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3 years ago

How many atoms are in 63.5 grams of Hg

1 answer:

6 0

Count the number of moles you have

25.8 g x (1 mole / 200.59g) = # mol since g cancels out

# mol x (6.02 * 10^23 atoms / 1 mol) = # atoms since mol cancels out

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Plz help all chemist!

solmaris [256]

Answer:

D. bromine

Explanation:

Highest electronegativity in the period has element closest to the *A group

K Ca Cu Br

6 0

4 years ago

Consider the conversion of 2-naphthol and 1-bromobutane into an ether via the williamson ether synthesis. list the procedural st

Lemur [1.5K]

Answer:

Following are the solution to this question:

Explanation:

Please find the complete question in the attachment.

Start of Laboratory

Dissolve 2-naphthol in the round bottom flask with ethanol.

Add pellets of sodium hydroxide and hot chips. Attach a condenser.

Heat for 20 minutes under reflux, until the put a burden dissolves.

After an additional hour, add 1-Bromobutane and reflux.

Pour the contents into a beaker with ice from a round bottom flask.

On a Bachner funnel, absorb the supernatant by vacuum filtration.

Utilizing cold water to rinse the material and dry that on the filter.

Ending of the Lab

6 0

3 years ago

Which of the following best describes the electron cloud model? A. It shows that electrons usually carry a negative charge. B. I

irakobra [83]

<span>D. It shows that the electrons within an atom do not have sharp boundaries.</span>

8 0

3 years ago

Need help on atoms and molecules

balu736 [363]

1-b

2-c

3-a

4-d

5-d

6-b

7-a

7 0

3 years ago

The ka of hypochlorous acid (hclo) is 3.0 ⋅ 10−8 at 25.0 °c. calculate the ph of a 0.0375m hypochlorous acid solution.

Scrat [10]

We can set up an ICE table for the reaction:
HClO H+ ClO-
Initial 0.0375    0 0
Change    -x     +x +x
Equilibrium 0.0375-x    x x

We calculate [H+] from Ka:
Ka = 3.0x10^-8 = [H+][ClO-]/[HClO] = (x)(x)/(0.0375-x)

Approximating that x is negligible compared to 0.0375 simplifies the equation to
3.0x10^-8 = (x)(x)/0.0375
3.0x10^-8 = x2/0.0375
x2 = (3.0x10^-8)(0.0375) = 1.125x10^-9
x = sqrt(1.125x10^-9) = 0.0000335 = 3.35x10^-5 = [H+]
in which 0.0000335 is indeed negligible compared to 0.0375.

We can now calculate pH:
pH = -log [H+] = - log (3.35 x 10^-5) = 4.47

6 0

4 years ago