Answer:
Explanation:
Incorrect use has resulted in accidents and disfiguring burns. Never leave a methylated spirit appliance unattended. Also make sure that the camping stove or appliance is on a flat surface and that the fuel cannot spill out.
1. The hypothesis for this is experiment is that the 50:50 of methanol-water mixture will not turn to solid when the temperature reaches to -40°C.
2. The procedure for this is measuring equal volumes of water and methanol using the graduated cylinder. You can measure 100 mL of water and 100 mL of methanol using the graduated cylinder. Then, mix them in the beaker. Next, measure 200 mL of water, and another 200 mL of methanol. Don't mix them. Also, make a 60:40 mixture by measuring 120 mL of water and 80 mL of methanol, then mix them together. Place them all in the refrigerator at the same time. Record the time when they would freeze to solid.
3. The controls for this experiment are the 200 mL water alone, and the 200 mL methanol alone.
4. The independent variable in here is the time, while the dependent variable is the temperature of the mixtures.
5. If the hypothesis turns out to be true, then all the mixtures prepared should freeze and become solid after a certain period of time, with the exception of the 50:50 mixture. The 50:50 mixture should still remain as a liquid even when left overnight.
Increase in Oxygen shift the equilibrium towards reactant side.
<u>Explanation:</u>
6CO₂ + 6H₂O ⇄ C₆H₁₂O₆ + 6O₂
This is the reaction occurs in the photosynthesis of plants by means of sunlight. In this case, if the concentration of Oxygen increases or adding more oxygen to the product side will shift the equilibrium towards the reactant side according to the Le Chatlier's principle, which adjusts the equilibrium by itself for any changes that is increase or decrease in pressure, temperature or concentration of reactants or products.
Answer:
1.
643.21g 1 mol 6.022^23
262.87 g 1 mol
= 1.4735E24 [Mg3(PO4)2]
2.
4.061x10^24 1mol 22.4 (L)
6.022^23 1mol
= 151 liters H2O2
3.
479.3g 1 mol 6.022^23
18.02g 1mol
= 1.60E25 H20 atoms
4.
80.34L 1mol 164.1
22.4L 1mol
588.6g Ca(NO3)2
5.
893.7g 1mol 22.4
44.01g 1mol
= 427 L CO2 or 427.4
6.
5.39 x 10^25 1mol 78.01
6.022^23 1mol
= 6980g Al(OH)3
hope this helps!! :)
