Which contain the greatest mass of oxygen 0. 75 mol of ethanol (C2H5OH),0. 60 mol of Formic acid (HCO2H)or 1. 0 mol of water (H2O) 2 explain why

Answer:

1.52atm is the pressure of the gas

Explanation:

To solve this question we must use the general gas law:

PV = nRT

Where P is pressure in atm = Our incognite

V is volume = 50.5L

n are moles of gas = 3.25moles

R is gas constat = 0.082atmL/molK

And T is absolute temperature = 288.6K

To solve pressure:

P = nRT / V

P = 3.25mol*0.082atmL/molK*288.6K / 50.5L

P = 1.52atm is the pressure of the gas

Option d. Essential amino acids cannot be formed from other amino acids but must be supplied in the diet. amino acids are used by cells to synthesize proteins.

Amino acids are made up of carbon, hydrogen, oxygen, nitrogen, and sometimes sulfur atoms. These molecules are first broken down by hydrolysis into simpler compounds and then used by cells to build new proteins or to produce energy.Amino acids are divided into two groups: essential and non-essential amino acids. Non-essential amino acids can be produced by the body from other amino acids or by using other metabolic processes. Essential amino acids, on the other hand, cannot be formed from other amino acids and must be supplied in the diet.

These amino acids include leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine.Excess dietary amino acids are either oxidized for energy or stored as fat or glycogen in the liver. Amino acids that cannot be oxidized are used to build new proteins, such as muscle tissue. In conclusion, essential amino acids cannot be formed from other amino acids but must be supplied in the diet. Excess dietary amino acids are either oxidized for energy or used to build new proteins or stored as fat or glycogen.

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Answer:

option C) Two liquids mixing to produce a precipitate.

this one is correct, because precipitate is only formed in a chemical reaction.

The temperature of the system in equilibrium is 25.51⁰ C.

The equilibrium temperature of the system depends on the heat released from both gold and water. The total heat received by the system will equal to total heat released by objects. It should follow

Q released = Q received

The heat can be defined by

Q = m . c . ΔT

where Q is heat, m is mass, c is the specific heat constant and ΔT is the change in temperature.

The given parameters are

m = 200 g = 0.2 kg

T1 = 24.5⁰ C

c = 4200 J/kg⁰ C

Q received = 850 J

Lets assume that the value of equilibrium temperature is T. Hence,

Q released = Q received

m . c . ΔT = 850

0.2 . 4200 . ΔT = 850

ΔT = 1.01

T - T1 = 1.01

T - 24.5 = 1.01

T =  25.51⁰ C

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Answer:

the argon because 40/4 = 10

The amount of heat absorbed by the water is 1606.6 J (383.99 Calories )

Initial temperature (T₁) = 21.5 °C

Final temperature (T₂) = 28.1 °C

ΔT = T₂ – T₁

ΔT = 28.1 – 21.5

Mass of water (M) = 58.18 g

Change in temperature (ΔT) = 6.6 °C

Specific heat capacity of water (C) = 4.184 J/gºC

Q = MCΔT

Q = 58.18 × 4.184 × 6.6

Divide by 4.184 to express in Calories

Q = 1606.6 / 4.184

Therefore, the amount of heat absorbed by the water is 1606.6 J (383.99 Calories )

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Answer:

120000drops

Explanation:

Average blood in human blood = 6L - 6*1000 = 6000ml

1 ml of blood is equal to 20 drops

6000ml of blood makes 20*6000 = 120000 drops

It is used for cleaning and other use. Higher pH means stronger base.

The percent yield of H₂O, if 87.0 g of H₂O is produced by combining 95.0 g of O₂ and 11.0 g of H₂ is 87.87%.

Mass of any substance will be calculated by using their moles as:

n = W/M, where

W = given or required mass

M = molar mass

Moles of 95g of Oxygen (O₂) = 95g / 32g/mol = 2.96 moles

Moles of 11g of hydrogen (H₂) = 11g / 2g/mol = 5.5 moles

Given chemical reaction is:
2H₂ + O₂ → 2H₂O

From the stoichiometry of the reaction, it is clear that:

1 moles of O₂ = reacts with 2 moles of H₂

2.96 moles of O₂ = reacts with 2×2.96=5.92 moles of H₂

Here hydrogen is the limiting reagent as it has lower moles and formation of water depends on this only.

2 moles of H₂ = produces 2 moles of water

5.5 moles of H₂ = produces 5.5 moles of water

Mass of 5.5 moles of water will be calculated as:

W = (5.5mol)(18g/mol) = 99g

Given theoretical yield of water = 87g

% yield of water will be calculated as:
% yield = (87 / 99)×100 = 87.87%

Hence required value is 87.87%.

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Answer:

D

Explanation:

For the pH after the addition of each volume of acid, we need to consider the reaction between methylamine (CH₃NH₂) and HNO₃. Methylamine is a weak base, and HNO3 is a strong acid. The reaction can be written as:

CH₃NH₂ + HNO₃ -> CH₃NH₃+ + NO₃-

First, let's calculate the initial moles of methylamine in the 100.0 ml sample:

moles CH₃NH₂ = volume (L) * concentration (mol/L)

moles CH₃NH₂ = 0.100 L * 0.100 mol/L

moles CH₃NH₂ = 0.010 mol

Since CH₃NH₂ is a weak base, it will react with HNO₃ in a 1:1 ratio. Therefore, the number of moles of CH₃NH₂ reacting will be equal to the number of moles of HNO₃ added.

Now let's calculate the moles of HNO₃ added for each case:

a) 0.0 ml (no HNO₃ added): 0.010 mol

b) 20.0 ml: moles HNO₃ = 0.020 L * 0.250 mol/L = 0.005 mol

c) 40.0 ml: moles HNO₃ = 0.040 L * 0.250 mol/L = 0.010 mol

d) 60.0 ml: moles HNO₃ = 0.060 L * 0.250 mol/L = 0.015 mol

Now we need to calculate the moles of CH₃NH₂ and CH₃NH₃+ remaining after the reaction.

For case a) 0.0 ml:

moles CH₃NH₂ remaining = 0.010 mol - 0.000 mol = 0.010 mol

moles CH₃NH₃+ formed = 0.000 mol

For case b) 20.0 ml:

moles CH₃NH₂ remaining = 0.010 mol - 0.005 mol = 0.005 mol

moles CH₃NH₃+ formed = 0.005 mol

For case c) 40.0 ml:

moles CH₃NH₂ remaining = 0.010 mol - 0.010 mol = 0.000 mol

moles CH₃NH₃+ formed = 0.010 mol

For case d) 60.0 ml:

moles CH₃NH₂ remaining = 0.010 mol - 0.015 mol = -0.005 mol (Excess acid)

moles CH₃NH₃₊ formed = 0.015 mol

Since methylamine is a weak base, we need to consider the Kb value to calculate the concentration of hydroxide ions (OH-) and then convert it to pH.

The Kb expression for methylamine is:

Kb = [CH₃NH₃+][OH-] / [CH₃NH₂]

We can assume that [OH-] ≈ [CH₃NH₃+], so the equation becomes:

Kb = [OH-]^2 / [CH₃NH₂]

Rearranging the equation:

[OH-] = sqrt(Kb * [CH₃NH₂])

Now, let's calculate the OH- concentration and convert it to pH for each case:

a) 0.0 ml:

[OH-] = sqrt(3.7x10^-4 * 0.010 mol) ≈ 0.00608 M

pOH = -log10(0.00608) ≈ 2.22

pH = 14

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Answer:

\(\large \boxed{\text{21.6 L}}\)

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:        180.16

         C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g:      24.5

(a) Moles of C₆H₁₂O₆

\(\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}\)

(b) Moles of CO₂

\(\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}\)

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37  °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

\(\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}\)

Answer: 2) Chloroform & Caustic potash

Explanation:

The carbylamine reaction is a kind of chemical test which is done to detect primary amines in an unknown solution. It cannot detect secondary and tertiary amines.

The reaction involves the heating with up of the unknown solution with alcoholic potassium hydroxide or caustic potash and the chloroform.

In the presence of primary amine, the production of isocyanide results.

The following solutions are buffers:

A solution made by mixing 100 ml of 0.100 M HCOOH and 50 ml of 0.100 M KCl, and a solution made by mixing 100 ml of 0.100 M HCOOH and 500 ml of 0.100 M NaOH.

Let's learn more about buffer solutions:

1. Buffer solutions can withstand or resist pH changes, even when adding a strong acid or base. A buffer solution comprises a weak acid and its conjugate base or a weak base and its conjugate acid. The pH change is insignificant when an acid or a base is added to a buffer solution.

2. The chemical equation for the ionization of HCOOH in water is as follows:

HCOOH + H2O ⇌ H3O+ + HCOO-

The salt made from the acid and the base in the buffer solution must be able to act as a source of ions to neutralize either the excess H3O+ ions or OH- ions introduced into the solution. For example, KCl and NaOH are used in the buffer solutions listed above, as they produce K+ and Na+ ions, which can interact with HCOO- ions in the first buffer and HCOO- ions and HCOOH molecules in the second buffer.

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When the electron in a hydrogen atom transitions from n, light with a wavelength of 486 nm is released.

Radio waves, light waves, and infrared (thermal) waves are examples of electromagnetic radiation that produce distinctive patterns as they move across space. Each wave is unique in its length and shape. The wavelength is the separation between peaks (high points). Therefore, when the electron in a hydrogen atom transitions from n, light with a wavelength of 486 nm is released. Radio waves, light waves, and infrared (thermal) waves are examples of electromagnetic radiation that produce distinctive patterns as they move across space. Each wave is unique in its length and shape.

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Answer:

A

Explanation:

The forward rate of a chemical reaction refers to the speed at which reactants are converted into products. The rate can be affected by various factors including temperature and concentration. The correct answer to the question is (1)

The concentration of the reactants decreases, and the temperature decreases. When the concentration of the reactants decreases, there are fewer reactant particles to react with each other, which leads to a decrease in the forward rate of the reaction. Similarly, when the temperature decreases, the  of the reactant particles decreases, which leads to a decrease in the number of successful collisions and a decrease in the forward rate of the reaction.

Overall, it is important to note that the forward rate of a chemical reaction can be affected by a variety of factors and conditions, and it is important to carefully consider each one in order to understand how they impact the reaction.

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Answer:

molecules heat and move faster, they are moving apart. So air, like most other substances, expands when heated and contracts when cooled.

Explanation:

Because there is more space between the molecules, the air is less dense than the surrounding matter and the hot air floats upward.

Answer:

the particles must 1. collide

2. with sufficient energy

3. in proper orientation.

Explanation:

hope this helps

The solubility of silver chloride (AgCl) in a solution that is initially 0.120 M in NH3 (ammonia) is approximately X mol/L.

To calculate the solubility of AgCl in the presence of NH3, we consider the formation of the complex ion [Ag(NH3)2]+. Using the equilibrium expression for the complex ion formation and the solubility product constant (Ksp) for AgCl, we determine that the concentration of Ag+ is equal to the solubility of AgCl. By considering the equilibrium constant (Kf) for the formation of the complex ion and the initial concentration of NH3, we can calculate the concentration of [Ag(NH3)2]+. Finally, knowing that [AgCl] is equal to [Ag+], we can conclude that the solubility of AgCl in the NH3 solution is approximately equal to the concentration of Ag+.

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Answer:

attracted towards each other

Explanation:

hope u like the ans

Answer:

The ions in metal behavior by repelled by electro

Answer:

Its C!

Explanation:

NH3 and OH- are both a Brønsted-Lowry base, while H2O and NH4 are acids!

Hope this helped!

The following is an example of a Bronsted-Lowry base will be

A material that receives protons is known as just a Bronsted-Lowry base.

Ammonia was pleased to receive a proton mostly from water's hydrogen (H2O) to form NH4. Therefore, NH3 will be considered as Bronsted-Lowry base.

Therefore, the correct answer will be NH3.

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Gene Vincent and Eddie Cochran were particularly popular with teenagers and young adults during the 1950s.

Gene Vincent and Eddie Cochran were American rock and roll musicians who gained popularity in the 1950s. They were part of the rockabilly movement, which combined elements of country music with rhythm and blues.

Gene Vincent was known for his hit song 'Be-Bop-A-Lula,' which became a rock and roll classic. Eddie Cochran was known for his energetic performances and songs like 'Summertime Blues' and 'C'mon Everybody.'

Both artists had a significant impact on the development of rock and roll music and were particularly popular with teenagers and young adults during their time.

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Gene Vincent and Eddie Cochran were particularly popular with the youth and rock and roll enthusiasts of the late 1950s and early 1960s.

Gene Vincent and Eddie Cochran were particularly popular with the youth and rock and roll music enthusiasts of the late 1950s and early 1960s. Their energetic performances and rebellious image resonated with the emerging teenage audience at the time.

They were influential figures in the rockabilly and rock and roll genres, known for their hits such as "Be-Bop-A-Lula" by Gene Vincent and "Summertime Blues" by Eddie Cochran. Their music and style captured the spirit of youthful rebellion and played a significant role in shaping the early rock and roll era.

Gene Vincent and Eddie Cochran were influential figures in the rock and roll music scene of the late 1950s and early 1960s. They were particularly popular with the teenage audience of the time, as their music and persona embodied the rebellious and energetic spirit of the youth culture.

Gene Vincent, born Vincent Eugene Craddock, rose to fame with his hit song "Be-Bop-A-Lula" in 1956. Known for his distinctive vocal style and wild stage presence, Vincent became a rockabilly icon. His music blended elements of rock and roll, rhythm and blues, and country, creating a unique sound that resonated with young listeners. Songs like "Bluejean Bop" and "Race with the Devil" further solidified his popularity.

Eddie Cochran, on the other hand, was a multi-talented musician, singer, and songwriter. He gained fame with his upbeat and catchy songs, such as "Summertime Blues" and "C'mon Everybody." Cochran's music was characterized by his skillful guitar playing, heartfelt lyrics, and a distinctive rock and roll sound. His contributions to the genre and his early death at the age of 21 in a tragic car accident solidified his status as a rock and roll legend.

Both Gene Vincent and Eddie Cochran were known for their electrifying live performances and their impact on the rock and roll genre. Their music resonated with young audiences who were seeking an outlet for their rebellious spirit and love for energetic, guitar-driven music. Their influence can still be felt in the development of rock music and the inspiration they provided to subsequent generations of musicians.

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Answer:

c2h6

Explanation:

molecular weight=vapour density*2=2*15=30

molecular formula=empirical formula*n

n=molecular weight/empirical formula weight=30/15=2

so molecular formula2 * (ch3) alkane =c2h6

The molecular formula of an alkane whose vapor density is 15 will be C₂H₆.

The alkanes are the carbon and hydrogen-containing compounds said to be hydrocarbons in which carbon is bounded with hydrogen with a single covalent bond by sharing electrons.

The molecular weight of alkane is =  vapor density × 2

The molecular weight of alkane is = 15  × 2 = 30

n = molar weight/ empirical weight

n =  30 / 15 = 2

molecular formula = ( alkane)ₙ

molecular formula = (CH₃)₂

molecular formula = C₂H₆

Therefore, C₂H₆ is the molecular formula of an alkane whose vapor density is 15.

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Determine The probality of offspring traits

Answer:

550 K

Explanation:

Given the following data;

Original Volume, V1 = 20mL

Original Temperature, T1 = 200K

New Temperature, V2 = 55mL

To find new temperature T2, we would use Charles' law.

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles is given by;

\( VT = K\)

\( \frac{V1}{T1} = \frac{V2}{T2}\)

Making T2 the subject of formula, we have;

\(T_{2} = \frac{V2}{V1} * T_{1}\\\)

Substituting into the equation;

\(T_{2} = \frac{55}{20} * 200\\T_{2} = \frac{11000}{20}\\T_{2} = 550K\)

Therefore, the new temperature is 550K.

This balanced redox reaction involves the transfer of electrons from Fe (s) to MnO4- (aq), resulting in the formation of Mn2+ (aq) and Fe3+ (aq) ions

The half-reaction occurring at the anode in the balanced reaction shown is the oxidation half-reaction, where electrons are lost. In this case, the anode is where Fe (s) is located. Therefore, the half-reaction at the anode is the oxidation of Fe (s) to Fe3+ (aq), which is represented as follows:

Fe (s) → Fe3+ (aq) + 3e-

This half-reaction involves the loss of electrons by Fe (s), resulting in the formation of Fe3+ (aq) ions. The electrons released in this half-reaction are then transferred through the external circuit to the cathode, where reduction occurs. Overall, this balanced redox reaction involves the transfer of electrons from Fe (s) to MnO4- (aq), resulting in the formation of Mn2+ (aq) and Fe3+ (aq) ions.

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Hess's Law states that the enthalpy change of a reaction is independent of the pathway between the initial and final states. In other words, the ΔH of a reaction can be calculated by adding or subtracting the enthalpy changes of other reactions that add up to the overall reaction of interest.

For example, if we have a reaction A → B with an experimental ΔH value of -100 kJ/mol, and a reaction B → C with an experimental ΔH value of +50 kJ/mol, then we can use Hess's Law to calculate the ΔH of the reaction A → C:

A → B (ΔH = -100 kJ/mol)

B → C (ΔH = +50 kJ/mol)

A → C (ΔH = -50 kJ/mol)

In this case, the ΔH of the overall reaction A → C is calculated by subtracting the ΔH of the reaction B → C from the ΔH of the reaction A → B.

To apply Hess's Law to the pre-lab exercise and calculate the ΔH for reaction 3, you would need to know the experimental ΔH values for other reactions that could be combined to give reaction 3. Once you have these values, you can add or subtract them to obtain the ΔH for reaction 3.

After obtaining the calculated value of ΔH using Hess's Law, you can compare it with the experimental value to see how well they agree. If the calculated value is within a reasonable range of the experimental value, it suggests that Hess's Law was a good approximation for the reaction. However, if the calculated and experimental values differ significantly, there may be some sources of error in the experiment, or there may be additional factors affecting the enthalpy change of the reaction that were not considered in the Hess's Law calculation.

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Answer

The formula form of Manganese (IV) perbromate is

Explanation

The formula of Manganese is Mn

The formula for perbromate is BrO₄⁻

Oxidation number of Manganese (IV) = +4, That is Manganese (IV) is Mn⁺⁴

Therefore, multiply the charge of manganese by 1 and perchlorate by 4 t