22.3°C-100°C)

Solve

-307.3 J = (-1320.9 g•°C) • c2

c2 = .233 J/g•°C or J/g•K (I'll explain later)

Question 3

%err = ((exp - actual)/actual) • 100%

Identify what you know

exp = .233 J/g•K

actual = .388 J/g•K

Plug in

%err = ((.233 J/g•K - .388 J/g•K)/ .388 J/g•K) • 100%

Solve

%err = -39.95 %

Take the absolute value

%err = 39.95%

Referring to earlier change in units:

The reason we can not use the K value of T2 (295.45K) is because the formula provided (T2-100) does not account for T2 being in K. It only accounts for T2 being in °C.

Menthol is a substance commonly used in cough drops. It has a molar mass of 156.3 g/mol and is comprised of 77.4% carbon and 12.9% hydrogen and 10.2% oxygen. What is the molecular formula for menthol. (25 points)

Answers

Menthol is a substance commonly used in cough drops. It has a molar mass of 156.3 g/mol and is comprised of 77.4% carbon and 12.9% hydrogen and 10.2% oxygen. The molecular formula is C[tex]_{10}[/tex]H[tex]_{20}[/tex]O.

The precise quantity and kind of atoms that make up a single chemical molecule are specified in its molecular formula. A compound's constituent elements are denoted by their chemical symbols, which are followed by numerical subscripts that indicate how many atoms of each constituent are contained in the molecule.

A molecular formula does not have a chemical name and is wordless. A molecular formula may suggest particular simple chemical structures, but it is not the same as a formula for the entire chemical structure. Compared to structural formulas, they are more restrictive.

moles of carbon =0.122/12=0.010

Moles of Hydrogen =0.0204/1=0.0204

moles of oxygen = 0.0171/16 =0.0017

Dividing by simple whole number ratio

The ratio of C : H : O = 10 : 20 : 1. The molecular formula is C[tex]_{10}[/tex]H[tex]_{20}[/tex]O.

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What volume will 5.00 mol of an ideal gas occupy at 25.0 C. and 153 kPa of pressure?

Answers

Answer:

5.00 mol of an ideal gas will occupy 103.6 L at 25.0 C and 153 kPa of pressure.

Explanation:

Using the ideal gas law, PV=nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin, we can solve for V.

First, we need to convert the temperature from Celsius to Kelvin by adding 273.15 K. Therefore, the temperature is 25.0 + 273.15 = 298.15 K.

Next, we can plug in the values we know:

PV = nRT

(153 kPa) V = (5.00 mol) (8.31 J/mol*K) (298.15 K)

Simplifying:

V = (5.00 mol) (8.31 J/mol*K) (298.15 K) / (153 kPa)

V = 103.6 L

Therefore, 5.00 mol of an ideal gas will occupy 103.6 L at 25.0 C and 153 kPa of pressure.

Luke needs to measure the iron (Fe) content of a 2.470 g steel sample. He starts by completely converting the iron in the steel to Fe2+ and dissolving it in 50.00 mL of water. Then, Luke determines the amount of Fe2+ by reacting it with a standardized aqueous 0.110 M potassium permanganate (KMnO4) solution.
MnO−4+8H++5Fe2+⟶Mn2++5Fe3++4H2O
The titration of the Fe2+ solution requires 39.23 mL KMnO4 solution to reach the end point. What is the percentage by weight of iron (Fe) in the steel sample?

Answers

The balanced equation for the reaction between [tex]Fe^+^2[/tex] and KMnO₄ is: MnO₄ + 8H+ + 5 [tex]Fe^+^2[/tex]⟶ [tex]Mn^+^2[/tex] + 5[tex]Fe^3^+[/tex] + 4H₂O. Therefore, the percentage by weight of iron in the steel sample is 1.94%.

Molarity of KMnO₄ = 0.110 M, Volume of KMnO₄ used = 39.23 mL = 0.03923 L

moles of  [tex]Fe^+^2[/tex] = 0.110 M x 0.03923 L / 5 = 0.000856 moles

The mass of iron in the steel sample can be calculated from the number of moles of  [tex]Fe^+^2[/tex]:

moles of Fe = 0.000856 moles  [tex]Fe^+^2[/tex] x 1 mole Fe / 1 mole  [tex]Fe^+^2[/tex]

= 0.000856 moles Fe

The mass of Fe can be calculated from the number of moles of Fe:

mass of Fe = 0.000856 moles Fe x 55.85 g/mol = 0.0479 g Fe

Finally, the percentage by weight of iron in the steel sample can be calculated as:

% Fe = (mass of Fe / mass of steel sample) x 100%

mass of steel sample = 2.470 g

% Fe = (0.0479 g / 2.470 g) x 100% = 1.94%

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Who has course hero?I really need the “lion king…Ecology science.” Answer key so I can print it.It wont let me print mines.

Answers

Note that the above prompt on Ecology draws it's analysis form a well known story which have been told visually called "The Lion King".

The answers are:

1) Biotic Factors, simply put are living things

2 examples of things from Lion King Introduction are:

The MonkeyThe LionThe Grass

3) Abiotic factors are Non -living things.

4) Examples from the introduction are:

MountainWaterDirt

5) the symbiotic relationship is called: commensalism.

What is commensalism?

Long-term biological interactions known as commensalism occur when individuals of one species benefit while those of the other species suffer neither advantages nor harm.

Ecology which is the study of the environment, allows  a person  to comprehend how different types of creatures coexist in various kinds of physical settings.

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Full Question:

Although part of your question is missing, you might be referring to this full question:

1) What is Biotic Factors

2) List three biotic factors from the Lion King introduction

3) What is Abiotic factors?

4) List three Abiotic factors from the Lion King introduction

5) The bird riding on the tusks of the elephant feed on insects the elephant stirs up. What kind of symbiotic relationship exists between the two?

Thank you in advance.

Answers

The rate expression is k [BF₃]²[NH₃], overall order is 3, rate constant k is 1.13 × 10⁻³ mol⁻² dm⁶ s⁻¹.

How to calculate rate expression?

The rate expression for this reaction can be written as:

Rate = k [BF₃]^m[NH₃]^n

The method of initial rates can be used to calculate the values of m and n. When we compare experiments 1 and 2, notice that halving the concentration of [NH3] reduces the initial rate of reaction. This suggests that the reaction is first order in terms of [NH3], implying that n = 1.

When comparing experiments 1 and 3, notice that increasing the concentration of [BF3] by a factor of 2.5 increases the initial rate of reaction by a factor of 9.39 (i.e., 2.13/0.227). This shows that the reaction is about second order in relation to [BF3], i.e., m 2.

Therefore, the rate expression for the reaction is:

Rate = k [BF₃]²[NH₃]

The overall order of the reaction is m + n = 3.

Using the data from experiment 4, substitute the values and solve for k:

Rate = k [BF₃]²[NH₃]

1.02 × 10-¹ = k (3.00 × 10⁻¹)² (1.00 x 10⁻¹)

k = 1.02 × 10-¹ / (3.00 × 10⁻¹)² (1.00 x 10⁻¹)

k = 1.13 × 10⁻³ mol⁻² dm⁶ s⁻¹

Therefore, the rate constant k is 1.13 × 10⁻³ mol⁻² dm⁶ s⁻¹.

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Determine E° for a galvanic (voltaic) cell if ∆G° = -6.3 kJ/mol and n = 3. (F = 96,500 J/(V・mol))

Answers

The for a galvanic cell is 0.000217 volts if  ∆G° = -6.3 kJ/mol and n = 3. (F = 96,500 J/(V・mol).

A galvanic cell or voltaic cell, named after the scientists Luigi Galvani and Alessandro Volta, respectively, is an electrochemical cell in which an electric current is generated from spontaneous Oxidation-Reduction reactions. A common apparatus generally consists of two different metals, each immersed in separate beakers containing their respective metal ions in solution that are connected by a salt bridge or separated by a porous membrane.

=ΔG°/-nF= -6.3/-3×96500=0.000217 V.

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What mass of oxygen gas, O2, from the air is consumed in the combustion of 702 g of octane, C8H18, one of the principal components of gasoline?
Select one:
A. 13.22 g
O B. 25.0 g/mol
O C. 32.0 g
O D. 33.22 g/mol
O E. 39.0 g

Answers

The principal components of gasoline is 702 g of octane.

Thus, Crude oil and other petroleum liquids are used to create gasoline, a fuel. The majority of gasoline is utilized in car engines.

For retail sale at gas stations, finished motor gasoline is produced at petroleum refineries and blending facilities.

In order to create finished motor gasoline, petroleum refineries primarily produce gasoline blendstocks, which must be blended with other liquids. In order to create finished motor gasoline in various grades and formulas, gasoline blendstocks, finished gasoline, and fuel ethanol are combined at blending terminals.

Thus, The principal components of gasoline is 702 g of octane.

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5 gallons to ? L
1gal = 3.785 L

Answers

A gallon is a common volume measurement unit for measuring liquids and occasionally dry things. 5 gallon is equal to 18.925 L.

Thus, The US liquid gallon, US dry gallon, and Imperial gallon are the three different sorts of gallons. To measure and store commodities like fuel, oil, milk, paint, and many other things, these three types are frequently employed.

The English Parliament created the gallon in 1696 to be used for measuring dry goods. After gaining independence, the US adopted the Winchester gallon, which is now known as the US dry gallon.

Both are now referred to as US liquid gallon and US dry gallon, respectively. On the other hand, the Imperial gallon was adopted by the British Empire in 1834.

Thus, A gallon is a common volume measurement unit for measuring liquids and occasionally dry things. 5 gallon is equal to 18.925 L.

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5. Which of the following are correct?
I. Electrons are attracted towards higher potential.
II. Protons are attracted towards higher potential.
III. Electrons are attracted towards lower potential.
IV. Protons are attracted towards lower potential.
(a) I & II
(b) I & III
(c) II & III
(d) I & IV

Answers

The statements that are correct is that electrons are attracted towards higher potential and protons are attracted towards lower potential. So the correct option is d. I & IV

What are electrons and protons?

Electrons are particles that are negatively charged so if it is placed in an electric field it will move from negative to positive, due to its negative charge, then it will attract a higher potential.

Protons are particles that are positively charged so if it is placed in an electric field it will move from a positive to a negative side, then due to its positive charge, it will be attracted to a lower potential.

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Help with this
A scientist discovers a new object in the Kuiper Belt and claims it is a dwarf planet. What would have to be true about the object for the scientist to be correct?

A The object would have to be irregularly shaped, have an atmosphere, and be traveling in a path with other objects.
B The object would have to be irregularly shaped, orbit the Sun, and be traveling in a path free from other objects.
C The object would have to be spherical, orbit the Sun, and be traveling in a path with other objects.
D The object would have to be spherical, have an atmosphere, and be traveling in a path free from other objects.

Answers

Answer:

C The object would have to be spherical, orbit the Sun, and be traveling in a path with other objects.

Explanation:

The sphere is a shape which allows for the most amount of volume with the least amount of surface. It is a very efficient way of enclosing objects and also allows for any point on its surface

[tex]\large \boxed{\mathrm{Correct \ answer \ Is \ C.}}[/tex]

[tex]\large \boxed{\mathrm{Sorry \ have \ no \ time \ for \ explanation \ so \ sorry!}}[/tex]

Step 1: A↽−−⇀B+C equilibrium Step 2: C+D⟶E slow Overall: A+D⟶B+E
Determine the rate law for the overall reaction, where the overall rate constant is represented as

Answers

The rate law for the reaction can be written as -

rate= k[C] [D]

The rate of reaction or reaction rate is the speed at which reactants are converted into products.

When we talk about chemical reactions, it is a given fact that rate at which they occur varies by a great deal. Some chemical reactions are nearly instantaneous, while others usually take some time to reach the final equilibrium.

The rate of the reaction is governed by the slow step.

This gives the rate equation-

rate= k[C] [D]

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Please help
Make sure to mark the answer as shown “question one= true/false”
Thank you

Answers

Answer:

Please mark me brainliest

Explanation:

The answer is false.

opaque materials do not reflect any light because they absorb or trap all light that comes in contact with them. No light is able to return from within an opaque material or reflect off its inner surface. Opaqueness is defined by a lack of light reflection, transmission or scattering. Light either disappears or is converted into other forms of energy within opaque substances.

Please let me know if this helps explain why opaque materials do not reflect light or if any part of the description is unclear. I can also provide any additional examples or details if needed.

Answer: 19) False. A totally opaque material does not let any light pass through it.

When 1500J of energy is lost from a 120 gram object, the temperature decreased from 45° C to 40° C. What is the specific heat of the object?

Answers

Specific Heat = 2.5 J/g•°C

We can set up an equation to solve this.

(Energy) = (mass)•(specific heat)•(change in temperature)

Plug in what you know

-1500J = 120g•(c)•(40°C-45°C)

[The energy is negative because the energy is lost, and the change in temperature can be found by subtracting the initial temperature from the final temperature (T(Final) - T(Initial))]

Solve

-1500J = 120g•(c)•(40°C-45°C)

-1500J = 120g•(c)•(-5°C)

-1500J = (-600g•°C)•(c)

(-1500J) / (-600g•°C) = c

c = 2.5 J/g•°C

If you mixed three plastic cup's contents containing 0,05 L of IM Kool-Aid, 0.05 L of 2.5M Kool-Aid, and 0.05 L of 0.5M Kool-Aid, what would the molarity be of the resulting solution?

Answers

The resultant solution has a molarity of 1 M.

How to determine molarity?

To find the molarity of the resulting solution, calculate the total amount of moles of Kool-Aid and the total volume of the solution.

The moles of Kool-Aid in each cup can be calculated as:

moles = concentration (M) × volume (L)

Cup 1: moles = 0.05 L × 0 M = 0 mol

Cup 2: moles = 0.05 L × 2.5 M = 0.125 mol

Cup 3: moles = 0.05 L × 0.5 M = 0.025 mol

The total amount of moles is:

total moles = moles in cup 1 + moles in cup 2 + moles in cup 3

total moles = 0 mol + 0.125 mol + 0.025 mol

total moles = 0.15 mol

The total volume of the solution is:

total volume = volume in cup 1 + volume in cup 2 + volume in cup 3

total volume = 0.05 L + 0.05 L + 0.05 L

total volume = 0.15 L

Now calculate the molarity of the resulting solution:

molarity = total moles ÷ total volume

molarity = 0.15 mol ÷ 0.15 L

molarity = 1 M

Therefore, the molarity of the resulting solution is 1 M.

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