Which of the following is not a property of water?

I need help with this, please :00000

mojhsa [17]

theoretical yield of the reaction is 121.38 g of NH₃ (ammonia)

limiting reactant is N₂ (nitrogen)

excess reactant is H₂ (hydrogen)

Explanation:

We have the following chemical reaction:

N₂ + 3 H₂ → 2 NH₃

Now we calculate the number of moles of each reactant:

number of moles = mass / molar weight

number of moles of N₂ = 100 / 28 = 3.57 moles

number of moles of H₂ = 100 / 2 = 50 moles

From the chemical reaction we see that 3 moles of H₂ are reacting with 1 moles of N₂, so 50 moles of H₂ are reacting with 16.66 moles of N₂ but we only have 3.57 moles of N₂ available, so the limiting reactant will be N₂ and the excess reactant will be H₂.

Knowing the chemical reaction and the limiting reactant we devise the following reasoning:

if 1 mole of N₂ produce 2 moles of NH₃

then 3.57 moles of N₂ produce X moles of NH₃

X = (3.57 × 2) / 1 = 7.14 moles of NH₃

mass = number of moles × molar weight

mass of NH₃ = 7.14 × 17 = 121.38 g

theoretical yield of the reaction is 121.38 g of NH₃

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limiting reactant

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4 0

4 years ago

The diameter of the He He atom is approximately 0.10 nm nm . Calculate the density of the He atom in g/cm 3 g/cm3 (assuming that

Sladkaya [172]

Answer:

Density of the He atom = 12.69 g/cm³

Explanation:

From the information given:

Since 1 mole of an atom = 6.022x 10²³ atoms)

1 atom of He = $1 \ atom \times (\dfrac{1 \ mole}{ 6.022 \times 10^{23} \ atoms}) \times ( \dfrac{4.003 \ grams}{ 1 \ mole})$

$=6.647 \times 10^{-24} \ grams$

The volume can be determined as folows:

since the diameter of the He atom is approximately 0.10 nm

the radius of the He = $\dfrac{0.10}{2}$ = 0.05 nm

Converting it into cm, we have:

$0.05 nm \times \dfrac{10^{-9} \ meters}{ 1 nm} \times \dfrac{ 1 cm }{10^{-2} \ meters}$

$=5 \times 10^{-9} \ cm$

Assuming that it is a sphere, the volume of a sphere is

= $\dfrac{4}{3}\pi r^3$

= $\dfrac{4}{3}\pi \times (5\times 10^{-9})^3$

= $5.236 \times 10^{-25} \ cm^3$

Finally, the density can be calcuated by using the formula :

$Density = \dfrac{mass}{volume}$

$D = \dfrac{6.647 \times 10^{-24} \ grams }{ 5.236 \times 10^{-25} \ cm^3}$

D = 12.69 g/cm³

Density of the He atom = 12.69 g/cm³

4 0

3 years ago

Read 2 more answers

The Doppler effect related change in wave frequency to what?

Irina18 [472]

Answer:

Relative motion of source and observer.

Explanation:

Doppler Effect : It is the change in the observed frequency of the wave due to relative motion of the source and object .

8 0

3 years ago

Which of the following would release the most heat? Assume the same mass of in each case. Specific heats of ice, liquid water, a

lesya692 [45]

Answer:

The process which releases most heat is E)

Explanation:

As we know that water freezes at 0ºC and vaporizes at 100ºC, we calculate the heat as follows:

Processes with temperatures < 0ºC : by using specific heat of ice (Sh ice) multiplied by the change in temperature (ΔT= Final Temperature - Initial Temperature)⇒ Sh ice x ΔT

Processes of ice melting (at 0ºC): by using heat of fusion of ice (ΔH fus) multiplied by a conversor factor (1 mol H20= 18 g)⇒ ΔHfus x 1mol/18g

Processes between 0ºC and 100ºC: by using specific heat of liquid water (Sh liq) multiplied by change in temperature ⇒ Sh liq x ΔT

Processes of water evaporation (at 100ºC): by using heat of vaporization (ΔH vap) multiplied by the conversor factor ⇒ ΔH vap x 1mol/18 g

Processes at a temperature >100ºC: by using specific heat of water vapor (Sh vap) multiplied by the change in temperature ⇒ Sh vap x ΔT

A) Water at -25ºC is ice. Ice is heated from -25ºC to 0ºC, then it melts at 0ºC (ice became liquid water) and then liquid water is heated from 0ºC to 70ºC. T

This is the only process in with the heat is absorbed (not releases), so it cannot be the right answer, but we calculate the heat involved to practice:

Heat= (Sh ice x ΔT) + (ΔH fus x 1/18 g) + Sh liq x ΔT

Heat= (2.05 J/g ºC x (0ºC -(-25ºC) ) + (6.01 x 10³ J/mol x 1 mol/18 g) + (4.18 J/g ºC x (70ºC-0ºC)

Heat= 51.25 J + 333,8 J +292.6 J

Heat= 677.65 J (heat is absorbed)

B) Water is cooled from 13ºC to 0ºC, then it is freezed at 0ºC and then the ice is cooled from 0ºC to -2.6 ºC

Heat= (Sh liq x ΔT) + (-ΔH melt x 1/18 g) + (Sh ice x ΔT)

Heat= 4.18 J/ºC x (0ºC- 13ºC) + (-6.01 x 10³ J/mol x 1mol/18 g) + (2.05 J/ºC x (-2.5ºc - 0ºC)

Heat= -54.34 J - 333.8 J + 5.33 J

Heat= -393.47 J (heat is released)

C) Liquid water is cooled from 74ºC to 95ºC

Heat= Sh liq x ΔT

Heat= 4.18 J/ºC x (74ºC - 95ºC)

Heat = -87.78 J (heat is released)

D) Water at 140ºC is in vapor state. Vapor at 140ºC is cooled to 110ºC (still vapor).

Heat = Sh vap x ΔT

Heat= 2.01 J/ºC x (110ºC - 140ºC)

Heat= -60.3 J (heat is released)

E) Vapor at 106ºC is cooled to 100ºC, then it condenses at 100ºC (convertion from gas to liquid), and liquid water is cooled from 100ºC to 88ºC.

Heat= (Sh vap x ΔT) + (-ΔHvap x 1mol/18g) + (Sh liq x ΔT)

Heat= (2.01 J/ºC x (100ºC-106ºC)) - (40.7 x 10³ J/mol x 1mol/18 g) + (4.18 J/ºC x (88ºC -100ºC)

Heat= -2323.32 J (heat is released) THIS IS THE RIGHT ANSWER (the more negative= the more released)

7 0

3 years ago

Which of the following is true of science? (2 points)

Tju [1.3M]

Answer:

It is based on testable and replicable evidence.

7 0

3 years ago