Answer:
Explanation:
A mole is defined as the number of molecules divided by 6.02×10^23
Mole = (2×10^19)/(6.02×10^23)
Mole = 3.32×10^-5 mole
Response:
The result is a pH of 7. I’m uncertain about the mechanics of Hydrochloric acid reactions
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.
Given:
Mass of the ionic compound = 10.00 g
Mass of water = 75.0 g
Initial temperature of water T1= 23.2 C
Final temperature of water T2 = 31.8 C
Specific heat of water c = 4.18 J/gC
To determine:
Enthalpy of dissolution of the ionic compound
Heat gained by water equation:
Q = mcΔT
m = mass of water
c = specific heat
ΔT = change in temperature (T2-T1)
Q = 75.0 g * 4.18 J/gC * (31.8-23.2)C = 2696 J
Thus, the heat gained by water equals heat lost by the ionic compound (enthalpy of dissolution)
Therefore, q(ionic) = 2696 J
ΔH = q(ionic)/mass of ionic compound = 2696 J/10.00 g = 2.7 *10² J/g
Answer: A) enthalpy change = 2.7*10² J/g
Answer:
The correct options include choice 2, 3, and 6.
Explanation:
Density is identified as the mass of a substance per unit volume occupied by that substance.
The density remains constant for a given substance, regardless of variations in mass and volume hence it is considered an intensive property.
2. 20.2 g of silver in 21.6 mL of water and 12.0 g of silver also in 21.6 mL of water.
3. 15.2 g of copper in 21.6 mL of water and 50.0 g of copper in 23.4 mL of water.
6. 11.2 g of gold in 21.6 mL of water and 14.9 g of gold in 23.4 mL of water.
The same metals in both instances will yield consistent densities due to the fixed density of the metal.
