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3 years ago

Are producers heterotrophs or autotrophs?

Chemistry

1 answer:

cupoosta [38]3 years ago

7 0

Answer:

Autotrophs

Explanation:

Autotrophs make their own food, or produce food for themselves through the process of photosynthesis. Therefore, we can say that autotrophs are also producers. Some examples of producers are plants, algae, and some types of bacteria.

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Solid ammonium chloride, NH4Cl, is formed by the reaction of gaseous ammonia, NH3, and hydrogen chloride, HCl. NH3(g)+HCl(g)⟶NH4

Mashutka [201]

9.41 atm is the pressure in atmospheres of the gas remaining in the flask

<h3>What is the pressure in atmospheres?</h3>

The equation NH3(g) + HCl(g) ==> NH4Cl(s) is balanced.

Divide the moles of each reactant by its coefficient in the balanced equation, and the limiting reagent is identified as the one whose value is less. With the issue we now have...

6.44 g NH3 times 1 mol NH3/17 g equals 0.3688 moles of NH3 ( 1 = 0.3688)

HCl: 6.44 g of HCl times one mole of HCl every 36.5 g equals 0.1764 moles ( 1 = 0.1764). CONTROLLING REAGENT

NH4Cl will this reaction produce in grams

0.1764 moles of HCl multiplied by one mole of NH4Cl per mole of HCl results in 9.44 g of NH4Cl (3 sig. figs.)

the gas pressure, measured in atmospheres, that is still in the flask

NH3(g) plus HCl(g) results in NH4Cl (s)

0.3688......0.1764............0..........

Initial

-0.1764....-0.1764........+0.1764...Change

Equilibrium: 0.1924.......0...............+0.1924

There are 0.1924 moles of NH3 and no other gases in the flask. This is at a temperature of 25 °C (+273 = 298 °K) in a volume of 0.5 L. After that, we may determine the pressure by using the ideal gas law (P).

PV = nRT

P = nRT/V = 0.1924 mol, 0.0821 latm/mol, and 298 Kmol / 0.5 L

P = 9.41 atm

9.41 atm is the pressure in atmospheres of the gas remaining in the flask

To learn more about balanced equation refer to:

brainly.com/question/11904811

#SPJ1

7 0

1 year ago

Given the following equation: 2K + Cl2 -> 2KCl How many grams of KCl is produced from 4.00 g of K and excess Cl2?

Thepotemich [5.8K]

Answer:

42.65g

Explanation:

Given parameters:

Mass of K = 4g

Unknown: Mass of KCl

Solution:

Complete equation of the reaction:

2K + Cl₂ → 2KCl

To solve this problem, we know that the reactant in short supply is potassium K and this dictates the amount of products that would be formed. The chlorine gas is in excess and we can't use it to determine the amount of product that would form.

Now, we work from the known to the unknown. Since we know the mass of K given in the reaction, we can simply find the molar relationship between the reacting potassium and the product. We simply convert the mass to mole and compare to the product. From there we can find the mass of KCl that would be produced.

Calculating number of moles of K

Number of moles = $\frac{mass}{molar mass}$

Number of moles of K = $\frac{4}{39}$ = 0.103mol

From the given reaction equation:

2 moles of K will produce 2 moles of KCl

Therefore 0.103mol of K will produce 0.103mol of KCl

To find the mass of KCl produced,

Mass of KCl = number of moles of KCl x molar mass

Molar mass of KCl = 39 + 35.5 = 74.5gmol⁻¹

Mass of KCl = 0.103 x 74.5 = 42.65g

4 0

3 years ago

Read 2 more answers

An atom (not a hydrogen atom) absorbs a photon whose associated wavelength is 300 nm and then immediately emits photon whose ass

KIM [24]

Answer:

The net energy is 2.196 eV

Explanation:

Basically, the energy of an atom increases when it absorbs a photon. In addition, the wavelength of the emitted photon is longer such that the atom absorbed a net energy in the process.

Using:

ΔE = h*c*(1/λ $_{1}$ - 1/λ $_{2}$ )

where:

ΔE is the net energy in eV (electron-volt). 1 eV is equivalent to 1.602* $10^{-19}$ J.

h = 4.135* $10^{-15}$ eVs

c = 3* $10^{8}$ m/s

λ $_{1}$ = 300 nm = 300* $10^{-9}$ m

λ $_{2}$ = 640 nm = 640* $10^{-9}$ m

Thus:

ΔE = 4.135* $10^{-15}$ eVs*3* $10^{8}$ m/s*( $\frac{1}{300*10^{-9}m } }-\frac{1}{640*10^{-9}m }$ )

ΔE = 4.135* $10^{-15}$ *3* $10^{8}$ *1.77* $10^{6}$ eV = 2.196 eV

6 0

3 years ago

A 0.0372-m3 container is initially evacuated. Then, 4.65 g of water is placed in the container, and, after some time, all of the

Montano1993 [528]

Answer:

18.3 kilopascals

Explanation:

We are given that the volume of this container is 0.0372 meters^3, that the mass of water is 4.65 grams, and that the temperature of this water vapor ( over time ) is 368 degrees Kelvins. This is a problem where the ideal gas law is an " ideal " application.

_______________________________________________________

First calculate the number of moles present in the water ( H2O ). Water has a mass of 18, so it should be that n, in the ideal gas law - PV = nRT, is equal to 4 / 18. It is the amount of the substance.

We now have enough information to solve for P in PV = nRT,

P( 0.0372 ) = 4 / 18( 8.314 )( 368 ),

P ≈ 18,276.9

Pressure ≈ 18.3 kilopascals

<u><em>Hope that helps!</em></u>

5 0

3 years ago

In the titration of HCl with NaOH, the equivalence point is determined

kondaur [170]

Answer:

In the titration of HCl with NaOH, the equivalence point is determined from the point where the phenolphthalein turns pink and then remains pink on swirling.

Explanation:

The equivalence point is the point at which exactly enough titrant (NaOH) has been added to react with all of the analyte (HCl). Up to the equivalence point, the solution will be acidic because excess HCl remains in the flask.

Phenolphtalein is chosen because it changes color in a pH range between 8.3 – 10. Phenolphthalein is naturally colorless but turns pink in alkaline solutions. It remains colorless throughout the range of acidic pH levels, but it begins to turn pink at a pH level of 8.3 and continues to a bright purple in stronger alkalines.

It will appear pink in basic solutions and clear in acidic solutions.

The more NaOH added, the more pink it will be. (Until pH≈ 10)

In strongly basic solutions, phenolphthalein is converted to its In(OH)3− form, and its pink color undergoes a rather slow fading reaction and becomes completely colorless above 13.0 pH

a. from the point where the pink phenolphthalein turns colorless and then remains colorless on swirling.

⇒ the more colorless it turns, the more acid the solution. (More HCl than NaOH)

b. from the point where the phenolphthalein turns pink and then remains pink on swirling.

The equivalence point is the point where phenolphtalein turns pink and remains pink ( Between ph 8.3 and 10). (

Although, when there is hydrogen ions are in excess, the solution remains colorless. This begins slowely after ph= 10 and can be noticed around ph = 12-13

c. from the point where the pink phenolphthalein first turns colorless and then the pink reappears on swirling.

Phenolphthalein is colorless in acid solutions (HCl), and will only turn pink when adding a base like NaOH

d. from the point where the colorless phenolphthalein first turns pink and then disappears on swirling

Phenolphthalein is colorless in acid or neutral solutions. Once adding NaOH, the solution will turn pink. The point where the solution turns pink, and stays pink after swirling is called the equivalence point. When the pink color disappears on swirling, it means it's close to the equivalence point but not yet.

3 0

3 years ago