Answer:
D. 0.3 M
Explanation:
NH4SH (s) <--> NH3 (g) + H2S (g)
Initial concentration 0.085mol/0.25L 0 0
Change in concentration -0.2M +0.2 M +0.2M
Equilibrium 0.035mol/0.25 L=0.14M 0.2M 0.2M
concentration
Change in concentration (NH4SH) = (0.085-0.035)mol/0.25L =0.2M
K = [NH3]*[H2S]/[NH4SH] = 0.2M*0.2M/0.14M ≈ 0.29 M ≈ 0.3M
Answer:
Element with 6s subshell
Explanation:
Reactivity of an element depends on the electronic configuration and position of element in the periodic table as reactivity increases as we go down the periodic table.
This is so because number of shell increases as move down the periodic table and the last electron is further away from the nucleus.
Element with 6s subshell is the largest among 3s and 4s subshell and has more number of shells so it will react more than 3s and 4s subshell.
Hence, the correct answer is "Element with 6s subshell".
Answer:CO2(g) will be formed at a faster rate in experiment 2 because more H+ particles can react per unit time
Explanation:
Answer:
Explanation:
a. Oxidation : 2O + 4e^- ------> 2O^2-
b. Reduction: 2Sr - 4e- -------> Sr^2+
c. Balanced redox reaction
2Sr + O2 ------------> 2Sr O
Oxidation and reduction can be defined by various means, addition of oxygen, removal of hydrogen, removal of electrons. For this reaction, this definition is used, oxidation is the loss of electrons while reduction is the gaining of electrons.
In (a) oxidation half reaction, the valency of oxygen is zero and then moves into lossing two electrons resulting into -2 valency.
In (b) reduction half reaction, the valency of Sr is zero and gains electrons resulting into valency of 2.
In the overall redox reaction, Sr and O2 with valency of 0 each reacts together and form SrO with valency of 2 and -2 respectively, which gives 0 and then balances the equation.
