Density=mass/volume
density=2/8
density=0.25g/ml
Answer:
The three statements are true
Explanation:
For the reaction:
I₂O₅(s) + 5CO(g) → I₂(s) + 5CO₂(g)
State oxidation of iodine in I₂O₅ is:
5 O²⁻ = 10⁻
As you have 2 I and the molecule has no charge, <em>oxidation state of I is +5</em>.
The carbon in CO has an oxidation state of +2 and in CO₂ is +4. That means <em>the carbon is oxidized</em>
<em />
An oxidizing agent is a substance that produce the oxidation of the agent that reacts with this one. CO is oxidized because of I₂O₅ is producing its oxidation being <em>the oxidizing agent</em>
<em></em>
Thus,<em> the three statements are true</em>.
Answer:
The pair co,Ni is out of order in terms of atomic mass
Explanation:
For co,Ni
Atomic mass of Co = 58.9331 u
Atomic mass of Ni = 58.6934
For Li,be
Atomic mass of Li = 6.941 u
Atomic mass of be = 9.012182 u
For I,xe
Atomic mass of I = 126.90447 u
Atomic mass of Xe = 131.293 u
Hence, the pair co,Ni is out of order in terms of atomic mass
In Lewis dot structures, you draw the atom in the center and then surround the atom with its valence electrons. The Lewis structure for O is as shown in the attached diagram.
<h3>What is the Lewis structure of O ?</h3>
Lewis Structure of an atom of oxygen contains 6 electrons in the valence shell. Four valence electrons exist in lone pairs. It implies that oxygen atom must participate in two single bonds or one double bond in order to have an octet configuration.
A simplified representation of the valence shell electrons in a molecule is called Lewis Structure. It shows how electrons are arranged around individual atoms in the molecule.
To know more about Lewis structure, refer
brainly.com/question/1525249
#SPJ4
The solution would be like
this for this specific problem:
<span>Moles of carbon = 58.8 /
12 = 4.9 </span><span>
<span>Moles of hydrogen = 9.8 / 1 = 9.8 </span>
<span>Moles of oxugen = 31.4 / 16 m= 1.96 </span>
<span>Ratio 4.9 / 1.96 = 2.5 9.8 / 1.96 = 5.0 1.96 / 1.96 = 1 </span></span>
Simplest
formula = C5H10<span>
</span><span>I hope this helps and if
you have any further questions, please don’t hesitate to ask again.</span>
