The MOST important biotic factor in this swamp would be the

Nina [5.8K]

It is C It is C It is C

3 0

3 years ago

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r

weqwewe [10]

Answer:

Therefore $\bigtriangledown H^\circ_{rxn}$ = -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by $\bigtriangledown H^\circ_{rxn}$

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s) $\bigtriangledown H^\circ_{rxn}$ =?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1) $\bigtriangledown H^\circ_{rxn}$ = +157KJ

P₄(g)+6Cl₂(g)→ 4PCl₃(g).............(2) $\bigtriangledown H^\circ_{rxn}$ = -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3) $\bigtriangledown H^\circ_{rxn}$ = -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4) $\bigtriangledown H^\circ_{rxn}$ =4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s) $\bigtriangledown H^\circ_{rxn}$ =( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s) $\bigtriangledown H^\circ_{rxn}$ = - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s) $\bigtriangledown H^\circ_{rxn}$ = -1835 KJ

Therefore $\bigtriangledown H^\circ_{rxn}$ = -1835 KJ

5 0

3 years ago

How does ionic radii increase

nalin [4]

<>"One such trend is closely linked to atomic radii -- ionic radii. Neutral atoms tend to increase in size down a group and decrease across a period. When a neutral atom gains or loses an electron, creating an anion or cation, the atom's radius increases or decreases, respectively."<>

3 0

3 years ago

How many moles of argon gas would be present in a 37.0 liter vessel at 45.00 °C at a pressure of 2.50 atm?

Pani-rosa [81]

Answer:

3.54 mol

Explanation:

Step 1: Given data

Volume (V): 37.0 L

Temperature (T): 45.00 °C

Pressure (P): 2.50 atm

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 45.00°C + 273.15 = 318.15 K

Step 3: Calculate the number of moles (n) of argon gas

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 2.50 atm × 37.0 L/(0.0821 atm.L/mol.K) × 318.15 K = 3.54 mol

8 0

3 years ago

t 745 K, the reaction below has an equilibrium constant (Kc) of 5.00 × 102. H2 (g) + I2 (g) ⇌ 2 HI (g) If a mixture of 0.10 mol

spayn [35]

Answer : The concentration of HI (g) at equilibrium is, 0.643 M

Explanation :

The given chemical reaction is:

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

Initial conc. 0.10 0.10 0.50

At eqm. (0.10-x) (0.10-x) (0.50+2x)

As we are given:

$K_c=5.00\times 10^2$

The expression for equilibrium constant is:

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

Now put all the given values in this expression, we get:

$5.00\times 10^2=\frac{(0.50+2x)^2}{(0.10-x)\times (0.10-x)}$

x = 0.0713 and x = 0.134

We are neglecting value of x = 0.134 because the equilibrium concentration can not be more than initial concentration.

Thus, we are taking value of x = 0.0713

The concentration of HI (g) at equilibrium = (0.50+2x) = [0.50+2(0.0713)] = 0.643 M

Thus, the concentration of HI (g) at equilibrium is, 0.643 M

8 0

3 years ago