Answer:
Specific heat of calcium carbonate(C) = 0.82 (Approx)
Explanation:
Given:
Energy absorbs (q) = 85 J
Change in temperature (Δt) = 34.9 - 21 = 13.9°C
Mass of calcium carbonate = 7.47 g
Find:
Specific heat of calcium carbonate(C)
Computation:
Specific heat of calcium carbonate(C) = q / m(Δt)
Specific heat of calcium carbonate(C) = 85 / (7.47)(13.9)
Specific heat of calcium carbonate(C) = 85 / 103.833
Specific heat of calcium carbonate(C) = 0.8186
Specific heat of calcium carbonate(C) = 0.82 (Approx)
Answer:
Glucose
Explanation:
Plants make glucose to store as energy.
Answer:
Explanation:
To solve this problem, we need to obtain the number of moles of the solute we desired to prepare;
Number of moles = molarity x volume
Parameters given;
volume of solution = 500mL = 0.5L
molarity of solution = 0.5M
Number of moles = 0.5 x 0.5 = 0.25moles
Now to know the volume stock to take;
Volume of stock = 
molarity of stock = 4M
volume = 
= 0.0625L or 62.5mL
The biological compounds that are nonpolar and insoluble in water are lipids. It is a group of molecules that are naturally occurring which includes sterols, waxes, fats, fat-soluble vitamins and the like. These molecules are nonpolar molecules so basically the cannot be dissolved in a polar solvent like water.
