Base anhydride


A base anhydride is an oxide of a chemical element from group 1 or 2 (the alkali metals and alkaline earth metals, respectively). They are obtained by removing water from the corresponding hydroxide base. If water is added to a base anhydride, a corresponding hydroxide salt can be re-formed.


Base anhydrides are not Brønsted–Lowry bases because they are not proton acceptors. However, they are Lewis bases, because they will share an electron pair with some Lewis acids, most notably acidic oxides.[1]



Examples


Group 1 oxides



  • Lithium oxide, which reacts with water forming lithium hydroxide


  • Sodium oxide, which reacts with water forming sodium hydroxide


  • Potassium oxide, which reacts with water forming potassium hydroxide


  • Rubidium oxide, which reacts with water forming rubidium hydroxide


  • Caesium oxide, which reacts with water forming caesium hydroxide


  • Francium oxide, which reacts with water forming francium hydroxide

Group 2 oxides



  • Magnesium oxide, which reacts with water forming magnesium hydroxide


  • Calcium oxide, which reacts with water forming calcium hydroxide


  • Strontium oxide, which reacts with water forming strontium hydroxide


  • Barium oxide, which reacts with water forming barium hydroxide


  • Radium oxide, which reacts with water forming radium hydroxide

Other



  • Thallium oxide, which reacts with water forming thallium hydroxide


See also


  • Acidic oxide


References




  1. ^ Principles of Modern Chemistry, 7th Edition. David Oxtoby, H. P. Gillis, Alan Campion. Published by Cengage Learning. Page 675-676. .mw-parser-output cite.citationfont-style:inherit.mw-parser-output qquotes:"""""""'""'".mw-parser-output code.cs1-codecolor:inherit;background:inherit;border:inherit;padding:inherit.mw-parser-output .cs1-lock-free abackground:url("//upload.wikimedia.org/wikipedia/commons/thumb/6/65/Lock-green.svg/9px-Lock-green.svg.png")no-repeat;background-position:right .1em center.mw-parser-output .cs1-lock-limited a,.mw-parser-output .cs1-lock-registration abackground:url("//upload.wikimedia.org/wikipedia/commons/thumb/d/d6/Lock-gray-alt-2.svg/9px-Lock-gray-alt-2.svg.png")no-repeat;background-position:right .1em center.mw-parser-output .cs1-lock-subscription abackground:url("//upload.wikimedia.org/wikipedia/commons/thumb/a/aa/Lock-red-alt-2.svg/9px-Lock-red-alt-2.svg.png")no-repeat;background-position:right .1em center.mw-parser-output .cs1-subscription,.mw-parser-output .cs1-registrationcolor:#555.mw-parser-output .cs1-subscription span,.mw-parser-output .cs1-registration spanborder-bottom:1px dotted;cursor:help.mw-parser-output .cs1-hidden-errordisplay:none;font-size:100%.mw-parser-output .cs1-visible-errorfont-size:100%.mw-parser-output .cs1-subscription,.mw-parser-output .cs1-registration,.mw-parser-output .cs1-formatfont-size:95%.mw-parser-output .cs1-kern-left,.mw-parser-output .cs1-kern-wl-leftpadding-left:0.2em.mw-parser-output .cs1-kern-right,.mw-parser-output .cs1-kern-wl-rightpadding-right:0.2em
    ISBN 978-0840049315









-

這個網誌中的熱門文章

Is there any way to eliminate the singular point to solve this integral by hand or by approximations?

圖片
Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite 1 $$int_0^1 (1+(x^-p-1)^1-p)^frac1pdx, pin mathbbR, pge 1$$ I find that $x=0,1$ are singular points(discontinuous points of this integral), is there a way to eliminate those singular points to make sure that this integral can be calculated by hand or by Taylor approximations. In either cases, what are those steps of calculating the integral? calculus real-analysis analysis approximation share | cite | improve this question edited Sep 11 at 5:07 asked Sep 11 at 4:48 Mclalalala 126 8 1 What is $p$? Please give more information to make it complete. – xbh Sep 11 at 5:02 @xbh already fixed it, thanks for reminding. – Mclalalala Sep 11 at 5:10 Seems that $0$ is not singular here… [I might compute wrong] – xbh Sep 11 at 5:30 @xbh can you show me how to get the close formula for the integral above? – Mclalalala Sep 11 at 5:33
閱讀完整內容

Why am i infinitely getting the same tweet with the Twitter Search API?

圖片
up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia
閱讀完整內容

Solve: $(3xy-2ay^2)dx+(x^2-2axy)dy=0$

圖片
Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite Solve: $(3xy-2ay^2)dx+(x^2-2axy)dy=0$ The given differential equation is not exact. The given differential equation is homogeneous. Therefore, $I.F=frac1Mx+Ny$ $=frac1(3xy-2ay^2)x+(x^2-2axy)y$ $=frac1xy(4x-4ay)$ With Integrating Factor of this form (ugly value), it is difficult to compute further integration, that is involved in solving the equation. So, can we not solve the equation using the rule: If in the differential eqn $Mdx+Ndy=0$, provided $ Mx+Ny ne 0$, andM and N ar eboth homogeneous then $I.F=frac1Mx+Ny$ differential-equations share | cite | improve this question edited Aug 8 at 9:19 asked Aug 8 at 9:13 Soumee 626 2 13 put $y=vx,$ ............. – Magneto Aug 8 at 9:44 add a comment  |  up vote 0 down vote favorite Solve: $(3xy-2ay^2)dx+(x^2-2axy)dy=0$ The given differential equation is not exact. The given differential eq
閱讀完整內容