Answer:
0.20M of nitric acid
0.00250M of KSCN
Explanation:
In the case of nitric acid, the solution's dilution changes from 10.00mL to 100.00mL, resulting in a 1/10 dilution. Given the original concentration of nitric acid is 2.0M, the updated concentration becomes: 2.0M×(1/10)=0.20M of nitric acid
Similarly, the dilution of KSCN extends from 50.00mL to 100.00mL, equal to a 1/2 dilution. Consequently, the new concentration of KSCN turns out to be:
0.00500M × (1/2) = 0.00250M of KSCN
I hope it aids you!
It must be a physical change unless a chemical substance interacts with the metal surface, resulting in a chemical change.
6.28 mol O2 × 2 mol H2 / 1 mol O2 = 12.56 moles H2
The enthalpy of hydration for copper sulfate is -1486.62 kJ/mol, indicating that 1486.62 kJ of energy is absorbed by a mole of copper sulfate during its hydration. Step 1: Calculate the energy released per mole of dissolved substance (Eq. 1). If 0.102 moles release 55.51 kJ, then 1 mole corresponds to 541.85 kJ/mol. Therefore, ΔH = -541.85 kJ/mol. Step 2: Identify the energy absorbed by dissolved substance (Eq. 2). When 0.101 moles absorb 95.31 kJ, 1 mole will absorb 944.77 kJ/mol, thus ΔH = 944.77 kJ/mol. Step 3: Subtract Eq. 2 from Eq. 1. Thus, ΔH = -541.85 kJ/mol (Eq. 1) and ΔH = 944.77 kJ/mol (Eq. 2), leading to ΔH = -541.85 - 944.77, so ΔH = -1486.62 kJ/mol.
Response: 670K
Rationale:
Provided data includes:
Initial volume of gas V1 = 1.22 L
Initial temperature T1 = 286 K
Final volume V2 = 2.86 L
Final temperature T2 =?
As temperature and volume are interrelated when pressure remains constant, apply Charles' law:
V1/T1 = V2/T2
1.22 L/286 K = 2.86 L/ T2
Cross multiplication yields:
1.22 L x T2 = 286 K x 2.86 L
1.22T2 = 817.96
Solving for T2:
1.22T2/1.22 = 817.96/1.22
T2 = 670.459 K (rounded to the nearest whole number is 670 K)
Therefore, the gas temperature is 670 Kelvin
