If we put 100 particles of heavy species of gas into a container and the pressure is 12
atmospheres, predict the new pressure of putting a total of 300 particles of this gas
into the container.

A - 12 atm

B- 36 atm

C- 24 atm

Answer:

Independent Variable: Number of Coils Wrapped Around

Dependent Variable: Number of Paperclips Magnet Picks Up

Possible Research Question: How does the number of wire wraps affect the strength of an electromagnet?

Trends noticed in the data: As the number of wire wraps (coils) around the nail increases, the magnetic strength of the electromagnet also increases. There is a positive correlation between the number of wire wraps and the number of paper clips the magnet picks up.

Claim based on the data: The magnetic strength of an electromagnet increases as the number of wire wraps (coils) around the nail increases.

Evidence from the chart or graph supporting the claim: The graph shows that as the number of wire wraps increases from 4 to 12, the number of paper clips the magnet picks up also increases from 6 to 25. This suggests a positive correlation between the two variables.

The options that affect the amount that the freezing point is lowered by the addition of the solute include :

The increasing of a solvent's boiling point as a result of the addition of a solute is known as boiling point elevation. Similar to freezing point depression, adding a solute lowers the freezing point of a solvent. In actuality, a solvent's freezing point drops as its boiling point rises.

Any solvent's freezing point will be lowered by the presence of a solute; this action is known as freezing-point depression. The fact that the solute is present in the liquid solution but not in the pure solid solvent is crucial to understanding this phenomenon.

This includes the compound being ionic or the solubility of the solvent.

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Answer:

I think you need to dissolve 62.116 (or 62.12g) of Li2SO4 .

The mass of lithium sulfate needed is 62g

Molality is also known as molal concentration. It is a measure of solute concentration in a solution. The solution is composed of two components; solute and solvent.

It is defined as the moles of the solute present in 1kg of the solvent. It is denoted by 'm'.

This term is useful in understanding the concentration of a solution because its formula is independent of temperature and pressure. Thus, the colligative properties are dependent on molality.

Given,

Molality of solution = 1.23m

Mass of solvent ( water) = 459g

Molality = number of moles of solute ÷ mass of solvent in kg

number of moles of solute = molality × mass of solvent in kg

                                       = 1.23 × 0.459

                                          = 0.564 moles

moles = mass / molar mass

mass= 0.564 × 109.94

            = 62 g

Therefore, the mass of lithium sulfate needed is 62g

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C₃H₈ has an empirical formula mass of 44 g/mol. Hence, C₃H₈ is also the molecular formula.

When twο οr mοre distinct elements are chemically mixed in a specific ratiο, the resultant substance is knοwn as a cοmpοund.  Chemical bοnds, such as cοvalent bοnds, iοnic bοnds, οr metallic bοnds, hοld the cοnstituent parts οf a cοmpοund tοgether.

We must first cοmpute the mοles οf each element present in the cοmpοund in οrder tο derive its empirical fοrmula.

We can figure οut hοw many mοles οf carbοn and hydrοgen there are in 100 g οf the chemical by dοing the fοllοwing calculatiοn:

81.71 g / 12.011 g/mol = 6.80 moles of carbon

18.29 g / 1.008 g/mol of hydrogen = 18.15 moles.

The simplest whole number ratio of carbon to hydrogen atoms must next be determined. In order to do this, we divide both values by the lesser amount, in this case 6.80 moles:

Carbon: 6.80 moles / 6.80 moles = 1.00 Hydrogen: 18.15 moles / 6.80 moles = 2.67

To get whole numbers, we must multiply both values by a factor. By multiplying each by 3, the smallest whole number ratio is obtained:

1.00 x 3 = 3 carbon

2.67 x 3 = 8.01 x 8 for hydrogen.

As a result, the compound's empirical formula is C₃H₈.

Knowing the compound's molecular weight is necessary to determine its molecular formula. The compound's MWt (molecular weight) is 44, as stated.

C₃H₈ has an empirical formula mass of 44 g/mol. Hence, C₃H₈ is also the molecular formula.

Any molecule with the empirical formula C₃H₈, like propane, might be the structural formula of the chemical (CH₃CH₂CH₃).

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The statement 'when charges pass through the bulb, light is produced' best describes how the charges move through the bulb.

An electrical charge is a type of positive or negative charge of the particles that move in a conductor.

At the atomic level, protons are positively charged particles, electrons are negative, and neutrons do not have a charge.

The electrical charges depend on the charge's initial time, the final time, and the net outward current.

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According to the information, we can infer that a 19.2-pound child would need to take approximately 2.49 tablets of 250.0 mg ampicillin per day.

To calculate the number of tablets the child needs to take, we first need to convert the weight of the child from pounds to kilograms. Since there are 2.20 pounds in 1 kg, we divide 19.2 pounds by 2.20 to get the weight in kilograms, which is approximately 8.73 kg.

Next, we calculate the daily dose of ampicillin for the child by multiplying the weight in kilograms by the recommended dosage of 115 mg/kg.

Since the available tablet size is 250.0 mg, we divide the total daily dose by the tablet size to find out how many tablets the child needs to take per day.

According to the above, we can infer that since we can't have a fraction of a tablet, we round the result to the nearest whole number. Therefore, a 19.2-pound child would need to take approximately 2.49 tablets (rounded to 2) of 250.0 mg ampicillin per day.

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Answer:

9.9 x 10^-2 g*cm²/s²

Explanation:

9.9 × 10^-9 kg*m²/s² =    g*cm²/s²

1 kg*m²/s² = 1 joule(s)

1 g*cm²/s² = 1 erg(s)

britannica

1 kg = 1000g = 1x10^3 g

1 m²= 10000 cm² = 1x10^4 cm²

add the exponents 3 and 4 which = 7

-9 + 7 = -2

9.9 × 10^-9 kg*m²/s² = 9.9 x 10^-2 g*cm²/s²

Answer:

4.0561×0.03÷63.5= 0.001916 mol cu

A coin is 3.00% Cu by mass. If the mass of the coin is 4.0561g, moles of Cu does it contain 0.001916 mole of Cu.

It is the most fundamental characteristic of matter and one of the fundamental quantities in physics.

Mass is a term used to describe how much matter is there in a body. The kilogramme is the SI unit of mass (kg). A body's mass does not alter at any point in time.

Given mass of coin = 4.0561g , moles =? , Cu by mass= 3.00%

4.0561×0.03÷63.5= 0.001916 mole Cu.

Thus, A coin is 3.00% Cu by mass. If the mass of the coin is 4.0561g, moles of Cu does it contain 0.001916 mole of Cu.

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The decreasing values of bond angle from 107º in NH3 to 91º in SbH3 down the group 15 elements of the periodic table owes to the decreasing electronegativity of the elements down the group.

The bond angle is the angle between any two bonds to a single atom.

The hydrides of group 15 elements form hydrides of the sort AH3 where A is the particular group 15 element.

Electronegativity is a periodic trend that decreases down the group and increases across the period.

As we move down group 15, atoms of elements become less electronegative and as such, repulsion between lone pair and bond pair and the repulsion between two bond pairs decreases. Ultimately, the bond angles  of the hydrides decrease also.

This accounts for the decreasing values of bond angle from 107º in NH3 to 91º in SbH3 down the group 15 elements of the periodic table.​

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Answer:

I think its D if I'm wrong I'm so sorry

Answer: b

Explanation:

I just took the test

The mass of hydrogen needed to produce 51.0 grams of ammonia is  ≈ 9.07 grams.

To determine the mass of hydrogen required to produce 51.0 grams of ammonia (NH3) using the Haber process, we need to calculate the stoichiometric ratio between hydrogen and ammonia.

From the balanced chemical equation:

3 H₂(g) + N₂(g) → 2 NH₃(g)

We can see that for every 3 moles of hydrogen (H₂), we obtain 2 moles of ammonia (NH₃).

First, we need to convert the given mass of ammonia (51.0 grams) to moles. The molar mass of NH₃ is 17.03 g/mol.

Number of moles of NH₃ = Mass / Molar mass

                     = 51.0 g / 17.03 g/mol

                     ≈ 2.995 moles

Next, using the stoichiometric ratio, we can calculate the moles of hydrogen required.

Moles of H₂ = (Moles of NH₃ × Coefficient of H₂) / Coefficient of NH₃

           = (2.995 moles × 3) / 2

           ≈ 4.493 moles

Finally, we can convert the moles of hydrogen to mass using the molar mass of hydrogen (2.02 g/mol).

Mass of H₂ = Moles × Molar mass

          = 4.493 moles × 2.02 g/mol

          ≈ 9.07 grams

Therefore, approximately 9.07 grams of hydrogen is needed to produce 51.0 grams of ammonia in the Haber process.

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The law of conservation of mass states that in a chemical reaction, the total mass of the reactants is equal to the total mass of the products. To provide evidence for this law, we can perform an experiment in which calcium carbonate (\(CaCO_3\)) is decomposed to produce calcium oxide (CaO) and carbon dioxide (\(CO_2\) ), and then measure the masses of the reactants and products.

Method:

Weigh a sample of \(CaCO_3\) using a balance.

Heat the \(CaCO_3\) in a crucible until it decomposes to CaO and \(CO_2\). The \(CO_2\) gas will escape, leaving only CaO in the crucible.

Allow the crucible to cool and then weigh it again to determine the mass of the CaO produced.

Collect the \(CO_2\) gas that is released during the reaction in a gas syringe or other collection device. Measure the volume of \(CO_2\) gas produced, and calculate its mass using its molecular weight.

Which measurements should be taken:

The following measurements should be taken:

The mass of the \(CaCO_3\) used as a reactant.

The mass of the CaO produced as a product.

The volume of \(CO_2\) gas produced during the reaction.

The temperature and pressure of the \(CO_2\) gas to allow for the calculation of its mass.

How the student could show evidence for the conservation of mass:

To show evidence for the law of conservation of mass, the student can compare the mass of the \(CaCO_3\) used as a reactant to the total mass of the products, which includes the mass of CaO produced and the mass of \(CO_2\) gas released.

The sum of the masses of CaO and \(CO_2\) should be equal to the mass of the \(CaCO_3\) used as a reactant, within experimental error. This will provide evidence that the mass of the reactants is conserved and equals the mass of the products, as required by the law of conservation of mass.

Additionally, the student could calculate the theoretical yield of CaO and CO2 based on the balanced equation for the reaction, and compare this to the actual yield obtained from the experiment. Any difference between the theoretical and actual yields could be due to experimental error, but the comparison can still provide additional evidence for the conservation of mass.

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Answer:

(a) Electrode reactions:

Pt/H+/H2: 2H+(aq) + 2e- -> H2(g) (reduction)

Ag/AgCl/Cl-: AgCl(s) + e- -> Ag(s) + Cl-(aq) (reduction)

Overall cell equation: 2AgCl(s) + H2(g) -> 2Ag(s) + 2HCl(aq)

(b) Direction of motion of ions and electrons:

In the Pt/H+/H2 half-cell, hydrogen ions (H+) move towards the platinum electrode and accept electrons to form hydrogen gas (H2). In the Ag/AgCl/Cl- half-cell, silver ions (Ag+) move towards the silver chloride (AgCl) electrode and accept electrons to form silver (Ag) metal while chloride ions (Cl-) move away from the electrode. Electrons move from the hydrogen electrode to the silver electrode through the external circuit.

(c) Electrode labeling:

The Pt/H+/H2 electrode is the cathode (-) and the Ag/AgCl/Cl- electrode is the anode (+).

(d) Standard cell potential (Eo):

The standard cell potential can be calculated using the standard reduction potentials for each half-cell:

Eo(cell) = Eo(reduction, Ag/AgCl/Cl-) - Eo(reduction, Pt/H+/H2)

Eo(reduction, Ag/AgCl/Cl-) = +0.222 V (from standard reduction potential tables)

Eo(reduction, Pt/H+/H2) = 0 V (by definition)

Eo(cell) = +0.222 V - 0 V = +0.222 V

(e) Actual cell potential (E):

E(cell) = Eo(cell) - (0.0592 V / n) * log[H3O+]

where n is the number of electrons transferred in the balanced equation (2 in this case)

E(cell) = +0.222 V - (0.0592 V / 2) * log(1.0 x 10^-4 M)

E(cell) = +0.222 V - (0.0296 V) = +0.1924 V

(f) Values of constants A and B:

E(cell) = A + B.pH

At pH 7 (neutral), E(cell) = Eo(cell) = +0.222 V

Therefore, A = +0.222 V and B = -0.0592 V/pH

When an alarm sounds in the laboratory, it is important to respond quickly and remain calm. Listen for any other instructions, turn off any equipment in use, and be prepared to leave immediately.

What is laboratory?

Laboratory is a place where scientific experiments and measurements are performed. It is equipped with various types of equipment and instruments to carry out experiments and tests. Laboratories are usually used by scientists and technicians to conduct experiments, analyze samples, and conduct research. It is an area that is designed to be kept clean and organized in order to ensure safety and accuracy when conducting experiments. Laboratories can be found in many different settings such as hospitals, universities, research and development centers, and manufacturing facilities.

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Answer:

D. If I brush my teeth, I will get fewer cavities than if I don't brush my

teeth.

Explanation:

Answer:

V₂ =1.99 L

Explanation:

Given data:

Initial volume = 1.8 L

Initial temperature = 28°C (28+273.15 K= 301.15 k)

Initial pressure = 102 KPa

Final volume = ?

Final pressure = 85 KPa

Final temperature = 4°C (4+ 273.15 K = 277.15 K)

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

V₂ = 102 KPa × 1.8 L × 277.15 K / 301.15 K × 85 KPa

V₂ = 50884.74 KPa .L. K / 25597.75 K.KPa

V₂ =1.99 L

Sucrose is a macromolecule from the given options.

Macromolecule is defined as any very large molecule, usually with a diameter ranging from about 100 to 10,000 angstroms (10−5 to 10−3 mm). The molecule is the smallest known of the substance that retains its characteristic properties.

The main 4 types of macromolecules are the carbohydrates, lipids (or fats), proteins, and nucleic acids. All of the major macromolecule classes are similar, in nature that, they are large polymers which are assembled from small repeating monomer subunits.

Sucrose isn't a type of macromolecule. Sucrose is the most common disaccharide, which consists of monomers of glucose and fructose. Hence, option D is correct.

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Action potential would not be possible because ions would not be able to move across the plasma membrane without the channels provided by proteins.

When the membrane potential of a particular cell site rapidly increases and decreases, an action potential happens.  Following this depolarization, nearby areas also experience depolarization. Excitable cells, which include neurons, muscle cells, and some plant cells, are a class of animal cells that exhibit action potentials.

Larger protein molecules are integrated within a lipid bilayer that makes up a cell membrane. The lipid bilayer serves as an insulator because it impedes the flow of electrically charged ions very effectively.

Action potentials are produced by channel proteins, which change their state from closed to open depending on the voltage differential between the cell's interior and exterior.

Therefore, action potential is not possible without proteins.

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1 mole is equal to 6.023 × 10 ²³ molecules. 15.05 × 10²³ atoms of cadmium are in 2.50 moles of cadmium.

The term mole is defined as the amount of substance of a system which contains as many elementary entities.

One mole of any substance is equal to 6.023 × 10²³ units of that substance such as atoms, molecules, or ions. The number 6.023 × 10²³ is called as Avogadro's number or Avogadro's constant.

The mole concept can be used to convert between mass and number of particles.

1 mole of any substance = 6.023 × 10²³ molecules

Therefore, 1 mole of cadmium = 6.023 × 10²³ molecules

Then 2.50 moles of cadmium = 2.50 × 6.023 × 10²³

= 15.05 × 10²³ molecules

Thus, 15.05 × 10²³ atoms of cadmium are in 2.50 moles of cadmium.

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How to express a number in scientific notation

Write the number in decimal form. Only a digit other than 0 must be before the period, that is, it must be a real number between 1 and 10 (example: 1.5).

Count how many decimal places the point has moved.

Put this number of places as an exponent of 10. You must be careful when walking with the point: if the number decreases, the exponent will be positive (example: 10^5). If the number increases, the exponent will be negative (example: 10^-6).

In this case:

You have the number 2385.9. The point will move 3 decimals to the left. The number will decrease, so the exponent will be positive and the value of it is 3.

So it will be: 2.3859 × 10^3

Answer: 2.3859 × 10^3

Answer:

A small amount of chemical splashes in Frank's eye. What should happen next? Frank should go to the eyewash station while his lab partner tells the teacher what happened.

Explanation:

Brainlist

I CANT DELETE THE ANSWER, SO HERE.

Answer: If you are talking about online school then it should be position X. Hope this helps!

Answer:B

Explanation: Pennfoster answer

Answer:

this si from google hope it helps

Explanation:

The Main Causes of Water Pollution in the U.S.

Runoff from Agricultural Operations. Agriculture represents one of the biggest sources of water pollution in the country. ...

Runoff and Nonpoint Source Pollution. ...

Industrial Activities. ...

Leakage from Underground Storage and Piping. ...

Leaking Sewers. ...

Vehicle Emissions. ...

Landfill Leakage. ...

Hazardous Waste.

Answer:

Runoff and Non-point Source Pollution.

Explanation:

it's caused by rainfall or snow-melt moving over and through the ground. As the runoff moves, it picks up and carries away natural and human-made pollutants, finally depositing them into lakes, rivers, wetlands, coastal waters and ground and NPS is it's abbreviation.

Answer:

56.04 g

Explanation:

Step 1: Given data

Moles of nitrogen (n): 2 mol

Step 2: Calculate the molar mass of nitrogen (M)

The molar mass of the molecule is the sum of the masses of the molecules that form it.

mN₂ = 2 × mN = 2 × 14.01 g/mol = 28.02 g/mol

Step 3: Calculate the mass corresponding to 2 moles of nitrogen

We will use the following expression.

m = n × M

m = 2 mol × 28.02 g/mol = 56.04 g