The energy released results in a kinetic energy of 92.2 keV for the products. We should convert keV into Joules, noting that 1 keV equals a kiloelectron volt. The required conversion is: 1.602×10⁻¹⁹ <span>joule = 1 eV
Kinetic energy = 92.2 keV * (1,000 eV/1 keV) * (</span>1.602×10⁻¹⁹ joule/1 eV) = 5.76×10²³ Joules
Next, we can determine the velocity of each He atom from the kinetic energy:
KE = 1/2*mv²
5.76×10²³ Joules = 1/2*(4)(v²)
This solves to give us: v = 5.367×10¹¹ m/s
Let's assume that the compound formula is as follows: Experiment 1: 1.00 g of the compound yields 1.95 g of AgCl. The molar mass of AgCl is 143.32 g/mol. Thus, the moles of AgCl for 1.95g are: The moles of Cl also equal 0.0136, considering that 1 mole of AgCl corresponds to 1 mole of Cl. Experiment 2: 1.00 g of the compound results in 0.900 g of CO2 and 0.735 g of H2O. The molar mass of CO2 is 44 g/mol, and for H2O, it's 18 g/mol. Therefore, the moles of C come to 0.0205 and the moles of H stand at 0.0816 (which is 2 times the moles of H2O). Now, from the provided details, it's derived that in 1.00 g of the compound, there are 0.0136 moles of Cl, 0.0205 moles of C, and 0.0816 moles of H. In terms of mass: Mass of Cl = 0.0136 * 35.5 = 0.4828 g. Mass of C = 0.0205 * 12 = 0.246 g. Mass of H = 0.0816 * 1 = 0.0816 g. Total mass = 0.4828 + 0.246 + 0.0816 + mass of N. Given that 1.00 = 0.8104 + Mass of N, it follows that Mass of N = 0.1896. Thus, upon dividing all moles by the smallest value, we find Cl = 0.0136 / 0.0135 = 1.0007; C = 0.0205 / 0.0135 = 1.52; H = 0.0816 / 0.0135 = 6.04; N = 0.0135 / 0.0135 = 1. Multiplying by 2 allows us to reach integer values: Cl = 2, C = 3, H = 12, N = 2.
Answer:
The response is provided below.
Explanation:
Numerous aspects can influence the actual results of titration. These factors vary from human error to misjudging measurements, a researcher's interpretation of color changes, and improper techniques during the experimental procedure.
Thus, to mitigate these errors, researchers must engage thoroughly throughout experimentation, and employing gross readings can assist in reducing mistakes when determining the final titre value.
The reaction can be described as follows: CO + 2H2 = CH3OH. Given the specified quantities of the reactants, we will identify the limiting reactant and compute the remaining excess amount. Calculating, 1.50 x 10^-6 g CO converts to 5.36 x 10^-8 mol CO, while 6.80 x 10^-6 g H2 equals 3.37 x 10^-6 mol H2. Thus, CO is fully consumed in the reaction, leaving 3.37 x 10^-6 - 5.36 x 10^-8 = 3.32 x 10^-6 moles of gas.
Clarification:
To obtain the specific element, you should multiply the grams provided by the ratio of grams of that particular element within its complete compound.
Since the query did not indicate the amount of NO2 produced, we can consider its mass to be negligible, thus assigning 1 mole to Nitrogen.