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
<h2>Answer:</h2><h3>Part 1:</h3>
Location the element zinc (Zn) on the periodic table:
- Group number : 12
- Period number : 4
- Block : d block
- Element : Transition elements.
<h3>Part 2:</h3>
Protons in an atom of Zn: 30
<h3>Part 3:</h3>
Electrons in a Zn atom: 30
<h3>Part 4 :</h3>
Neutron in an atom of Zn: 35
<h3 />
Answer:
18.22874999999973
I recommend you to round the nearest 1 d.p
Explanation:
<em>h</em><em>a</em><em>v</em><em>e</em><em> </em><em>a</em><em> </em><em>g</em><em>r</em><em>e</em><em>a</em><em>t</em><em> </em><em>d</em><em>a</em><em>y</em><em>!</em>
Answer:
74.4 ml
Explanation:
C₆H₈O₇(aq) + 3NaHCO₃(s) => Na₃C₆H₅O₃(aq + 3CO₂(g) + 3H₂O(l)
Given 15g = 15g/84g/mol = 0.1786mole Sodium Bicarbonate
From equation stoichiometry 3moles NaHCO₃ is needed for each mole citric acid or, moles of citric acid needed is 1/3 of moles sodium bicarbonate used.
Therefore, for complete reaction of 0.1786 mole NaHCO₃ one would need 1/3 of 0.1786 mole citric acid or 0.0595 mole H-citrate.
The question is now what volume of 0.8M H-citrate solution would contain 0.0595mole of the H-citrate? This can be determined from the equation defining molarity. That is => Molarity = moles solute / Liters of solution
=> Volume (Liters) = moles citric acid / Molarity of citric acid solution
=> Volume needed in liters = 0.0.0595 mole/0.80M = 0.0744 Liters or 74.4 ml
