Answer:
<u>Our beaches would be unprotected</u>
In the short-term, these artificial sand hills will be destroyed by the elements. Because sand dunes protect inland areas from swells, tides, and winds, they must be protected and defended like national treasures. ... The ocean and the wind can have an unpredictable, destructive force on coastal regions.
- surfertoday
Natural sand dunes play a vital role in protecting our beaches, coastline and coastal developments from coastal hazards such as erosion, coastal flooding and storm damage. Sand dunes protect our shorelines from coastal erosion and provide shelter from the wind and sea spray.
- Waikato Regional Council
Your answer is radio waves
Answer:
1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹
2. 0.58 mol
Explanation:
1.Given ΔO₂/Δt…
2H₂O₂ ⟶ 2H₂O + O₂
-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt
d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹
d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹
2. Moles of O₂
(a) Initial moles of H₂O₂
(b) Final moles of H₂O₂
The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.
(c) Moles of H₂O₂ reacted
Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol
(d) Moles of O₂ formed
answer:
This allows us to tell alkenes apart from alkanes using a simple chemical test. Bromine water is an orange solution of bromine. It becomes colourless when it is shaken with an alkene. Alkenes can decolourise bromine water, but alkanes cannot.
explanation:
Explanation:
Moles of N2 = 35.0g / (28g/mol) = 1.25mol
Moles of H2 = 60.0g / (2g/mol) = 30.0mol
Since 1.25mol * 3 < 30.0mol, nitrogen is limiting.
Moles of NH3 = 1.25mol * 2 = 2.50mol.
Mass of NH3 = 2.50mol * (17g/mol) = 42.5g.
30.0mol - 1.25mol * 3 = 26.25mol.
Excess mass of H2
= 26.25mol * (2g/mol) = 52.5g.
