Answer:
concave lens
Explanation:
it's concave lens because it diverges the ray/beam of light.
Answer:
Elecric charge/ Electricity
Explanation:
Electric charge is the net gain or loss of electrons
I hope im right!!
C) medium.
The medium of a wave is any substance that carries the wave, or through which the wave travels.Ocean waves are carried by water, sound waves are carried by air, and. the seismic waves of an earthquake are carried by rock and soil.
The answer is: H₃PO₄.
A phosphoric acid is three protic acid, which means that in water release tree protons.
Phosphoric acid ionizes in three steps in water.
First step: H₃PO₄(aq) ⇄ H₂PO₄⁻(aq) + H⁺(aq).
Second step: H₂PO₄⁻(aq)⇄ HPO₄²⁻(aq) + H⁺(aq).
Third step: HPO₄²⁻(aq) ⇄ PO₄³⁻(aq) + H⁺(aq).
Species that are present: H₃PO₄, H₂PO₄⁻, HPO₄²⁻, PO₄³⁻ and H⁺.
A buffer can be defined as a substance that prevents the pH of a solution from changing by either releasing or absorbing H⁺ in a solution.
Buffer is a solution that can resist pH change upon the addition of an acidic or basic components and it is able to neutralize small amounts of added acid or base, pH of the solution is relatively stable.
Answer:
The answer to your question is below
Explanation:
Data
mass of CaCO₃ = 155 g
mass of HCl = 250 g
mass of CaCl₂ = 142 g
reactants = CaCO₃ + HCl
products = CaCl₂ + CO₂ + H₂O
1.- Balanced chemical reaction
CaCO₃ + 2HCl ⇒ CaCl₂ + CO₂ + H₂O
2.- Limiting reactant
molar mass of CaCO₃ = 40 + 12 + 48 = 100 g
molar mass of HCl = 2[1 + 35.5 ] = 73 g
theoretical proportion CaCO₃ /HCl = 100 / 73 = 1.37
experimental proportion CaCO₃ /HCl = 155 / 250 = 0.62
As the experimental proportion was lower than the theoretical proportion the limiting reactant is CaCO₃
3.-
Calculate the molar mass of CaCl₂
CaCl₂ = 40 + 71 = 111 g
100 g of CaCO₃ ------------------ 111 g of CaCl₂
155 g of CaCO₃ ----------------- x
x = (155 x 111) / 100
x = 17205 / 100
x = 172.05 g of CaCl₂
4.- percent yield
Percent yield = 142 / 172.05 x 100 = 82.5 %
5.- Excess reactant
100 g of CaCO₃ -------------------- 73 g of HCl
155 g of caCO₃ ------------------- x
x = (155 x 73)/100
x = 133.15 g
Mass of HCl = 250 - 133.15
= 136.9 g
