A balloonist begins a trip in a helium-filled balloon in early morning when the temperature is 15°C. By mid-afternoon, the temperature is 30.°C. Assuming the pressure remains at 1.00 atm, for each mole of helium, calculate the following:
(f)  Explain the relationship between the answers to parts (d) and (e).

The density of the substance having a mass of 2.0 g is 0.4 g/mL (Option D)

First, we shall obtain the volume of the substance. This can be obtained as follow:

Volume of substance = (Volume of water + substance) - (Volume of water)

Volume of substance = 75 - 70

Volume of substance = 5 mL

Finally, we shall determine the density of the substance. This is illustrated below:

Density = mass / volume

Density of substance = 2 / 5

Density of substance = 0.4 g/mL

Thus, the density is 0.4 g/mL (Option D)

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The volume of the KOH solution needed to make the buffer is also 780 mL. To prepare an acetate buffer of pH 5.54, we need to calculate the amount of acetic acid and potassium hydroxide required.

The Henderson-Hasselbalch equation can be used to determine the ratio of the concentration of the acid (HA) to its conjugate base (A⁻) in the buffer solution:

pH = pKa + log([A⁻]/[HA])

Given that the pKa of acetic acid is 4.76 and the desired pH is 5.54, we can rearrange the equation to solve for the ratio [A⁻]/[HA]:

5.54 = 4.76 + log([A⁻]/[HA])

0.78 = log([A⁻]/[HA])

By taking the antilog of both sides, we can find the ratio [A⁻]/[HA]:

[A⁻]/[HA] = 10^0.78

[A⁻]/[HA] = 6.02

Now, let's consider the initial concentration of acetic acid (HA) in the acetic acid solution. We have 0.659 M acetic acid, which means that the concentration of the acetate ion (A⁻) will be 0 M initially.

To achieve the desired ratio of [A⁻]/[HA] = 6.02, we need to add potassium hydroxide (KOH) to convert some of the acetic acid to acetate ion. The reaction between acetic acid and KOH is as follows:

CH3COOH + KOH ⇌ CH3COOK + H2O

The stoichiometry of the reaction is 1:1, meaning that equal volumes of acetic acid and KOH solutions are needed to achieve the desired ratio. Since we have 780 mL of the acetic acid solution, we will need an equal volume of the KOH solution.

In summary, to prepare an acetate buffer of pH 5.54, you will need to add 780 mL of the 2.42 M KOH solution to 780 mL of the 0.659 M acetic acid solution. This will result in a buffer solution with the desired pH, maintaining the appropriate ratio of acetate ion to acetic acid as required by the Henderson-Hasselbalch equation.

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The final temperature is 280.59 degrees Celsius.

The term atmospheres (atm), defined as follows: A mercury column one millimetre high would impose a pressure of one atm, or 1.01325 x 105 Pa. millimetres of mercury (mmHg). One atmosphere of pressure is generated by a mercury column that is 760 mm high. Because pressure and force are connected, you may use the physics equation to compute one if you know the other.

P1V1/T1 = P2V2/T2

Where P1, V1, and T1 are the initial pressure, volume, and temperature, and P2, V2, and T2 are the final pressure, volume, and temperature.

Given:

Initial pressure (P1) = 1.015 atm

Initial volume (V1) = 9.00 L

Initial temperature (T1) = -211.76 °C (62.39 K)

Mass of hydrogen gas initially (m1) = 2.035 g

Additional mass of hydrogen gas added (m2) = 2.099 g

Final pressure (P2) = 20.15 atm

PV = nRT

n1 = m1/MH2 = 2.035 g / 2.016 g/mol = 1.011 mol

n2 = m1/MH2 + m2/MH2 = 2.035 g / 2.016 g/mol + 2.099 g / 2.016 g/mol = 2.059 mol

We can assume that the volume of the container is constant, so we can set V1 = V2 in the combined gas law equation:

P1/T1 = P2/T2

Solving for T2:

T2 = P2T1V1 / (P1*V2)

Substituting the values:

V2 = V1 = 9.00 L

T2 = (20.15 atm)(62.39 K)(9.00 L) / (1.015 atm)(2.059 mol)(8.314 J/(mol*K))

T2 = 553.75 K or 280.59 °C

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Question:

A container holds 2.035 grams of hydrogen gas at a pressure of 1.015 atm and a temperature of -211.76 °C. What will be the temperature in °C if an additional 2.099 grams of hydrogen gas is added to the container and the pressure increases to 20.15 atm?

The boiling point of a solution is related to the molality of the solution (the number of moles of solute per kilogram of solvent) by the equation: The Correct option is \(Na_{3} PO_{4}\)

ΔTb = Kb x molality

where ΔTb is the change in boiling point, Kb is the boiling point elevation constant, and molality is the molality of the solution.

We can calculate the molality of compound X using the given information:

ΔTb = Tb - Tb°

where Tb is the boiling point of the solution and Tb° is the boiling point of the pure solvent, which is 100°C for water at standard pressure.

ΔTb = 101.4°C - 100°C = 1.4°C

molality = ΔTb / Kb

For water, Kb = 0.52°C/m, so:

molality = 1.4°C / 0.52°C/m = 2.7 m

Now we need to identify which of the given compounds could form a 1.35 m solution with water, resulting in a boiling point elevation of 1.4°C.

ΔTb = Kb x molality = 0.52°C/m x 2.7 m = 1.4°C

Therefore, the compound that could be X is \(Na_{3} PO_{4}\).

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Answer:

Leucippus and Democritus

Explanation:

Hope this helps~ :D

Thank you so much! :D

The correct statement about Dunite is that it is formed when magma cooled slowly below the ground; Option B.

Rocks are large aggregates of minerals that occur in the earth's crusts as a result of the cooling ad solidification of molten magma or the deposition of the remains of dead organic matter which changes from pressure ad heat.

There are three types of rocks, and they are:

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The half-life of phosphorus-32 is 14.3 days, which means that after each 14.3-day period, the amount of phosphorus-32 remaining will be reduced by half. We can use the following equation to calculate the amount of phosphorus-32 remaining after a certain number of half-lives:

N = N0 * (1/2)^(t/t1/2)

where:

N = the amount of radioactive material remaining

N0 = the initial amount of radioactive material

t = the time elapsed

t1/2 = the half-life of the radioactive material

In this case, we know that:

N0 = 1 kg (the initial amount)

t1/2 = 14.3 days (the half-life)

t = 100.1 days (time elapsed)

Plugging these values into the equation, we get:

N = 1 kg * (1/2)^(100.1/14.3)

Simplifying this expression yields:

N = 1 kg * 0.0684

Thus, the amount of phosphorus-32 remaining after 100.1 days is:

N = 0.0684 kg

Therefore, at the end of the 100.1-day period, approximately 68.4 grams of phosphorus-32 remains.

Answer:

a free-falling object is an object that is falling under the sole influence of gravity. That is to say that any object that is moving and being acted upon only be the force of gravity is said to be "in a state of free fall." Such an object will experience a downward acceleration of 9.8 m/s/s.

Answer:

a. Oxidising agent: Cl₂

b. Reducing agent: NaBr

c. Oxidised: NaBr

d. Reduced: Cl₂

e. Oxidation numbers before reaction: Cl= 0, Na= +1, Br= -1

f. Oxidation numbers after reaction: Cl= -1, Na= +1, Br= 0

Explanation:

Oxidising agents reduces themselves, oxidising other elements/compounds.

Reducing agents oxidise themselves, reducing other elements/compounds.

Oxidation is the loss of electrons or an increase in oxidation number.

Reduction is the gain of electrons or decrease in oxidation number.

The molar mass of dinitrogen triphosphide is  121 grams per mole.

The molar mass of dinitrogen triphosphide (N₂P₃) can be calculated by adding up the atomic masses of its constituent elements:

2 x atomic mass of nitrogen (N) + 3 x atomic mass of phosphorus (P)

The atomic mass of nitrogen is 14 g/mol and the atomic mass of phosphorus is 31 g/mol.

The molar mass of dinitrogen triphosphide is calculated as;

2(14 g/mol) + 3(31 g/mol)

= 121 g/mol

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The empirical formula of tartaric acid is C₄H₆O₆.

Tartaric acid is a compound that contains carbon (C), hydrogen (H), and oxygen (O). To determine its empirical formula, we need to analyze the given information about the production of carbon dioxide (CO₂) and water (H₂O). From the data, we can calculate the mole ratios of carbon, hydrogen, and oxygen in the compound.

Using the molar masses of carbon (12.01 g/mol), hydrogen (1.008 g/mol), and oxygen (16.00 g/mol), we can convert the given masses of CO₂ and H₂O to moles. From the balanced chemical equation of the reaction, we know that 1 mole of butyric acid produces 4 moles of CO₂ and 2 moles of H₂O.

By dividing the number of moles of carbon and hydrogen by the smallest number of moles obtained, we find that the empirical formula of tartaric acid is C₄H₆O₆. This means that tartaric acid consists of four carbon atoms, six hydrogen atoms, and six oxygen atoms in its simplest, empirical form.

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the element is in group 2

Answer:

harder

Explanation:

Answer:

The increased distance weakens the nuclear attraction to the outer-most electron, and is easier to remove

Explanation:

Answer:

The answer would be 15, letter D.

Explanation:

Since it has 5 electrons in its outermost shell, it will be located in group 15 in the periodic table.

The electron configuration of an atom of an element for the electrons in the s and p orbitals 15th group number of the element in the periodic table option D is correct.

Tha common electronic configuration of 15th group elements in ns2 np3, Group 15 elements in the modern periodic table are known as the pnictogens which means suffocation in Greek language.

This group consists of nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb) and bismuth (Bi).

The group 15 elements consist of five valence electrons. Due to this the elements can either lose five electrons or gain three electrons in order to attain the stable configuration. The general electronic configuration of nitrogen family is ns 2 np 3.

therefore , because of ns2 np3 option D is correct.

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We know that 1 mole of sodium contains Avogadro number of atoms. - But we have to find the number of atoms in 0.2 mole of sodium. - Therefore 0.2 moles of sodium (Na) contains 12.046×1023atoms in it.

the strength of the solution formed is 3.8.

The amount of solute dissolved in grammes per litre of the solution is used to determine the solution's strength. It stands for the solution's potency or concentration. It uses grammes per litre of expression.

Concentration of NaOH = 19 M

Initial volume of NaOH = 25 mL= 0.025 L

Moles of NaOH = 19*0.025

                        = 0.475 moles

Final volume of NaOH solution = 5 L

Final concentration = 0.475/5

                              = 0.095 M

so, the strength is 19/5

                            = 3.8

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For the given data , (a) the process cannot reach State 3. Thus, it is impossible to determine the pressure and volume at this state ; (b) Work done is : W1-2 = 0.1005 kJ ; W2-3 = 0.1435 kJ ; W3-1 = 2.019 kJ.

(a) To determine the pressure and volume at State 3 :

For, process 1-2: P1V1 = P2V2

At state 2 :

V2 = 0.004m

P1V1/P2 = V2 = 0.004m

=> P2 = P1V1/V2

=> P2 = 100 kPa * 0.002 m / 0.004 m = 50 kPa

For process 3-1: It is Polytropic compression process

PVn = C where n = 1.4

State 3 = Initial state and State 1 = final state

Let the pressure and volume at state 3 be P3 and V3.

P3V3n = P1V1n

=> P3 = P1(V1/V3)^n= 100 kPa(0.002 m / V3)^1.4

Now, to find V3: V1 = V2 + V3

∴ V3 = V1 − V2

=> V3 = 0.002 m − 0.004 m = -0.002 m (unacceptable)

Volume at state 3 is negative. It is an unacceptable value.

Therefore, the process cannot reach State 3.

Therefore, it is not possible to determine the pressyre and volume at this state.

(b) To determine : work for each process :

We know, Process 1-2: constant pressure process

Q1-2 = ΔH1-2 = H2 − H1

=>  Q1-2 = CpΔT = Cp(T2 − T1)

where Cp = specific heat at constant pressure = 1.005 kJ/kgK

ΔT = T2 − T1

For a constant pressure process, work done, W1-2 = Q1-2 = Cp(T2 − T1)

Now, P1V1 = P2V2 = mRT1

P2/P1 = T2/T1 = V1/V2 = 0.5

At state 2, P2 = 50 kPa

Process 2-3: constant volume process

Work done during a constant volume process is :

W2-3 = Q2-3 = ΔU2-3 =

=> W2-3 = U3 − U2= CvΔT = Cv(T3 − T2)

where Cv = specific heat at constant volume = 0.718 kJ/kgK

ΔT = T3 − T2

Now, Process 3-1: Polytropic compression process

Work done during a polytropic compression process is :

W3-1 = Q3-1 = ΔH3-1

=> W3-1 = H1 − H3 = Cp

ΔT + V(P1 − P3)= Cp(T1 − T3) + V(P1 − P3)

where ΔT = T1 − T3 ;  V = (V1V3)^0.5

State 3 = Initial state and State 1 = final state

As, PVn = CP1V1n = C,

P3V3n = CP3V3n = C

∴ P1V1n = P3V3n

V1/V3 = (P3/P1)^(1/n) = (50 kPa / 100 kPa)^(1/1.4) = 0.883

At state 3, P3V3 = P1V1(V1/V3)^n= 100 kPa * 0.002 m * (0.002 m / 0.004 m)^1.4= 3.925 kJ/kgK

=> W3-1 = Cp(T1 − T3) + V(P1 − P3)

= 1.005 kJ/kgK(T1 − T3) + 0.002 m(100 kPa − 3.925 kPa)

= 2.019 kJ

The pressure and volume at this state is not possible to determine.

Therefore, the work done by all the processes is :

W1-2 = 0.1005 kJ

W2-3 = 0.1435 kJ

W3-1 = 2.019 kJ.

Thus, the required answers are given above.

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Answer:

Sandstone, limestone, and shale

Explanation:

The rocks are most likely to start as sediments carried in rivers and end up in lakes and oceans

Answer:

D - materials dissolved in solutions

Explanation:

EDGE 2021

Carbon dioxide molecules can absorb and emit infrared radiation, and they are one of the most abundant constituents of Earth's atmosphere.

Thus, the correct options are:d) Are one of the most abundant constituents of Earth's atmospheree) Can move in many ways, thus absorbing and emitting infrared radiation

Carbon dioxide is a trace gas present in the Earth's atmosphere. It's a vital component of Earth's carbon cycle, which helps to regulate Earth's temperature and support life as we know it. Carbon dioxide molecules are one of the most common gases in the atmosphere, accounting for around 0.04% of the Earth's atmosphere.

The greenhouse effect is caused by carbon dioxide, methane, and other greenhouse gases. When the Sun's energy reaches the Earth's surface, it is absorbed and then radiated back into space as infrared radiation. Greenhouse gases absorb this radiation and trap it in the atmosphere, which causes the Earth's temperature to rise and the climate to change.

Carbon dioxide molecules are capable of absorbing and emitting infrared radiation due to their molecular structure, which consists of one carbon atom and two oxygen atoms. This property of carbon dioxide is the main reason it's classified as a greenhouse gas.

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The correct option is:D. CH2BrCH2Br.  The product that results from the reaction of CH2=CH2 (ethylene) with Br2 (bromine) is CH2BrCH2Br (1,2-dibromoethane). The reaction of CH2==CH2 with Br2 is a halogenation reaction, which involves the addition of a halogen to an unsaturated organic compound.

The product that results from this reaction is CH2BrCH2Br, which is option D. This product is formed by the addition of one Br atom to each of the carbon atoms in the double bond of ethene. The resulting molecule is a dibromoalkane, which is a type of organic compound that contains two bromine atoms attached to adjacent carbon atoms. This reaction is an example of an addition reaction, where the unsaturated organic compound undergoes a reaction with a halogen to form a saturated organic compound. In summary, the correct answer is D, CH2BrCH2Br. This reaction is an example of an addition reaction, in which the bromine atoms are added to the carbon atoms in the double bond, resulting in a single bond between the carbon atoms and a bromine atom attached to each carbon.

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Answer:

I believe the Answer is B, as time is not dependent on growth, but growth is dependent on time.

Explanation:

Answer:

B.) An atom of arsenic has one more valence electron and more electron shells than an atom of silicon, so the conductivity decreases because the arsenic atom loses the electron.

Explanation:

Silicon is located in the 3rd row and 14th column in the periodic table. Arsenic is located in the 4th row and 15th column in the periodic table. This means that arsenic has one more valence electron than silicon. Since arsenic is located one row down from silicon, its valence electrons occupy higher energy orbitals.

Silicon maintains a crystal-like lattice structure. Each silicon atom is covalently connected to assume this shape. When silicon gains one extra electron from arsenic, it experiences n-type doping. This new electron is not tightly bound in the lattice structure. This allows it to move more freely and conduct more electricity. This can also be explained using band gaps. Silicon, which previously had an empty conduction band, now has one electron in this band. This lowers the band gap between the conduction and valence bands and increases conductivity.

Answer:

its d

Explanation:

One of the problems that occurs as a consequence of chlorofluorocarbon (CFC) pollution is ozone depletion. CFCs are synthetic compounds that were commonly used in refrigerants, aerosol propellants, and foam-blowing agents.

When released into the atmosphere, CFCs rise to the stratosphere, where they are broken down by ultraviolet (UV) radiation. This process releases chlorine atoms that catalytically destroy ozone molecules, leading to a reduction in the ozone layer.

Ozone depletion has significant environmental consequences, including increased exposure to harmful UV radiation, which can have detrimental effects on human health, ecosystems, and agricultural productivity.

Chlorofluorocarbons (CFCs) are chemical compounds that were widely used in various industries due to their stability, non-toxicity, and non-reactivity.

However, when CFCs are released into the atmosphere, they eventually reach the stratosphere, where they undergo photodissociation by high-energy UV radiation. This photodissociation process breaks down CFC molecules and releases chlorine atoms.

The released chlorine atoms are highly reactive and act as catalysts in the destruction of ozone molecules. Each chlorine atom can participate in a series of reactions that lead to the destruction of thousands of ozone molecules before it is eventually deactivated. This process is known as ozone depletion.

Ozone depletion is a critical environmental issue because the ozone layer in the stratosphere plays a vital role in protecting life on Earth from harmful UV radiation.

UV radiation can cause various health problems in humans, including skin cancer, cataracts, and weakened immune systems. It can also have adverse effects on marine ecosystems, agricultural productivity, and the overall balance of ecosystems.

To address the problem of ozone depletion, the international community came together and took action through the Montreal Protocol in 1987. This agreement aimed to phase out the production and use of CFCs and other ozone-depleting substances.

As a result, the production and consumption of CFCs have significantly decreased, leading to a gradual recovery of the ozone layer. However, it will take several more decades for the ozone layer to fully heal.

In conclusion, the release of chlorofluorocarbons (CFCs) into the atmosphere causes ozone depletion, leading to a reduction in the protective ozone layer in the stratosphere.

This depletion increases the levels of harmful UV radiation reaching the Earth's surface, posing risks to human health, ecosystems, and agriculture.

International efforts to reduce CFC production and consumption have been successful in mitigating ozone depletion, but continued vigilance and adherence to protocols are necessary to ensure the full recovery of the ozone layer.

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Understanding the properties of acids and bases is critical for everyday life and several career fields.

Topic: Acids and Bases

Acids and bases are two major branches of chemistry that deal with the behavior of acids and bases in solutions and the properties of their aqueous solutions. Acids and bases can be found all around us, from the foods we eat to the products we use in our daily lives. The pH scale is used to measure the acidity or basicity of solutions.

Solutions with pH less than 7 are acidic, while solutions with pH greater than 7 are basic. A solution with a pH of 7 is neutral.

A few examples of acids and bases in everyday life are:

Acids

Vinegar: Acetic acid is a weak acid found in vinegar.

Citrus fruits: Citric acid is a weak acid found in citrus fruits.

Carbonated drinks: Carbonic acid is a weak acid found in carbonated drinks.

Stomach acid: Hydrochloric acid is a strong acid found in the stomach.

Base

Soap: Sodium hydroxide is a strong base found in soaps.

Ammonia: Ammonia is a weak base found in cleaning products.

Antacids: Antacids are basic compounds used to neutralize stomach acid.

So, in everyday life, people can observe and apply the properties of acids and bases while cooking, cleaning, and consuming food and drinks. The use of acids and bases is also critical in several professions, including healthcare, agriculture, and manufacturing.

In healthcare, pH regulation is critical for maintaining homeostasis in the body. In agriculture, pH regulation is critical for soil fertility and plant growth. In manufacturing, acids and bases are used in the production of various products.

Therefore, understanding the properties of acids and bases is critical for everyday life and several career fields.

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