Cu(NO3)2 --> MM187.5558
NiNO3 *COEF2* --> 120.6983
Answer:
3.816 × 10⁻³ M
Explanation:
A stock solution of Cu²⁺(aq) is made by dissolving 0.8875 g of solid Cu(NO₃)₂∙2.5H₂O in a 100.0-mL volumetric flask, and then brought up to volume with water. What is the molarity (in M) of Cu²⁺(aq) in this stock solution?
We can derive the following relations:
- The molar mass of Cu(NO₃)₂∙2.5H₂O is 232.59 g/mol.
- Each mole of Cu(NO₃)₂∙2.5H₂O yields one mole of Cu²⁺.
The moles of Cu²⁺ present in 0.8875 g of Cu(NO₃)₂∙2.5H₂O are:
The molarity of Cu²⁺ is:
The resulting temperature, following the change in volume and pressure, is -27.26°C. To find this temperature, we apply the combined gas law equation—a formulation where initial and final pressures, volumes, and temperatures are compared. Given the initial conditions and transformations, when we input the stipulated values, we reach the conclusion that the resultant temperature is -27.26°C.
<span>The partial pressure of oxygen is 438.0 mmHg. The ideal gas equation is expressed as PV = nRT where P represents pressure, V denotes volume, n is the number of moles, R is the ideal gas constant (8.3144598 (L*kPa)/(K*mol)), and T signifies absolute temperature. To convert from Celsius to Kelvin, we have 43.4 + 273.15 = 316.55 K. For the pressure conversion from mmHg to kPa: 675.9 mmHg * 0.133322387415 = 90.11260165 kPa. When solving for n using the ideal gas equation, we derive n = PV / (RT) which provides n = 90.11260165 kPa * 16.2 L / (8.3144598 (L*kPa)/(K*mol) * 316.55 K)= 1459.824147 L*kPa / 2631.94225 (L*kPa)/(mol), resulting in n = 0.554656603 mol. Thus, we have 0.554656603 moles of gas particles. Next, we determine the contribution from oxygen. The atomic weight of oxygen is 15.999 g/mol, while argon is 39.948 g/mol, and the molar mass of O2 is 31.998 g/mol. We establish the relationships where M is the number of moles of O2, and 0.554656603 - M gives the number of moles of Ar. Setting up the equation: M * 31.998 + (0.554656603 - M) * 39.948 = 19.3, we solve for M resulting in 0.359424148 moles of oxygen out of 0.554656603 total moles. This leads to oxygen providing 0.359424148 / 0.554656603 = 0.648012024 or 64.8012024% of the total pressure of 675.9 mmHg. The partial pressure therefore calculates to 675.9 * 0.648012024 = 437.9913271 mmHg, rounded to 438.0 mmHg</span>
What is being removed during the wash is the solvent.
