Given the following equivalents, make the following conversion: 1.00 knop = ? knips

4 clips = 5 blips
1 knop = 6 bippy
3 blip = 18 pringle
1 clip = 10 knip
10 bippy = 8 pringle

Answer:

c) will never be spontaneous at any pressure.

Explanation:

Hello,

In this case, we need to remember that the second law of thermodynamics states that heat flow is transferred from a hot object to a cold object only, never the other way around, therefore, the answer is c) will never be spontaneous at any pressure. This is supported by driving force that in this case is the temperature difference which must be negative for the hot object (it gets eventually cold) and positive for the cold object (it gets eventually hot) until they reach the equilibrium temperature.

Best regards.

Answer:

A precipitate will form, AgI

Explanation:

When Ag⁺ and I⁻ ions are in an aqueous media, AgI(s), a precipitate, is produced or not based on its Ksp expression:

Ksp = 8.3x10⁻¹⁷ = [Ag⁺] [I⁻]

Where the concentrations of the ions are the concentrations in equilibrium

For actual concentrations of a solution, you can define Q, reaction quotient, as:

Q = [Ag⁺] [I⁻]

If Q > Ksp, the ions will react producing BaCO₃, if not, no precipitate will form.

Actual concentrations of Ag⁺ and I⁻ are:

[Ag⁺] = [AgNO₃] = 2.0x10⁻⁵ × (300mL / 500.0mL) = 1.2x10⁻⁵M

[I⁻] = [NaI] = 2.5x10⁻⁹ × (200mL / 500.0mL) = 1.0x10⁻⁹M

500.0mL is the volume of the mixture of the solutions

Replacing in Q expression:

Q = [Ag⁺] [I⁻]

Q = [1.2x10⁻⁵M] [1.0x10⁻⁹M]

Q = 1.2x10⁻¹⁴

As Q > Ksp

Answer:

Half life = 13.197 hour

Explanation:

Given:

Old amount (A₀) = 3.2

New amount (A) = 0.4

Radiation decay time (t) = 39.6 hour

Half life = T(1/2)

Find:

Half life = T(1/2) = T

Computation:

A = A₀[tex]e^{-(\frac{0.693t}{T} )}[/tex]

[tex]e^{-(\frac{0.693t}{T} )}[/tex] = 0.4 / 3.2

-[27.4428 / T] = In (0.125)

-[27.4428 / T] = -2.0794

[27.4428 / T] = 2.0794

T = 13.197

Half life = 13.197 hour

Answer:

[tex]Kc=0.14[/tex]

Explanation:

Hello,

In this case, for the given reaction, the equilibrium expression is:

[tex]Kc=\frac{[NO]^2[O_2]}{[NO_2]^2}[/tex]

That in terms of the reaction extent [tex]x[/tex] is written as (initial concentration of NO2 is 0.24 M for 0.24 mol in 1.00 L):

[tex]Kc=\frac{(2*x)^2(x)}{(0.24M-2*x)^2}[/tex]

Moreover, since at the equilibrium 0.14 moles of NO are present (a 0.14-M concentration), we can compute the reaction extent as shown below:

[tex][NO]=2*x=0.14M[/tex]

[tex]x=0.14M/2=0.07M[/tex]

In such a way, knowing [tex]x[/tex], we compute Kc as shown below:

[tex]Kc=\frac{(2*0.07)^2(0.07)}{(0.24M-2*0.07)^2}\\\\Kc=0.14[/tex]

Regards.

The kc is 0.14.

Since

[tex]Kc = \frac{[NO]^2[O_2]}{[NO_2]^2}\\\\[/tex]

Here the reaction extent x should be written like the initial concentration of NO2 is 0.24 M for 0.24 mol in 1.00 L.

Now

[tex]Kc = \frac{(2\times\ x)^2}{(0.24M - 2\times x)^2}[/tex]

Since at the equilibrium 0.14 moles of NO are presented so the reaction should be

NO = 2*x = 0.14m

x = 0.07

Now kc should be

[tex]= \frac{(2\times 0.07)^2 (0.07)}{(0.24M - 2\times 0.07)^2}[/tex]

= 0.14

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Answer:

there are 37,8541 liters in 10 gallons

Answer:

0.27 V

Explanation:

Given that the both half cells contain 1.0 molar solutions of their respective electrolytes.

E°Pb= -0.13 V

E°Cd = -0.40 V

Since it is a galvanic cell, the electrode having a more negative electrode potential will serve as the anode and the electrode having a less negative electrode potential will serve as the cathode.

Hence cadmium will serve as the anode and lead will serve as the cathode.

E°cell = E°cathode - E°anode

E°cell = -0.13 - (-0.40)

E°cell = 0.27 V

Answer:

Molar mass is the mass of one mole of a substance and is ontained by adding atomic mases of all the constituent atoms and group of atoms.

Atomic masses: Cu = 63.55 g/mol,

S = 32.07 g/mol

O x 4 = 16.00 g/mol x 4 = 64.00 g/mol

H = 1.008 g/mol,

So, H2O = (2*1.008+16.00)*5 = 90.10 g/mol

Molar mass of CuSO4,5H2O = 249.72 g/mol.

Out of that 249.72 grams, 90.10 grams is water,

So, percentage by mass of water in CuSO4,5H2O

= 90.10*100/249.72 = 36.08.

I think this is what you are asking. hope this helps

Answer:

2

Explanation:

A single covalent bond is formed when two electrons are shared between the same two atoms, one electron from each atom.

Answer:

the answer is 2

Explanation:

Explanation:

The answer is directly proportional, because when there is more concentration their will more reactants to react fast diring the chemical reaction which increases the rate of chemical reaction.

So, we can state that the relationship between them are directly proportional.

Hope it helps...

Answer:

1.84 × 10⁻³

Explanation:

Step 1: Write the balanced equation

2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)

Step 2: Calculate the initial concentration of NOBr

0.143 moles of NOBr(g) are introduced into a 1.00 liter container. The molarity is:

M = 0.143 mol / 1.00 L = 0.143 M

Step 3: Make an ICE chart

         2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)

I             0.143               0           0

C              -2x               +2x        +x

E          0.143-2x            2x          x

Step 4: Find the value of x

The equilibrium concentration of NOBr(g) was 0.108 M. Then,

0.143-2x = 0.108

x = 0.0175

Step 5: Calculate the concentrations at equilibrium

[NOBr] = 0.108 M

[NO] = 2x = 0.0350 M

[Br₂] = x = 0.0175 M

Step 6: Calculate the equilibrium constant (Kc)

Kc = [0.0350]² × [0.0175] / [0.108]²

Kc = 1.84 × 10⁻³

Answer:

A : At the end of the hour, isotope B has a greater decay constant λ than isotope A

Explanation:

Firstly, we need to understand that radioactive decay follows a first order rate law.

What this means is that we can calculate the radioactive decay constant using the following formula from the half-life

Mathematically;

[tex]t_{1/2}[/tex]  = 0.693/λ

where λ represents the radioactive decay constant.

Rearranging the equation, we can have

λ = 0.693/[tex]t_{1/2}[/tex]

Now, to have a fair level playing ground, it is best that the half-life of both isotopes are in the same unit of time(seconds)

For A, the half-life = 2.3 minutes which is same as 2.3 × 60 = 138 seconds

For B, the half-life is 24 seconds

Thus, at the end of the hour, the decay constant for isotope A will be;

λ = 0.693/138 = 0.0050 [tex]s^{-1}[/tex]

For isotope B, the decay constant will be;

λ = 0.693/24 = 0.028875  [tex]s^{-1}[/tex]

We can see that the decay constant of isotope B is higher than that of A at the end of the experiment

The given question is incomplete.

The complete question is:

When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many grams of methane were needed for the reaction?

Answer: 4 grams of methane were needed for the reaction

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

{tex]CH_4+2O_2\rightarrow CO_2+H_2O[/tex]

Given:  mass of oxygen = 16 g

Mass of carbon dioxide = 11 g

Mass of water = 9 g

Mass of products = Mass of carbon dioxide + mass of water = 11 g  +9 g = 20 g

Mass or reactant = mass of methane + mass of oxygen = mass of methane + 16 g

As mass of reactants = mass of products

mass of methane + 16 g= 20 g

mass of methane  = 4 g

Thus 4 grams of methane were needed for the reaction

Answer:

False

Explanation:

Reaction rates is a field under chemical kinetics that deals with the measure of speed of a chemical reaction. It is the change in the concentration of a reactant or product per unit time.

Activation Energy is a theory been  put forward to explain why different chemical reactions proceed at different rates.

Activation Energy theory postulates that for a reactant to transform into a product , the colliding particles or molecules of the reactant must possess a certain amount of energy so as to overcome the reaction barrier.

An important factor which may influence the attainment of activation energy by colliding particles of reactants is the temperature at which the reaction is carried out. The higher the temperature, the  greater is the fraction of the reactant particles which possess the activation energy and thus the faster the reaction becomes. SO , in essence increasing the temperature of a reaction system do not decreases the activation energy of the reaction but rather also increases  the activation energy of the reaction.

Answer:

ELECTROLYTIC CELL

An electrochemical cell in which a non-spontaneous chemical reaction takes place when electric current is passed through the solution,is called an electrolytic cell.

EXAMPLE

Nelsons cell and Downs cell

VOLTAIC CELL OR GALVANIC CELL

The electrochemical cell in which a spontaneous chemical reaction takes place and generates electric current is called galvanic and voltaic cell.

EXAMPLE

Daniel cell

Explanation:

Electrolytic cell by the non-spontaneous reactions covert the chemical energy into electical and volataic cells are those in which spontaneous redox reaction takes place.

Cell is a device which converts chemical energy into electrical energy.

Electrolytic Cell: An electrolytic cell is an electrochemical cell in which a non-spontaneous redox chemical reaction occurs when an electric current is conducted through the solution.

Example- Electrolysis of sodium chloride, by which formation of sodium metal and chlorine gas takes place.

Voltaic cell: Voltaic cells are electrochemical cells in which a spontaneous redox chemical reaction occurs and creates electric current. These cells are also known by the name of Galvanic cells.

Example- It is used in the form of batteries which can be portable easily.

Hence, in electrolytic cells non - spontaneous reaction occur and in voltaic cell spontaneous reaction is occured.

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Answer:

ΔS surrounding (entropy change of the reservoir) = -1 J/K

Explanation:

Given:

Change in heat (ΔH) = 150 joules

Temperature (T) = 150 K

Find:

ΔS surrounding (entropy change of the reservoir)

Computation:

ΔS surrounding (entropy change of the reservoir) = - ΔH / T

ΔS surrounding (entropy change of the reservoir) = - 150 / 150

ΔS surrounding (entropy change of the reservoir) = -1 J/K

Answer:

pH of solution B is 5

Explanation:

A weak acid, HA, is in equilibrium with water as follows:

HA(aq) + H₂O(l) ⇄ A⁻(aq) + H₃O⁺(aq)

Where Ka (10^-pKa = 1x10⁻⁹) is:

Ka = 1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]

Where concentrations of this species are equilibrium concentrations

As initial concentration of HA is 0.1M, the equilibrium concentrations of the species are:

[HA] = 0.1M - X

[A⁻] = X

[H₃O⁺] = X

Where X is the amount of HA that reacts until reach the equilibrium, X is reaction coordinate.

Replacing in Ka expression:

1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]

1x10⁻⁹ = [X] [X] / [0.1 - X]

1x10⁻¹⁰ - 1x10⁻⁹X = X²

1x10⁻¹⁰ - 1x10⁻⁹X - X² = 0

Solving for X:

X = -0.00001 → False solution, there is no negative concentrations.

X = 1x10⁻⁵ → Right solution.

As [H₃O⁺] = X

[H₃O⁺] = 1x10⁻⁵M

And pH = -log[H₃O⁺]

pH = 5

Answer:

[tex]\huge\boxed{A \ noble \ gas}[/tex]

Explanation:

A noble gas always exists in a mono-atomic form. This is because the valency of noble gas is zero and thus they don't tend to lose, gain or share electrons. They are inert gases and do not react.

Answer:

a noble gas

Explanation:

the the noble gases exist as mono atomic elements that is helium, neon ,argon, Krypton,etc.

mono atomic elements exist as stable single atoms

Answer:

The binding energy = 8.64972649×10¹⁰ kJ/mole

Explanation:

Given that:

An atom of 108Te has a mass of 107.929550 amu.

In a 108 Te atom, there are 52 protons and 56 neutrons

where;

mass of proton= 1.007825 amu

mass of neutron= 1.008665 amu

Similarly; The atomic number of Te = 52

the mass of 52 protons = 52 ×  1.007825  amu

the mass of 52 protons = 52.4069 amu

the mass of 56 neutrons = 56 ×  1.008665 amu

the mass of 56 neutrons = 56.48524 amu

The total mass can now be = the mass of 52 protons + the mass of 56 neutrons

The total mass = 52.4069 amu +  56.48524 amu

The total mass = 108.89214  amu

Recall  : it is given that An atom of 108Te has a mass of 107.929550 amu.

Therefore, the mass defect will be = 108.89214  amu - 107.929550 amu

the mass defect = 0.96259amu

where 1 amu = 1.66× 10⁻²⁷ kg

Therefore, 0.96259amu = (0.96259  × 1.66× 10⁻²⁷) kg

= 1.5978994 × 10⁻²⁷kg

The binding energy = mass defect × (speed of light)²

where;

speed of light c = 2.99792 × 10⁸ m/s

The binding energy = 1.5978994 × 10⁻²⁷kg  ×  2.99792 × 10⁸ m/s

The binding energy = 1.43611597 × 10⁻¹⁰  J

The binding energy =  1.43611597 × 10⁻¹³ kJ/atom

since 1 mole = 6.023 × 10²³ atom (avogadro's constant)

Then;

The binding energy = ( 1.43611597 × 10⁻¹³ )× (6.023 × 10²³)  kJ/mole

The binding energy = 8.64972649×10¹⁰ kJ/mole

Answer: Km = 10μM

Explanation: Michaelis-Menten constant (Km) measures the affinity a enzyme has to its substrate, so it can be known how well an enzyme is suited to the substrate being used. To determine Km another value associated to an eznyme is important: Turnover number (Kcat), which is the number of time an enzyme site converts substrate into product per unit time.

Enzyme veolcity is calculated as:

[tex]V_{0} = \frac{E_{t}.K_{cat}.[substrate]}{K_{m}+[substrate]}[/tex]

where Et is concentration of enzyme catalitic sites and has to have the same unit as velocity of enzyme, so Et = 20nM = 0.02μM;

To calculate Km:

[tex]V_{0}*K_{m} + V_{0}*[substrate] = E_{t}.K_{cat}.[substrate][/tex]

[tex]K_{m} = \frac{E_{t}.K_{cat}.[substrate]-V_{0}*[substrate]}{V_{0}}[/tex]

[tex]K_{m} = \frac{0.02*600*40-9.6*40}{9.6}[/tex]

Km = 10μM

The Michaelis-Menten for the substrate SAD is 10μM.

Answer:

1, C, Xe 2, B,He

Explanation:

1, cause as u go down a group the boiling point increases.

2, boiling point of single element is greater than a compound

According to  periodic trends in periodic table boiling point increases down the  group and hence Xe has highest boiling point and more amount of heat is required to boil an element hence He has highest boiling point.

Periodic table is a tabular arrangement of elements in the form of a table. In the periodic table, elements are arranged according to the modern periodic law which states that the properties of elements are a periodic function of their atomic numbers.

It is called as periodic because properties repeat after regular intervals of atomic numbers . It is a tabular arrangement consisting of seven horizontal rows called periods and eighteen vertical columns called groups.

Elements present in the same group have same number of valence electrons and hence have similar properties while elements present in the same period show gradual variation in properties due to addition of one electron for each successive element in a period.

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Answer:

0.5 mol MgCl₂

Explanation:

Step 1: Write the balanced equation

Mg + 2 HCl → MgCl₂ + H₂

In words, 1 mole of Mg reacts with 2 moles of HCl to form 1 mole of MgCl₂ and 1 mole of H₂.

Step 2: Establish the appropriate molar ratio

The molar ratio of HCl to MgCl₂ is 2:1.

Step 3: Calculate the moles of MgCl₂ produced from 1 mole of HCl

1 mol HCl × (1 mol MgCl₂/2 mol HCl) = 0.5 mol MgCl₂

Answer:

it is 2.0, the above one is wrong

Explanation:

I did the test :

Answer:

E) NaF and SrO

Explanation:

The ionic bonding occurs between atoms with a great difference in electronegativity. This usually happens between a metal and a non-metal.

In which pair do both compounds exhibit predominantly ionic bonding?

A) KCl and CO₂. NO. C and O are non-metals and present covalent bonding.

B) SO₂ and BaF₂. NO. S and O are non-metals and present covalent bonding.

C) F₂ and N₂O. NO. Both compounds contain non-metals and present covalent bonding.

D) N₂O₃ and Rb₂O. NO. N and O are non-metals and present covalent bonding.

E) NaF and SrO. YES. Na and Sr are metals while F and O are non-metals.

Answer:

Oxidation half equation;

3Cd(s) -------> 3Cd^2+(aq) + 6e

Reduction half equation;

2In^3+(aq) + 6e -----> 2In(s)

Explanation:

Since the reduction potentials of Indium and Cadmium are -0.34 V and - 0.40 V respectively, we can see that cadmium will be oxidized while indium will the reduced.

We arrived at this conclusion by examining the reduction potential of both species. The specie with more negative reduction potential is oxidized in the process.

Oxidation half equation;

3Cd(s) -------> 3Cd^2+(aq) + 6e

Reduction half equation;

2In^3+(aq) + 6e -----> 2In(s)

Answer:

0.500 moles of oxygen

Explanation:

Full question says: "Calculate the moles of oxygen needed to produce 0.500 moles of acetic acid.

The reaction of ethanol (C₂H₅OH) with oxygen (O₂) is:

C₂H₅OH + O₂ → CH₃COOH + H₂O

Where 1 mole of ethanol reacts per mole of oxygen to produce 1 mole of acetic acid (CH₃COOH) and 1 mole of water

Based on the chemical equation (1 mole of oxygen produce 1 mole of acetic acid; Ratio 1:1). Thus, if you want to produce 0.500 moles of acetic acid you will need:

Answer:

quartz

Explanation:

The correct option would be quartz.

Allotropy is a phenomenon that describes the natural existence of the same element in different forms with different physical characteristics. Allotropes are therefore different forms of the same element.

Carbon as an element has several allotropes which include diamond, graphite, graphene, amorphous carbon, and fullerenes. Quartz is a crystalline solid that is composed of silicon dioxide and not carbon.

Hence, all the options are carbon allotropes except quartz.

Answer:

3

Explanation:

Resonance is a valence bond concept put forward by Linus Pauling to explain the fact that the observed properties of a molecule may be as a result of the fact that its actual structure lie somewhere between a given number of structural extremes called canonical structures or resonance structures.

There are three resonance structures for SO3 that obey the octet rule. All the S-O bonds in SO3 are equivalent in these resonance structures.

Seven equivalent resonance structures for the molecular of SO3 can be drawn without breaking the octet rule.

We can arrive at this answer because:

In an SO3 molecule, electrons are shared between atoms. This sharing can be done with seven resonance structures.

These structures are shown in the figure below.

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Answer:

I believe the answer The case study was influenced by bias, and led to incorrect conclusions being drawn. plz correct me if I am wrong

Explanation:

Answer: options B

Explanation:

Answer:

Reaction of Chlorine with Hydrogen Chlorine and Hydrogen mixed together explodes when exposed to sunlight, which produces Hydrogen Chloride. In the dark away from sunlight, no reaction occurs, so light energy is required for a reaction. Cl2 + H2 = 2 HCl Reaction of Chlorine with Non-Metals Chlorine directly combines with most non-metals.

Explanation:

I hope this helps bro

Answer:

[tex]m_{H_3PO_4}=31.5gH_3PO_4[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]P_2O_5(s)+3H_2O(l)\rightarrow 2H_3PO_4(aq)[/tex]

Thus, since the diphosphorus pentoxide to water molar ratio is 1:3 and we are given the mass of both of them, for the calculation of the maximum mass phosphoric acid that is yielded, one could first identify the limiting reactant, for which we compute the available moles of diphosphorus pentoxide (molar mass 142 g/mol):

[tex]n_{P_2O_5}=22.8gP_2O_5*\frac{1molP_2O_5}{142gP_2O_5}=0.161molP_2O_5[/tex]

And the moles of diphosphorus pentoxide that are consumed by 13.5 g of water (molar mass 18 g/mol):

[tex]n_{P_2O_5}^{consumed}=13.5gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molP_2O_5}{3molH_2O} =0.25molP_2O_5[/tex]

Hence, since less moles of diphosphorus pentoxide are available, we sum up it is the limiting reactant, therefore, the maximum mass of phosphoric acid (molar mass 98 g/mol) is computed by considering the 1:2 molar ratio between them as follows:

[tex]m_{H_3PO_4}=0.161molP_2O_5*\frac{2molH_3PO_4}{1molP_2O_5} *\frac{98gH_3PO_4}{1molH_3PO_4} \\\\m_{H_3PO_4}=31.5gH_3PO_4[/tex]

Regards.

Answer:

3BaO + Al₂(SO₄)₃  →  Al₂O₃+ 3BaSO₄

Explanation:

Yes! A reactiin occurs between barium hydroxide and auminium sulphate.

barium sulfate (BaSO4) and aluminum hydroxide (Al(OH)3) are the products obtained in this reaction.

The reaction is given by the equation below;

3BaO + Al₂(SO₄)₃  →  Al₂O₃+ 3BaSO₄