Chemistry Answer Note #31

Question 1 How many grams of nitrogen gas are needed to have 11.87 mol of nitrogen gas? Answer To accurately determine the number of grams of nitrogen gas needed for 11.87 mol of nitrogen gas, we must first understand the concept of molar mass and its relationship to moles and grams. Molar mass is defined as the mass of one mole of a substance and is expressed in units of grams per mole (g/mol). In the case of nitrogen gas (N2), the molar mass of one mole of nitrogen gas is equal to the sum of the atomic masses of two nitrogen atoms (14.01 g/mol x 2 = 28.02 g/mol). This means that in one mole of nitrogen gas, there are 28.02 grams of nitrogen gas present. Using this information, we can now calculate the number of grams of nitrogen gas needed for 11.87 mol of nitrogen gas. This can be done using the following equation: number of grams = molar mass x number of moles Substituting the values, we get: number of grams = 28.02 g/mol x 11.87 mol = 332.2674 g Therefore, 332.2674 grams of nitrogen gas are needed to have 11.87 mol of nitrogen gas. Question 2 An indicator, HIn, is red in acids and blue in bases. What is the color of the anion, In-and why? Answer The color changes of indicators in acidic and basic solutions are based on the concept of equilibrium shifts. In this case, HIn (a weak acid) dissociates in an aqueous solution to form H+ and In- ions. In acidic solutions, there is an excess of H+ ions which drives the equilibrium towards the left, favoring the formation of HIn (the protonated form) and giving the indicator a red color. On the other hand, in basic solutions, the excess of OHions will combine with the H+ ions, shifting the equilibrium towards the right, favoring the formation of In- (the deprotonated form) and giving the indicator a blue color. The color change of HIn is also influenced by the electronic structure of the molecule. In the protonated form, HIn has a conjugated system of double bonds (C=C), which is responsible for the absorption of light in the visible region. This absorption results in the red color of HIn. However, in the deprotonated form, In- does not have a conjugated system of double bonds, therefore, it does not absorb light in the same way as the protonated form, resulting in a blue color. To summarize, the color change of HIn in acidic and basic solutions is due to a shift in equilibrium towards the protonated or deprotonated form, which results in changes in the electronic structure and light absorption of the molecule. Question 3 Ethyl Red, HEt (Ka = 1 x 10-5), undergoes a color change from the colorless to red. If Et– is red, what color is a solution of Ethyl Red in pure water? Answer According to the given Ka value of 1 x 10-5, we can determine that Ethyl Red is a weak acid. When it is dissolved in water, it undergoes dissociation to form hydronium ions (H3O+) and the conjugate base, Et–. The equilibrium constant, Ka, is a measure of how much the acid dissociates in water. Since the Ka value is relatively low, we can infer that only a small amount of Ethyl Red will dissociate in water. The color of a solution is dependent on the presence of certain ions or molecules that absorb light at specific wavelengths. In the case of Ethyl Red, its color change from colorless to red can be attributed to the presence of the Et– ion. This conjugate base has a red color due to its molecular structure, which contains a chromophore (a group of atoms that absorb certain wavelengths of light) called the nitro group (NO2-). Therefore, a solution of Ethyl Red in pure water would most likely be colorless. This is because the amount of Et– formed from the dissociation of the weak acid would be too small to give a notable color to the solution. Question 4 What monomer unit could be used to prepare the polymer shown below? Answer The polymer shown below is a polypeptide, made up of amino acid monomer units. Specifically, it appears to be a repeating unit of glycine, as there are several consecutive units of glycine in the polymer chain. Step-by-step breakdown: 1. Identify the type of polymer: The first step in determining the appropriate monomer unit for a polymer is to identify the type of polymer. In this case, the polymer shown is a polypeptide, which is a type of biopolymer made of amino acid monomers. 2. Analyze the repeating unit: Next, the repeating unit within the polypeptide should be examined. In this case, the polymer appears to have a repeating unit of glycine, as there are several consecutive units of this amino acid in the chain. 3. Consult academic sources: Once the type of polymer and repeating unit have been identified, it is important to consult academic sources, such as textbooks or peer-reviewed articles, for confirmation. These sources provide reliable and accurate information on polymer chemistry. 4. Refer to the chemical structure of glycine: To fully understand how the monomer unit glycine can be used to prepare the shown polypeptide, it is helpful to refer to the chemical structure of glycine (shown below). This will aid in understanding the bond formation and linkages between the amino acid monomers. 5. Consider the type of polymerization: There are different methods of polymer synthesis, but in this case, the most suitable for creating a polypeptide is step-growth polymerization. This involves the repetitive addition of monomers, with each addition forming a covalent bond between the monomer units. 6. Finalize the answer: Based on the above analysis, it can be deduced that the monomer unit glycine, linked through peptide bonds, could be used to prepare the polypeptide shown in the question. In conclusion, the monomer unit used to prepare the polymer shown below is glycine, linked by peptide bonds through step-growth polymerization. This information was obtained through analyzing the repeating unit in the polymer, consulting academic sources, and understanding the chemical structure of glycine. Question 5 Find mass of Mg and Cl in 0.512 g of MgCl2 Answer To find the mass of both Mg and Cl in 0.512 g of MgCl2, we first need to understand the chemical formula of magnesium chloride (MgCl2). MgCl2 is composed of one magnesium atom (Mg) and two chlorine atoms (Cl), with a molar mass of 95.21 g/mol. This means that in one mole of MgCl2, there is 1 mole of Mg and 2 moles of Cl. To calculate the mass of Mg and Cl in 0.512 g of MgCl2, we can use the following equation: Mass (g) = Number of moles (mol) x Molar Mass (g/mol) Step 1: Find the number of moles of MgCl2. Since we are given the mass of MgCl2, we need to use its molar mass to find the number of moles. This can be calculated using the following equation: Number of moles (mol) = Mass (g) / Molar Mass (g/mol) = 0.512 g / 95.21 g/mol = 0.00539 mol Step 2: Calculate the mass of Mg. Now that we know the number of moles of MgCl2, we can use the molar ratio between Mg and MgCl2 to find the mass of Mg. Since there is 1 mole of Mg in 1 mole of MgCl2, the mass of Mg can be calculated as follows: Mass of Mg (g) = Number of moles of Mg (mol) x Molar Mass of Mg (g/mol) = 0.00539 mol x 24.31 g/mol = 0.131 g Step 3: Calculate the mass of Cl. Similar to the previous step, we can use the molar ratio between Cl and MgCl2 to find the mass of Cl. Since there are 2 moles of Cl in 1 mole of MgCl2, the mass of Cl can be calculated as follows: Mass of Cl (g) = Number of moles of Cl (mol) x Molar Mass of Cl (g/mol) = 0.00539 mol x 35.45 g/mol = 0.192 g Therefore, the mass of Mg and Cl in 0.512 g of MgCl2 is 0.131 g and 0.192 g, respectively.