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Andrews [41]
4 years ago
12

Simple Chem question - help!

Chemistry
1 answer:
Lunna [17]4 years ago
6 0

d, one atom of oxygen and two atoms of hydrogen

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I need filling the blank spots ​
AnnZ [28]

Answer:

have you tried c

Explanation:

the chicken and I don't know if you can make it

4 0
3 years ago
What is the chemical formula for ammonia sulfide
Mandarinka [93]

Answer:

(NH4)2S

Explanation:

5 0
3 years ago
If it takes 25 mL of 0.05 M HCl to neutralize 345 mL of NaOH solution, what is the concentration, M, of the NaOH solution?
Rashid [163]

Answer:

400ml

Explanation:

4 0
3 years ago
This set contains the spectra of benzil, biphenyl, and bibenzyl, three compounds used in the melting point experiment. Using you
murzikaleks [220]

Answer:

Spectrum 1- Biphenyl

Spectrum 2 - Benzil

Spectrum 3 - Bibenzyl

Explanation:

<u>For Benzil: </u>

<u> </u>The Spectrum -  2  shows this compound.The C=O and C-H bond stretching observed at following values.

C=O \,at\,1657cm^{-1}

C=H \,at\,3063cm^{-1}

<u> For Biphenyl: </u>

<u> </u>Spectrum -  1  shows this compound.

There are stretching vibrations from 3061, 3032\,and \,1597 -1428cm^{1} due to C-H stretching and C-C stretching respectively of aromatic ring only.

<u>For Bibenzyl: </u>

Spectrum - 3  shows this compound

There are stretching vibrations at 3057,3026 and 1598 - 1449cm^{-1} due to C-H stretching and C-C stretching of aromatic ring.

Along with this, stretching vibrations at 2943 - 2917 cm^{-1}due to C-H of alkyl.

8 0
4 years ago
The heat of vaporization for liquid zinc is 1.76 kj/g. How much heat is needed to boil 11.2 g of liquid zinc already at its boil
sladkih [1.3K]
<h3>Answer:</h3>

= 19.712 kJoules

<h3>Explanation:</h3>
  • Heat of vaporization refers to the amount of heat required to change a unit mass of a substance from liquid to gaseous state without change in temperature.

To calculate the amount of heat, we use,

Amount of heat = Mass × Heat of vaporization

Q = m×Lv

Given;

Mass of liquid Zinc = 11.2 g

Lv of liquid Zinc = 1.76 kJ/g

Therefore;

Q = 11.2 g × 1.76 kJ/g

  = 19.712 kJ

Thus, the amount of heat needed to boil 11.2 g of zinc is 19.712 kilo-joules.

6 0
3 years ago
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