+2, because Beryllium is in the Group II of the periodic table.
Hope this helps!
The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵
<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>
It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.
Hence, the equilibrium constant of the reaction in discuss is;
K = [5.6]²/[0.10]³[0.10]
k = 5.6² × 10⁴
k = 3.136 × 10⁵
K = 3.1 × 10⁵.
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Answer:
Equation: - 2Li + H2O = Li2O + H2 Uncoated lithium metal reacts with water to form a colorless lithium hydroxide solution and hydrogen gas.
Explanation:
Answer:
NO2- is the reducing agent.
Cr2O7_2- is the oxidizing agent.
H+ is neither
Explanation:
Reduction is the gain in electron. A chemical specie that undergoes reduction is called the oxidizing agent.
Oxidation is simply the loss in electrons. A chemical specie that undergoes oxidation is called the reducing agent.
Let us look at the species.
The first specie is the NO2-. In this specie, the oxidation number of nitrogen changed from +3 to +5 in NO3-. Thus we can see that there is more loss of electron to have caused an increase in the oxidation number positively. This shows an oxidation. Hence, NO2- is the reducing agent.
Let us look at the chromium. We can see that the oxidation number of chromium changed from +7 to +3.
Now we can see that it is a decrease and hence, it is a gain of electron and thus it is reduction. This means the first chromium specie is the oxidizing agent.
The hydrogen ion is simply placed there to balance the ions and hence it is neither the oxidizing nor the reducing agent.
Answer: -64.1 kJ.
Explanation:
According to first law of thermodynamics:
=Change in internal energy
q = heat absorbed or released
w = work done or by the system
w = work done by the system=
{Work is done by the system is negative as the final volume is greater than initial volume}
w = -855 Joules = 0.855 kJ (1kJ=1000J)
q = -65.0 kJ {Heat released by the system is negative}
Thus the change internal energy (ΔE) for a system that is giving off 65.0 kJ of heat and is performing 855 J of work on the surroundings is -64.1 kJ.
