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Tomtit [17]
3 years ago
5

If 25.0 g of NH3 and 38.9g of O2 react in the following reaction, how

Chemistry
1 answer:
Elan Coil [88]3 years ago
8 0

Answer:

There will be formed 36.49 grams of NO

Explanation:

<u>Step 1</u>: The balanced equation

4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H20 (g)

<u>Step 2:</u> Data given

Mass of NH3 = 25.0 grams

Mass of O2 = 38.9 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of 02 = 32 g/mol

Molar mass of NO = 30.01 g/mol

<u>Step 3:</u> Calculate number of moles

Moles of NH3 = 25.0 grams / 17.03 g/mol

Moles of NH3 = 1.47 moles

Moles of O2 = 38.9 grams / 32 g/mol

Moles of O2 = 1.216 moles

<u>Step 4:</u> Calculate limiting reactant

For 4 moles of NH3 consumed, we need 5 moles of O2 to produce 4 moles of NO and 6 moles of H2O

O2 is the limiting reactant. It will complete be consumed (1.216 moles).

NH3 is the limiting reactant. There will 4/5 * 1.216 = 0.9728 moles of NH3 consumed.

There will remain 0.4972 moles of NH3.

<u>Step 5: </u>Calculates moles of NO

If there is consumed 4 moles of NH3, there is produced 4 moles of NO

For 1.216 moles of NH3 consumed, we will have 1.216 moles of NO

<u>Step 6:</u> Calculate mass of NO

Mass NO = moles NO * Molar mass NO

Mass NO = 1.216 moles * 30.01 g/mol = 36.49 grams

There will be formed 36.49 grams of NO

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Explanation:

Given is that, the atmospheric pressure on the surface of Venus is

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To convert divide the pressure value by 101325.

Pressure in atm = \frac{6.84 \times 10^{4} }{101325}

= 0.675055 atm

Rounding it off to 3 significant digits: 0.675 atm

Now,  one Torr is 133.322 Pa. For conversion, divide the pressure value by 133.322.

Pressure in Torr = \frac{6.84 \times 10^{4} }{133.322}

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3 years ago
Which of the following transitions represent the emission of a photon with the largest energy?
s2008m [1.1K]

Answer:

n=6 to n=3 (B)

Explanation:

Energy of an electron present in the n^{th} orbit is directly proportional to n^{2} .Hence a transistion from one orbit to another orbit emits an energy proportional to the difference of their squares of the orbits. that is if an electron travels from orbit n1 to orbit n2 then it emits an energy corresponding to n_{1} ^{2} -n_{2} ^{2}.So in the above question the highest energy emission occurs when an electron moves from n=6 to n=3.(Highest difference of energy levels).

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Hello!

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Looking at our period table, sodium has a +1 charge, written as Na 1+, and sulfate has a charge of -2, and it is written as SO4 2-.

Now, we need to make the charges equivalent. To do this, we need to "criss-cross" the charges. This means that sodium will need to additional atoms to make the charges equal, and sulfate will need one.

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How the calculation of the [OH-], pH and % ionization for 0.619 M ammonia (NH3) NH3 + H2O (liq) rightwards harpoon over leftward
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Answer:

[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52

Explanation:

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Is:

Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]

<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>

<em>[NH₄⁺] = [OH⁻] = X</em>

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The amount of heat required to convert the 1 kg solid  into 1 kg liquid at constant temperature.  

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