Advantages
less wasted energy
Reduces fossilisation
Disadvantages
- Expensive
- doesn't look nice ( apparently)
Gneiss is a sedimentary rock and granite is a metamorphic rock.
<h3>What are rocks?</h3>
Rocks are geological hard materials that are made up of various types which include:
- Sedimentary rocks: These are rocks that made up of various layers formed from sediments. Example is the gnesis.
- Metamorphic rocks: These are rocks that are form from pre existing rocks that undergoes some transformation. Example is granite
Therefore, Gneiss is a sedimentary rock and granite is a metamorphic rock.
Learn more about rocks here:
brainly.com/question/398139
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Answer:
<h3>The precipitate is MgCl2</h3>
Explanation:
The reaction that is described goes as follows:
2AgCl + Mg(OH)2 ---> MgCl2 + 2AgOH
The precipitate here is the MgCl2 salt.
I hope it helped!
1) Chemical reaction
HCl + NaOH ---> NaCl + H2O
25.0 ml
0.150 M 0.250M
2) 50% completion => 0.025 l * 0.150 M * (1/2) = 0.001875 mol HCl consumed and 0.001875 mol HCl in solution
0.001875 mol HCl => 0.001875 mol H(+)
Volume = Volume of HCl solution + Volumen of NaOH solution added
Volume of HCl solution = 0.0250 l
Volume of NaOH = n / M = 0.001875 mol / 0.250M = 0.0075 l
Total volume = 0.0250 l + 0.0075 l = 0.0325 l
[H+] = 0.001875 mol / 0.0325 l = 0.05769 M
pH = - log [H+] = - log (0.05769) = 1.23
Answer: 1.23
3) Equivalence point
0.02500 l * 0.150 M = 0.250M * V
=> V = 0.02500 * 0.150 / 0.250 = 0.015 l
4) 1.00 ml NaOH added beyond the equivalence point
1.00 ml * 1 l / 1000 ml * 0.250 M = 0.00025 mol NaOH in excess
0.00025 mol NaOH = 0.00025 mol OH-
Volume of the solution = 0.02500 l + 0.015 l + 1.00/1000 l = 0.041 l
[OH-] = 0.00025 mol / 0.041 l = 0.00610 M
pOH = - log (0.00610) = 2.21
pH + pOH = 14 => pH = 14 - pOH = 14 - 2.21 = 11.76
Answer: 11.76
So we know that the equation for density is:
where D is the density, m is the mass in grams, and V is the volume in mL.
So since we know two of the variables, mass and density, we can solve for the volume:
Therefore, the volume of this urine sample is 144.12mL.
