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Valentin [98]
2 years ago
8

Help asapp plzz!!!!

Chemistry
1 answer:
tino4ka555 [31]2 years ago
7 0

Answer:

Determining what data to collect.

Explanation:

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Agata [3.3K]
<span>11.2G is the answer to this problem.
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3 years ago
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The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In t
Sloan [31]

Answer:

Mass =  42.8g

Explanation:

4 NH 3 ( g ) + 5 O 2 ( g ) ⟶ 4 NO ( g ) + 6 H 2 O ( g )

Observe that every 4 mole of ammonia requires 5 moles of oxygen to obtain 4 moles of Nitrogen oxide and 6 moles of water.

Step 1: Determine the balanced chemical equation for the chemical reaction.

The balanced chemical equation is already given.

Step 2: Convert all given information into moles (through the use of molar mass as a conversion factor).

Ammonia = 63.4g × 1mol / 17.031 g = 3.7226mol

Oxygen = 63.4g × 1mol / 32g = 1.9813mol

Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.

If all of the 1.9831 moles of oxygen were to be used up, there would need to be 1.9831 × 4 / 5 or 1.5865 moles of Ammonia. We have 3.72226 moles of ammonia - Far excess. Because there is an excess of Ammonia, the Oxygen amount is used to calculate the amount of the products in the reaction.

Step 4: Use the amount of limiting reactant to calculate the amount of H2O produced.

5 moles of O2  = 6 moles of H2O

1.9831 moles = x

x = (1.9831 * 6 ) / 5

x = 2.37972 moles

Mass of H2O = Molar mass * Molar mass

Mass = 2.7972 * 18

Mass =  42.8g

6 0
3 years ago
When most liquids freeze, explain what happens to the motion and the space between the atoms.
KiRa [710]

Answer:

atoms relative motion slow down and begin to vibrate in place

8 0
3 years ago
The first excited vibrational energy level of diatomic chlo- rine (Cl2) is 558 cm^-1 above the ground state. Wave- numbers, the
Usimov [2.4K]

Answer:

The answer is "0.0000190 and 2.7 J".

Explanation:

\to P_4=\frac{e^{-\beta}(4+\frac{1}{2}) hev}{2vb}\\\\

         =\frac{e^{-9(1.35)}}{0.278v}\\\\=\frac{e^{-12.5}}{0.278}\\\\=\frac{0.000005285}{0.278}\\\\=0.0000190

Given:

h=6.626 \times 10^{-34}\\\\c=3\times 10^{10}\\\\v=558\ cm^{-1}\\\\K=1.38 \times 10^{-23}\\\\ T=298\\\\

E=\frac{hcv}{KT}\\\\

by putting the value into the above formula so, the value is 2.7 J  

6 0
3 years ago
Robert has pure samples of both D-ribose and D-arabinose, but he forgot to label them. He only has some nitric acid, the reagent
Sauron [17]

Answer:

Following are the responses to the given question:

Explanation:

Since HN03 is an oxidation substance D-ribose u.ith oxidized to form in rubric acid Ribose is chiral, but rubric acid is achiral because of its symmetry mirror level, Hence no infrared roster in the sample holder is observed.

Please find the attached file.

D-Arabinose, on either hand, gives optical aldaric acid with such a net optical rotation observed inside the polarimeter for diagnosis with HN03.

4 0
3 years ago
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