How many moles of O₂ are needed to react completely with 35.0 mol of FeCl₃?


4FeCl3+3O2->2Fe2O3+3Cl2


The 2 for the second half of the equation is not a coefficient it’s supposed to be for 2Fe2 Not 2O3.

26.3 mol

46.7 mol

23.3 mol

10.0 mol

Answer:lmk

Explanation:

Answer:

steel, plastics

Hope it helps!

Answer:

Explanation:

To solve this problem we have to find the moles of iron(II) chloride that react. Using the chemical equation, we can kknow moles FeCl2 = Moles FeCO3. Thus, we can find the moles of FeCO3. Converting these moles to grams using its molar mass -Molar mass FeCO3: 115.854g/mol-

Moles FeCl2 = Moles FeCO3:

1.24L * (2.00mol / L) = 2.48 moles FeCl2

Mass FeCO3:

2.48mol * (115.854g / mol) =

Answer:

its 287

Explanation:

i got it right on edm

A car engine produces 604 kJ of mechanical energy from fuel capable of producing 2416 kJ of energy, then efficiency of the engine is 25%.

We can calculate the efficiency by using the below equation as:

Efficiency = (Actual production / Theoretical production) × 100%

Given that,

Actual production = 604 kJ

Theoretical production = 2,416 kJ

On putting values in given equation, we get

Efficiency = (604/2,416) × 100% = 25%

Hence efficiency of the engine is 25%.

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Explanation:

Here is why: If the atmosphere contains too much of these gases, the whole Earth becomes a hotter and hotter greenhouse. The atmosphere holds onto too much of the heat at night instead of letting it escape into space. Then, the next day, the Sun heats Earth's surface even more.

Answer:

2.14 Na2S + 1.99 HCl ----> 3.20 NaCl + 0.935 H2S

Explanation:

The Balanced equation is

Na2S + 2HCl = 2NaCl + H2S

As we can see -

78.0452 g/mol of sodium sulfide reacts with 2 mole of HCl.

Mass of one mole of HCl = 36.458 g/mol

So, 78.0452 g/mol of sodium sulfide reacts with 2* 36.458 g/mol of HCL to produce 2* 58.44 g of NaCl and 34.1 g/mol of H2S

Thus,

2.14g of sodium sulfide will react with (2* 36.458/78.0452)* 2.14 of HCl

2.14g of sodium sulfide will react with 1.99 g of HCl

2.14 Na2S + 1.99 HCl ----> (2* 58.44/78.0452) *2.14 NaCl + (34.1/78.0452) * 2.14 H2S

2.14 Na2S + 1.99 HCl ----> 3.20 NaCl + 0.935 H2S

Answer:

Q = 1455.12 Joules.

Explanation:

Given the following data;

Mass = 300 grams

Initial temperature = 22.3

Final temperature = 59.9°C

Specific heat capacity = 0.129 J/gºC.

To find the quantity of energy;

[tex] Q = mcdt[/tex]

Where,

Q represents the heat capacity.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt represents the change in temperature.

dt = T2 - T1

dt = 59.9 - 22.3

dt = 37.6°C

Substituting the values into the equation, we have;

[tex] Q = 300*0.129*37.6 [/tex]

Q = 1455.12 Joules.

Answer:

Acoustic survey data can help inform scientists about the locations of the rainforest with the highest biodiversity. ... By working with these companies, The Nature Conservancy can help inform them of where they might try to create wildlife corridors or connected habitats on their lands

Answer:

Density= 3.68g/cm³

The object will sink in water

Explanation:

D=m/v

=350/95

=3.684

≈3.68g/cm³

Water= 1g/cm³

The object will sink in water because the object is denser than the water.

3.684g/cm³ is the density of an object that has a mass of 350 g and a volume of 95 cm³. The object will sink in water.

The density of a material (volumetric density and specific mass) is its mass per unit volume. The most common sign for density is however the Roman letter D can be used as well. Density is defined mathematically as mass divided with volume.

The density of a pure material has the same numeric values just like its mass concentration. Various materials have varied densities, and density can be important in terms of buoyancy, purity, and packing.

D=m/v

=350/95

=3.684g/cm³

The object will sink in water

Therefore, 3.684g/cm³ is the density of an object that has a mass of 350 g and a volume of 95 cm³. The object will sink in water.

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Answer:

C. Transpiration, evaporation, and sublimation.

Explanation:

thank you to the internet.

Answer:

Hydrologists study how water interacts with the earth's crust. For example, they may study how rainfall and snowfall cause erosion, create caves, percolate through soil and rock to become groundwater, or eventually reach the sea.Hydrologists strive to improve water quality and increase our access to water so that we can continue to make use of it in all the ways that are necessary to our lives. A hydrologist measures the stream flow in a tributary to the Coeur d'Alene River. layers of gases surrounding a planet or other celestial body

Answer:

Tetrahedral

Explanation:

Answer:

question two is most likely you can describe the shape as a twisted ladder, thats what I learned in school.

Explanation:

YW

Answer:

The empirical formula of the compound is [tex]C_{0.504}HO_{1.008}[/tex].

Explanation:

We need to determine the empirical formula in its simplest form, where hydrogen ([tex]H[/tex]) is scaled up to a mole, since it has the molar mass, and both carbon ([tex]C[/tex]) and oxygen ([tex]O[/tex]) are also scaled up in the same magnitude. The empirical formula is of the form:

[tex]C_{x}HO_{y}[/tex]

Where [tex]x[/tex], [tex]y[/tex] are the number of moles of the carbon and oxygen, respectively.

The scale factor ([tex]r[/tex]), no unit, is calculated by the following formula:

[tex]r = \frac{M_{H}}{m_{H}}[/tex] (1)

Where:

[tex]m_{H}[/tex] - Mass of hydrogen, in grams.

[tex]M_{H}[/tex] - Molar mass of hydrogen, in grams per mole.

If we know that  [tex]M_{H} = 1.008\,\frac{g}{mol}[/tex] and [tex]m_{H} = 0.2\,g[/tex], then the scale factor is:

[tex]r = \frac{1.008}{0.2}[/tex]

[tex]r = 5.04[/tex]

The molar masses of carbon ([tex]M_{C}[/tex]) and oxygen ([tex]M_{O}[/tex]) are [tex]12.011\,\frac{g}{mol}[/tex] and [tex]15.999\,\frac{g}{mol}[/tex], then, the respective numbers of moles are: ([tex]r = 5.04[/tex], [tex]m_{C} = 1.2\,g[/tex], [tex]m_{O} = 3.2\,g[/tex])

Carbon

[tex]n_{C} = \frac{r\cdot m_{C}}{M_{C}}[/tex] (2)

[tex]n_{C} = \frac{(5.04)\cdot (1.2\,g)}{12.011\,\frac{g}{mol} }[/tex]

[tex]n_{C} = 0.504\,moles[/tex]

Oxygen

[tex]n_{O} = \frac{r\cdot m_{O}}{M_{O}}[/tex] (3)

[tex]n_{O} = \frac{(5.04)\cdot (3.2\,g)}{15.999\,\frac{g}{mol} }[/tex]

[tex]n_{O} = 1.008\,moles[/tex]

Hence, the empirical formula of the compound is [tex]C_{0.504}HO_{1.008}[/tex].

The type of molecule which is shown in the given question is alcohol, due to presence of hydroxyl group.

Funcional groups are those specific groups which after attaching to any compound changes the property of that compound and then that compound will known by the propeties of that functional group.

Given molecule has the molecular structure CH₃-CH₂-OH, means in this molecule hydroxyl group is present which defines the general properties of alcohol.

Hence given molecule is an alcohol.

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Answer:

2,3-dimetylpentane

Explanation:

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Answer:

one

Explanation:

Answer:

[tex]P \approx 225atm[/tex]

Explanation:

From the question we are told that:

Moles of sample n=3.20_mol

Volume V=350mL

Temperature T=300k

Generally the equation for ideal gas is mathematically given by

 [tex]PV=nRT[/tex]

 [tex]P=\frac{nRT}{V}[/tex]

 [tex]P=\frac{3.20*0.08206*300}{350*10^{-3}}[/tex]

 [tex]P=225.079atm[/tex]

 [tex]P \approx 225atm[/tex]

Answer:

C, when the ball lands it will have the least amount potential energy.

Explanation:

here is a picture that should help you.

Answer:

27.5°C = Final Temperature of the metal

Explanation:

The change in temperature using the specific heat of a material could be obtained using the formula:

Q = m*S*ΔT

Where Q is heat absorbed in Joules = 128J in the problem

m is the mass of the substance = 10.5g

S is Specific Heat of the substance = 0.449J/g°C for Iron

ΔT is change in temperature = Final T - Initial T

Replacing:

128J = 10.5g*0.449J/g°C*ΔT

2.5°C = ΔT

2.5°C = Final T - 25°C

Answer:

47730 grams.

Explanation:

First we convert 1.752x10²⁶ formula units of Ca(NO₃)₂ into moles, using Avogadro's number:

Then we convert 290.88 moles of Ca(NO₃)₂ into grams, using the molar mass of Ca(NO₃)₂: