For the answer to the question above asking, h<span>ow many moles of glucose (C6H12O6) are in 1.5 liters of a 4.5 M C6H12O6 solution?
The answer to your question is the the third one among the given choices which is 6.8 mol.
</span><span>moles glucose = 1.5 x 4.5 = 6.8 </span>
Using the chart that has been provided, we may determine water temperature. We do this by drawing a straight line form the bottom scale which has the ppm of oxygen dissolved to the middle scale which has the percentage saturation.
The line starts from 11.5 ppm on the bottom scale and goes to 90% on the middle scale. Next, we continue this line, without changing its slope, to the third scale showing temperature. We see that it crosses the temperature scale at 4°C.
The temperature of the water is 4 °C.
1.0 mole ---------- 6.02x10²³ molecules
4.5 moles -------- ?
4.5 * 6,02x10²³ / 1.0
= 2.709x10²⁴ molecules units
Explanation:
Let us take the volume of block is x.
Since, the block is floating this means that it is in equilibrium. Formula to calculate net force will be as follows.
Also, buoyancy force 
= (volume submerged in water × density of water) + (volume in oil × density of oil)
= 
= 
g
As, W = V × density of graphite × g
It is given that density of graphite is 
or 2160 
.
So, W = 2160 V g
= (0.592 V \rho + 408 V) g - 2160 V g = 0
= 1752
= 2959.46 or 2.959 
is the density of oil.
It is given that mass of flask is 124.8 g.
Mass of 35.3 
oil = 
104.7 g
Hence, in second weighing total mass will be calculated as follows.
(124.8 + 104.7) g
= 229.27 g
Thus, we can conclude that in the second weighing mass is 229.27 g.
