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3 years ago

How can you obtain hydrogen from a mixture of ethene, hydrogen and ammonia gases?

Chemistry

1 answer:

irga5000 [103]3 years ago

3 0

Answer:

We normally separate unreacted hydrogen from ammonia (product) in Haber process. The reaction mixture contains some ammonia, plus a lot of unreacted hydrogen and nitrogen. The mixture is cooled and compressed, causing the ammonia gas to condense into a liquid.

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Determine the percent yield forthe reaction between 82.4 g of Rband 11.6 g of O2 if 39.7 g of Rb2Ois produced

Mazyrski [523]

Step 1

The reaction is written and balanced:

4 Rb + O2 =>2 Rb2O

-----------

Step 2

Define % yield of product (Rb2O) = (Actual yield/Theoretical yield) x 100

The actual yield is provided by the exercise = 39.7 g

----------

Step 3

Determine the limiting reactant. The molar masses are needed to solve this:

For Rb) 85.4 g/mol

For O2) 32 g/mol

Procedure:

4 Rb + O2 =>2 Rb2O

4 x 85.4 g Rb ----- 32 g O2

82.4 g Rb ----- X = 7.72 g O2 are needed

For 82.4 g Rb, 7.72 g O2 is needed, but there is 11.6 g O2. Therefore, O2 is the excess agent. Rb is the limiting reactant.

--------

Step 4

Determine the theoretical yield from the limiting reactant:

The molar mass Rb2O) 187 g/mol

Procedure:

4 x 85.4 g Rb ------ 2 x 187 g Rb2O

82.4 g Rb ------ X = 90.2 g Rb2O = Theoretical yield

---------

Step 5

% yield = Actual y./Theoretical y. x 100 = (39.7 g/90.2 g) x 100 = 44 % approx.

Answer: % yield = 44 %

4 0

1 year ago

It takes 945. kJ/mol to break a nitrogen-nitrogen triple bond. Calculate the maximum wavelength of light for which a nitrogen-ni

kolezko [41]

Answer: 1.274 * 10^ -7 meter (same as 127.4 nanometers

Explanation:

It's given that the energy

required to break the N N triple bond is 945 * 10^3 joules per mole.

One mole contains 6.02 * 10^ 23 molecules, so the energy required per molecule

= 945 * 10^3 / 6.023 * 10^23, or 1.56 * 10^-18 joules.

Then we need a photon whose energy (E) is at least that amount.

The energy E of a photon is related to its frequency f by PLANCK'S EQUATUON,

E = hf,

where h is Planck's constant (6.625 * 10^-34 joule-sec)

and the wavelength w is inversely proportional to the frequency by w = c/f, where c is the speed of light, 2.998 * 10^8 meters per sec.

If h & c are both constants, their product hc is constant, so we can say E = hc/w,

or if we know E and want to find w, a little algebra gives: w = hc/E.

The product hc = 1.9875 * 10^-25 joule-meters,

so w = 1.9875 * 10^-25 / 1.56 * 10^-18, or 1.274 * 10^ -7 meter (same as 127.4 nanometers

6 0

3 years ago

The black, granular material that fills a dry cell in a common flashlight (between the carbon rod and the zinc shell) is mangane

mestny [16]

Answer: c. +4

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

$Mn:25:1s^22s^22p^63s^23p^64s^23d^{5}$

$Mn^{4+}:21:1s^22s^22p^63s^23p^63d^1$

$O:8:1s^22s^22p^4$

$O^{2-}:10:1s^22s^22p^6$

Here manganese is having an oxidation state of +4 called as cation and oxygen forms an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged to give $Mn_2O_4$ which written in simplest whole number ratios to give neutral $MnO_2$