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2 years ago

Hi no Link please

Chemistry

1 answer:

d1i1m1o1n [39]2 years ago

5 0

Answer:

Tell him u like him and be respect his idea of what he say it’s a 50/50 chance he likes u. I really believe u can do it.

Explanation:

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I was checking my answers for a chem test and came across a solution for a problem where the ∆S was pos, ∆H was neg and ∆G was n

nevsk [136]

Answer:

there is only 15 points

Explanation:

The second law of thermodynamics says that the entropy of the universe always increases for a spontaneous process: \Delta \text {S}_{\text{universe}}=\Delta \text {S}_{\text{system}} + \Delta \text {S}_{\text{surroundings}} > 0ΔS

universe

=ΔS

system

+ΔS

surroundings

>0delta, start text, S, end text, start subscript, start text, u, n, i, v, e, r, s, e, end text, end subscript, equals, delta, start text, S, end text, start subscript, start text, s, y, s, t, e, m, end text, end subscript, plus, delta, start text, S, end text, start subscript, start text, s, u, r, r, o, u, n, d, i, n, g, s, end text, end subscript, is greater than, 0

At constant temperature and pressure, the change in Gibbs free energy is defined as \Delta \text G = \Delta \text H - \text{T}\Delta \text SΔG=ΔH−TΔSdelta, start text, G, end text, equals, delta, start text, H, end text, minus, start text, T, end text, delta, start text, S, end text.

When \Delta \text GΔGdelta, start text, G, end text is negative, a process will proceed spontaneously and is referred to as exergonic.

The spontaneity of a process can depend on the temperature.

Spontaneous processes

In chemistry, a spontaneous processes is one that occurs without the addition of external energy. A spontaneous process may take place quickly or slowly, because spontaneity is not related to kinetics or reaction rate. A classic example is the process of carbon in the form of a diamond turning into graphite, which can be written as the following reaction:

8 0

3 years ago

The Haber Process synthesizes ammonia at elevated temperatures and pressures. Suppose you combine 1580 L of nitrogen gas and 351

ikadub [295]

Answer : The volume of reactant measured at STP left over is 409.9 L

Explanation :

First we have to calculate the moles of $N_2$ and $H_2$ by using ideal gas equation.

$PV_{N_2}=n_{N_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $N_2$ gas = 1580 L

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 1580L=n_{N_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{N_2}=70.49mole$

$PV_{H_2}=n_{H_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $H_2$ gas = 3510 L

n = number of moles $H_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 3510L=n_{H_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{H_2}=156.6mole$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

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

From the balanced reaction we conclude that

As, 3 mole of $H_2$ react with 1 mole of $N_2$

So, 156.6 moles of $H_2$ react with $\frac{156.6}{3}\times 1=52.2$ moles of $N_2$

From this we conclude that, $N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the excess moles of $N_2$ reactant (unreacted gas).

Excess moles of $N_2$ reactant = 70.49 - 52.2 = 18.29 moles

Now we have to calculate the volume of reactant, measured at STP, is left over.

$PV=nRT$