Consider the process of diluting 1.2 M NH4Cl solution to prepare 1.4 L of a 0.25 M NH4Cl solution. Determine if each of the foll

Answer:

1 M

Explanation:

The reaction equation is as follows:

CH3COOH + CH3COONa -------------------> 2CH3COO^- + NaH

One mole from each of the reactants yields 2 moles of acetate ions.

According to the problem, 2.00 mL, which is (2÷1000)L, of 0.50 M acetic acid reacts with 8.00 mL, equating to (8/1000)L, of 0.50 sodium acetate.

From the equation, we use n = CV -------------------------------------------(1).

Here, n = number of moles, V = volume, C = concentration.

The number of moles, n, of acetic acid = 0.50M × 2/1000L.

n(acetic acid) = 0.001 moles.

The number of moles, n, of sodium acetate = 0.50M × (8/1000)L.

n(sodium acetate) = 0.004 moles.

0.001 moles of acetic acid reacts with 0.004 moles of sodium acetate.

Thus, acetic acid acts as the limiting reagent.

One mole of acetic acid generates 2 moles of acetate ions.

0.001 mole of acetic acid results in = 0.002 moles of acetate ions.

According to the formula (1), n = CV.

0.002 = C × 2/1000

C = 0.002/0.002

C = 1 M

Answer:

The amount of NaHCO3 needed is 9.6 E-3 moles.

Explanation:

According to the balanced equation:

• 2 NaHCO3(s) + H2SO4(ac) ↔ Na2SO4(ac) + 2 CO2(g) + 2 H2O(l)

• assumed concentration of H2SO4 is 6M, a common lab setup.

Thus, n H2SO4 = 0.0008 L * 6 mol/L = 0.0048 mol of H2SO4.

For every mole of H2SO4, two moles of NaHCO3 are required, leading to:

⇒ moles of NaHCO3 = 0.0048 mol H2SO4 * (2 mol NaHCO3 / mol H2SO4)

⇒ moles of NaHCO3 = 0.0096 mol.

This indicates that 9.6 E-3 mol of NaHCO3 is the minimum needed to neutralize the acid.

Answer:

The kinetic energies of all the gases are equal.

Explanation:

The average kinetic energy of gas molecules is described by the equation:

K.E = (3/2)KT

where,

K.E = Average Kinetic Energy of the molecule

K = Boltzmann's Constant = 1.38 x 10^-23 J/K

T = Absolute temperature of the gas

As all gases are mixed together, their temperatures will be uniform and equal to that of the mixture.

Thus, all variables in the equation remain constant, leading to the Average Kinetic Energies of all involved gases (specifically, Neon Ne, Krypton Kr, Radon Rn), being equal.

Unbalanced half-reaction:

<span>NO(g) ---> HNO2(aq) </span>

<span>adjust to equal the oxygen: </span>

<span>NO(g) + H2O(l) ---> HNO2(aq) </span>

<span>equal the hydrogen (as in acid solution): </span>

<span>NO(g) + H2O(l) ---> HNO2(aq) + H^+ </span>

<span>equal the charge: </span>

<span>NO(g) + H2O(l) ---> HNO2(aq) + H^+ + e- </span>

<span>modify for basic solution: </span>

<span>OH^-(aq) + NO(g) + H2O(l) ---> HNO2(aq) + H2O(l) + e- </span>

<span>eliminate water: </span>

<span>OH^-(aq) + NO(g) ---> HNO2(aq) + e- </span>

My answer is MgCl2 because the bases of group one elements (alkali metals) such as sodium, lithium, and cesium are all known to be strong bases, unlike those from groups two and three. The bases of group one elements, like NaOH, LiOH, and CsOH, easily dissolve in water, creating strong soluble bases. Comparatively, when assessing the bases of Ba (Ba(OH)2) and Mg (Mg(OH)2), Ba's base is stronger due to its higher reactivity. Mg(OH)2, being less soluble in water, does not easily dissociate to yield OH- ions.