Calculate the empirical formula 9.1 g of lithium and 10.4 g of oxygen

A characteristic feature of any form of chromatography is the ________.a. calculation of an Rf value for the molecules separated

castortr0y [3046]

Answer: The right choice is (c) application of both a mobile phase and a stationary phase.

Explanation:

Chromatography: This refers to a technique for separating a mixture where the mixture is distributed between two phases at varying rates, one being stationary and the other moving.

Mobile phase: The component in which the mixture is dissolved is referred to as the mobile phase.

Stationary phase: This is an adsorbent medium that remains in place while a liquid or gas passes over its surface, thus remaining stationary.

Consequently, a key characteristic of any chromatography technique involves utilizing both a mobile and a stationary phase.

4 0

2 months ago

In a titration experiment, H2O2(aq) reacts with aqueous MnO4-(aq) as represented by the equation above. The dark purple KMnO4 so

Alekssandra [3086]

Respuesta:

El oxígeno en H2O2 es la especie que se reduce a H2O y se oxida a O2.

Explicación:

5 H2O2(aq) + 2 MnO4-(aq) + 6 H+(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g)

La oxidación se define como la pérdida de electrones. La oxidación provoca un aumento en el número de oxidación de un elemento.

Si se descompone esta reacción en sus mitades de reducción y oxidación

Se observa que, de los reactivos mencionados anteriormente,

H202 se convierte en H2O y O2

MnO4- + H+ se convierte en Mn2+ y H2O

El número de oxidación de Mn cambia de +7 en MnO4- a +2 en Mn2+ (lo que indica evidentemente una reducción)

El oxígeno en MnO4- no cambia su número de oxidación, ya que se mantiene en -2

El número de oxidación del oxígeno cambia de -1 en H2O2 a -2 en H2O y 0 en O2

El hidrógeno en H2O2 no cambia su número de oxidación, y su número de oxidación se mantiene en +1 tanto en H2O2 como en H2O.

Esto indica que H2O2 sufre tanto oxidación como reducción; más específicamente, el oxígeno en H2O2 es la especie que se reduce a H2O y se oxida a O2.

Espero que esto ayude

7 0

2 months ago

A chamber with a fixed volume is shown above. The temperature of the gas inside the chamber before heating is 25.2 C and it’s pr

KiRa [2933]

Answer:

Explanation:

Given data:

Initial temperature T₁ = 25.2°C = 298.2K

Initial pressure P₁ = 0.6atm

Final temperature = 72.4°C = 345.4K

What we need to find:

Final pressure = ?

To determine this, we apply a modified version of the combined gas law with constant volume. This simplifies our calculations to:

$\frac{P_{1} }{T_{1} } = \frac{P_{2} }{T_{2} }$

Here, P and T signify pressure and temperatures, 1 refers to initial and 2 to final temperatures.

Now we can substitute the known variables:

$\frac{0.6}{298.2} = \frac{P_{2} }{345.4}$

P₂ = 0.7atm

3 0

3 months ago

If 3.18 x 10^23 atoms of iron react with 67.2 L of chlorine gas at STP, what is the maximum

Tems11 [2777]

84.34 grams of iron (III) chloride is the maximum produced since iron is the limiting reagent, and chlorine gas is in excess.

Explanation:

Balanced equation:

2 Fe + 3 Cl2 → 2 FeCl3

DATA PROVIDED:

iron = atoms

mass of chlorine = 67.2 liters

mass of FeCl3 =?

The number of moles of iron will be calculated as

number of moles = $\frac{total number of atoms}{Avagaro's number}$

number of moles = $\frac{3.18 x 10^23}{6.022x 10^23}$

number of moles = 0.52 mol of iron

moles of chlorine gas

number of moles = $\frac{mass}{molar mass of 1 mole}$

Substituting the values into the equation:

n = $\frac{67200}{70.96}$ (molar mass of chlorine gas = 70.96 g/mol)

= 947.01 moles

As iron is the limiting reagent therefore

2 moles of Fe lead to 2 moles of FeCl3

0.52 moles of Fe will yield

$\frac{2}{2}$ = $\frac{x}{0.52}$

0.52 moles of FeCl3 is produced.

To express this in grams:

mass = n x molar mass

= 0.52 x 162.2 (molar mass of FeCl3 is 162.2g/mol)

= 84.34 grams

3 0

2 months ago