Answer:
The concentration of this sodiumhydroxide solutions is 0.50 M
Explanation:
Step 1: Data given
Mass of sodium hydroxide (NaOh) = 8.0 grams
Molar mass of sodium hydroxide = 40.0 g/mol
Volume water = 400 mL = 0.400 L
Step 2: Calculate moles NaOH
Moles NaOH = mass NaOH / molar mass NaOH
Moles NaOH = 8.0 grams / 40.0 g/mol
Moles NaOh = 0.20 moles
Step 3: Calculate concentration of the solution
Concentration solution = moles NaOH / volume water
Concentration solution = 0.20 moles / 0.400 L
Concentration solution = 0.50 M
The concentration of this sodiumhydroxide solutions is 0.50 M
The correct answer is - deflation.
The process of deflation can be caused by the winds. It is an erosive process in which the main role has the wind that is carrying lot of sediment in the shape of very small particles with it.
Through this process, the winds manage to erode large areas, especially in the drier places where the vegetation is very sparsely distributed. By this type of erosion, the winds manage to make lot of hollows that can range significantly in size. The hollows made by the deflation can be anywhere from few cm deep and several meters long, up to several km long and 50-60 meters of depth.
This is the process that is responsible for the creation of most of the oasis in the largest desert in the world, Sahara, some even being lowered enough to be under the sea level.
Answer:
60 g/100 g water
Explanation:
Find 5 °C on the horizontal axis.
Draw a line vertically from that point until you reach the solubility curve for CaCl₂.
Then draw a horizontal line from there to the vertical axis.
The solubility of CaCl₂ is 60 g/100 g water.
The correct answer is approximately 11.73 grams of sulfuric acid.
The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.
The balanced equation is:
2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O
Each mole of Al(OH)3 corresponds to 3/2 moles of H₂SO₄. The molecular mass of Al(OH)3 is 78.003 g/mol. There are 15/78.003 = 0.19230 moles of Al(OH)3 in the five grams of Al(OH)3 available. Al(OH)3 is in limiting, which means that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g
40 grams of sulfuric acid is available, out of which 28.289 grams is consumed. The remaining 40-28.289 = 11.711 g is in excess, which is closest to the first option, that is, 11.73 grams of H₂SO₄.
If the sun was not there the earth would travel in a straight line
