How many calories of heat are necessary to raise the temperature of 378.2 g of water from 32.2 oC to the boiling point?

Compare the solubility of silver chromate in each of the following aqueous solutions: Clear All 0.10 M AgCH3COO 0.10 M Na2CrO4 0

slavikrds [6]

Solution :

Comparing the solubility of silver chromate for the solutions :

$0.10 \ M \ AgCH_3COO$ ----- Less soluble than in pure water.

$0.10 \ M \ Na_2CrO_4$ ----- Less soluble than in pure water.

$0.10 \ M \ NH_4NO_3$ ----- Similar solubility as in the pure water

$0.10 \ M \ KCH_3COO$ ----- Similar solubility as in the pure water

The silver chromate dissociates to form :

$AgCrO_4 (s) \rightleftharpoons 2Ag^+ (aq) +CrO_4^{2-}(aq)$

When 0.1 M of $AgCH_3COO^-$ is added, the equilibrium shifts towards the reverse direction due to the common ion effect of $Ag^+$ , so the solubility of $Ag_2CrO_4$ decreases.

Both $AgCH_3COO$ and $KCH_3COO$ are neutral mediums, so they do not affect the solubility.

4 0

3 years ago

1. Unas de las formas de producir nitrógeno gaseoso (N2) es mediante la oxidación de metilamina (CH3NH2), tal como se muestra en

Maslowich

Answer:

a) 4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂

b) m = 5,043 g

c) % = 69,4 %

Explanation:

a) La ecuación balanceada es la siguiente:

4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂

En el balanceo, se tiene en la relación estequiométrica que 4 moles de metilamina reacciona con 9 moles de oxígeno para producir 4 moles de dióxido de carbono, 10 moles de agua y 2 moles de nitrógeno.

b) Para determinar la masa de nitrógeno se debe calcular primero el reactivo limitante:

$n_{O_{2}} = \frac{m}{M} = \frac{25,6 g}{31,99 g/mol} = 0,800 moles$

$n_{CH_{3}NH_{2}} = \frac{4}{9}*0,800 moles = 0,356 moles$

De la ecuación anterior se tiene que la cantidad de moles de metilamina necesaria para reaccionar con 0,800 moles de oxígeno es 0,356 moles, y la cantidad de moles iniciales de metilamina es 0,5 moles, por lo tanto el reactivo limitante es el oxígeno.

Ahora, podemos calcular la masa de nitrógeno producida:

$n_{N_{2}} = \frac{2}{9}*n_{O_{2}} = \frac{2}{9}*0,8 moles = 0,18 moles$

$m_{N_{2}} = n_{N_{2}}*M = 0,18 moles*28,014 g/mol = 5,043 g$

Por lo tanto, se pueden producir 5,043 g de nitrógeno.

c) El redimiento de la reacción se puede calcular usando la siguiente fórmula:

$\% = \frac{R_{r}}{R_{T}}*100$

<u>Donde</u>:

$R_{r}$ : es el rendimiento real

$R_{T}$ : es el rendimiento teórico

$\% = \frac{3,5}{5,043}*100 = 69,4$

Entonces, el procentaje de rendimiento de la reacción es 69,4%.

Espero que te sea de utilidad!

5 0

3 years ago

Please help asap :)))))

deff fn [24]

Answer:

um i dont know anything about music sorry

6 0

2 years ago

List three different forms of potential energy

Wewaii [24]

Answer:

Types of Potential Energy

Elastic Potential Energy. Anything that can act like a spring or a rubber band can have elastic potential energy. ...

Gravitational Potential Energy. There is a constant attractive force between the Earth and everything surrounding it, due to gravity. ...

Chemical Potential Energy.

(IF THIS HELPED CAN YOU GIVE ME A BRAINYLEST PLEASE?)

3 0

3 years ago

How much calcium hypochlorite (65% strength) is needed to make 200 L of 2% hypochlorite solution? (Assume that a 1% solution is

g100num [7]

To solve this we use the equation,

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

65 x V1 = 2 x 200 L

V1 = 6.15 L

5 0

3 years ago