Electric Potential is the work done per unit charge in order to bring the charge from infinity to a point in electric field while Electric potential difference is the Potential developed while moving a charge from one point to another in the field itself.
Answer:
CHAPTER 2 ATOMS, MOLECULES, AND IONS Questions 16. ... A molecule has no overall charge (an equal number of electrons and protons are present). Ions ... The first sample of chloroform has a total mass of 12.0 g C + 106.4 g Cl + 1.01 g H ... SrF2 f. Al2Se3 g. K3N h. Mg3P2 72. a. mercury(I) oxide b. iron(III) bromide c.
Explanation:
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Answer:
They represent it by ensuring that the number of atoms of each element (matter) in the reactant side is the same as the product side
Explanation:
The law of conservation of matter stated that matter can neither be created nor destroyed. Chemical equations involve combining atoms of elements. The compounds combined by chemists are called REACTANTS while the produced compounds are called PRODUCTS.
In order to conform to the law of conservation of matter, the same quantity of matter present in the reactants must be present in the products. This means that the number of atoms of each element (matter) in the reactant side must be the same as the product side. For example;
C6H12O6 + 6O2 → 6CO2 + 6H2O
In this chemical equation for photosynthesis, number of atoms in the reactant side (6 carbon, 12 hydrogen, 18 oxygen) are the same as that in the product side (6 carbon, 12 hydrogen, 18 oxygen), hence, this obeys the law of conservation of mass.
In a nutshell, chemists chemists properly represent the law of conservation of matter in their chemical equations by making sure that same number of atoms of reactants is present in the products.
1. base
2. neutral
3. acid
4. base
If this is what you meant, hope it helps
Answer:
solubility is 1.984x10⁻⁹M
Explanation:
When CaCO₃ is in water, the equilibrium that occurs is:
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Kps = [Ca²⁺] [CO₃²⁻] = 4.96x10⁻⁹
If you have a 0.250M solution of Na₂CO₃, [CO₃²⁻] = 0.250M:
[Ca²⁺] [0.250M] = 4.96x10⁻⁹
Assuming you are adding an amount of CaCO₃:
[X] [0.250 + X] = 4.96x10⁻⁹
<em>Where X is the amoun of CaCO₃ you can add, that means, solubility</em>
X² + 0.250X - 4.96x10⁻⁹ = 0
Solving for X:
X = -0.25M → False answer, there is no negative concentrations.
X = 1.984x10⁻⁹M.
That means, <em>solubility is 1.984x10⁻⁹M</em>
