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:
THE VOLUME OF 0.200M CALCIUM HYDROXIDE NEEDED TO NEUTRALIZE 35 mL of 0.050 M NITRIC ACID IS 43.75 mL.
Explanation:
Using
Ca VA / Cb Vb = Na / Nb
Ca = 0.0500 M
Va = 35 mL
Cb = 0.0200 M
Vb = unknown
Na = 2
Nb = 1
Equation for the reaction:
Ca(OH)2 + 2HNO3 --------> Ca(NO3)2 + 2H2O
So therefore, we make Vb the subject of the equation and solve for it
Vb = Ca Va Nb / Cb Na
Vb = 0.0500 * 35 * 1 / 0.0200 * 2
Vb = 1.75 / 0.04
Vb = 43.75 mL
The volume of 0.02M calcium hydroxide required to neutralize 35 mL of 0.05 M nitric acid is 43.75 mL
Answer:
ΔH = 2.68kJ/mol
Explanation:
The ΔH of dissolution of a reaction is defined as the heat produced per mole of reaction. We have 3.15 moles of the solid, to find the heat produced we need to use the equation:
q = m*S*ΔT
<em>Where q is heat of reaction in J,</em>
<em>m is the mass of the solution in g,</em>
<em>S is specific heat of the solution = 4.184J/g°C</em>
<em>ΔT is change in temperature = 11.21°C</em>
The mass of the solution is obtained from the volume and the density as follows:
150.0mL * (1.20g/mL) = 180.0g
Replacing:
q = 180.0g*4.184J/g°C*11.21°C
q = 8442J
q = 8.44kJ when 3.15 moles of the solid react.
The ΔH of the reaction is:
8.44kJ/3.15 mol
= 2.68kJ/mol
