Reaction: 2K₍s₎ + 2H₂O₍l₎ → 2KOH₍aq₎ + H₂₍g₎.
K - potassium.
H₂O - water.
KOH - potassium-hydroxide.
H₂ - hydrogen.
s - solid phase.
l - liquid.
aq - disolves in water.
g - gas.
Reaction is exothermal (release of energy) and potassium burns a purple flame. H<span>ydrogen released during the reaction reacts with </span>oxygen<span> and ignites.</span><span>
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These are two questions and two answers
Question 1.
Answer:
Explanation:
<u>1) Data:</u>
a) m = 9.11 × 10⁻³¹ kg
b) λ = 3.31 × 10⁻¹⁰ m
c) c = 3.00 10⁸ m/s
d) s = ?
<u>2) Formula:</u>
The wavelength (λ), the speed (s), and the mass (m) of the particles are reltated by the Einstein-Planck's equation:
- h is Planck's constant: h= 6.626×10⁻³⁴J.s
<u>3) Solution:</u>
Solve for s:
Substitute:
- s = 6.626×10⁻³⁴J.s / ( 9.11 × 10⁻³¹ kg × 3.31 × 10⁻¹⁰ m) = 2.20 × 10 ⁶ m/s
To express the speed relative to the speed of light, divide by c = 3.00 10⁸ m/s
- s = 2.20 × 10 ⁶ m/s / 3.00 10⁸ m/s = 7.33 × 10 ⁻³
Answer: s = 7.33 × 10 ⁻³ c
Question 2.
Answer:
Explanation:
<u>1) Data:</u>
a) m = 45.9 g (0.0459 kg)
b) s = 70.0 m/s
b) λ = ?
<u>2) Formula:</u>
Macroscopic matter follows the same Einstein-Planck's equation, but the wavelength is so small that cannot be detected:
- h is Planck's constant: h= 6.626×10⁻³⁴J.s
<u>3) Solution:</u>
Substitute:
- λ = 6.626×10⁻³⁴J.s / ( 0.0459 kg × 70.0 m/s) = 2.06 × 10 ⁻³⁴ m
As you see, that is tiny number and explains why the wave nature of the golf ball is undetectable.
Answer: 2.06 × 10 ⁻³⁴ m.
Ba 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s² → [Xe]6s²
Ba - 2e⁻ → Ba⁺² [Xe]
The metal is aluminium
<u>Explanation:</u>
Given:
Heat, q = 4680 J
Mass, m = 100g = 0.1kg
ΔT = 52°C
sample = ?
We know:
q = mcΔT
On substituting the value we get:
Thus, the metal is aluminium which has a specific heat capacity of 900 J/kg°C
The balanced chemical reaction is:
2HCl + Ca = CaCl2 + H2
We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.
100 g HCl ( 1 mol HCl / 36.46 g HCl ) = 2.74 mol HCl
100 g Ca ( 1 mol Ca / 40.08 g ) = 2.08 mol Ca
From the reaction, the mole ratio of the reactants is 2:1 where every 2 moles of hydrochloric acid, 1 mole of calcium is required. Therefore, the limiting reactant for this case is calcium.
