Two examples of energy transformations are shown. The energy transformations are similar because they both involve transformations that begin with chemical energy. Begin with electrical energy. Result in radiant energy. Result in mechanical energy.

Answer:

b. 2 mol of KI in 500. g of water

Explanation:

We have to apply the colligative property of freezing point depression.

The formula is: ΔT = Kf . m . i

As the (Kf . m . i) is higher, then the freezing temperature will be lower.

i refers to the Van't Hoff factor (number of ions dissolved in the solution)

KI → K⁺ + I⁻    (i =2)

Kf is constant so, we have to search for the highest m (molality)

Molality means the moles of solute in 1kg of solvent.

The highest m is option b → 2 mol of KI / 0.5 kg = 4 mol/kg

a. 1 mol of KI / 0.5 kg = 2 mol/kg

c. 1 mol of KI / 1kg = 1 mol/kg

d. 2 mol of KI / 1kg = 2 mol/kg

1000 g = 1kg. In order to determine molality we need to convert the mass (g) of solvent to kg

Answer:

D. Nitrogen Fixation

Explanation:

Nitrogen fixation is a process by which molecular nitrogen in the air is converted into ammonia (NH

3) or related nitrogenous compounds in soil.

Answer:

The answer to the question is

The equilibrium temperature T = 22.016 °C

Explanation:

To solve the question, we note the given variables thus

Mass of copper block =235 grams

Temperature of the copper block = 255 °C

Mass of aluminium calorimeter cup = 155 g

Mass of water in calorimeter cup = 16 °C

Also we note the specific heat capacities of the materials involved in the question

Specific heat capacity of water = 4.186 joule/gram

Specific heat capacity of copper= 0.385 joule/gram

Specific heat capacity of aluminium = 0.900  joule/gram

There for from the first law of thermodynamics, energy is neither created nor destroyed but it changes from one form to another, we have

Heat lost by copper = heat gained by water and the calorimeter cup

Therefore we have

The equation for heat capacity =

mass * specific heat capacity * Temperature change  = m·c·ΔT

therefore

m·c·ΔT for copper = m·c·ΔT for aluminium + m·c·ΔT for water

where ΔT on the left of the equation = Initial temperature - final temperature

while on the right  ΔT = Final temperature - Initial temperature

and the final temperature in this case = the equilibrium temperature

255*0.385*(255 -T) = 155*0.9*(T-16) +875*4.186*(T-16)

Which gives the equilibrium temperature T = 22.016 °C

The equilibrium temperature when a 235 g copper block at 255 °C is placed in a 155 g aluminum calorimeter cup containing 875 g of water at 16.0 °C can be found by applying conservation of energy. The heat lost by the hot copper block is equal to the heat gained by the water and the aluminum calorimeter. By setting up and solving this equation, we can determine the final equilibrium temperature.

In this calorimeter scenario, we have a hot copper block transferring its heat to the colder aluminum calorimeter and the water it contains. As per the law of conservation of energy, the heat lost by the hot copper block will be equal to the heat gained by the water and the aluminum calorimeter. We can express this mathematically as:
     (mass of copper)x(specific heat of copper)x(temperature change of copper) = (mass of water)x(specific heat of water)x(temperature change of water) + (mass of aluminum)x(specific heat of aluminum)x(temperature change of aluminum).

Given the specific heat of aluminum is 0.897 J/g °C, water is 4.184 J/g °C and copper is 0.390 J/g °C, we can substitute these values, as well as the rest of given values, into the equation and solve for the common final temperature, which represents the equilibrium temperature.

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Answer:

N2= 193.25mmHg

Cl2= 579.75mmHg

Explanation:

Total number of moles=4

For N2 mole fraction=1/4

Partial pressure of N2= 1/4× 773= 193.25mmHg

For Cl2, mole fraction= 3/4

Partial pressure of Cl2= 3/4 × 773 = 579.75 mmHg

The correct partial pressures of each gas in the container are:

[tex]\[ P_{\text{N}_2} = \frac{2}{3} \times 773 \text{ mmHg} \] \[ P_{\text{Cl}_2} = \frac{1}{3} \times 773 \text{ mmHg} \][/tex]

To find the partial pressures of each gas, we use Dalton's Law of Partial Pressures, which states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each individual gas. The chemical equation for the decomposition of nitrogen trichloride (NCl3) is:

[tex]\[ 2\text{NCl}_3(l) \rightarrow 2\text{N}_2(g) + 3\text{Cl}_2(g) \][/tex]

From the stoichiometry of the balanced equation, we see that 2 moles of NCl3 decompose to form 2 moles of N2 and 3 moles of Cl2. Therefore, for every mole of NCl3 that decomposes, 1 mole of N2 and 1.5 moles of Cl2 are produced. This gives us a molar ratio of N2 to Cl2 of 2:3 or, simplified, 2 parts N2 for every 3 parts Cl2, which corresponds to a ratio of 2/5 for N2 and 3/5 for Cl2 in the mixture.

Since the total pressure of the mixture is 773 mmHg, we can calculate the partial pressure of each gas by multiplying the total pressure by the respective molar ratios:

For nitrogen (N2):

[tex]\[ P_{\text{N}_2} = \frac{2}{5} \times 773 \text{ mmHg} \] \[ P_{\text{N}_2} = \frac{2}{3} \times 773 \text{ mmHg} \][/tex]

For chlorine (Cl2):

[tex]\[ P_{\text{Cl}_2} = \frac{3}{5} \times 773 \text{ mmHg} \] \[ P_{\text{Cl}_2} = \frac{1}{3} \times 773 \text{ mmHg} \][/tex]

Thus, the partial pressure of nitrogen gas (N2) is[tex]\( \frac{2}{3} \times 773 \text{ mmHg} \)[/tex] , and the partial pressure of chlorine gas (Cl2) is [tex]\( \frac{1}{3} \times 773 \text{ mmHg} \)[/tex]. These are the partial pressures of each gas in the container after the decomposition of liquid nitrogen trichloride.

On a molecular level, the author means that no compound is grosser than any other, as all compounds are made up of the same fundamental building blocks: atoms and molecules.

The author means that, on a molecular level, no one compound is inherently more disgusting or repulsive than another. This statement suggests that all compounds, regardless of their odor, appearance, or taste, are made up of the same fundamental building blocks: atoms and molecules. For example, two compounds may have different smells, but they are both composed of the same elements and bond together in similar ways. So, while certain compounds may be perceived as gross or unpleasant to our senses, from a molecular standpoint, they are all equally fascinating and interconnected.

Answer:

The solution is shown in the attached image to this answer.

Explanation:

The first attached image is the complete question.

The second attached image is the solution to the question.

The major organic product formed when the compound undergoes a reaction with HNO3 and H2SO4 is the corresponding nitro compound, specifically 2-nitrobenzoic acid.

When the given compound reacts with a mixture of concentrated nitric acid (HNO3) and concentrated sulfuric acid (H2SO4), it undergoes a nitration reaction. This reaction results in the substitution of a nitro group (-NO2) at the meta position (position 2) of the benzene ring, yielding 2-nitrobenzoic acid as the major product.

The concentrated sulfuric acid serves as a catalyst and helps in the generation of the nitronium ion (NO2+), which is the electrophile responsible for attacking the benzene ring. The nitro group is introduced as a substituent on the aromatic ring, and the carboxylic acid group (-COOH) remains intact. Overall, this reaction is a common method for introducing nitro groups onto aromatic rings and is widely used in organic synthesis for the preparation of various nitroaromatic compounds.

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Answer:

The initial temperature is 300 K (The temperature doesn't change)

Explanation:

Step 1: Data given

Initial volume = 21L

Final volume = 14L

Initial pressure = 100 kPa = 0.986923 atm

Final pressure = 150 kPa = 1.48038 atm

The final temperature = 300K

Step 2: Calculate the initial temperature

Calculate the initial temperature

(P1*V1)/T1 = (P2*V2)/T2

⇒with P1 = the initial pressure = 0.986923 atm

⇒with V1 = the initial volume = 21 L

⇒ with T1 = the initial temperature = ?

⇒with P2 = the final pressure = 1.48038 atm

⇒with V2 = the final volume = 14 L

⇒with T2 = the final temperature = 300 K

(0.986923 * 21)/T1 = (1.48038*14)/300

T1 = 300 K

The initial temperature is 300 K (The temperature doesn't change)

Answer : The balanced net ionic equation for the reactions are:

(A) [tex]2Ag^{+}(aq)+2Br^{-}(aq)\rightarrow AgBr(s)[/tex]

(B) [tex]H^{+}(aq)+OH^{-}(aq)\rightarrow H_2O(l)[/tex]

(C) [tex]Co^{2+}(aq)+S^{2-}(aq)\rightarrow CoS(s)[/tex]

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

Part A :

The balanced molecular equation will be,

[tex]2AgNO_3(aq)+MgBr_2(aq)\rightarrow Mg(NO_3)_2(aq)+2AgBr(s)[/tex]

The complete ionic equation in separated aqueous solution will be,

[tex]2Ag^+(aq)+2NO_3^{-}(aq)+Mg^{2+}(aq)+2Br^{-}(aq)\rightarrow Mg^{2+}(aq)+2NO_3^{-}(aq)+2AgBr(s)[/tex]

In this equation the species [tex]Mg^{2+}\text{ and }NO_3^-[/tex] are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

[tex]2Ag^{+}(aq)+2Br^{-}(aq)\rightarrow AgBr(s)[/tex]

Part B :

The balanced molecular equation will be,

[tex]HClO_4(aq)+KOH(aq)\rightarrow H_2O(l)+KClO_4(aq)[/tex]

The complete ionic equation in separated aqueous solution will be,

[tex]H^+(aq)+ClO_4^{-}(aq)+K^{+}(aq)+OH^{-}(aq)\rightarrow K^{+}(aq)+CLO_4^{-}(aq)+H_2O(l)[/tex]

In this equation the species [tex]K^{+}\text{ and }ClO_4^-[/tex] are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

[tex]H^{+}(aq)+OH^{-}(aq)\rightarrow H_2O(l)[/tex]

Part C :

The balanced molecular equation will be,

[tex](NH_4)_2S(aq)+CoCl_2(aq)\rightarrow 2NH_4Cl(aq)+CoS(s)[/tex]

The complete ionic equation in separated aqueous solution will be,

[tex]2NH_4^+(aq)+S^{2-}(aq)+Co^{2+}(aq)+2Cl^{-}(aq)\rightarrow 2NH_4^{+}(aq)+2Cl^{-}(aq)+CoS(s)[/tex]

In this equation the species [tex]NH_4^{+}\text{ and }Cl^-[/tex] are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

[tex]Co^{2+}(aq)+S^{2-}(aq)\rightarrow CoS(s)[/tex]

Final answer:

The net ionic equations are: 2Ag+ + 2Br- -> 2AgBr(s); H+ + OH- -> H2O(l); Co2+ + S2- -> CoS(s).

Explanation:

The net ionic equation for the reaction between silver nitrate (AgNO3) and magnesium bromide (MgBr2) is:

2Ag+ + 2Br- -> 2AgBr(s)

The net ionic equation for the reaction between perchloric acid (HClO4) and potassium hydroxide (KOH) is:

H+ + OH- -> H2O(l)

The net ionic equation for the reaction between ammonium sulfide ((NH4)2S) and cobalt(II) chloride (CoCl2) is:

Co2+ + S2- -> CoS(s)

Answer: noble gas

Explanation:

Chemical bond is formed between atoms by losing gaining, or sharing electrons.

When the number of electrons and protons differ, it leads to the formation of ionic species. When an atom gains electrons, it will lead to the formation of negatively charged ion known as anion and when an atom looses electrons, it will lead to the formation of positively charged ion known as cation.

The atoms lose , gain or share electrons to achieve nearest noble gas configuration or to get stable.

Example : Sodium (Na) with atomic number 11 loses one electron to form [tex]Na^+[/tex] to get electronic configuration of neon with 1 0 electrons.

Answer:

Empirical formula (which matches the molecular formula) is = PbC₈H₂₀

Explanation:

Our sample: 60 g of tetraethyl lead

In order to determine the compound empirical formula we need the centesimal composition:

(Mass of element / Total mass) . 100 =

(38.43 g lead / 60g ) . 100 = 64.05%

(17.83 g C / 60g) . 100 = 29.72%

(3.74 g H / 60g) . 100 = 6.23 %

These % are the mass of the elements in 100 g of compound. Let's find out the moles of them:

64.05 g / 207.2 g/mol = 0.309 moles

29.72 g / 12 g/mol = 2.48 moles

6.23 g/ 1 g/mol = 6.23 moles

Next, we divide the moles, by the lowest value of them (0.309)

0.309 / 0.309 = 1 mol Pb

2.48 / 0.309 = 8 mol C

6.23 / 0.309 = 20 mol H

There, we have our formula PbC₈H₂₀

Answer:

K = 6.5 × 10⁻⁶

Explanation:

C₅H₆O₃(g) → C₂H₆(g) + 3 CO(g)

The formula for the ideal gas law

PV = nRT

where

P = total pressure

V = volume

n = total moles

R = gas constant

T = absolute temperature

P(C₅H₆O₃) = nRT/V

where

n = 5.63 /114

= 0.049

[tex]= \frac{0.049 * 0.0821 * 473}{2.5}[/tex]

= [tex]0.78atm[/tex]

C₅H₆O₃(g) ⇄ C₂H₆(g) + 3 CO(g)

0.78atm           0              0

0.78 - x            x              3x

P(total) = 1.63atm

0.78atm - x + x + 3x⇒x = 0.288atm

P(C₅H₆O₃)  = 0.78 - 0.288

                  = 0.489 atm

P(C₂H₆)  = 0.288 atm

P(CO) = 0.846 atm

[tex]Kp = \frac{0.288 * 0.864^3}{0.489}[/tex]

= 0.379

K = Kp/(RT³)

[tex]= \frac{0.379}{(0.0821 * 473)^3} \\= 6.5 * 10^-^6[/tex]

K = 6.5 × 10⁻⁶

The value of K for this reaction is : 6.5 * 10⁻⁶

The decomposition equation

C₅H₆O₃ ----> C₂H₆ (g) + 3 CO (g)

Given data :

mass of pure C₅H₆O₃ = 5.63 g

volume of flask = 2.50 L

Temperature = 200°C = 473 K

Pressure in flask = 1.63 atm

we will apply Ideal gas law formula

PV = nRT

therefore P( C₅H₆O₃ ) = nRT / V  ---- ( 1 )

where : n = 5.63 / 114 = 0.049,  V = 2.5 L,  R = 0.0821,  T = 473 k

Insert values into equation ( 1 )

P( C₅H₆O₃ ) = 0.78 atm - x  ---- ( 2 )

P(total) = 1.63 atm.

Considering the decomposition equation

0.78 - x + x + 3x = 1.63 atm

therefore ; x = 0.288

back to equation ( 2 )

P( C₅H₆O₃ ) = 0.78 - 0.288 = 0.489 atm

Given that :

P(C₂H₆) = 0.288 atm

P(CO) = 0.846 atm

Find K using the relationship below

K = Kp / ( RT)³  ------ ( 3 )

While Kp = ( 0.288 * 0.864³ ) / 0.489

               = 0.379

Back to equation ( 3 )

K = 0.379 / ( 0.0821 * 473 )³

  = 6.5 × 10⁻⁶.

Hence we can conclude that the value of K for this reaction is : 6.5 * 10⁻⁶

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Answer:

This tells us the radial velocity of the object and that the object is approaching or coming towards us.

Explanation:

Certain chemicals radiate with particular wavelengths or colors  when their temperature is raised or when they are charged electrically. Also observable are dark strokes separating the spectrum known as absorption lines

These spectral lines of chemicals are well known as stated above and from the phenomenon of Doppler effect, spectroscopy can be used to detect the movement of a distant object by the change of the emitted frequency of the wavelength

The Doppler effect is used in calculating the radial velocity of a distant object due to the fact that an approaching object compresses its emitted signal wavelength while a receding object has a longer wavelength than normal

Answer:

3.7g

Explanation:

First we calculate the molar mass of the drug. Secondly, we now obtain the mass of the drug solution required at the given concentration to obtain a 1.5 milligram dose of the drug. The substitution is now properly done and the mass is obtained as 3.7g of the nalorphine drug which is a relative or morphine used to combat withdrawal symptoms in narcotic users.

Answer:

2. Option B.

Explanation:

H₂SO₄ + Ba(OH)₂ → BaSO₄ + 2H₂O

You can count 2H in sulfuric acid and 2 H in the barium hyrdoxide, so the coefficient for water must be 2.

You will have 4 H on both sides of the reaction.

Try with the dissociations of each reactant

Sulfuric acid ⇒ H₂SO₄ →  2H⁺  + SO₄⁻²

Barium hydroxide ⇒ Ba(OH)₂ → Ba²⁺  +  2OH⁻

Sulfate anion bonds to barium cation to produce the salt, therefore the 2 protons will bond the 2 hydroxide in order to produce, 2 moles of H₂O

2H⁺ + 2OH⁻ → 2H₂O

Answer:

P₂ = 5.55 atm

Explanation:

Mathematically, Boyle's law can be stated as:

[tex]{\displaystyle P\propto {\frac {1}{V}}}[/tex]

Pressure is inversely proportional to the volume

[tex]{\displaystyle PV=k}[/tex] Pressure multiplied by volume equals some constant [tex]{\displaystyle k}[/tex]

where P is the pressure of the gas, V is the volume of the gas, and k is a constant.

The equation states that the product of pressure and volume is a constant for a given mass of confined gas and this holds as long as the temperature is constant. For comparing the same substance under two different sets of conditions, the law can be usefully expressed as:

[tex]{\displaystyle P_{1}V_{1}=P_{2}V_{2}[/tex]

The  final  pressure   of a  gas  is calculated  using  Bolyes law  formula  

that is  P1V1 =P2V2

Because the temperature and number of moles remained constant, we can use the formula

P₁V₁=P₂V₂

3.00 L * 1.85 atm = P₂ * 1.00 L

P₂ = 5.55 atm

Answer:

B. Relative Humidity

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

[tex]\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9[/tex]

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = [tex]164.5\times 10^{-9} g[/tex]

[tex]1 ng=10^{-9} g[/tex]

Volume of the sample = V = [tex]15.3 cm^3[/tex]

Density of the lake water sample ,d= [tex]1.00 g/cm^3[/tex]

Mass of sample =  M = [tex]d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g[/tex]

[tex]ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75[/tex]

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in [tex]1 cm^3[/tex]  of lake water = [tex]\frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g[/tex]

Mass of arsenic in [tex]0.710 km^3[/tex] lake water be m.

[tex]1 km^3=10^{15} cm^3[/tex]

Mass of arsenic in [tex]0.710\times 10^{15} cm^3[/tex] lake water :

[tex]m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g[/tex]

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

[tex]\frac{2.74}{41.90} days[/tex]

Then days required to remove 7,633.66 kg of arsenic from the lake water :

[tex]=7,633.66\times \frac{2.74}{41.90} days=499.19 days[/tex]

1 year = 365 days

499.19 days = [tex]\frac{499.19}{365} years = 1.367 years\approx 1.37 years[/tex]

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

[tex]\frac{2.74}{41.90} days[/tex]

Then days required to remove 7,633.66 kg of arsenic from the lake water :

[tex]=7,633.66\times \frac{2.74}{41.90} days=499.19 days[/tex]

1 year = 365 days

499.19 days = [tex]\frac{499.19}{365} years = 1.367 years\approx 1.37 years[/tex]

It will take 1.37 years to remove all of the arsenic from the lake.

The concentration of arsenic in the lake water sample is 10.5 ppb. The total mass of arsenic in the entire lake is 7445 kg. However, the time required to remove all arsenic from the lake cannot be calculated as the daily removal rate is not specified.

To compute the concentration of arsenic in the sample in parts per billion (ppb), we first need to convert the mass of arsenic from ng to g. This gives us 164.5 ng  x 10^-9 = 1.645 x 10^-7 g. Given that the density of the water is 1 g/cm^3 = 1 g/mL, a sample of 15.7 cm^3 would weigh 15.7 g. so, the concentration in ppb would be: (1.645 x 10^-7 g / 15.7 g) x 10^9 = 10.5 ppb.

Next, to find the total mass of arsenic in the whole lake, first convert the volume of the lake to cm^3. 0.710 km^3 = 0.710 x 10^15 cm^3. Using the concentration we found earlier, the total mass of arsenic (in g and then in kg) is (10.5 ppb x 0.710 x 10^15 g) x 10^-9 = 7445 kg of arsenic.

Finally, the time to remove all of the arsenic will depend on how much arsenic the company can remove per day, which is not mentioned in the problem. Thus, we can't calculate it based on the information given.

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Answer: covalent bond

Explanation:

An ionic bond is formed when an element completely transfers its valence electron to another element. The element which donates the electron is known as electropositive element or the metal and the element which accepts the electrons is known as electronegative element or non metal. Example: [tex]NaCl[/tex]

A covalent bond is formed when an element shares its valence electron with another element. This bond is formed between two non metals. Example: [tex]F_2[/tex] which is formed by sharing of one electron each from flourine atom.

A covalent bond is formed when one pair of electrons is shared between two atoms. This bond helps the atoms to complete their valence shell. Multiple pairs of electrons can also be shared, resulting in double and triple bonds.

When one pair of electrons is shared between two atoms, a covalent bond is formed. This type of bond is a strong one that can be formed between two atoms of the same or different elements. The sharing of electrons fulfills each atom's need to complete its valence shell, or outer ring of electrons.

For example, hydrogen gas (H-H) is an instance of a single covalent bond where two atoms of hydrogen each share their solitary electron.

Further, it is important to note that more than one pair of electrons can be shared between a pair of atoms, leading to what we call double and triple bonds. For instance, in formaldehyde (CH₂O), a double bond forms as two pairs of electrons are shared between the carbon and oxygen atoms.

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Answer:

The rate of formation of CO2 = 1.43 mol L-1s-1.

The rate of formation of H2O = 1.43 mol L-1s-1.

The rate of reaction of O2 = 1.43 mol L-1s-1.

Explanation:

2C6H14(g) + 19O2(g) -------> 12CO2(g) + 14H2O(g)

The rate of reaction of C6H14 is 1.43 mol L-1s-1.

The rate of reaction is theoretically defined as the speed of a chemical reaction. It is how fast or how slow the reactants are being used up or products are being formed.

It is the rate of change of concentration, amount etc., of any of the reactant or any of the product with time. The rate of reaction is the same and uniform for any of the products and reactants.

For a chemical reaction, A + 2B -----> 2C

r = -(dA/dt) = -(dB/dt) = (dC/dt)

Hence, the rate of reaction = rate of disappearance of C6H14 = rate of formation of CO2 = rate of formation of H2O = rate of reaction of O2 = 1.43 mol L-1s-1.

Hope this Helps!!!

Answer:

lattice parameter = 5.3355x10^-8 cm

atomic radius = 2.3103x10^-8 cm

Explanation:

known data:

p=0.855 g/cm^3

atomic mass = 39.09 g/mol

atoms/cell = 2 atoms

Avogadro number = 6.02x10^23 atom/mol

a) the lattice parameter:

Since potassium has a cubic structure, its volume is equal to:

v = [(atoms/cell)x(atomic mass)/(p)x(Avogadro number)]

substituting values:

v =[(2)x(39.09)/(0.855x6.02x10^23)]=1.5189x10^-22 cm^3

but as the cell volume is

a^3 =v

[tex]a=\sqrt[3]{v}=\sqrt[3]{1.5189x10^{-22} } = 5.3355x10^-8[/tex] cm

for a BCC structure, the atomic radius is equal to

[tex]r=\frac{ax\sqrt{3} }{4}=\frac{5.3355x10^{-8}x\sqrt{3} }{4}=2.3103x10^{-8}cm[/tex]

Answer:

The answer to your question is letter A

Explanation:

Data

Empirical formula CH₂

Molecular weight = 70 u

Process

1.- Calculate the molecular weight of the empirical formula (CH₂)

CH₂ = (12 x 1) + (1 x 2) = 12 + 2 = 14 g

2.- Divide the molecular weight by the molecular weight of the empirical formula

                      70/14 = 5

3.- Write the empirical formula

                       5(C₁H₂) = C₅H₁₀

1. C₅H₁₀

Given:

Empirical formula = CH₂

Molecular weight = 70 u

To find Molecular formula of the compound:

1.- Calculate the molecular weight of the empirical formula (CH₂)

CH₂ = [tex](12 * 1)+(1*2)=12+2=14 u[/tex]

(As molecular weight of C=12u and H= 1u)

2.- Divide the molecular weight by the molecular weight of the empirical formula

[tex]\frac{\text{Molecular weight}}{\text{ Molecular weight of Empirical formula}} = \frac{70}{14} =5[/tex]

3.- Write the empirical formula

5(C₁H₂) = C₅H₁₀

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Answer: The kilogram is the SI unit of mass and it is the almost universally used standard mass unit.

Explanation: When we use kilograms to measure weight, we are actually referring to kgf or kilogram-force.

Answer:

the awnser is 20dggrees

Explanation:

The effects of changes in concentrations or volume on an equilibrium are predicted by Le Chatelier's principle. The reaction will typically shift to restore equilibrium in response to such changes. The exact direction of the shift depends on the specifics of the change in conditions.

This question refers to Le Chatelier's principle, which allows predictions about how a change in conditions will affect a chemical equilibrium. (a) By doubling the concentration of B and halving the concentration of C, you are disrupting the equilibrium. The reaction will shift to the right (towards C) to reestablish equilibrium. (b) Doubling the concentrations of both products and quadrupling the container volume will decrease pressure, and the reaction will shift towards the side with more moles of gas, which is the right in this case. (c) Doubling the container's volume also reduces the pressure, again causing the reaction to shift to the right. Likewise, (d) the increase of products will shift the equilibrium to the left to minimize the effect. (e) Adding more A will cause the reaction to shift to the right, resulting in the formation of more C. (f) Doubling the concentrations of both the products and the container volume has two opposing effects. The added products would shift the reaction to the left, while the increased volume would shift it to the right. Thus, the final shift in equilibrium is dependent on which effect is greater.

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The equilibrium of the reaction A(g) + B(s) \<=> 2C(g) will shift to counteract changes such as concentration and volume adjustments, consistent with Le Chatelier's principle.

The response of the reaction A(g) + B(s) \<=> 2C(g) at equilibrium to various changes can be explained by Le Chatelier's principle, which states that a system at equilibrium will adjust to minimize any changes applied to it.

Answer: 60 ml of the 65% mixture will you need to add to obtain the desired solution.

Explanation:

According to the dilution law,

[tex]C_1V_1+C_2V_2=C_3V_3[/tex]

where,

= concentration of given alcohol solution = 25 %

[tex]V_1[/tex] = volume of given alcohol solution = 100 ml

[tex]C_2[/tex] = concentration of another alcohol solution=65 %

[tex]V_2[/tex] = volume of another acid solution= x ml

[tex]C_3[/tex]  = concentration of resulting alcohol solution = 40 %

[tex]V_3[/tex] = volume of resulting acid solution = (100+x)

[tex]25\times 100+65\times x=40\times (100+x)[/tex]

[tex]x=60ml[/tex]

Thus 60 ml of the 65% mixture will you need to add to obtain the desired solution.

Answer: [tex]2 \cdot CH_{4} O (l)+3 \cdot O_{2}(g) \rightharpoonup 2 \cdot CO_{2}(g) + 4 \cdot H_{2}O(l)[/tex]

Explanation:

Let consider that one mole of diatomic oxygen is used. So, the stoichometric can be modelled by using three variables:

[tex]x \cdot CH_{4} O (l)+O_{2}(g) \rightharpoonup y \cdot CO_{2}(g) + z \cdot H_{2}O(l)[/tex]

Where [tex]x,y,z[/tex] are the required variables.

Now, three equations are constructed from the number of elements involved (Carbon, Hydrogen and Oxygen):

Carbon

[tex]x=y[/tex]

Oxygen

[tex]x+2=2\cdot y + z[/tex]

Hydrogen

[tex]4\cdot x = 2 \cdot z[/tex]

The coefficients can be found by solving the abovementioned 3 x 3 Linear System:

[tex]x = \frac{2}{3}, y = \frac{2}{3}, z = \frac{4}{3}[/tex]

The whole-number coefficients are determined by multiplying every coefficient by 3, then:

[tex]2 \cdot CH_{4} O (l)+3 \cdot O_{2}(g) \rightharpoonup 2 \cdot CO_{2}(g) + 4 \cdot H_{2}O(l)[/tex]

The stoichiometric coefficient for oxygen in the balanced chemical equation for the combustion of methanol (CH4O) is 2, as it takes two molecules of O2 to provide enough oxygen atoms to form one molecule of CO2 and two molecules of H2O.

The stoichiometric coefficient of oxygen (O2) when the equation involving CH4O (methanol) combusting to form CO2 (carbon dioxide) and H2O (water) is balanced, can be determined through following a step-by-step approach:

The balanced equation is therefore CH4O(l) + 2 O2(g) -> CO2(g) + 2 H2O(l). The stoichiometric coefficient for oxygen is 2.

Answer:

Molarity of stock HCl solution is 0.89 M

Explanation:

The given problem can be solved using laws of dilution.

According to laws of dilution-    [tex]C_{1}V_{1}=C_{2}V_{2}[/tex]

Where, [tex]C_{1}[/tex] and [tex]C_{2}[/tex] are initial and final concentration of a solution

            [tex]V_{1}[/tex] and [tex]V_{2}[/tex] are initial and final volume of a solution

Here, [tex]V_{1}=35.0mL[/tex], [tex]C_{2}=0.062M[/tex] and [tex]V_{2}=500mL[/tex]

So, [tex]C_{1}=\frac{C_{2}V_{2}}{V_{1}}[/tex] = [tex]\frac{(0.062M)\times (500mL)}{35.0mL}[/tex] = 0.89 M

Hence, molarity of stock HCl solution is 0.89 M

Answer:

We need 19.3 mL of HCl

Explanation:

Step 1: Data given

Molarity HCl = 5.00 M

Mass Zn = 3.15 grams

Step 2: The balanced equation

Zn (s) + 2HCl (aq) ⟶ ZnCl2 (aq) + H2 (g)

Step 3: Calculate moles Zn

Moles Zn = mass Zn / molar mass Zn

Moles Zn = 3.15 grams  / 65.38 g/mol

Moles Zn = 0.0482 moles

Step 4: Calculate moles HCl needed

For 2 moles HCl we need 1 mol Zn to produce 1 mol ZnCl2 and 1 mol H2

For 0.0482 moles Zn we need 2*0.0482 = 0.0964 moles HCl needed

Step 5: Calculate volume needed

Molarity = moles / volumes

Volumes = moles / molarity

Volume HCl needed = 0.0964 moles HCl /5.00 M

Volume HCl needed = 0.01928 L = 19.3 mL

We need 19.3 mL of HCl

Answer:

The answer to your question is No, is not enough

Explanation:

I attached the problem because it says that my answer has bad words.

Answer:

Explanation:

Each orbital may contain a maximum of two electrons.

The energy level is the principal quantum number: 1, 2, 3, 4, 5, 6, 7.

The principal quantum number limits the kind of orbital (ℓ ) and the number of orbitals (mℓ ).

I. Principal quantum number, n = 1

  Angular or orbital quantum number : ℓ = 0 ⇒ ortibal s

  Magnetic quantum number:  mℓ = 0

                                 ⇒ number of orbitals 1 ⇒ number of electrons: 2

II. Principal quantun number, n = 2

  Angular or orbital quantum number : ℓ = 0 and 1 ⇒ ortibal s and p

  Magnetic quantum number:  

                                            for ℓ:  mℓ = 0: 1 s ortibal

                                            for ℓ = 1: mℓ = -1, 0, and +1:  3 p ortitals

                                 ⇒ number of orbitals 4 ⇒ number of electrons: 8

III. Principal quantun number, n = 3

    Angular or orbital quantum number : ℓ = 0, 1, and 2 ⇒ ortibal s, p and d

    Magnetic quantum number:  

                                            for ℓ:  mℓ = 0: 1 s ortibal

                                            for ℓ = 1: mℓ = -1, 0, and +1:  3 p ortitals

                                            for ℓ = 2: mℓ = -2, -1, 0, +1, and +2:  5 d ortitals

                                 ⇒ number of orbitals 9 ⇒ number of electrons: 18

IV. Principal quantun number, n = 4

    Angular or orbital quantum number : ℓ = 0, 1, 2, and 3 ⇒ ortibal s, p, d and f

    Magnetic quantum number:  

                                            for ℓ:  mℓ = 0: 1 s ortibal

                                            for ℓ = 1: mℓ = -1, 0, and +1:  3 p ortitals

                                            for ℓ = 2: mℓ = -2, -1, 0, +1, and +2:  5 d ortitals

                                            for ℓ = 3: mℓ = ±3, ±2, ±1, and 0: 7 f ortitals

                                 ⇒ number of orbitals 16 ⇒ number of electrons: 32.

Thus, you have:

energy level, n    maximum number of     maximum number of

                             orbitals                           electrons

      1                           1                                      2

      2                          4                                     8

      3                          9                                    18

      4                          16                                   32

Those numbers follow a rule: n² and 2n². You can verify that the previous numbers are in accordance with those formulas.