subject
Chemistry, 30.01.2020 07:04 jskdkfjf

How many atoms are present in 179.0 g of iridium

ansver
Answers: 2

Another question on Chemistry

question
Chemistry, 21.06.2019 17:30
You are performing an experiment in a lab to attempt a new method of producing pure elements from compounds. the only problem is that you do not know what element will form. by your previous calculations you know that you will have 6.3 moles of product. when it is complete, you weigh it and determine you have 604.4 grams. what element have you produced?
Answers: 1
question
Chemistry, 21.06.2019 22:40
How many electrons does silver have to give up in order to achieve a sido noble gas electron configuration
Answers: 1
question
Chemistry, 22.06.2019 08:20
What is the formula for the compound dinitrogen pentoxide? a. n4o5 b. n5o4 c. n4o6 d. n5o2 e. n2o5
Answers: 3
question
Chemistry, 23.06.2019 16:00
Be sure to answer all parts. the catalytic destruction of ozone occurs via a two-step mechanism, where x can be any of several species: (1) x + o3 → xo + o2 [slow] (2) xo + o → x + o2 [fast] (a) write the overall reaction. o3 + o → 2 o2 o3 + xo → x + 2 o2 xo + x + o3 + o → xo + x + 2 o2 x + o3 → xo + o2 (b) write the rate law for each step (using k for the rate constant). reaction 1: reaction 2: k[x][o2] k[xo][o2] k[x][o3] k[xo][o3] k[xo][o2] k[x][o] k[x][o2] k[xo][o] (c) x acts as a and xo acts as a catalyst intermediate intermediate catalyst (d) high-flying aircraft release no into the stratosphere, which catalyzes this process. when o3 and no concentrations are 3 ă— 1012 molecule/cm3 and 9.9 ă— 109 molecule/cm3, respectively, what is the rate of o3 depletion? the rate constant k for the rate-determining step is 6 ă— 10â’15 (cm3)2/moleculeâ·s. give your answer in scientific notation. use one significant figure in your answer. ă— 10 molecule/s
Answers: 2
You know the right answer?
How many atoms are present in 179.0 g of iridium...
Questions
question
Computers and Technology, 08.10.2021 17:20
question
Mathematics, 08.10.2021 17:30
Questions on the website: 13722363