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Nana76 [90]
2 years ago
13

Suppose you had 1 mole of oxygen (O2). How many moles of hydrogen (H2) would react completely with the oxygen, and how many mole

s of H2O would b produced? Show the balance equation with explanations.
Chemistry
1 answer:
Lostsunrise [7]2 years ago
7 0
First, we write an equation for the reaction of Hydrogen and Oxygen to produce water:
H₂ + O₂ ⇒ H₂O
This equation is not balanced, we then balance the Oxygens:
H₂ + O₂ ⇒ 2H₂O
And now the Hydrogens:
2H₂ + O₂ ⇒ 2H₂O
Therefore, if one mole of Oxygen reacts, it will require two moles of Hydrogen and produce two moles of water.
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Identify the reactants 4 A g + 2 H 2 S + O 2​
Arturiano [62]
2Ag2S + 2H2O—>4Ag+2H2S+O2

The reactants are silver sulphide (Ag2s) and water (H2O)
8 0
3 years ago
Does mass change when you add alka seltzer to water??
Gennadij [26K]

Answer & Explanation:

As an Alka-Seltzer tablet dissolves in water, it liberates carbon dioxide. ... This carbon dioxide gas has mass, but since it is a gas it escapes from the container and diffuses into the atmosphere. The loss of mass from the container is measured directly with the analytical balance.

5 0
3 years ago
Read 2 more answers
A non-stoichiometric compound is a compound that cannot be represented by a small whole-number ratio of atoms, usually because o
ikadub [295]
<span>Average oxidation state = VO1.19
Oxygen is-2. Then 1.19 (-2) = -2.38
Average oxidation state of V is +2.38 

Consider 100 formula units of VO1.19
There would be 119 Oxide ions = Each oxide is -2. Total charge = -2(119) = -238 
The total charge of all the vanadium ions would be +238. 
Let x = number of of V+2 
Then 100 – x = number of V+3 
X(+2) + 100-x(+3) = +238 
2x + 300 – 3x = 238 
-x = 238-300 = -62 
x = 62
 
Thus 62/100 are V+2 
62/100 * 100 = 62%

</span>62 % is the percentage of the vanadium atoms are in the lower oxidation state. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help. 
7 0
3 years ago
Which procedure must be completed before switching to high power objective on a microscope
Elis [28]
1.turn  the  revolving  turret  so  that low  power  lens  come  into  position
2 place  the  microscope  slide  on  the  stage
3 look  at  the  objective  lens  and the  stage  from  the   side  as  you  turn   adjust  knob  so  that  stage   can  move  upward
4 look  through  the  eyepiece  and  move  the    focus  knob  until  image  come  into  focus
5.adjust  the   condenser   to regulate  thee  amount of  light
6 move  the  microscope  slide  around  until  the  sample  is  in  the   center of  field  of  view







7 0
3 years ago
2 A(g) + B(g) ---&gt; 2 C(g)Rate = k [A][B]At the beginning of one trial of this reaction, [A] = 3.0 M and [B] = 1.0 M. The obse
Orlov [11]

Answer:

Lmol⁻¹s⁻¹

Explanation:

The rate law of the given reaction is:-

Rate=k[A][B]

Wherem, k is the rate constant.

Given that:-

Rate = 0.36 mol/Lsec = 0.36 M/sec

[A] = 3.0 M

[B] = 1.0 M

Thus,

Applying in the equation as:-

0.36 M/sec =k × 3.0 M× 1.0 M

k = 0.12 (Ms)⁻¹ = 0.12 Lmol⁻¹s⁻¹

<u>The units of k = Lmol⁻¹s⁻¹</u>

6 0
3 years ago
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