300 mL of an aqueous solution of HCl 0.1 M are diluted with water to 1.5 L. Calculate (to two decimal places) the molar concentration (mol L-1) of the resulting aqueous solution of HCl.
A positive reaction for Molisch's test is given by almost all carbohydrates (exceptions include tetroses & trioses). It can be noted that even some glycoproteins and nucleic acids give positive results for this test (since they tend to undergo hydrolysis when exposed to strong mineral acids and form monosaccharides).
Answer: 7.8125 g
Explanation: Given:
Original amount (N₀) = 500 g
Number of half-lives (n) = 9612/1602 = 6
Amount remaining (N) = ?
N = 1/2ⁿ × N₀
N = 1/2^6 × 500
N = 0.015625 × 500
N = 7.8125 g
Answer: The product formed is potassium chloride.
Explanation:
Precipitation reaction is defined as the chemical reaction in which an insoluble salt is formed when two solutions are mixed containing soluble substances. The insoluble salt settles down at the bottom of the reaction mixture.
The chemical equation for the reaction of potassium phosphate and magnesium chloride follows (look at the picture)
2 moles of aqueous solution of potassium phosphate reacts with 3 moles of aqueous solution of magnesium chloride to produce 1 mole of solid magnesium phosphate and 6 moles of aqueous solution of potassium chloride.
Answer: B. carbon tetrachloride, CCI4
Explanation: The other options are incorrect. Let's write the correct formulas:
A. Diarsenic pentoxide - As2O5
C. Sodium dichromate - Na2Cr2O7
D. magnesium phosphide - Mg3P2
Explanation: Acids and bases can be classified in terms of hydrogen ions or hydroxide ions, or in terms of electron pairs. (look at the picture)
Let us note that from the pH scale, a pH of;
0 - 6.9 is acidic
7 is neutral
8 - 14 is basic
a. [H+] = 6.0 x 10-10M
pH= 9.22 is basic
b. [OH-] = 30 × 10-2M
pH = 13.5 is basic
C. IH+1 = 20× 10-7M
pH = 0.56 is acidic
d. [OH-] = 1.0 x 10-7M
pH = 7 is neutral
Answer: 306 L
Explanation: Using the ideal gas law,
PV = nRT
where R = 0.08206 L•atm/(mol•°K), solving for n gives
n = PV/(RT)
n = (845 mmHg) (270 L) / ((0.08206 L•atm/(mol•°K)) (24 °C))
Convert the given temperature to °K and the given pressure to atm:
24 °C = (273.15 + 24) °K ≈ 297.2 °K
(845 mmHg) × (1/760 atm/mmHg) ≈ 1.11 atm
Then the balloon contains
n = (1.11 atm) (270 L) / ((0.08206 L•atm/(mol•°K)) (297.2 °K))
n ≈ 12.3 mol
of He.
Solve the same equation for V :
V = nRT/P
Convert the target temperature to °K:
-50 °C = (273.15 - 50) °K = 223.15 °K
Then the volume under the new set of conditions is
V = (12.3 mol) (0.08206 L•atm/(mol•°K)) (223.15 °K) / (0.735 atm)
V ≈ 306 L
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