Total solar eclipse where only the corona is visible is called?

Calcule a variação da entalpia dessa reação ( 2 NH3 (g) ---> CO(NH2)2 (s) + H2O (L) ) a partir das seguintes equações termoqu

Nitella [24]

ΔH = +438 kJ

We have three equations:

(I) N₂ + 3H₂ → 2NH₃; ΔH = -92 kJ

(II) H₂ +½O₂ → H₂O; ΔH = -286 kJ

(III) CO(NH₂)₂ + ³/₂O₂ → CO₂ + 2H₂O + N₂; ΔH = -632 kJ

From these, we must devise the target equation:

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; ΔH = ?

_________________________________

The target equation has 2NH₃ on the left, so you reverse equation (I).

When you reverse an equation, you reverse the sign of its ΔH.

(V) 2NH₃ → N₂ + 3H₂; ΔH = +92 kJ

Equation (V) has 1N₂ on the right, and that is not in the target equation.

You need an equation with 1N₂ on the left.

Reverse Equation (III).

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; ΔH = +632 kJ

Equation (VI) has ³/₂O₂ on the right, and that is not in the target equation.

You need ³/₂O₂ on the left.

Multiply Equation (II) by three.

When you multiply an equation by three, you multiply its ΔH by three.

(VII) 3H₂ +³/₂O₂ → 3H₂O; ΔH = -286 kJ

Now, you add equations (V), (VI), and (VII), cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

_______________________________________

We get the target equation (IV):

(V) 2NH₃ → N₂ + 3H₂; ΔH = + 92 kJ

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; ΔH = +632 kJ

(VII) 3H₂ +³/₂O₂ → 3H₂O; ΔH = -286 kJ

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; ΔH = +438 kJ

7 0

2 years ago

Consider the following reaction where Kc = 34.5 at 1150 K:2SO2(g) + O2(g) 2SO3(g)A reaction mixture was found to contain 4.77×10

sukhopar [10]

Answer:

Explanation:

[ so₃] = 4.37 x 10⁻²

[so₂] = 4.77 x 10⁻²

[ o₂] = 4.55 x 10⁻²

Qc = (4.37)²x10⁻⁴ /(4.77)².(4.55) x 10⁻⁶ =18.44

Qc is less than Kc hence in order to reach equilibrium more of so₃ will be produced . Statement 1 is true.

Kc is always constant . Statement 2 is false.

Statement 3 is false because statement 1 is true.

Qc Is smaller than Kc . So statement 4 is false.

The reaction is not in equilibrium. Statement 5 is false.

3 0

3 years ago

) The cooling or warming of air that occurs because air is allowed to expand or is compressed, not because it is added or subtra

Pani-rosa [81]

Answer:

Adiabatic temperature changes.

Explanation:

Temperature changes related to changes of pressure without external gain or loss of heat. If no heat is added or lost to the surroundings, then when an air parcel rises and expands, its temperature drops. Conversely, when the parcel is compressed, its temperature rises. So the term adiabatic, implies a change in temperature of the air parcel without gain or loss of heat from outside the air parcel.

3 0

3 years ago

A 30.g sample of Al(s) is heated to 50°C and placed in a calorimeter containing 150g of water at 20°C. As the system approaches

SpyIntel [72]

Energy is transferred from the Al (solid) to water and the temperature of the water increases while the temperature of the Al (solid) decreases.

<h3>Heat energy</h3>

Heat energy is a form of energy transferred from one body to another due to a temperature difference between the bodies.

Heat energy is also known as thermal energy.

In heat transfer, heat is transferred from a body at higher temperature to a body at lower temperature.

In the heat transfer between the heated aluminium and the water, the solid aluminium is at a higher temperature than the water, hence, heat energy is transferred from the aluminium to water.

Therefore heat energy is transferred from the Al (solid) to water and the temperature of the water increases while the temperature of the Al (solid) decreases.

Learn more about about heat transfer at: brainly.com/question/12072129

5 0

1 year ago

If wind is eroding an area at a rate of 2 mm per year and depositing it in a smaller area at a rate of 7 mm per year, how much l

amid [387]

Answer:

After 2000 years, the first after will be 4m lower.

After 2000 years, the second area will be 14m higher.

Explanation:

From the given information:

At the first area:

if the wind is eroding at 2 mm/year, then after 2000 years; we have:

2 mm/year × 2000 year = 4000 mm

4000 mm to meters, we have:

$=4000 \ mm \times \dfrac{1 \ m}{ 1000 \ mm}$

= 4 m lower

At the second smaller area:

Given that the wind is depositing it at the rate of 7 mm per year;

Then:

7 mm/year × 2000 year = 14000 mm

$=14000 \ mm \times \dfrac{1 \ m}{ 1000 \ mm}$

=14 m higher

6 0

3 years ago