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NikAS [45]
3 years ago
13

Answer number 1 only please and thank you

Chemistry
2 answers:
I am Lyosha [343]3 years ago
7 0
The answer is d because team 1 is pulling with more force
bulgar [2K]3 years ago
5 0
Number 1 is D because 124-110 is 14n in team 1s favor so it moves towards team 1
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What are the limitations of bohr's model of atom​
nataly862011 [7]

Answer:

The Bohr Model is very limited in terms of size. Poor spectral predictions are obtained when larger atoms are in question. It cannot predict the relative intensities of spectral lines. It does not explain the Zeeman Effect, when the spectral line is split into several components in the presence of a magnetic field.

Explanation:

7 0
3 years ago
Read 2 more answers
Calculate the volume of each of the following gases at STP
kherson [118]
For Ar :

1 mol ------------ 22.4 L ( at STP )
7.6 mol ---------- x L 

x = 7.6 * 22.4

x = 170.24 L
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For C2H3:

1 mol ------------ 22.4 ( at STP)
0.44 mol --------- y L

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hope this helps !.



7 0
3 years ago
N2 + 3 H2 → 2 NH3
sweet [91]

Answer:

B

Explanation:

5 0
3 years ago
I have asked this before, to get more points, just put the same answer on the other questions
aleksklad [387]
The standard atomic weight is the average mass of an element in atomic mass units ("amu"). Though individual atoms always have an integer number of atomic mass units, the atomic mass on the periodic table is stated as a decimal number because it is an average of the various isotopes of an element.
8 0
3 years ago
G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu
Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄Ag^{+} + Br^{-1}.

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains \frac{0.6485}{45}×1000 = 14.411 gm-mole. Thus the solubility product K_{sp}of AgBr = [Ag^{+}]×Br^{-}.

Or, 5.0×10^{-13} = S^{2} (the given value of solubility product of AgBr is 5.0×10^{-13} and the charge of the both ions are same).

Thus S = (5.00×10^{-13})^{1/2} = 7.071×10^{-7} g/mL.

Thus the concentration of Br^{-1} or HBr is 7.071×10^{-7} g/mL.

4 0
4 years ago
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