From the provided data, the unknown mixture was composed of salt, salicylic acid, and sand. It is understandable that the student suspected the presence of sand, yet scientific experimentation must verify such assumptions. The test involving salt and salicylic acid reveals that salt dissolves in water, while salicylic acid is only slightly soluble, and sand does not dissolve at all. By introducing the unknown into water, the salt would dissolve first, followed by the partial dissolution of salicylic acid. Heating the mixture could allow for the evaporation of salicylic acid, resulting in the remaining salt. If traces of sand were observed in the dissolved sample, it could suggest contamination.
The stated condition has been verified. Construct the resonance structure for CSO, where the central atom carries a +2 formal charge and the oxygen atom has a +1 charge. We need to create the resonance structure for CSO as shown in the figure. According to the problem, there is a +2 formal charge on the central atom and a +1 charge on the oxygen atom. The central atom in this structure is sulfur. We will calculate the formal charge of sulfur based on the information presented, demonstrating that it aligns with the necessary formal charges.
Answer: The correct option is (1).
Explanation:
Group 16 is the third-from-last column in the periodic table and is known as the oxygen family.
Members of this group include oxygen, sulfur, selenium, tellurium, and polonium.
Elements in Group 16 have 6 valence electrons in their outermost shell.
The electronic configuration of sulfur is
.
Because sulfur belongs to Group 16, it has 6 valence electrons.
The conversion of -70 degrees Celsius results in 9/5 x C + 32 = 158 degrees Fahrenheit. CO2 transitions to solid state at -108.5 degrees Fahrenheit, indicating that at -70 degrees Celsius, it will be in the state commonly known as 'dry ice'.
At 30°C, glucose has a solubility of 1.25 g per gram of water. Given that the density of water at this temperature is 1 g/mL, the mass corresponding to 400 mL of water is also 400 g. Therefore, the concentration of the solution is calculated as 550 g divided by 400 g of water, which gives 1.375 g of glucose per gram of water. Since this concentration exceeds the solubility limit for glucose at this temperature, the solution can be classified as SATURATED.