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3 years ago

What are the symbol and units for Enthaply

Chemistry

1 answer:

NISA [10]3 years ago

3 0

Answer:

The SI unit for specific enthalpy is joule per kilogram. It can be expressed in other specific quantities by h = u + pv, where u is the specific internal energy, p is the pressure, and v is specific volume, which is equal to 1ρ, where ρ is the density.

Explanation:

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How does an open system differ from a closed system

Len [333]

Explanation:

A system can be either closed or open; A closed system is a system that is completely isolated from its environment. An open system is a system that has flows of information, energy, and/or matter between the system and its environment, and which adapts to the exchange.

7 0

3 years ago

Of the following equilibria, only ________ will shift to the right in response to a decrease in volume. N2 (g) + 3H2 (g) 2NH3 (g

igomit [66]

Answer:

Of the following equilibria, only one will shift to the right in response to a decrease in volume.

$N_2 (g) + 3H_2 (g)\rightleftharpoons 2NH_3 (g)$

On decreasing the volume the equilibrium will shift in right direction due to less number of gaseous moles on product side.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Decrease the volume

If the volume of the container is decreased , the pressure will increase according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place. So, the equilibrium will shift in the direction number of gaseous moles are less.

$N_2 (g) + 3H_2 (g)\rightleftharpoons 2NH_3 (g)$

On decreasing the volume the equilibrium will shift in right direction due to less number of gaseous moles on product side.

$2 Fe_2O_3 (s) \rightleftharpoons 4 Fe (s) + 3O_2 (g)$

On decreasing the volume the equilibrium will shift in left direction due to less number of gaseous moles on reactant side.

$2 SO_3 (g) \rightleftharpoons 2 SO_2 (g) + O_2 (g)$

On decreasing the volume the equilibrium will shift in left direction due to less number of gaseous moles on reactant side.

$H_2 (g) + Cl_2 (g) \rightleftharpoons 2 HCl (g)$

On decreasing the volume the equilibrium will shift in no direction due to same number of gaseous moles on both sides.

$2HI (g) \rightleftharpoons H_2 (g) + I_2 (g)$

On decreasing the volume the equilibrium will shift in no direction due to same number of gaseous moles on both sides.

8 0

3 years ago

What adaptations of the harpy eagle allow it to be such a successful rainforest predator?

IgorC [24]

The ranking of most eagles in the food chain is at the top. They are called apex predator.

The harpy eagle is one of the largest eagle there is. Common to most eagles are its sharpness of their vision. Most eagles use it to sight their predator while circling above the ground at a distance unseen by the prey. However, harpy eagles adapted a way of hunting prey which is called perch-hunting.

Perch-hunting is a way of looking for predator by scanning the trees for prey by flying in short distances from a tree to another tree. The harpy eagle developed this skill since it is hard to sight a prey in a rain forest because of the leaves and trees blocking its vision. Hence, perch-hunting is very apt for the scenario of a rain forest eagle.

3 0

3 years ago

If 30.0 mL of a 12.0 M hydrochloric acid (HCI) stock solution is diluted to a volume of 500.0 mL,

klemol [59]

Answer:

The concentration of dilute solution is 0.72M

Explanation:

According to the given question

V1=30 ml

S1 = 12 M

V2= 500 ml

then S2 =?

we all know that V1S1=V2S2

or S2= V1S1÷V2

or S2 = 30×12÷500

or S2= 360÷500

or S2 =0.72 M.

The concentration of dilute solution is 0.72 M

5 0

3 years ago

PLEASE ANSWER!

Marianna [84]

Answer:

0.7atm

Explanation:

Given parameters:

Initial temperature = 25.2°C in Kelvin; 25.2 + 273 = 298.2K

Initial pressure = 0.6atm

Final temperature = 72.4°C in kelvin = 72.4 + 273 = 345.4K

Unknown:

Final pressure = ?

Solution:

Since we are dealing with pressure temperature relationships under a fixed volume, we use a simplification of the combined gas law to solve this problem.

At fixed volume;

$\frac{P_{1} }{T_{1} } = \frac{P_{2} }{_T{2} }$

where P and T are temperature values

1 and 2 are the initial and final states

Input the parameters and solve for P₂

$\frac{0.6}{298.2} = \frac{P_{2} }{345.4}$

P₂ = 0.7atm

3 0

3 years ago