Cover echo de cobalto

Place these ions in order of preferential discharge, 1-4,where 1 is the most and 4 is least perfered.

egoroff_w [7]

Answer:

Silver ions 1 Hydrogen ions 2 Iron ions 3 Sodium ions 4

Explanation:

6 0

3 years ago

A tank at is filled with of dinitrogen monoxide gas and of boron trifluoride gas. You can assume both gases behave as ideal gase

blsea [12.9K]

Answer:

(1). Mole fraction = 0.152 for sulfur tetrafluoride gas.

Mole fraction = 0.848 For dinitrogen monoxide gas.

(2). Partial Pressure for dinitrogen monoxide gas = 187 kPa

Partial Pressure for sulfur tetrafluoride gas = 33.4 kpa.

(3). Total Partial Pressure = 220.4 kpa.

Explanation:

So, we are given the following data or parameters or information in the question above;

• Volume of the tank = 5.00L per tank;

• Temperature of the tank = 7.03°C;

• The mass of the content in the tank =

17.7g of dinitrogen monoxide gas and

7.77g of sulfur tetrafluoride gas.

So, we will be making use of the formulae below to calculate the MOLE FRACTION:

Moles, n= mass/molar mass and mole fraction = n(1)/ n(1) + n(2) per each constituents.

Moles, n1 = 17.7g of dinitrogen monoxide gas/ 44 grams per mole. =0.4023 moles.

Moles, n2 = 7.77g of sulfur tetrafluoride gas/ 108.1 grams per mole. = 0.07188 moles.

Total numbers of moles = n1 + n2 = 0.47415 moles

Mole fraction =0.4023 / 0.47415 = 0.848 of dinitrogen monoxide gas.

Mole fraction = 0.07188/0.47415 = 0.152 of sulfur tetrafluoride gas.

PART TWO: CALCULATE THE PARTIAL PRESSURE AND TOTAL PRESSURE BY USING THE FORMULA BELOW;

pressure × volume = number of moles × gas constant, R × temperature.

Pressure = n × R × T/ V.

For dinitrogen monoxide gas. ;

Partial Pressure = 0.4023 × 8.314 × 280.03 / 5 × 10^-3 = 187 kPa.

For sulfur tetrafluoride gas

Partial Pressure = 0.07188 × 8.314 ( × 280.03 / 5 × 10^-3. = 33.4 kpa.

(3). Total pressure = (187 + 33.4)kpa = 220.4 kpa

6 0

3 years ago

i have an unknown volume of gas at a pressure of 0.50 atm a temperature of 325 k. If i raise the pressure to 1.2 atm, decrease t

Ronch [10]

Assuming that the number of mols are constant for both conditions:
$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
Now you plug in the given values. V_1 is the unknown.
$\frac{0.50 atm*V_1}{325K} = \frac{1.2 atm* 48 L}{230K}
$
Separate V_1
$V_1= \frac{1.2 atm* 48 L * 325K }{230K*0.50 atm }$
V= 162.782608696 L

There are 2 sig figs

V= 160 L

3 0

3 years ago

In completing a Fischer Esterification experiment using acetic acid and unknown alcohol catalyzed by sulfuric acid a student sta

masha68 [24]

Answer:

The equilibrium law

Explanation:

According to the equilibrium law , when $\text{concentration}$ of any of the reactants or the products in the reaction at an equilibrium state, is changed, then it changes the composition of equilibrium mixture in order to minimize the effect of

Acetic acid (in excess) + $\text{alcohol}$ ⇄ $\text{ester}$ + water

$\text{Fischer esterification}$ is a reaction which involves the equilibrium. So to force the reaction towards the side of $\text{ester}$ , one of the reactant is taken in excess. All this can be explained easily by applying the equilibrium law.

3 0

3 years ago

a car traveling at 20 m/s starts to decelerate steadily it comes to a complete stop in 10s .what is its acceleration

Anna007 [38]

A car traveling at 20 m/s starts to decelerate steadily it comes to a complete stop in 10s. The equation for the acceleration is velocity divided by time. So divide 20 m/s by 10 s and you will get an acceleration of 2 m/s2.

4 0

3 years ago