How did the work of Dmitri Mendeleev differ from that of John Newlands in the development of the periodic table?

In a study of the following reaction at 1200 K it was observed that when the equilibrium partial pressure of water vapor is 15.0

sp2606 [1]

Answer:

The value of $K_p$ for this reaction at 1200 K is 4.066.

Explanation:

Partial pressure of water vapor at equilibrium = $p^o_{H_2O}=15.0 Torr$

Partial pressure of hydrogen gas at equilibrium = $p^o_{H_2}=?$

Total pressure of the system at equilibrium P = 36.3 Torr

Applying Dalton's law of partial pressure to determine the partial pressure of hydrogen gas at equilibrium:

$P=p^o_{H_2O}+p^o_{H_2}$

$p^o_{H_2}=P-p^o_{H_2O}=36.3 Torr- 15.0 Torr = 21.3 Torr$

$3 Fe(s) 4 H_2O(g)\rightleftharpoons Fe_3O_4(s) 4 H_2(g)$

The expression of $K_p$ is given by:

$K_p=\frac{(p^o_{H_2})^4}{(p^o_{H_2O})^4}$

$K_p=\frac{(21.3 Torr)^4}{(15.0 Torr)^4}=4.066$

The value of $K_p$ for this reaction at 1200 K is 4.066.

6 0

3 years ago

Dissolution of KOH, ΔHsoln:

swat32

Using Hess's law we found:

1) By adding reaction 10.2 with the reverse of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) ΔH (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy (ΔH).

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq) ΔHsoln (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq) ΔHneut (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from differents changes in various reactions can be calculated as the sum of all the enthalpies of all those reactions.

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

We need to add reaction 10.2 to the reverse of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

Canceling the KOH(s) from both sides, we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

2) The change in enthalpy for reaction 10.3 can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

$\Delta H = \Delta H_{soln} + \Delta H_{neut}$

The enthalpy of reaction 10.1 (ΔHsoln) changed its sign when we reversed reaction 10.1, so:

$\Delta H = \Delta H_{neut} - \Delta H_{soln}$

Therefore, the ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy ΔH.

Learn more here:

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I hope it helps you!

6 0

2 years ago

3 waves are shown with a line through their center. The bottom of the first wave is labeled C. A bracket labeled D connects the

Likurg_2 [28]

Answer:Label the parts of this wave.

A:

✔ crest

B:

✔ amplitude

C:

✔ trough

D:

✔ wavelength

Explanation:

8 0

2 years ago

Read 2 more answers

Part iv. Is the neutralization reaction enthalpy favored?

Burka [1]

Yes, it is a special case of enthalpy of neutralization.

The enthalpy of neutralization (ΔHn) is the change in enthalpy that occurs when one equivalent of an acid and one equivalent of a base undergo a neutralization reaction to form water and a salt.

The standard enthalpy change of neutralization is the enthalpy change when solutions of an acid and an alkali react together under standard conditions to produce 1 mole of water.

5 0

2 years ago

A sample of an unknown gas at STP has a density of 0.630 gram per liter. What is the gram molecular mass of this gas?

jeyben [28]

STP means standard temperature and pressure at 0°C (273K) and 1 atm (atmosphere). The density of the unknown gas is 0.63 gram per liter. The deal gas equation is PV = nRT. The n is the numer of moles and can be represented as mass of the gas, m, divided by the molar mass, c. so we have,

PV = nRT
PV = (m/c)RT
Since the density is d = m/V
Pc = (m/V)RT
Pc = dRT
c = drT/P

substitute the values into the equation,
c = [(0.63g/L)(0.08206 L-atm/mol-K)(273K)]/(1atm)
c = 14.11 g/mol

3 0

3 years ago