Answer: HCl
Explanation:
calcium carbonate dissolves in HCl acid producing CO 2 gas. It will not dissolve in pure water. The Ksp for calcium carbonate in water is 3.4 x 10-9 moldm-3 which is very low. What takes place here is actually a chemical reaction:
CaCO 3 (s) + 2HCl(aq) → CaCl 2 (aq) + H 2CO 3(aq)
This reaction accounts for the solubility of the Calcium carbonate in HCl and not in pure water.
1 is D - double-replacements do not make solid metals
2 is A - to have complete combustion the original compound must ONLY have C, H and O
3 is B - the elemental Mg replaces the H in the HCl
Answer:
(a). 4°C, (b). 2.4M, (c). 11.1 g, (d). 89.01 g, (e). 139.2 g and (f). 58 g/mol.
Explanation:
Without mincing words let's dive straight into the solution to the question.
(a). The freezing point depression can be Determine by subtracting the value of the initial temperature from the final temperature. Therefore;
The freezing point depression = [ 1 - (-3)]° C = 4°C.
(b). The molality can be Determine by using the formula below;
Molality = the number of moles found in the solute/ solvent's weight(kg).
Molality = ( 11.1 / 58) × (1000)/ ( 90.4 - 11.1) = 2.4 M.
(c). The mass of acetone that was in the decanted solution = 11.1 g.
(d). The mass of water that was in the decanted solution = 89.01 g.
(e). 2.4 = x/ 58 × (1000/1000).
x = 2.4 × 58 = 139.2 g.
(f). The molar mass of acetone = (12) + (1 × 3) + 12 + 16 + 12 + (1 x 3) = 58 g/mol.
<span>1. Which variable is the independent variable and which is the dependent variable? Density vs. ethylene glycol
The independent variable would be ethylene glycol and dependent variable would be density.
A. A 25-mL volumetric flask with its stopper has a mass of 32.6341 g. The same flask filled to the line with ethylene glycol (C2H6O2, automotive antifreeze) solution has a mass of 58.0091 g. What is the density of the ethylene glycol solution?
Density = 58.0091 - 32.6341 / .025 = 1015 g/L
B. What is the molarity of the ethylene glycol solution, if the mass of ethylene glycol in the solution is 12.0439 g?
Molarity = 12.0439 ( 1 mol / 62.07 g) / 0.025 = 7.8 M</span>
