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Question #85ba3 - Socratic The fundamental frequency of a streched string is given by the follwing formula n=1 (2l)sqrt(T m) Where l->"Length of the string" T->"Tension of the string" m->"Mass per unit length of the string" So T and m remaining constant nprop1 l let the frequency of the string is n_1 when its length is 20cm and its frequency is n_2 when its length is 21cm Si by the law of length n_1 n_2=21 20 Let" " n
Question #cbdf9 - Socratic K_text (c) = 1 2 × 10^"-8" > The chemical equation is "N"_2 + "3H"_2 ⇌ "2NH"_3 Step 1 Calculate the initial concentrations of "N"_2 and "H"_2 If nitrogen and
Question #3fa48 - Socratic Here's what I got It is absolutely vital that you work with a balanced chemical equation, so 8always* make sure that your equation is balanced before doing any stoichiometric calculations In this case, the balanced chemical equation is color (blue) (3)"H"_ (2 (g)) + "N"_ (2 (g)) -> color (purple) (2)"NH"_ (3 (g)) Notice that you have a color (blue) (3):color (purple) (2) mole ratio between
Question #aa818 - Socratic Explanation: #-10+log (n+3)=-10# #log (n+3)=0# #n+3=10^0=1# #n=1-3=-2# Answer link You can reuse this answer
Question #81dd7 - Socratic Explanation: When you "add together" two or more chemical equations, it goes similarly as algebra: equal quantities on opposite sides of either reaction can cancel out
Question #09aa6 - Socratic Here's what I got Your equilibrium reaction looks like this "N"_ (2(g)) + "O"_ (2(g)) rightleftharpoons color(red)(2)"NO"_ ((g)) You know that the equilibrium constant for this reaction is K_c = 1 7 * 10^(-3) Right from the start, the fact that K_c < 1 should tell you that when equilibrium is reached, the reaction vessel will contain significantly higher concentrations of nitrogen gas and
If sum_ (n=1)^oo abs (a_ (n+1)-a_n) converges then does the sequence a . . . Essentially yes Given the convergence of: sum_ (n=1)^oo abs (a_ (n+1) - a_n) Note that for any M <= N we have: abs (a_N - a_M) <= abs (a_ (M+1)-a_M) + abs (a_ (M+2)-a_ (M+1)) + + abs (a_N - a_ (N-1)) So: abs (a_N - a_M) <= sum_ (n=M)^oo abs (a_ (n+1) - a_n) But if the sum to oo converges, then for any epsilon > 0 there is some M_1 such that for all M >= M_1: sum_ (n=M)^oo abs (a_ (n+1)-a_n