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Stells [14]
2 years ago
13

Calculate the number of atoms of each element. 2,6,15,3,8,5?

Chemistry
1 answer:
katen-ka-za [31]2 years ago
6 0

Answer:

P 6 atoms

O 15 atoms

Explanation:

P = 2×3= 6

O = 5×3= 15

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2,3,7,10,13 i did this yesterday can u mark me brainliest

3 0
3 years ago
What is the highest occupied molecular orbital in the O2 molecule?
Aleonysh [2.5K]

H2 is known to exist. For dihydrogen, H2, we can identify the frontier molecular orbitals (FMOs). The highest occupied molecular orbital (or HOMO) is the σ (sigma) 1s MO. The lowest unoccupied MO (LUMO) is the σ* (sigma star) 1s MO which is antibonding.


3 0
3 years ago
What is the half-life in minutes of a compound if 75.0 percent of a given sample decomposes in 40.0 minutes? Assume first-order
yaroslaw [1]

Answer:

See below

Explanation:

.75 = 1/2^(40/h)      

log .75 / ( log 1/2) = 40 / h

<u>h = half life =   96.37683 min</u>

8 0
1 year ago
Can you guys help me with this one part?
Pani-rosa [81]

Answer: Do want the experiment done or help of the experiment

3 0
2 years ago
Read 2 more answers
Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N
Montano1993 [528]

The question is incomplete, here is the complete question:

Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.

Determine the limiting reactant. (express your answer as a chemical formula)

<u>Answer:</u> The limiting reactant is ammonia (NH_3)

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For ammonia:</u>

Given mass of ammonia = 135.9 kg = 135900 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia}=\frac{135900g}{17g/mol}=7994.12mol

  • <u>For carbon dioxide gas:</u>

Given mass of carbon dioxide gas = 211.4 kg = 211400 g

Molar mass of carbon dioxide gas = 44 g/mol

Putting values in equation 1, we get:

\text{Moles of carbon dioxide gas}=\frac{211400g}{44g/mol}=4804.54mol

The given chemical reaction follows:

2NH_3(aq.)+CO_2(aq,)\rightarrow CH_4N_2O(aq.)+H_2O(l)

By Stoichiometry of the reaction:

2 moles of ammonia reacts with 1 mole of carbon dioxide

So, 7994.12 moles of ammonia will react with = \frac{1}{2}\times 7994.12=3997.06mol of carbon dioxide

As, given amount of carbon dioxide is more than the required amount. So, it is considered as an excess reagent.

Thus, ammonia is considered as a limiting reagent because it limits the formation of product.

Hence, the limiting reactant is ammonia (NH_3)

5 0
2 years ago
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