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zalisa [80]
3 years ago
8

Please help me #6!!!!!!!!

Chemistry
1 answer:
Vesnalui [34]3 years ago
3 0
I believe that the answer is 12 because there is already 3 O molecules and since its in parentheses with 3 outside it that means that there are 3 of those CO molecules meaning that for every 1 CO there will be 3 O’s so 3, four times Is 12
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Due to the small and highly electronegative nature of fluorine, the oxyacids of the this element are much less common and less s
Molodets [167]

Answer:

HOFO = (0, 0, +1, -1)

Explanation:

The formal charge (FC) can be calculated using the following equation:

FC = V - N - \frac{1}{2}B

<u>Where:</u>

V: are the valence electrons

N: are the nonbonding electrons

B: are the bonding electrons

The arrange of the atoms in the oxyacid is:

H - O₁ - F - O₂

Hence, the formal charge (FC) on each of the atoms is:

H: FC = 1 - 0 - 1/2*(2) = 0            

O₁: FC = 6 - 4 - 1/2*(4) = 0        

F: FC = 7 - 4 - 1/2*(4) = +1

O₂: FC =  6 - 6 - 1/2*(2) = -1

We can see that the negative charge is in the oxygen instead of the most electronegative element, which is the F. This oxyacid is atypical.  

I hope it helps you!

3 0
2 years ago
Acetone major species present when dissolved in water
jek_recluse [69]

Answer: acetone molecule ( CH₃-CO-CH₃)


Explanation:


1) Acetone is CH₃-CO-CH₃


2) That is a molecule (build up of covalent bonds).


3) When dissolved, covalent bonded compounds remain as separate molecules, then it is said that the major species present in the solution is the molecule. The molecules of acetone are surrounded (sovated) by the molecules of water.


This as opposed to the case of ionic compounds that ionize. When a compound as NaCl dissolves in water, it ionizes completely, so the major speceies are not NaCl formulas, but the ions Na⁺ and Cl⁻, not molecules.


That leads to the answer: the major species present when acetone is dissolved in water is the molecules of acetone (you do not need to state the fact that the molecules of water are part of the solution, because that is not the target of the question).



3 0
3 years ago
3. What mass of copper could be deposited from a copper(II) Sulphate solution using a current of 5.0 A over 100 seconds? ( F =96
arsen [322]

Mass of copper : 0.165 g

<h3>Further explanation</h3>

Given

5.0 A over 100 seconds

Required

Mass of copper

Solution

Faraday's law:

<em>The mass of the substance formed at each electrode is proportional to the electric current flowing in the electrolysis</em>

<em />\tt W=\dfrac{e.i.t}{96500}<em />

e = Ar / valence = eqivalent weight

i = current

t = time

W = weight

CuSO₄ ----> Cu²⁺ + SO₄²⁻

Cu ----> Cu²⁺ + 2e

e = Ar/2

= 63,5/2 = 31,75

\tt W=\dfrac{31.75\times 5\times 100}{96500}=0.165~g

8 0
2 years ago
Water from a riverbed carries sediment downstream. The water pressure cuts deep into the riverbed, creating a deep, narrow chann
atroni [7]

Answer:

A canyon has formed.

Explanation:

5 0
2 years ago
n a coffee cup calorimeter, 100.0 mL of 1.0M NaOH and 100.0 mL of 1.0M HCl are mixed. Both solutions were originally at 24.6 deg
RoseWind [281]

Answer:

Explanation:

mass of the solution = volume x density = 200 x 1 = 200 gm

heat absorbed = m x s x Δ t , s is specific heat , Δt is rise in temperature

= 200 x 4.18 x ( 31.3 - 24.6 )

= 5601 J .

This is the enthalpy change required.

3 0
3 years ago
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