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
Answer:
Polyester
Acrylic
Nylon
Are all made of fossil fuels.
<em>malai</em><em> </em><em>aaudaina</em><em> </em><em>aati</em><em> </em><em>jabo</em><em> </em><em>tapai</em><em> </em>
<em>lai</em><em> </em><em>aaudaina</em><em> </em>
Answer:
24x10³
Explanation:
2CO₂(g) + 4H₂O(g) → 2CH₃OH(l) + 3O₂ (g)
The equilibrium constant for this reaction is:
Kc = ![\frac{[O_2]^3}{[CO_2]^2[H_2O]^4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BO_2%5D%5E3%7D%7B%5BCO_2%5D%5E2%5BH_2O%5D%5E4%7D)
The expression of [CH₃OH] is left out as it is a pure liquid.
Now we <u>convert the given masses of the relevant species into moles</u>, using their <em>respective molar masses</em>:
- CO₂ ⇒ 3.28 g ÷ 44 g/mol = 0.0745 mol CO₂
- H₂O ⇒ 3.86 g ÷ 18 g/mol = 0.214 mol H₂O
- O₂ ⇒ 2.80 g ÷ 32 g/mol = 0.0875 mol O₂
Then we calculate the concentrations:
- [CO₂] = 0.0745 mol / 7.5 L = 0.0099 M
- [H₂O] = 0.214 mol / 7.5 L = 0.0285 M
- [O₂] = 0.0875 mol / 7.5 L = 0.0117 M
Finally we <u>calculate Kc</u>:
- Kc =
= 24x10³
