4 NH3 (g) + 7 O2 (g) 4 NO2 (g) + 6 H2O (g)
The combustion of 28.8 g of ammonia consumes __________ g of oxygen.

Answers

Answer 1

Answer:

21

Explanation:


Related Questions

Please answer I will give you brainliest!!

Answers

Answer:

Warm front

Explanation:

A warm front forms when a warm air mass pushes into a cooler air mass, shown in the image to the right (A). Warm fronts often bring stormy weather as the warm air mass at the surface rises above the cool air mass, making clouds and storms. Warm fronts move more slowly than cold fronts because it is more difficult for the warm air to push the cold, dense air across the Earth's surface. Warm fronts often form on the east side of low-pressure systems where warmer air from the south is pushed north.

You will often see high clouds like cirrus, cirrostratus, and middle clouds like altostratus ahead of a warm front. These clouds form in the warm air that is high above the cool air. As the front passes over an area, the clouds become lower, and rain is likely. There can be thunderstorms around the warm front if the air is unstable.

On weather maps, the surface location of a warm front is represented by a solid red line with red, filled-in semicircles along it, like in the map on the right (B). The semicircles indicate the direction that the front is moving. They are on the side of the line where the front is moving. Notice on the map that temperatures at ground level are cooler in front of the front than behind it.

How will an increase in wind speed affect soil erosion?

O A Soil erosion will increase.

O B. Soil erosion will decrease.

O C. Soil erosion will completely stop.

OD Soil erosion will remain the same.​

Answers

Answer:

A. Soil erosion will increase.

Option A is correct. Increase in wind speed will increase the rate of erosion.

The process of eroding, transferring, and depositing tiny particles as well as nutrients from the top layer of soil is known as soil erosion by wind.

In arid and semiarid regions, wind-driven soil erosion is a serious issue that impedes the environmental sustainability of animal husbandry and agriculture and threatens ecological security.

A wind that is blowing at 30 mph will erode at a rate that is more than 3 times faster than a wind that is blowing at 20 mph. As soil moisture rises, wind erosion declines. For instance, dry soil erodes approximately 1.3 times more quickly than soil with just enough moisture to support plant life.

Learn more about soil erosion;

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7. Which object would be the least dense in a tub of
water?
A. Golf ball
B. Glass marble
C. Rubber ball
D. Beach ball

Answers

The most dense one out of all of the answers would be the glass marble with a higher density than that of the golf ball and the rubber ball
Beach ball because it has less mass causing it to float


8. Which of the following elemental gases would have properties closest to an ideal gas?
a. hydrogen
argon
b. helium
d. fluorine

Answers

Answer:

Helium

Explanation:

Please tell me if it was right

At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1·M−1s−1 : →2SO3g+2SO2gO2g Suppose a vessel contains SO3 at a concentration of 1.44M . Calculate the concentration of SO3 in the vessel 0.240 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits.

Answers

Answer:

[tex][SO_3]=0.25M[/tex]

Explanation:

Hello there!

In this case, since the integrated rate law for a second-order reaction is:

[tex][SO_3]=\frac{[SO_3]_0}{1+kt[SO_3]_0}[/tex]

Thus, we plug in the initial concentration, rate constant and elapsed time to obtain:

[tex][SO_3]=\frac{1.44M}{1+14.1M^{-1}s^{-1}*0.240s*1.44M}\\\\[/tex]

[tex][SO_3]=0.25M[/tex]

Best regards!

During the workup portion of the reaction of alkenes with HBr as described in the experiment provided, a student transferred the reaction mixture to a separatory funnel, rinsed the reaction flask with diethyl ether, and added the ether rinses to the separatory funnel. The student then added sodium bicarbonate to the separatory funnel. Extremely vigorous bubbling occurred. What did the student do wrong

Answers

Answer:

Explanation:

Because of the acid-base reaction, as sodium bicarbonate is introduced to the separatory funnel, the additional or unreacted HBr reacts vigorously to yield CO2 gas, which exits the separatory funnel together with any dissolved compound(s) in the ether layer. This is due to a wrong and incorrect selection of the solvent mixture and the addition of sodium bicarbonate to an acidic solution.

Nothing to do with it until it has leaked out of the separatory funnel. Even then, the student may separate the components from the remaining reaction mixture by washing the ether coating layer several times with brine water, then running it into a dry sodium sulfate bed and evaporating solvent ether under decreased pressure.

PLZ HELP *NO LINKS*
1) How many moles of gaseous arsine (AsH3) occupy 0.372 L at STP?
In sig fig 4
2) What is the density of gaseous arsine?
In sig fig 4
Thanks!

Answers

Answer: (1). There are  0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP.

(2). The density of gaseous arsine is 3.45 g/L.

Explanation:

1). At STP the pressure is 1 atm and temperature is 273.15 K. So, using the ideal gas equation number of moles are calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

[tex]PV = nRT\\1 atm \times 0.372 L = n \times 0.0821 L atm/mol K \times 273.15 K\\n = 0.0165 mol[/tex]

2). As number of moles are also equal to mass of a substance divided by its molar mass.

So, number of moles of Arsine [tex](AsH_{3})[/tex] (molar mass = 77.95 g/mol) is as follows.

[tex]No. of moles = \frac{mass}{molar mass}\\0.0165 mol = \frac{mass}{77.95 g/mol}\\mass = 1.286 g[/tex]

Density is the mass of substance divided by its volume. Hence, density of arsine is calculated as follows.

[tex]Density = \frac{mass}{volume}\\= \frac{1.286 g}{0.372 L}\\= 3.45 g/L[/tex]

Thus, we can conclude that 0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP and the density of gaseous arsine is 3.45 g/L.

State the methods of preparing salts​

Answers

Answer:

Preparation of Salts by The Action of An Acid Upon a Metal.

Preparation of Salts by Double Decomposition.

Preparation of Salts by Neutralization.

Using A Soluble Base (alkali)

(b). Using An Insoluble Base.

Preparation of Salts by the Action of An Acid on The Trioxocarbonate (IV) of A Metal.

Explanation:


Genes influence an organism's traits by coding for:
A
Cells

Answers

Answer:

First, the protein may be a structural protein, contributing to the physical properties of cells or organisms. ... Second, the protein may be an enzyme that catalyzes one of the chemical reactions of the cell. Therefore, by coding for proteins, genes determine two important facets of biological structure and function.

Explanation:

i think it will help you

Thorium-236 has a half life of 10.0 minutes. If I have 250 grams of Thorium-236, How many half lives will have taken place after 50.0 minutes?

Answers

My 6.00 aped disks. Aosksbsd osidndd did eodnfnfb

What's you're favorite year?

Answers

Answer:

2020 cause of the lockdown

Explanation:

Why are some chemical substances, like oil and coal, considered nonrenewable?

the process that forms them stops working after a decade
the process that forms them has not worked for millions of years
the process that forms them is very fast
the process that forms them is very slow

Answers

Answer:

The process of the formation of coal and oil is a very gradual and slow process.

Explanation:

The process of the formation of coal and oil is a very gradual process that takes up to millions of years.

HELPP ASAP I WILL MARK BRAINIST


Answers

Answer:

I'm thinking ethier D or A

Explanation:

How many grams of potassium carbonate are needed to make 300ml of a 4.5M solution?

Answers

Answer:

186.3g

Explanation:

4.5moles of K₂CO₃  is in 1000ml

? moles of K₂CO₃ is in 300 ml

(4.5 × 300)/ 1000 = 1.35 moles of K₂CO₃

1 mole of K₂CO₃ = (39 × 2) + 12 + (16 × 3) = 78 + 12 + 48 = 138g

1.35 moles of K₂CO₃  = ?

= (1.35 × 138)/1 = 186.3g

Which of the following is an advantage of asexual reproduction compared to sexual reproduction?

Both will produce genetically identical offspring from the parent.

Sexual reproduction will increase genetic variability within a species.

Asexual reproduction requires less energy and will produce more offspring over time.

Sexual reproduction has minimal changes of mutations compared to asexual reproduction.

Answers

Explanation:

sexual reproduction has minimal changes of mutations compared to asexual reproduction

Answer:

sexual reproduction has minimal changes of mutations compared to asexual reproduction

I hope this helps

15.
A tank containing 173 grams of methane, CH (9), registers 15.1 atmospheres at 298 Kelvin. What is the volume of the tank (assuming the entire volume is available to the gas)?
A 3,410L
B. 213
C.O 175L
D. 280.
Chemistry 4/28 5454
Copyright © 2021 Illuminate Education, Inc. All Rig

Answers

Answer: The volume of tank is 17.5 L.

Explanation:

Given: Mass of methane = 173 g

Pressure = 15.1 atm

Temperature = 298 K

Molar mass of methane is 16.04 g/mol.

Therefore, moles of methane are calculated as follows.

[tex]No. of moles = \frac{mass}{molar mass}\\= \frac{173 g}{16.04 g/mol}\\= 10.78 mol[/tex]

Now, ideal gas equation is used to calculate the volume as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

[tex]PV = nRT\\15.1 atm \times V = 10.78 mol \times 0.0821 L atm/mol K \times 298 K\\V = \frac{10.78 mol \times 0.0821 L atm/mol K \times 298 K}{15.1 atm}\\= 17.5 L[/tex]

Thus, we can conclude that volume of the tank is 17.5 L.

What volume will 28 grams of nitrogen gas occupy at 27 Celsius and a
pressure of 785 mm Hg?

Answers

Answer: [tex]2.49\ m^3[/tex]

Explanation:

Given

Mass of nitrogen present [tex]m=28\ g[/tex]

Temperature [tex]T=27^{\circ}C\equiv 300\ K[/tex]

Pressure [tex]P=785\ mm\ \text{of}\ Hg\ \text{or}\ 1.032\ atm[/tex]

The molar mass of Nitrogen [tex]M=28\ g/mol[/tex]

No of moles of nitrogen present

[tex]n=\dfrac{m}{M}\\\\n=\dfrac{28}{28}\\\\n=1[/tex]

Using [tex]PV=nRT[/tex]

[tex]\Rightarrow 1.032\times V=1\times 8.314\times 300\\\\\Rightarrow V=\dfrac{2494.2}{1.032}\\\\\Rightarrow V=2494.2\ L\ \text{or}\ 2.49\ m^3[/tex]

How many grams are there in 1.8055 x 10^25 molecules of sodium sulfate? Hint: Convert to moles first

Answers

Answer:

4258.82 g of Na₂SO₄

Explanation:

From the question given above, the following data were obtained;

Number of molecules of Na₂SO₄ = 1.8055x10²⁵ molecules.

Number of mole of Na₂SO₄ =?

From Avogadro's hypothesis,

6.02×10²³ molecules = 1 mole

Therefore,

6.02×10²³ molecules = 1 mole of Na₂SO₄

Next, we shall determine the mass of 1 mole of Na₂SO₄. This can be obtained as follow:

1 mole of Na₂SO₄ = (23×2) + 32 + (16×4)

= 46 + 32 + 64

= 142 g

Thus,

6.02×10²³ molecules = 142 g of Na₂SO₄

Finally, we shall determine the mass of Na₂SO₄ that contains 1.8055x10²⁵ molecules. This can be obtained as follow:

6.02×10²³ molecules = 142 g of Na₂SO₄

Therefore,

1.8055x10²⁵ molecules

= (1.8055x10²⁵ × 142) / 6.02×10²³

= 4258.82 g of Na₂SO₄

Thus, 4258.82 g of Na₂SO₄ contains 1.8055x10²⁵ molecules

What is the volume, in liters, of 0.350 mol of nitrogen gas at 32°C and
0.980 atm of pressure? *
A. 9.85 L
B. 8.94 L
C. 104.6 L
D. 0.94 L

Answers

Answer: The volume is 8.94 L.

Explanation:

Given: no. of moles = 0.350 mol,   Pressure = 0.980 atm

Temperature = [tex]32^{o}C = (32 + 273) K = 305 K[/tex]

Formula used to calculate the volume is as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

[tex]PV = nRT\\0.980 atm \times V = 0.350 mol \times 0.0821 L atm/mol K \times 305 K\\V = 8.94 L[/tex]

Thus, we can conclude that the volume is 8.94 L.

a) How do you prepare %3 (w/v) Na2CO3 solution from Na2CO3⸱2H2O? (15p) Na2CO3 MW=106 g/mol

Answers

Answer:

4.02g of Na2CO3⸱2H2O must be added completing the volume of the solution to 100mL

Explanation:

A 3%(w/v) solution contains 3g of solute (In this case, Na2CO3) in 100mL of solution.

Assuming we require 100mL of solution we must add 3g of Na2CO3. The reactant that is available is its dihydrate, with molar mass:

106g/mol + 2*MW H2O

106g/mol + 2*18g/mol = 142g/mol

That means the mass of Na2CO3.2H2O that must be added to prepare the solution is:

3g Na2CO3 * (142g/mol Na2CO3.2H2O / 106g/mol Na2CO3) =

4.02g of Na2CO3⸱2H2O must be added completing the volume of the solution to 100mL

How to pass chem!!!!!!!!!!

Answers

Answer:

refer to below

Explanation:

- read your textbook! especially the topics that are most confusing

- as you read, make sure to take notes/make an outline. write the words from the text in your own words. In chem, the textbook explanations are usually very detailed, so putting it in your own words will help you remember and understand what you write.

- make sure you really understand the concepts. don't just memorize what you read.

- draw diagrams and structures. seeing how things are structured will help you visualize what you learn

- ask questions to your teacher. if there is anything confusing, make sure to refer to your teacher as they will hopefully be able to explain it so  it won't be confusing

These tips have helped me pass chem with an A. Hope all of this helped!

Answer:

Explanation:

Study material, ask for help, read textbook and do practice problems, get a tutor, beg your teacher for marks,

Introduction: Reaction rates are also influenced by surface area and concentration. The surface area of a solid is a measure of how much of the solid is exposed to other substances. The concentration of a substance is a measure of how many molecules of that substance are present in a given volume. Question: How do surface area and concentration affect reaction rates

Answers

Answer:

See explanation

Explanation:

Surface area has to do with the number of solid particles that are exposed at a given time and is capable of colliding with other reactant particles. When more surface area is exposed for reaction, then it means that more particles are likely to collide with each other leading to faster chemical reaction rates. When few particles are exposed for reaction (low surface area) then less collisions occur and the rate of reaction is decreased.

Similarly, concentration refers to the amount of substance present. The greater the amount of substance present, the greater the likelihood of collision between particles and the greater the rate of reaction and vice versa.

At 47c a gas has a pressure of 140kpa. The gas is cooled until the pressure decreases to 105kpa. If the volume remains constant, what will the final temperature be in kelvin’s? In degrees Celsius

Answers

Answer:

The final temperature is equal to 240 K or -33.15°C

Explanation:

Given that,

Initial temperature of the gas, T₁ = 47°C = 320 K

Initial pressure, P₁ = 140 kpa

Final pressure, P₂ = 105 kpa

We need to find the final temperature if the volume remains constant.  The relation between temperature and pressure is given by :

[tex]P\propto T[/tex]

or

[tex]\dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}\\\\T_2=\dfrac{P_2T_1}{P_1}\\\\T_2=\dfrac{105\times 320}{140}\\\\T_2=240\ K\\\\T_2=-33.15^{\circ} C[/tex]

So, the final temperature is equal to 240 K or -33.15°C.

3. How many times does earth rotate on its axis in one year?

Answers

Answer:

There are approximately 366.25 sidereal days in a year so that the Earth spins 366.25 times with respect to distant stars in a year

hope this helps

have a good day :)

Explanation:

Jerry is trying to classify cells by their physical characteristics. He discovers a multicellular organism containing cells that have a nucleus and a cell wall as well as the ability to conduct photosynthesis. Into which of the three domains would this organism most likely fit? A. Archaea B. bacteria C. Eukarya D. Viral​

Answers

Turn on will turn on will turn on both turn on mobile

Domains eukarya  would this organism most likely fit.

The domain eukarya comprised of eukaryotes or organisms whose cells contain true nucleus.

What is a domain?

It is  the largest of all groups in the classification of life. There are three domains:-

Archaea domainBacteria domainEukarya domainWhat is Eukarya?

It is the domain of organism called eukaryotes. These are the organism who have a well defined nucleus and membrane bound organelles.

Hence, C) option is correct.

To know more about domain here

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What happens to the entropy when a solution is made?
A. The entropy increases.
B. The entropy decreases.
C. The entropy goes to zero.
D. The entropy is unaffected.

Answers

Answer:

The entropy increases

Explanation:

Just took the quiz

is scandium a transition metal?​

Answers

Answer:no

Explanation:

Answer:

Scandium is a transition metal

Explanation:

Consider the balanced equation Zn + 2HCl ZnCl2 + H2 How many
moles of ZnCl2 will be produced if 2 moles of HCl are used?

Answers

Answer:

1 mole of ZnCl₂

Explanation:

Just from the stoichiometric equation/ balanced equation:

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

            1 mole   2 moles    1 mole       1 mole

Therefore: 2 moles of 2HCl produce 1 mole of ZnCl₂

t-Butyl alcohol was produced by the liquid-phase hydration (using water, W) of isobutene (I) over an Amberlyst-15 catalyst.

a. True
b. False

Answers

Answer:

True

Explanation:

This is because, The hydroboration oxidation of an alkene which is isobutene in the presence of a catalyst will result to alcohol as the product . Therefore, the OH group will attach or link itself to the carbon which is less obstructed. Thus this reaction is in accordance to Anti-Markownikoff's rule.

So isobutene on hydroboration oxidation will produce ter isobutyl alcohol.

Bears stop coming to a river ecosystem where they have been eating many fish each day. The fish the bears eat normally eat smaller fish, which eat plants along the river bottom.

What happens to the ecosystem?


Both the larger and the smaller fish populations grow quickly but then die out because the plant life is insufficient for them all to eat.

The larger fish population will drop first, and the smaller fish population will grow quickly. The plants will die off because too many of the smaller fish are eating them.

The larger fish population explodes at first, and the smaller fish population begins to drop. Eventually, the river runs out of smaller fish so larger fish die out, and the plant population grows.

The smaller fish population begins to eat more plants and to grow. The larger fish have more food to eat so their population is able to grow, too.

Answers

Answer:

The larger fish population explodes at first, and the smaller fish population begins to drop. Eventually, the river runs out of smaller fish so larger fish die out, and the plant population grows.

Explanation:

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