It would be copper and iron -A
<span>The
surface of a mirror is smoothly and therefore the rays of light are
reflected with the same angle they hit the surface. The surface of a
piece of wood on the other hand has a lot of microscopical
imperfections and hence the rays of light are reflected in multiple
directions.</span>
Answer:
Option d is correct = Negative ion
Explanation:
We know that an atom consist of electrons, protons and neutrons. Neutrons and protons are present inside the nucleus while electrons are present out side the nucleus. Electron has a negative charge and is written as e⁻. The mass of electron is 9.10938356×10⁻³¹ Kg
. While mass of proton and neutron is 1.672623×10⁻²⁷Kg and 1.674929×10⁻²⁷ Kg respectively.
Symbol of proton= P⁺
Symbol of neutron= n⁰
The number of electron or number of protons are called atomic number while mass number of an atom is sum of protons and neutrons. The umber of protons and electrons are always equal to make the atom electrically neutral and when an atom loses its valance electron the number of protons increases and thus positive charge increased and atom form cation.
When an atom gain electron negative charge increase because of more number of electron thus atom form negative ion or anion. For example,
Anion formation:
X + e⁻ → X⁻
Cation formation:
X → X⁺ + e⁻
Thus option d is correct option.
Answer:
B
Explanation:
try your best and dont guess <3
Answer:
1. 35 mg of H₃PO₄
2. 27 mol AlF₃; 82 mol F⁻
3. 300 mL of stock solution.
Explanation:
1. Preparing a solution of known molar concentration
Data:
V = 80 mL
c = 4.5 × 10⁻³ mol·L⁻¹
Calculations:
(a) Moles of H₃PO₄
Molar concentration = moles of solute/litres of solution
c = n/V
n = Vc = 0.080L × (4.5 × 10⁻³ mol/1 L) = 3.60 × 10⁻⁴ mol
(b) Mass of H₃PO₄
moles = mass/molar mass
n = m/MM
m = n × MM = 3.60 × 10⁻⁴ mol × (98 g/1 mol) = 0.035 g = 35 mg
(c) Procedure
Dissolve 35 mg of solid H₃PO₄ in enough water to make 80 mL of solution,
2. Moles of solute.
Data:
V = 4900 mL
c = 5.6 mol·L⁻¹
Calculations:
Moles of AlF₃ = cV = 4.9 L AlF₃ × (5.6 mol AlF₃/1L AlF₃) = 27 mol AlF₃
Moles of F⁻ = 27 mol AlF₃ × (3 mol F⁻/1 mol AlF₃) = 82 mol F⁻.
3. Dilution calculation
Data:
V₁= 750 mL; c₁ = 0.80 mol·L⁻¹
V₂ = ? ; c₂ = 2.0 mol·L⁻¹
Calculation:
V₁c₁ = V₂c₂
V₂ = V₁ × c₁/c₂ = 750 mL × (0.80/2.0) = 300 mL
Procedure:
Measure out 300 mL of stock solution. Then add 500 mL of water.
