Is the aqueous solution of each of these salts acidic, basic, or neutral? (a) cr(no3)3 acidic basic neutral (b) nahs acidic basi

Respuesta:

D

Explicación:

Utilizamos la relación de moles para calcular la presión parcial. El número total de moles es 0.2 + 0.2 + 0.1 = 0.5 moles

Ahora, sabemos que la fracción molar del gas argón es 0.2/0.5

La presión parcial se determina así. Para calcular esto, simplemente multiplicamos el número de moles por la presión total.

0.2/0.5 * 5 = 1.0/0.5 = 2.00atm

D

Answer:

The molar mass of the gas is 36.25 g/mol.

Explanation:

• To determine this, we utilize the mathematical relationship:

ν =

Where, ν represents the speed of light in a gas (ν = 449 m/s),

R denotes the universal gas constant (R = 8.314 J/mol.K),

T stands for the temperature of the gas in Kelvin (T = 20 °C + 273 = 293 K),

M is the molar mass of the gas in (Kg/mol).

ν =

(449 m/s) = √(3(8.314 J/mol.K)(293 K)/M,

by squaring both sides:

(449 m/s)² = (3(8.314 J/mol.K)(293 K))/M,

thus M = (3(8.314 J/mol.K)(293 K)/(449 m/s)² = 7308.006/201601 = 0.03625 Kg/mol.

Thus, the molar mass of the gas is 36.25 g/mol.

The correct answer is: c. N2O, N2O4, N2O5.

According to the law of multiple proportions, also referred to as Dalton's Law, when two elements form compounds, the mass ratios of the second element that combine with a specific mass of the first yield small whole number ratios.

1) For NO, the mass ratio m(N): m(O) is 14: 16, simplified to 7: 8.

2) In N₂O, the ratio m(N): m(O) equates to 2·14: 16, which simplifies to 7: 4.

3) For NO₂, the masses yield m(N): m(O) = 14: 2·16, simplifying to 7: 16.

4) In N₂O₅, the ratio is (2·14): (5·16), which simplifies to 7: 20.

5) For NO₄, the mass ratio is m(N): m(O) = 14: (4·16), which simplifies to 7: 32.

6) N₂O₄ gives a ratio of m(N): m(O) as (2·14): (4·16), simplifying to 7: 16.

A) This means m(N): m(O) = 5.6 g: 3.2 g, simplifying results in 1.75: 1, which further translates to m(N): m(O) = 7: 4.

B) Here, m(N): m(O) is 3.5 g: 8.0 g. Dividing reveals a ratio of 1: 2.285, which alters to m(N): m(O) = 7: 16.

C) Lastly, for m(N): m(O) = 1.4 g: 4.0 g, then adjustments yield a ratio of m(N): m(O) = 7: 20.

The enthalpy change associated with the precipitation reaction is 84 kJ/mole

Why?

The chemical equation for the reaction can be written as

AgNO₃(aq) + NaCl (aq) → AgCl(s) + NaNO₃(aq)

To determine the enthalpy change, the following equation applies

To calculate the heat (Q):

Next, we need to calculate the number of moles involved in the reaction (n):

With these two values, we can substitute them into the first equation:

Have a great day!