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lapo4ka [179]
3 years ago
15

Orbital notation of h

Chemistry
1 answer:
shutvik [7]3 years ago
8 0

Answer: in the box, put an arrow facing up. Below the box, write 1s

Explanation:

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What type of animal might become preserved in amber
goldfiish [28.3K]
Insect's or Bugs preserved in Amber
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3 years ago
The pH at the equivalence point will be a. less than 7.00 b. equal to 7.00 Allohalineup of a sto quivalence point of a titration
True [87]

Answer: c. greater than 7.00

Explanation: The equivalence point of a titration is when all the base is consumed by the acid. When a strong base and a strong acid react, the medium is neutralized because is produced water and salt (which won't suffer hydrolysis). How water's pH is 7, in this type of titration the pH of the equivalence point will be at pH=7. But on titration of a weak acid with a strong base, the reaction of the equivalence point produces water and the conjugate base of the acid. Because the acid is weak, their conjugate base will be strong and will suffer hydrolysis, producing hydroxyl ions, elevating the pH of the water and making it greater than 7.

8 0
3 years ago
Instructions
Marysya12 [62]

Answer:

-104.7?

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Explanation:

7 0
3 years ago
Consider the incorrectly balanced combustion equation:
Ulleksa [173]
Oxygen is balanced incorrectly.
The correct balanced equation: 2C6H6+15O2=12CO2+6H2O
6 0
2 years ago
Consider the following reaction: 2{\rm{ N}}_2 {\rm{O(}}g)\; \rightarrow \;2{\rm{ N}}_2 (g)\; + \;{\rm{O}}_2 (g)
fredd [130]

Answer:

a. 5.9 × 10⁻³ M/s

b. 0.012 M/s

Explanation:

Let's consider the following reaction.

2 N₂O(g) → 2 N₂(g) + O₂(g)

a.

Time (t): 12.0 s

Δn(O₂): 1.7 × 10⁻² mol

Volume (V): 0.240 L

We can find the average rate of the reaction over this time interval using the following expression.

r = Δn(O₂) / V × t

r = 1.7 × 10⁻² mol / 0.240 L × 12.0 s

r = 5.9 × 10⁻³ M/s

b. The molar ratio of N₂O to O₂ is 2:1. The rate of change of N₂O is:

5.9 × 10⁻³ mol O₂/L.s × (2 mol N₂O/1 mol O₂) = 0.012 M/s

4 0
3 years ago
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