Answer:
identify and name the landforms in the following diagrams
Answer:
First, let's determine how many moles of oxygen we have.
Atomic weight oxygen = 15.999
Molar mass O2 = 2*15.999 = 31.998 g/mol
We have 3 drops at 0.050 ml each for a total volume of 3*0.050ml = 0.150 ml
Since the density is 1.149 g/mol,
we have 1.149 g/ml * 0.150 ml = 0.17235 g of O2
Divide the number of grams by the molar mass to get the number of moles 0.17235 g / 31.998 g/mol = 0.005386274 mol
Now we can use the ideal gas law. The equation PV = nRT where P = pressure (1.0 atm) V = volume n = number of moles (0.005386274 mol) R = ideal gas constant (0.082057338 L*atm/(K*mol) ) T = Absolute temperature ( 30 + 273.15 = 303.15 K)
Now take the formula and solve for V, then substitute the known values and solve.
PV = nRT V = nRT/P V = 0.005386274 mol * 0.082057338 L*atm/(K*mol) * 303.15 K / 1.0 atm V = 0.000441983 L*atm/(K*) * 303.15 K / 1.0 atm V = 0.133987239 L*atm / 1.0 atm V = 0.133987239 L
So the volume (rounded to 3 significant figures) will be 134 ml.
Reaction of Cu(NO₃)₂ with each salt is as follow,
1) with KNO₃;
Cu(NO₃)₂ + KNO₃ → Cu(NO₃)₂ + KNO₃
Both salt products are water soluble.
2) With CuSO₄;
Cu(NO₃)₂ + CuSO₄ → CuSO₄ + Cu(NO₃)₂
Again both Salt products are water soluble.
3) With K₂SO₄;
Cu(NO₃)₂ + K₂SO₄ → CuSO₄ + 2 KNO₃
Again both salt products are water soluble.
4) With K₂S;
Cu(NO₃)₂ + K₂S → CuS + 2 KNO₃
In this case CuS is water insoluble, hence precipitates out.
Result:
Option-4 is the correct answer.
<span>An insect would have an easier time walking on the surface of water than on the surface of ethanol. Water's stronger intermolecular forces lead to higher surface tension. Higher surface tension allows water to support the insect. I hope this helps.</span>
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