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Ronch [10]
4 years ago
7

According to valence bond theory, which orbitals on n and h overlap in the nh3 molecule? 2s on n overlaps with 1s on h 2p on n o

verlaps with 1s on h sp3 on n overlaps with 1s on h sp3 on n overlaps with sp on h 2p on n overlaps with 2s on h
Chemistry
1 answer:
Tcecarenko [31]4 years ago
7 0
Answer is: <span>sp3 on nitrogen overlaps with 1s on hydrogen.
</span>Electron configuration for nitrogen: ₇<span>N 1s² 2s² 2p³.
</span>Electron configuration for hydrogen: ₁H 1s¹.
<span>The central atom in ammonia, nitrogen, is </span><span>sp3 </span>hybridized (<span>one </span>s<span> and three </span>p-orbitals<span> will combine for a four sp3 </span><span>hybridized orbitals), because sp3 orbitals have lower energy</span>.
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The equation representing the solubility of PbI_{2}:

PbI_{2}(aq)Pb^{2+}(aq) + 2 I^{-}(aq)

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What happens when you heat iodine and ammonium chloride?
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None of the above I believe it is what can form a tornado??
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3 years ago
I am using all my points to ask this. please help
ELEN [110]

Answer:

C5H7N

C10H14N2

Explanation:

ok so lets lay out this nicely

elements                                                         C              H         N                    

percentage                                                    74             8.7       17.3

mass number                                                 12              1          14

divided percentage by mass                        6.167          8.7      1.2357

now divide the smallest number with the 4.99            7.04    1

rest for instance 1,237 is the smallest.    

so 6.167÷1,237

and 8.7 ÷1.237

and 1,237 divide 1.237    

now the ratio is                              C5H7N

now for molecular formula find Mr of C5H7N which is (12×5)+7+14=81

next divide 162.1÷81=2

so now for molecular formula just multipily everythng by 2

C10H14N2

5 0
3 years ago
How many grams of carbon dioxide will be produced if 76.4 grams of
goldfiish [28.3K]

Answer:

Assuming that all of the oxygen is used up, 1.53×4111.53×411 or 0.556 moles of C2H3Br3 are required. Because there are only 0.286 moles of C2H3Br3 available, C2H3Br3 is the limiting reagent.

Limiting Reagent What is the limiting reagent if 76.4 grams of C2H3Br3 were reacted with 49.1 grams of O2? C2H3Br3 + 11O2 → 8CO2 + 6H2O + 6Br2 SOLUTION Using Approach 1: A. 76.4g &times; (1 mol/ 266.72 g) = 0.286 moles C2H3Br3 49.1g &times; (1 mole/ 32 g) = 1.53 moles O2 B.

Explanation:

MRK ME BRAINLIEST PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Map%3A_Introductory_Chemistry_(Tro)/08%3A_Quantities_in_Chemical_Reactions/8.04%3A_Limiting_Reactant_and_Theoretical_Yield

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3 years ago
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