Answer:
Sliding friction
Explanation:
SLiding friction produces force
Answer:
1.427x10^-3mol per L
Explanation:
I could use ⇌ in the math editor so I used ----
from the question each mole of Y(IO3)3 is dissolved and this is giving us a mole of Y3+ and a mole of IO3^3-
Ksp = [Y^3+][IO3-]^3
So that,
1.12x10^-10 = [S][3S]^3
such that
1.12x10^-10 = 27S^4
the value of s is 0.001427mol per L
= 1.427x10^-3mol per L
so in conclusion
the molar solubility is therefore 1.427x10^-3mol per L
There is a missing portion of this question which shows the reaction that needs balancing:
"In a balanced equation, the same number of each kind of atom is shown on each side of the equation. Calculate the number of iron (Fe), oxygen (O), and carbon atoms (C).
Fe2O3+ 3CO --> 2Fe + 3CO<span>2
</span><span>Based on these values, is the equation balanced?</span><span>"
</span>
To check if this equation is balanced we simply compare the number of each element on each side of the equation.
On the reactant side of the equation we have:
2 Fe atoms
6 O atoms
3 C atoms
On the product side of the equation we have:
2 Fe atoms
6 O atoms
3 C atoms
Therefore, both side of the reaction have the correct and equal number of each atom, so the equation is balanced.
<span>Made up of particles packed relatively close together, having an indefinite shape but a definite volume</span>
Answer:
Explanation:
Cu²⁺ + 2e⁻ → Cu ( copper gets reduced )
Cu → Cu²⁺ + 2e⁻ ( copper gets oxidized )
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O
The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
