Find the mass of nitrogen trioxide (NO2) that occupies 156 L at STP.

At standard pressure, the volume is 490 ml.

Boyle's law is used in this situation: P1V1 = P2V2. Volume and pressure need not be converted to standard forms. Even the pressure in mmHg might be used.

1 atm equals 760 mmHg

V2 = P1V1V1 / P2 = 745 x 500 / 760 = 490 ml

Keep in mind that we are assuming a constant temperature here.

An object that has neither a set size nor shape is a gas. A gas will expand to fill the space inside a tightly closed container. The air that you breathe is a type of gas.

The main component of natural gas, methane, is a colorless, gaseous combination of hydrocarbons (CH4). It makes up around 30% of the energy consumed in the US.

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The pressure inside the vessel would be 15 atm if we increased the temperature of the vessel to 450 K at constant volume. As a result, Option D is right.

According to the Ideal Gas Law, the pressure inside the vessel is directly proportional to the temperature (Kelvin) and the number of moles of gas present in the vessel, if the volume is kept constant. In other words, if we increase the temperature of the gas, the pressure inside the vessel will also increase.

The Ideal Gas Law equation is:

Where:

Given that V is constant, we can rearrange the equation to solve for P:

Assuming that n and V are constants, then P is directly proportional to T. So, if the temperature increases by a factor of T2/T1, the pressure will increase by the same factor.

Let's say the initial pressure was 10 atm, and the initial temperature was 400 K. Then P1/T1 = 10 atm/400 K.

When the temperature is increased to 450 K, the pressure will become:

So, when the temperature of the vessel is increased to 450 K at constant volume, the pressure inside the vessel will increase, and the answer is "d. 15 atm".

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The volume of O₂ required to react with 4.8 L of CO, measured at same temperature and pressure is 2.4 L

To obtain the volume of O₂ required, we shall consider the equation for the reaction, in order to obtain relevant information. This is shown below:

2CO(g) + O₂(g) -> 2CO₂(g)

From the balanced equation above,

2 L of CO required 1 L of O₂

With the above information, we can obtain the volume of O₂ required to react with 4.8 L of CO as follow:

From the balanced equation above,

2 L of CO required 1 L of O₂

Therefore

4.8 L of CO will require = (4.8 × 1) / 2 = 2.4 L of O₂

THus, the volume of O₂ required is 2.4 L

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Complete question:

2CO(g) + O₂(g) -> 2CO₂(g)

for the given reaction, what volume of o2 would be required to react with 4.8 l of co , measured at the same temperature and pressure?

There will be a decrease in the concentration of\(NH_4HS\)(s) as the reactant. Therefore, the forward reaction is favored by the system to compensate for the removal of\(H_2S\) gas.

\(H_2S\) gas is removed from the system at equilibrium. How does the system adjust to reestablish equilibrium?The chemical reaction is:

\(NH_4HS(s)\) ⇌ \(NH_3\)(g) + \(H_2S\)(g)When the \(H_2S\)

gas is removed from the system at equilibrium, the equilibrium shifts to the right-hand side to compensate for the loss. Since the H2S gas is one of the products, the forward reaction will be favored to compensate for the removal of \(H_2S\) gas. In other words, to reestablish the equilibrium, the equilibrium shifts to the right side to produce more\(H_2S\) gas in the forward reaction. The shift of equilibrium to the right side would result in an increase in the concentration of \(NH_3\)(g) and \(H_2S\)(g).

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The half-life for the transmutation of radon-222 to lead-214 is 3.8 days. If there is an initial mass of 100.0 g of radon-222, we can calculate how much radon-222 would remain after 7.6 days.

After one half-life (3.8 days), half of the radon-222 would decay. So, we are left with 50.0 g of radon-222. Now, after another 3.8 days (a total of 7.6 days), another half of the remaining radon-222 would decay. Therefore, we would have half of 50.0 g remaining, which is 25.0 g of radon-222 after 7.6 days.

The decay process follows an exponential decay model, where the remaining amount decreases by half with each half-life. By understanding the concept of half-life, we can determine the amount of radon-222 that would remain after a given time period. In this case, with an initial mass of 100.0 g and a half-life of 3.8 days, we calculate that 25.0 g of radon-222 would be left after 7.6 days.

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Answer:

The purpose of all seed is reproduction and is to grow a new plant, thus propagating the species.

Red (Na) = 5, White (H) = 5, Black (C) = 5, Blue (O) = 15

In this case, for a single molecule of NaHCO3, you would need 1 red jellybean (Na), 1 white jellybean (H), 1 black jellybean (C), and 3 blue jellybeans (O).

It is important to note that the molecular formula of sodium bicarbonate is NaHCO3, which is made up of 1 sodium atom, 1 hydrogen atom, 1 carbon atom and 3 oxygen atoms, so each model of sodium bicarbonate would require 1 red jelly bean, 1 white jelly bean, 1 black jelly bean and 3 blue jelly beans.

Sodium bicarbonate, also known as baking soda, is a chemical compound with the molecular formula NaHCO3. It is composed of one sodium atom, one hydrogen atom, one carbon atom, and three oxygen atoms. The compound is an ionic compound, meaning that it is composed of positively charged ions (sodium ions, Na+) and negatively charged ions (bicarbonate ions, HCO3-).

The compound is commonly used in baking as a leavening agent, as it reacts with acids, such as those found in buttermilk, yogurt or vinegar, to produce carbon dioxide gas, which causes the dough to rise. Additionally, it can be used as an antacid to neutralize stomach acid and relieve indigestion.

The jellybean model is a simple way to illustrate the molecular structure of a compound, and can be helpful for visualizing how atoms are bonded together to form a molecule.

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The identity of the daughter nuclide from the positron emission of c-11/6  is option c)11/5-b

What is beta decay?

In the radioactive decay process known as beta decay, a parent nucleus emits an electron (or positron). An electron and a neutron can separate from certain neutrons in the parent nucleus. The parent nucleus will emit an electron, followed by the production of a further proton, increasing the atomic number. In terms of nuclear physics, some elements in group A can form daughter nuclei of elements that are next to them in the periodic table. Let's call it B in this situation.:

NZA→ 0−1e+NZ+1BZNA→ −10e+Z+1NB

where:Z is the atomic number,N is the mass number

If the emission is a positron, then the process is:

NZA→ 0+1e+NZ−1BZNA→ +10e+Z−1NB

So from the following equation we learn

11(top) 6(bottom) C ---> 0(top) 1(bottom) e + 11(top) 5(bottom) B

Therefore the identity of the daughter nuclide from the positron emission of c-11/6  is option c)11/5-b

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Yo,

In Geranium plant life red vegetation is given by using the allele R this is dominant (so RR or Rr ) genotype will produce crimson plant life.

If we communicate approximately the connected punnet square which i took from the workout e book, you could see that there was a cross made among vegetation with heterozygous genotype for red flora.

P1: Rr X Rr

Gametes: R: r : R : r

F1: RR: Rr: Rr: rr

so there is three: 1 in F1 offspring for red to white plant life.

Now the query is set f2 technology, while we cross F1 generation suppose we do a move of Rr and rr from F1 generations allow us to see what takes place.

F1: Rr: rr

Gametes: R: r: r: R

F2 : Rr : rr: Rr: rr

Rr: purple flowers

rr: White flora:

So the ratio in F2 era might be 2 crimson : 2 white. Please see connected punnet squares (first one for query) and the second of F1 move for better know-how.

I'm hoping that this will benefit you, so from my assumption and study it'd be 2 crimson : 2 white, if this is indeed the one i'm thinking of since nothing else was provided, if not please correct me.

Answer: 2 red : 2 white

Explanation:

I know

Answer:

191 days =16,502, 400 seconds

Answer:

What are the trends of ionization energy?

Ionization energy is the minimum energy required to remove an electron from an atom or ion in the gas phase.

The most common units of ionization energy are kilojoules per mole (kJ/M) or electron volts (eV).

Ionization energy exhibits periodicity on the periodic table.

Explanation:

please add me as brainliest..........

Answer:

Ca - calcium

Cu - copper

Cl - chlorine

Cs - cesium

Explanation:

70.9-gram sample of \(Cl_{2}\) gas will occupy Opton C. 22.4 liters at STP.

To determine the volume occupied by the sample of \(Cl_{2}\) (g) at STP, we can use the ideal gas law equation, PV = nRT

where P = pressure

V = volume

n = number of moles

R = ideal gas constant

T = temperature.

At STP (Standard Temperature and Pressure), the pressure is 1 atmosphere (atm) and the temperature is 273.15 Kelvin (K).

First, calculate the number of moles of \(Cl_{2}\) (g) using its molar mass. The molar mass  \(Cl_{2}\) is 70.9 grams/mol.

Number of moles (n) = mass (m) / molar mass (M)

n = 70.9 g / 70.9 g/mol

n = 1 mol

Now, we can calculate the volume using the ideal gas law:

V = (nRT) / P

V = (1 mol * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm

V ≈ 22.4 L

Therefore, the correct answer is C. 22.4 L.

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Change in volume is 0 m3 when three moles of a monatomic ideal gas undergoes an isochoric process in which the temperature changes from 500 k to 300 k.  

An isochoric process, also known as a constant-volume process, isovolumetric process, or isometric process in thermodynamics, is a thermodynamic process where the volume of the closed system having undergone such a method remains constant.

In everyday life, we see an isochoric process once we boil the water in a pressure cooker. Because the temperature increase is accomplished at a constant volume, we encounter an isovolumetric process when we transfer heat to the container.

In this problem we have T1=500K T2=300K

Since in isochoric process volume in constant, change in volume = 0

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Answer:

17.5moles

Explanation:

The number of moles in a substance can be calculated by using the formula;

Number of moles (n) = mass (m) ÷ molar mass (MM)

According to this question, mass of ammonia (NH3) = 297g

Molar Mass of NH3 = 14 + 1(3)

= 17g/mol

n = 297/17

n = 17.47

Number of moles of NH3 = 17.5moles

Answer:

C.

Explanation:

The correct name of the compound whose chemical formula is Na₂SO₄ is sodium sulfate.

Chemical  formula is a way of representing the number of atoms present in a compound or molecule.It is written with the help of symbols  of elements. It also makes use of brackets and subscripts.

Subscripts are used to denote number of atoms of each element and brackets indicate presence of group of atoms. Chemical formula does not contain words. Chemical formula in the simplest form  is called empirical formula.

It is not the same as structural formula and does not have any information regarding structure.It does not provide any information regarding structure of molecule as obtained in structural formula.

There are four types of chemical formula:

1)empirical formula

2) structural formula

3)condensed formula

4)molecular formula

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Answer:

This is thermodynamics.

Using simple thermodynamics operation equation

Answer:

Aluminium (A1) -Used in the production of duralumin alloy for use in the construction of aeroplane bodies. 3. Silicon (Si) -To make microchips 4. Sulphur (S) -To make matches, fireworks and for the manufacture of sulphuric acid 5.

Explanation:

Hope this helped!

Answer:

yes

Explanation:

we use different elements and compounds in our life ex salt and iron but most of the substances found in our environment are mixture btw scientists need to classified substances to work with our substances and to study pure substances hope it helps u

Answer:

uh i think this is the answer

Explanation:

The common way is the use of active voice when writing an abstract because passive voice make an indirect writing of the action and indication of a person in the action.

Abstract is the first phase of a thesis or a publication in a journal. It briefly summarizes about what we are going tp communicate through that paper or thesis what are the major methods used and also tell about the major findings and conclusions.

Thus, abstract must be concise but sharp and clear to the point about each step in the experiment of the scientific investigation whatever. It must be following the same tense of action that is in past or present. All sentence should follow the same tense.

Now, it is commonly better to avoid passive voice to write the actions and if possible write it in active voice. Direct indication of the person in charge and the action is good to read.

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Answer:

1.4 L.

Explanation:

Applying Charles law,

V/T = V'/T'....................... Equation 1

Where V = Initial volume, V' = Final volume, T = Initial Temperature in Kelvin, T' = Final Temperature in Kelvin.

Make V' the subject of the equation

V' = (V/T)T'..................... Euqation 2

Given: V = 1.5 L, T = 295 K, T' = 2 °C = (2+273) K = 275 K

Substitute these values into equation 2

V' = (1.5/295)275

V' = 1.398

V' ≈ 1.4 L

Answer:

7249.1

Explanation:

Multiply 7.1x1021= 7249.1

Answer:

should be half wich is 750

Explanation:

The electrode classification that identifies a filler metal that can be used in the overhead position is known as an AWS E6010 classification electrode.

AWS E6010 classification electrodeAWS E6010 classification electrode is an electrode classification that identifies a filler metal that can be used in the overhead position. The E6010 electrode classification is ideal for welding thinner sheets of metal, usually, less than 1/8" thick. The welding process is often used in pipes, fuel tanks, and other non-structural welds. The electrodes have a thick and cellulose-based coating, making them ideal for a variety of welding applications.

The E6010 classification electrode is commonly used for root passes and pipe welding. The classification of the electrode indicates how much current is required for proper welding. It also specifies the type of metal that can be welded with the electrode. The classification is also used to identify the filler metal that is used in the electrode, making it easier to determine which electrode is appropriate for a particular welding task.

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Answer:

It's the second one down.

Explanation:

Gold

Mass 197

Number of Protons: 79

Number of Neutrons: 197 - 79 = 118

Number of electrons: = number of protons = 97

Answer:

Mass number

Explanation:

All atoms of the same element have the same number of protons, but some may have different numbers of neutrons, which would cause a different mass number

Answer:

The answer is mass number

Answer:

Only 3 atoms

Explanation:

The 3 came from the coefficient.

The mass of oxygen gas in the unknown mixture is approximately 5.868 g. The mass of ethane gas in the unknown mixture is approximately 0.812 g. The mass percent of oxygen gas in the unknown mixture is approximately 71.83%.

To solve this problem, we need to determine the masses of oxygen gas and ethane gas in the unknown mixture, as well as the mass percent of oxygen gas.

Let's start by calculating the mass of water produced;

Mass of water = 1.48 g

Since water is composed of hydrogen and oxygen in a 2:1 ratio, the molar mass of water is:

Molar mass of water = 2(1.00784 g/mol) + 15.999 g/mol = 18.015 g/mol

Now we can calculate the number of moles of water produced;

Number of moles of water = Mass of water/Molar mass of water

= 1.48 g / 18.015 g/mol

≈ 0.082 mol

From balanced chemical equation for the combustion of ethane;

C₂H₆ + 7/2 O₂ → 2 CO₂ + 3 H₂O

We can see that 1 mole of ethane produces 3 moles of water. Therefore, the number of moles of ethane can be calculated as;

Number of moles of ethane = (Number of moles of water) / 3

≈ 0.082 mol / 3

≈ 0.027 mol

The molar mass of ethane (C₂H₆) is;

Molar mass of ethane = 2(12.011 g/mol) + 6(1.00784 g/mol)

= 30.0708 g/mol

Finally, we can calculate the mass of ethane;

Mass of ethane = Number of moles of ethane × Molar mass of ethane

≈ 0.027 mol × 30.0708 g/mol

≈ 0.812 g

Now, to determine the mass of oxygen gas, we subtract the mass of ethane and water from the total mass of the unknown mixture:

Mass of oxygen gas = Total mass of unknown mixture - Mass of ethane - Mass of water

= 8.16 g - 0.812 g - 1.48 g

≈ 5.868 g

Finally, we can calculate the mass percent of oxygen gas in the unknown mixture;

Mass percent of oxygen gas = (Mass of oxygen gas / Total mass of unknown mixture) × 100%

= (5.868 g / 8.16 g) × 100%

≈ 71.83%

Therefore;

The mass of oxygen gas in the unknown mixture is approximately 5.868 g.

The mass of ethane gas in the unknown mixture is approximately 0.812 g.

The mass percent of oxygen gas in the unknown mixture is approximately 71.83%.

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Answer:

The light microscope bends a beam of light at the specimen using a series of lenses to provide a clear image of the specimen to the observer.

Answer:

If someone is suffering from the problem of acidity after overeating Baking soda solution would be suggested as a remedy as it is basic in nature, it neutralises excess acid in the stomach.