Calculate the amount of energy (in kJ) required to increase the temperature of 255 g of water from 25.2 C to 90.5 C. Specific heat of water is 4.184J/g C.

Answers

Answer 1

Answer:

70.91 kJ

Explanation:

The amount of energy (in kJ) required to increase the temperature of 255 g of water from 25.2 C to 90.5 C can be calculated using the formula:

Q = m * c * ΔT

Where Q is the amount of energy, m is the mass of the substance, c is the specific heat, and ΔT is the change in temperature.

Substituting the given values:

m = 255 g

c = 4.184 J/g C

ΔT = (90.5 - 25.2) C = 65.3 C

Q = 255 g * 4.184 J/g C * 65.3 C

Q = 70905.564 J

Q = 70.91 kJ (rounded to two decimal places)

Therefore, the amount of energy required to increase the temperature of 255 g of water from 25.2 C to 90.5 C is 70.91 kJ.


Related Questions

If the average speed of an oxygen molecule is 4.37 ✕ 104 cm/s at 25°C, what is the average speed of a CO2 molecule at the same temperature?

Answers

The average speed of a gas molecule is proportional to the square root of its temperature and inversely proportional to the square root of its molar mass. Therefore, we can use the following equation to find the average speed of a CO2 molecule at the same temperature:

v2/v1 = sqrt(M1/M2)

where v1 and v2 are the average speeds of the oxygen and CO2 molecules, respectively, M1 and M2 are the molar masses of oxygen and CO2, respectively.

The molar mass of oxygen (O2) is 32 g/mol, and the molar mass of CO2 is 44 g/mol.

We are given that the average speed of an oxygen molecule is 4.37 × 10^4 cm/s at 25°C. We can convert the temperature to Kelvin by adding 273.15 to get:

T = 25°C + 273.15 = 298.15 K

Now we can solve for v2:

v2 = v1 * sqrt(M1/M2)

v2 = 4.37 × 10^4 cm/s * sqrt(32 g/mol / 44 g/mol)

v2 = 3.67 × 10^4 cm/s

Therefore, the average speed of a CO2 molecule at the same temperature is 3.67 × 10^4 cm/s.

According to the Law of Conservation of Mass, Matter cannot be created or destroyed.
Given that, if 15 grams of reactant went into the reaction, then how many grams of products are formed?
NH,NO,
N₂ +
H₂O

Answers

Answer:

Explanation:

According to the Law of Conservation of Mass, matter cannot be created or destroyed in a chemical reaction. Therefore, the mass of the reactants must be equal to the mass of the products.

If 15 grams of reactant went into the reaction, then the mass of the products formed must also be 15 grams. This assumes that the reaction is complete and no reactants are left unreacted.

It is important to note that this applies to closed systems where there is no loss or gain of mass. In real-world situations, some mass may be lost due to factors such as evaporation or incomplete reactions, which can affect the accuracy of the calculations.

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A student has a 2.97 L
bottle that contains a mixture of O2
, N2
, and CO2
with a total pressure of 5.68 bar
at 298 K
. She knows that the mixture contains 0.225 mol N2
and that the partial pressure of CO2
is 0.309 bar
. Calculate the partial pressure of O2
.

Answers

To calculate the partial pressure of O2, we can use the ideal gas law equation:

PV = nRT

where:
P = pressure
V = volume = 2.97 L
n = number of moles
R = gas constant = 0.08314 L bar K^-1 mol^-1
T = temperature = 298 K

We can start by calculating the total number of moles of gas in the bottle:

n_total = PV/RT

n_total = (5.68 bar)(2.97 L)/(0.08314 L bar K^-1 mol^-1)(298 K)

n_total = 0.725 mol

We know that the mixture contains 0.225 mol N2, so we can calculate the number of moles of the other gases:

n_other = n_total - n_N2

n_other = 0.725 mol - 0.225 mol

n_other = 0.500 mol

We also know that the partial pressure of CO2 is 0.309 bar, so we can calculate the number of moles of CO2:

n_CO2 = P_CO2 V/RT

n_CO2 = (0.309 bar)(2.97 L)/(0.08314 L bar K^-1 mol^-1)(298 K)

n_CO2 = 0.0112 mol

Now we can use the mole fractions of O2 and N2 to calculate the partial pressure of O2:

X_O2 = n_O2/n_other

X_N2 = n_N2/n_other

We know that the mole fraction of N2 is 0.225/0.500 = 0.450, so:

X_N2 = 0.450

Therefore:

X_O2 = 1 - X_N2

X_O2 = 1 - 0.450

X_O2 = 0.550

Now we can use the ideal gas law to calculate the partial pressure of O2:

P_O2 = n_O2 RT/V

P_O2 = X_O2 n_other RT/V

P_O2 = (0.550)(0.500 mol)(0.08314 L bar K^-1 mol^-1)(298 K)/(2.97 L)

P_O2 = 0.876 bar

Therefore, the partial pressure of O2 in the mixture is 0.876 bar.

PLEASE ACTUALLY ANSWER THE WHOLE ASSIGNMENT FOR BRAINLIEST

Answers

The results of the lab activity showed that the larger the mass of the sun, the more likely at least one planet will fall into the habitable zone.

What effect does the mass of the Sun have on the orbits of Planets?

The mass of the sun affects the orbits of planets in a solar system. When the mass of the sun is larger, the gravitational force between the sun and the planets is stronger, causing the planets to move at a slower pace around the sun.

Conversely, when the mass of the sun is smaller, the gravitational force is weaker, causing the planets to move at a faster pace.

Additionally, when Earth is closer to the sun, the gravitational force is stronger, causing its orbit to become faster, while a farther distance from the sun results in a slower orbit.

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Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6 , and an unknown amount of propane, C3H8 ) were added to the same 10.0- L container. At 23.0 ∘C, the total pressure in the container is 3.70 atm. Calculate the partial pressure of each gas in the container.

Answers

The partial pressure of each gas are:

Partial pressure of CH₄ is 1.22 atmPartial pressure of C₂H₆ is 1.46 atmPartial pressure of C₃H₈ is 1.02 atm

How do i determine the partial pressure of each gas?

First, we shall determine the mole of 8.00 g of methane, CH₄ and 18.0 g of ethane, C₂H₆. Details below:

For methane, CH₄

Mass of CH₄ = 8 g Molar mass of CH₄ = 16 g/mol Mole of CH₄ =?

Mole = mass / molar mass

Mole of CH₄ = 8 / 16

Mole of CH₄ = 0.5 mole

For ethane, C₂H₆

Mass of C₂H₆ = 18 g Molar mass of C₂H₆ = 30 g/mol Mole of C₂H₆ =?

Mole = mass / molar mass

Mole of C₂H₆ = 18 / 30

Mole of C₂H₆ = 0.6 mole

Next, we shall determine the total mole. Details below:

Volume (V) = 750 mL = 10 LTemperature (T) = 23 °C = 23 + 273 = 296 KPressure (P) = 3.70Gas constant (R) = 0.0821 atm.L/mol KTotal of mole (n) =?

PV = nRT

3.70 × 10 = n × 0.0821 × 293

Divide both sides by (0.0821 × 293)

n = (3.70 × 10) / (0.0821 × 293)

n = 1.52 mole

Finally, we shall determine the partial pressure of each gas. Details below:

For methane, CH₄

Mole of CH₄ = 0.5 moleTotal mole = 1.52 moleTotal pressure = 3.70 atmPartial pressure of CH₄ =?

Partial pressure = (Mole / total mole) × total pressure

Partial pressure of CH₄ = (0.5 / 1.52) × 3.70

Partial pressure of CH₄ = 1.22 atm

For ethane, C₂H₆

Mole of C₂H₆ = 0.6 moleTotal mole = 1.52 moleTotal pressure = 3.70 atmPartial pressure of C₂H₆ =?

Partial pressure = (Mole / total mole) × total pressure

Partial pressure of C₂H₆ = (0.6 / 1.52) × 3.70

Partial pressure of C₂H₆ = 1.46 atm

For propane, C₃H₈

Partial pressure of CH₄ = 1.22 atmPartial pressure of C₂H₆ = 1.46 atmTotal pressure = 3.70 atmPartial pressure of C₃H₈ =?

Partial pressure of C₃H₈ = Total pressure - (Partial pressure of CH₄ + Partial pressure of C₂H₆)

Partial pressure of C₃H₈ = 3.7 - (1.22 + 1.46)

Partial pressure of C₃H₈ = 1.02 atm

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A bag of potato chips is sealed in a factory near seal level. The atmospheric pressure is 99.82 kPa. What is the difference in Pa between the pressure in the bag and the atmospheric pressure?

Answers

The difference in Pa between the pressure in the bag and the atmospheric pressure is 1.505 kPa.

How to obtain the difference in pressure

To obtain the difference in pressure, we first need to know the atmospheric pressure near sea level. This is 760 mm Hg. When we convert this to pascals, we will have,  101.32472 kPa.

Now, the difference in pressure will be obtained by subtracting the atmospheric pressure in the bag from the atmospheric pressure near sea level and this is:

101.32472 kPa -  99.82 kPa

= 1.505 kPa.

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The Environmental Protection Agency was assigned which task? A. setting standards and governing the cleanliness of water used by Americans B. setting standards and governing the highways and interstates in the United States C. setting standards and governing the use of national parks and monuments in the United States. D. setting standards and governing the civil and commercial air transportation throughout America​

Answers

Answer:

A. setting standards and governing the cleanliness of water used by Americans

Explanation:

The responsibilities of the Environmental Protection Agency (EPA) is to make sure that:

People in America should have clean air, water, and good quality soil so that land is fertile.Using Scientific information to come up with efforts on a national level and reduce risks to the environment.A fair and effective administration of federal laws centered around the protection of human health and our environment.

Identify the reagent that is used to confirm the presence of each of the following:
a. CO32-: [C]
b. S2-: [S]
c. I-: [I]

Answers

a. [C]: [tex]HCl[/tex] or any other strong acid b. [S]: Lead acetate or any other heavy metal salt c. [I]: Lead nitrate or silver nitrate

a. To confirm the presence of [tex]CO_32[/tex]-, a solution of dilute [tex]HCl[/tex] (hydrochloric acid) is added. If [tex]CO_32[/tex]- is present, it will react with the [tex]HCl[/tex] to produce [tex]CO_2[/tex] gas, which can be identified by bubbling it through limewater [tex](Ca(OH)_2)[/tex].

b. To confirm the presence of [tex]S_2[/tex]-, a solution of lead acetate [tex](Pb(CH_3COO)_2)[/tex] is added. If [tex]S_2[/tex]- is present, it will react with the lead acetate to produce a black precipitate of lead sulfide ([tex]PbS[/tex]).

c. To confirm the presence of I-, a solution of chlorine water ([tex]Cl_2[/tex] in water) is added. If I- is present, it will react with the chlorine to produce a brown color, which is due to the formation of iodine (I2).

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Explain how your model is different from the model in the picture.

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My model is distinct from the model in the image in that it takes a more thorough and all-encompassing approach to comprehending the fundamental parts of a system.

It considers the interactions between various system elements as well as the connections between those elements and their surroundings. It also looks at how the system changes over time, and how different components interact with each other.

As a result, the system may be understood more precisely, and management choices can be made with more knowledge. In order to offer a more precise and current picture of the system, my model also integrates the most recent research and technological advancements.

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Dimensional analysis with shapes

Answers

The surface area of the rectangular prism is 0.034 square meters.

For a rectangular prism with length l, width w, and height h, the surface area is:

Surface area = 2lw + 2lh + 2wh

Substituting the given values, we get:

Surface area = 2(10 cm x 5 cm) + 2(10 cm x 8 cm) + 2(5 cm x 8 cm)

Surface area = 100 cm² + 160 cm² + 80 cm² = 340 cm²

We can use dimensional analysis. So the conversion factor is:

1 m² / 10,000 cm²

Multiplying the surface area by this conversion factor, we get:

Surface area = 340 cm² x (1 m² / 10,000 cm²)

Surface area = 0.034 m²

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--The complete Question is, What is the surface area of a rectangular prism that has a length of 10 cm, a width of 5 cm, and a height of 8 cm? Use dimensional analysis to convert the answer to square meters--

Chemistry. . . Reaction: AB₂C (g) → B₂ (g) + AC (g), find the value of K
At equilibrium [AB₂C]=0.0168 M, [B₂]= 0.007 M, and [AC] = 0.0118 M

Answers

The value of K at equilibrium, for the reaction is 0.0049

How do i determine the value of K at equilibrium?

First, we shall list out the given parameters from the question. This is shown below:

AB₂C (g) ⇌ B₂(g) + AC(g) Concentration of AB₂C, [AB₂C] = 0.0168 MConcentration of B₂, [B₂]= 0.007 MConcentration of AC, [AC] = 0.0118 MEquilibrium constant (K) =?

Equilibrium constant is defined as:

Equilibrium constant = [Product]ᵐ / [Reactant]ⁿ

Where

m is the coefficient of productsn is the coefficient of reactants

With the above formula, we can obtain the equilibrium constant, K as follow:

Equilibrium constant, K = [B₂][AC] / [AB₂C]

K = (0.007 × 0.0118) / 0.0168

K = 0.0049

Thus, the equilibrium constant, K for the reaction is 0.0049

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Predict which of the following reactions has a positive change in entropy.
I. 2N2(g) + O2(g) → 2N2O(g)
II. CaCO3(s) → CaO(s) + CO2(g)
III. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)​

Answers

Answer:

Explanation:

The change in entropy of a system can be determined by comparing the entropy of the reactants to the entropy of the products. The reaction that leads to an increase in the number of moles of gas or particles will generally have a positive change in entropy.

I. 2N2(g) + O2(g) → 2N2O(g)

The reactants have 3 moles of gas, while the product also has 3 moles of gas. Therefore, there is no change in the number of moles of gas, and the change in entropy is likely to be small.

II. CaCO3(s) → CaO(s) + CO2(g)

The reactant is a solid, while the products are a solid and a gas. The formation of a gas from a solid leads to an increase in the number of moles of particles, and therefore an increase in entropy.

III. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

The reactants consist of a solid and a liquid, while the products consist of an aqueous solution and a gas. The formation of a gas leads to an increase in the number of moles of particles, and therefore an increase in entropy.

Therefore, reactions II and III have a positive change in entropyentropy

The following first-order reaction occurs in CCL4(l) at 45°C: N2O5》N2O4+1÷2O2. The rate consast is k=6.2×10^-4 s^-1 an 80.0 g sample of N2O5 in CCL4 is allowed to decompose at 45°C
a) how long does it take for the quantity of N2O5 to be reduced yo 2.5 g ?
b) how many liters of O2 measured at 745 mmHg and 45°C, are produced up to this point ?

Answers

a) The amount of N₂O₅ is lowered to 2.5 g during the course of around 4.41 × 10⁴  seconds or 12.25 hours.

b) 9.71 L of O₂ are generated at 745 mmHg and 45 °C.

How to find quantity?

a) To solve for the time required for the quantity of N₂O₅ to be reduced to 2.5 g, use the first-order integrated rate law:

ln[N₂O₅]t/[N₂O₅]0 = -kt

where [N₂O₅]t = concentration of N₂O₅ at time t, [N₂O₅]0 = initial concentration of N₂O₅, k = rate constant, and t = time.

Find the initial concentration of N₂O₅:

n(N₂O₅) = m/M = 80.0 g / 108.01 g/mol = 0.7413 mol

[N₂O₅]0 = n/V = 0.7413 mol / 0.153 L = 4.846 M

where M = molar mass of N₂O₅ and V = volume of the solution.

Substituting the given values into the equation:

ln([N₂O₅]t / 4.846 M) = -6.2×10⁻⁴ s⁻¹ × t

When the quantity of N₂O₅ is reduced to 2.5 g, the concentration is:

n(N₂O₅) = m/M = 2.5 g / 108.01 g/mol = 0.02314 mol

[N₂O₅]t = n/V = 0.02314 mol / 0.153 L = 0.151 M

Substituting this concentration into the equation and solving for t:

ln(0.151 M / 4.846 M) = -6.2×10⁻⁴ s⁻¹ × t

t = 4.41 × 10⁴ s

Therefore, it takes approximately 4.41 × 10⁴ seconds or 12.25 hours for the quantity of N₂O₅ to be reduced to 2.5 g.

b) The balanced equation for the reaction shows that 1 mole of N₂O₅ produces 1/2 mole of O₂:

N₂O₅ → N₂O₄ + 1/2 O2

Therefore, the number of moles of O₂ produced can be calculated using the stoichiometry:

n(O₂) = 1/2 × n(N₂O₅) = 1/2 × 0.7413 mol = 0.3707 mol

The ideal gas law can be used to calculate the volume of O₂ produced at 745 mmHg and 45°C:

PV = nRT

where P = pressure, V = volume, n = number of moles, R = gas constant, and T = temperature in Kelvin.

Convert the pressure to atm and the temperature to Kelvin:

P = 745 mmHg / 760 mmHg/atm = 0.980 atm

T = 45°C + 273.15 = 318.15 K

Substituting the values and solving for V:

V = nRT/P = (0.3707 mol) × (0.08206 L·atm/mol·K) × (318.15 K) / (0.980 atm) = 9.71 L

Therefore, the volume of O₂ produced at 745 mmHg and 45°C is 9.71 L.

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100 grams of 4 degree celsius water is heated until its temperature is 37 degrees celsius. If the specific heat of water is 4.18 J/g degrees celsius, calculate the amount of heat energy needed to cause this rise in temperature.

Answers

To calculate the amount of heat energy needed to cause the rise in temperature, you can use the formula:
Q = mcΔT

Where Q represents the heat energy, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Given:
m = 100 grams
c = 4.18 J/g°C
Initial temperature (T1) = 4°C
Final temperature (T2) = 37°C
First, find the change in temperature (ΔT):
ΔT = T2 - T1 = 37°C - 4°C = 33°C
Now, plug the values into the formula:
Q = (100 g) × (4.18 J/g°C) × (33°C)
Q = 13794 J

So, the amount of heat energy needed to cause this rise in temperature is 13,794 Joules.

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The total pressure of gas collected over water is 725.0 mmHg and the temperature is 18.0 C what is the pressure of hydrogen gas formed in mmHg

Answers

The pressure of hydrogen gas formed is 709.5 mmHg.

Partial pressure is the pressure exerted by a single gas component in a mixture of gases, assuming all other gases are held constant.

In this case, the hydrogen gas is formed by a chemical reaction.

To calculate the partial pressure of hydrogen gas, we need to subtract the vapor pressure of water from the total pressure of the gas collected.

The vapor pressure of water at 18.0 °C is 15.5 mmHg.

Therefore, the partial pressure of hydrogen gas can be calculated as:

Partial pressure of hydrogen gas = Total pressure - Vapor pressure of water

Partial pressure of hydrogen gas = 725.0 mmHg - 15.5 mmHg = 709.5 mmHg

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Ocean water contains 3.3 % NaCl by mass.
How much salt can be obtained from 234g of seawater?

Answers

Answer:

Ans: 8.9 NaCl

Explanation:

Ocean water contains 3.5 nacl by mass how much salt can be obtained from 254 g of seawater

Question: Ocean water contains 3.5% NaCl by mass. How much salt can be obtained from 254g of seawater?

A solution contains 0.0400 M Ca2+ and 0.0990 M Ag+. If solid Na3PO4 is added to this mixture, which of the phosphate species would precipitate out of solution first? Ca3(PO4)2
Ag3PO4
Na3PO4
When the second cation just starts to precipitate, what percentage of the first cation remains in solution?

Answers

When the second cation first starts to precipitate, 80.8% of Ca²⁺ will still be in solution.

What is cation?

A cation is an ion with a positive charge. It is formed when an atom loses one or more of its electrons, resulting in a net positive charge. Cations are attracted to anions (ions with a negative charge) due to their opposite charges. Cations are found in many different substances, including acids, bases, and salts.

Ca₃(PO₄)₂ will be the first species that separates out of solution when solid Na₃PO₄ is introduced to the mixture. This is due to Ca3(PO4)2 having a substantially lower solubility than Ag₃PO₄ and Na₃PO₄.

The proportion of the first cation (Ca²⁺ ) still in solution when the second cation (Ag⁺) is just beginning to precipitate will depend on the starting concentrations of the two cations. In this instance, the starting concentrations of Ca²⁺  and Ag⁺ are 0.0400 M and 0.0990 M, respectively. Therefore, 80.8% of Ca²⁺  will still be in solution when its second cation first begins to precipitate.

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Question
How many moles of Na₂S₂O3 are needed to dissolve 0.65 mol of AgBr in a solution volume of
1.0 L, if Ksp for AgBris 3.3 x 10-13 and K for the complex ion [Ag(S₂03)2] is 4.7 × 10¹3?
Remember to use correct significant figures in your answer (round your answer to the nearest
tenth). Do not include units in your response.

Answers

The precipitation of an ionic substance from solution occurs when the ionic product exceeds the value of its solubility product at that temperature. Here the moles of Na₂S₂O₃ needed is

The solubility product of a sparingly soluble salt is defined as the product of the molar concentrations of its ions in a saturated solution of it at a given temperature.

Here the concentration of Ag⁺ ions = √Ksp = √3.3 × 10⁻¹³ = 1.81 × 10⁻¹³.

Moles of Ag⁺ ions: (1.82 x 10⁻¹³ M) x 1.0 L = 1.82 x 10⁻¹³  mol Ag⁺

Use the stoichiometry of the reaction to find the moles of Na₂S₂O₃ needed: 1 mol Na₂S₂O₃ / 2 mol Ag⁺ = 0.5 mol Na₂S₂O₃/mol Ag⁺

Moles of Na₂S₂O₃ required: 0.5 mol Na₂S₂O₃/mol Ag⁺ x 1.82 x 10⁻¹³ mol Ag⁺ = 9.1 x 10⁻¹⁴ mol Na₂S₂O₃

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A platinum ring is composed of 2.35×1023 atoms. Calculate the mass of the ring in grams.

Answers

The mass of the platinum ring is 76.0 grams.

To calculate the mass of the platinum ring

We need to know the molar mass of platinum and the number of platinum atoms in the ring.

The molar mass of platinum (Pt) is 195.08 g/mol.

The number of platinum atoms in the ring is 2.35×10^23.

Now we can use the following formula to calculate the mass of the ring:

mass = (number of atoms) x (atomic mass) / Avogadro's number

where Avogadro's number is 6.022 x 10^23 mol^-1.

Substituting the values:

mass = (2.35×10^23 atoms) x (195.08 g/mol) / (6.022 x 10^23 mol^-1)

mass = 76.0 g

Therefore, the mass of the platinum ring is 76.0 grams.

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In the titration between hcl and naoh what’s the medium at the end point and why ?

Answers

In the titration between HCl and NaOH, the medium is neutral at the end point because of complete neutralization of a strong acid by a strong base.

Neutralization is a chemical reaction in which acid and base react to form salt and water. Hydrogen (H⁺) ions and hydroxide (OH⁻ ions) react with each other to form water.

The strong acid and strong base neutralization have a pH value of 7.

The beaker gets warm which indicates that the reaction between acid and base is an exothermic reaction releasing heat energy into the surroundings.

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Which of these would be the least dense?Marble
Feather
Coin
Phone

Answers

The feather is the least thick of the bunch. This is due to the fact that density is defined as mass per unit volume.

A feather has a relatively low mass compared to its volume, due to its porous and lightweight nature. Marble, coin, and phone all have substantially higher densities than a feather since they are constructed of denser materials such as stone, metal, and plastic/electronics.

As a result, when the density of these things is compared, the feather is the least dense.

Because it has a significantly smaller mass than the other objects listed, the feather would be the least dense. Because density is defined as mass per unit volume, an object with a lower mass and the same or greater volume has a lower density.

The stone, coin, and phone all have greater masses and thus higher densities than the feather. However, because density varies based on the exact material used, the relative densities of these things may change if they are made of different materials.

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50cm³ of 1.0M hydrochloric acid reacted with excess zinc. i) Write the equation for the reaction. ii) How many mole of aqueous hydrogen ions were present in the acid solution? iii) Calculate the volume of gas evolved at s.t.p. [Molar volume = 22.4 dm³ at s.t.p. of gas].​

Answers

i) The equation for the reaction between hydrochloric acid and zinc is:

[tex]Zn + 2HCl → ZnCl2 + H2[/tex]

ii) n(HCl) = C × V = 1.0M × 0.05 L = 0.05 moles

iii) The volume of gas evolved at STP is 0.544 L or 544 mL.

The concentration of hydrochloric acid is 1.0M, which means that there is 1 mole of hydrochloric acid in 1 liter (1000 cm³) of solution. The volume of the hydrochloric acid used is 50 cm³, which is 0.05 liters.

According to the stoichiometry of the reaction, each mole of hydrochloric acid produces one mole of hydrogen ions, so the number of moles of hydrogen ions in the solution is also 0.05 moles.

The volume of gas evolved can be calculated from the ideal gas law:

PV = nRT

where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the gas constant (0.0821 L·atm/K·mol), and T is the temperature of the gas in Kelvin. At standard temperature and pressure (STP), the pressure is 1 atm and the temperature is 273 K. The molar volume of a gas at STP is 22.4 L/mol.

From the equation for the reaction, we know that one mole of hydrogen gas is produced for every two moles of hydrochloric acid used. Therefore, the number of moles of hydrogen gas produced is:

n(H2) = 0.5 × n(HCl) = 0.5 × 0.05 moles = 0.025 moles

Using the ideal gas law, we can calculate the volume of hydrogen gas produced at STP:

V(H2) = n(H2) × RT/P = 0.025 mol × 0.0821 L·atm/K·mol × 273 K/1 atm = 0.544 L

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The volume of a sample of air in a cylinder with
a movable piston is 2.0 L at a pressure P1 , as
shown in the diagram above. The volume is
increased to 5.0 L as the temperature is held
constant. The pressure of the air in the cylinder is
now P2 . What effect do the volume and pressure
changes have on the average kinetic energy of the
molecules in the sample?
(A) The average kinetic energy increases.
(B) The average kinetic energy decreases.
(C) The average kinetic energy stays the same.
(D) It cannot be determined how the kinetic
energy is affected without knowing P1
and P2 .

Answers

Answer:

I used Chat GPT to answer the question here is the answer

Assuming the gas behaves ideally, the answer is (C) The average kinetic energy stays the same.

According to the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature. If the temperature is held constant, then nR is also constant. Therefore, for a given amount of gas, if V increases, P must decrease (and vice versa) to maintain the same value of PV.

The average kinetic energy of gas molecules is proportional to temperature, so if the temperature is held constant, the average kinetic energy of the gas molecules stays the same. The changes in volume and pressure only affect the density and distribution of the gas molecules, but not their average kinetic energy.

The average kinetic energy of a gas is directly proportional to its temperature, according to the kinetic theory of gases. This means that if the temperature is held constant, the average kinetic energy of the gas molecules will also be constant, regardless of any changes in volume or pressure.

Therefore, the correct answer is (C) the average kinetic energy stays the same.

How many hydrogen molecules (h2) are needed to convert the triacylglycerol shown to saturated fat

Answers

We would need about 16 hydrogen atoms so that we can convert the compound to a saturated fat.

What is a saturated fat?

In animal products like meat and dairy, saturated fat is a form of dietary fat that is normally solid at room temperature. It is known as being "saturated" because each molecule of fat has the most hydrogen atoms possible, giving it a stable structure.

We can see this by counting the number of double bonds in the fat and there are eight of them so sixteen hydrogen atoms are needed for saturation.

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A container of hellum has 4.3 moles of gas in a container with a volume of 3.9 liters and a pressure of 201.6kPa at 298K. A container of xenon has a volume of 3.9 liters
and a pressure of 201.6kPa at 298K. How many moles of xenon gas is present?

Answers

The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin. Rearranging the equation, we get:

n = PV/RT

For the container of helium:

n = (201.6 kPa) x (3.9 L) / [(8.31 J/mol*K) x (298 K)] = 0.0688 mol

Now, using the same equation for the container of xenon:

n = (201.6 kPa) x (3.9 L) / [(8.31 J/mol*K) x (298 K)] = 0.0688 mol

Therefore, there are also 0.0688 moles of xenon gas present in the container.

What best describes the energy in light?
A. It increases as it is absorbed by an atom.
B. It increases as the light moves from violet toward red.
C. It is absorbed and emitted in discrete chunks.
D. It is absorbed when it comes into contact with an object.

Answers

C. It is absorbed and emitted in discrete chunks.

The energy in light is carried by particles called photons, which behave both like waves and like particles. According to the theory of quantum mechanics, photons can only be absorbed or emitted in discrete amounts of energy, known as quanta. This means that the energy in light is not continuous, but rather comes in specific packets or chunks. This phenomenon is known as quantization, and it has important implications for many areas of physics, including atomic and molecular physics, as well as the study of electromagnetic radiation.

Answer: C it is absorbed and emitted in descrete chunks.

Explanation:

photons of light are emitted or absorbed as electrons change energy levels

Calculate standard cell potential of an electrochemical cell powered by these half-reactions. (Write values to two decimal places. If a value is less than 1, be sure to write a 0 before the decimal.)

 Pb4+ + 2e− → Pb2+

 Co3+ + e− → Co2+

E°cell = V
Is the reaction spontaneous

Answers

The standard cell potential is found as +1.95 V and is a  spontaneous  reaction.

What is  standard cell potential ?

The standard cell potential (E°cell) of an electrochemical cell is given by the difference between the standard reduction potentials of the two half-cells involved.

E°cell = E°reduction (cathode) - E°reduction (anode)

The half-reactions given are:

Pb4+ + 2e− → Pb2+ (reduction)

Co3+ + e− → Co2+ (reduction)

The standard reduction potentials for these half-reactions are:

E°reduction(Pb4+/Pb2+) = -0.13 V

E°reduction(Co3+/Co2+) = +1.82 V

We then calculate as:

E°cell = E°reduction (Co3+/Co2+) - E°reduction (Pb4+/Pb2+)

E°cell = (+1.82 V) - (-0.13 V)

E°cell = +1.95 V

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using the equation PCI5(g) PCI3(g) + CI2(g), if CI2 is added, what way will the euilibeium shift

Answers

When an equilibrium system is put under stress, Le Chatelier's principle can be used to forecast changes in equilibrium concentrations.

Thus, The adjustments required to reach equilibrium might not be as obvious if we have a mixture of reactants and products that have not yet reached equilibrium.

In this situation, we can compare the Q and K values for the system to forecast changes.

By adding or withdrawing one or more of the reactants or products, an equilibrium chemical system can be momentarily moved out of equilibrium. After that, additional adjustments are made to the reactant and product concentrations in order to bring the system back to equilibrium.

Thus, When an equilibrium system is put under stress, Le Chatelier's principle can be used to forecast changes in equilibrium concentrations.

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How much H2 gas at STP can be produced by
the reaction
2 Na(s) + 2 H2O(ℓ) −→ H2(g) + 2 NaOH(aq)
of 3.60 g of Na and excess water?
Answer in units of L.

Answers

1.71 L of [tex]H^2[/tex] gas can be produced at STP from the given reaction.

To solve this problem

According to the equation, 1 mole of hydrogen gas [tex]H^2[/tex]  is created for every 2 moles of sodium (Na) that react with extra water.

Using the molar mass of Na, we can get the number of moles from the given amount of sodium (3.60 g):

3.60 g Na × (1 mol Na / 22.99 g Na) = 0.157 mol Na

Since the reaction requires 2 moles of Na to produce 1 mole of [tex]H^2[/tex]  the number of moles of [tex]H^2[/tex] produced is

0.157 mol Na × (1 mol [tex]H^2[/tex] / 2 mol Na) = 0.079 mol [tex]H^2[/tex]

Now, to calculate the volume of hydrogen gas produced at STP (standard temperature and pressure), we can use the ideal gas law

PV = nRT

Where

The ideal gas constant R = 0.08206 L atm/(mol K) P = 1 atm (since it's at STP) V is the volume we're looking forn = 0.079 mol (from above)T = 273 K (since it's at 0°C)

Solving for V, we get:

V = nRT/P = (0.079 mol)(0.08206 L atm/(mol K))(273 K)/(1 atm) = 1.71 L

Therefore, 1.71 L of % [tex]H^2[/tex] gas can be produced at STP from the given reaction.

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Lab: Limiting Reactant and Percent Yield
Step 7: Determine the Limiting Reactant (Trial 2)
Analysis: aluminum
there is no aluminum left
yes
Convert Mass:
2.50g=.019
.25g=.0093
The limiting reactants is/are aluminum.
Are these answers correct?
Yes they are I did the lab.

Answers

The given answer statement  "there is no aluminum left" and " limiting reactants is aluminum" are correct.

In the analysis of Trial 2, it was found that there was no aluminum left after the reaction had taken place. This indicates that all of the aluminum had reacted with the copper (II) chloride and that it was the limiting reactant in the reaction.

To confirm this, the mass of each reactant was converted to moles using their respective molar masses. It was found that the aluminum had a smaller number of moles than the copper (II) chloride, indicating that it would be used up first and thus be the limiting reactant.

Therefore, the limiting reactant in Trial 2 was aluminum.

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