Answer:
The displacement reactions are regarded as redox reactions because they involve the formal transfer of electrons from one chemical specie to the other
2) The series of reactivity of the metals in the order of increasing reactivity are;
↓
↓
Explanation:
1) Redox (oxidation-reduction) reaction is a chemical reaction involving the loss and gain of electrons from one chemical species to another, where the chemical species that undergoes oxidation, loses electrons and is termed the reducing agent, while the other chemical species that undergoes reduction, gains electrons, and is termed the oxidizing agent
2) The redox reaction can being based on the affinity for electrons depends on the positions of the reactants in the electrochemical series as well as the chemical reactivity of the metals with zinc being a stronger reducing agent and more chemically reactive than copper and magnesium being a stronger reducing agent and more chemically reactive than zinc
Least reactive (Cu) < (Zn) < Mg Most reactive
Copper < Zinc < Magnesium.
The formula is SrCl2. hope this helps
Complete question:
ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.
Answer:
The magnitude of q for the process 568 J.
Explanation:
Given;
change in internal energy of the gas, ΔU = 475 J
work done by the gas, w = 93 J
heat added to the system, = q
During gas expansion process, heat is added to the gas.
Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.
ΔU = q - w
q = ΔU + w
q = 475 J + 93 J
q = 568 J
Therefore, the magnitude of q for the process 568 J.
A mole of any gas occupied 22.4 L at STP. So, the number of moles of nitrogen gas at STP in 846 L would be 846/22.4 = 37.8 moles of nitrogen gas.
Alternatively, you can go the long route and use the ideal gas law to solve for the number of moles of nitrogen given STP conditions (273 K and 1.00 atm). From PV = nRT, we can get n = PV/RT. Plugging in our values, and using 0.08206 L•atm/K•mol as our gas constant, R, we get n = (1.00)(846)/(0.08206)(273) = 37.8 moles, which confirms our answer.
H^+(aq) + OH^-(aq) ---> H2O(l)
<span>Na^+ and ClO4^- are the spectator ions.</span>
