Can anyone help me with these poblems need help?
6. Which ion in each pair has a smaller ionic radius?KCu7. Which ion in each pair has a smaller ionic radius?PoO8. Which ion in each pair has a smaller ionic radius?SeSc9. Which ion in each pair has a smaller ionic radius?NBi10. Which ion in each pair has a smaller ionic...
Asked by Adriene Frausto 1 year ago.
6. Which ion in each pair has a smaller ionic radius? KCu 7. Which ion in each pair has a smaller ionic radius? PoO 8. Which ion in each pair has a smaller ionic radius? SeSc 9. Which ion in each pair has a smaller ionic radius? NBi 10. Which ion in each pair has a smaller ionic radius? AtHf Answered by Gus Livington 1 year ago.
6. Cu 7. O 8. Se 9. Bi 10.At As you go across a period the pull gets stronger cause you have the same of shell being pulled IN by a greater positively charged nucleus. As You go down a group the radius gets bigger because you have more shells so the pull Decreases Answered by Bradly Thim 1 year ago.
Which are redox reactions, and for all redox reactions write the red and ox agents and half ionic equations?
6) Ni + CuSO4 = NiSO4 + Cu 7) Cl2O7 + H2O = 2HCLO4
Asked by Arie Norg 1 year ago.
Ni goes from 0 to +2 => reducing agent Cu goes from +2 to 0 => oxidizing agent Cl2O7 + H2O >> 2 HClO4 is NOT a redox Answered by Mayola Hurtt 1 year ago.
What is the mass of 7 mol Cu?
Asked by Lu Degrella 1 year ago.
? g Cu = (7 mol Cu) / (63.5 g Cu / 1 mol Cu) = 444.5 g Cu. Since you gave me one sig fig, it should be rounded to 400 g Cu! Hope this helps! Answered by Sandra Litano 1 year ago.
IUD users, please give me your experiences. Would you recommend them over the Pill?
I'm looking into birth control options and don't want the hassle of taking the Pill everyday or the high dosage. The IUD sounds like a good option but am unfamiliar with them. I'm 35 years old. Any information you can share, that would be great. Thanks!
Asked by Ericka Coutino 1 year ago.
I had a CU-7 for seveal years without problems...I think it was much better than birth control and I didnt have tobother with all the creams and crap...I would not have changed it..I will warn you they are alot easier going in than out..but I got pregnant shortly after removal...then I got my tubes tied...hope it helps Answered by Esta Potaczala 1 year ago.
The IUD is a good option if you don't get bad period cramps. If you do have bad period cramps, the IUD will make the pain MUCH worse. Have you considered the Nuvaring? You only have to put it in one time per month. Answered by Laticia Denney 1 year ago.
If you are married have your husband get a vasectomy. My wife used iuds for several years. she had those hideous menstral periods with big clots and masses.. i think the way IUDs work it to cause abortions. Not to prevent conception. Live your conscience. A vasectomy is irreversal, but doesn't hurt and is 100% effective unless you cheat. woops! Answered by Claretha Puhl 1 year ago.
an IUD is, for a man, like making love to a mousetrap fyi Answered by Sharyl Blasko 1 year ago.
Gas Stoichiometry, limiting reagent?
Calculate the volume of nitrogen dioxide produced at 1.029 atm and 28.3°C by the reaction of 7.45 cm3 copper (d = 8.95 g/cm3) with 222.8 mL nitric acid (d = 1.42 g/cm3, 68.0% HNO3 by mass): Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
Asked by Valery Stea 1 year ago.
Calculate the moles of Cu: 7.45 cm^3 x 8.95 g/cm^3 / 63.54 g/mol = 1.049 moles Calculate the moles of HNO3: 222.8 mL x 1.42 g/mL x 0.68 / 63.0 = 3.415 moles 1.049 mole of Cu requires 4x1.049 or 4.196 mole of HNO3 for complete reaction. HNO3 is the limiting reactant 3.415 moles of HNO3 yield 3.415/2 or 1.71 moles of NO2 V = nRT/P V = 1.71x0.08206x301.4/1.029 = 41.1 L Answered by Matilda Crusinberry 1 year ago.
I need some Chemistry help please, 10 points?
1. What do the coefficients in a balanced chemical equation tell you? A. amount of energy produced B. number of moles which react C. physical state of the compounds reacting D. elements involved in the reaction 2. The coefficients of a reaction only give the ratio in which substances...
Asked by Loni Cyfers 1 year ago.
1. What do the coefficients in a balanced chemical equation tell you? A. amount of energy produced B. number of moles which react C. physical state of the compounds reacting D. elements involved in the reaction 2. The coefficients of a reaction only give the ratio in which substances react. They do not in any way tell you HOW MUCH is reacting. A. true B. false 3. WHERE would you find the mass of 1 mole of magnesium? A. Periodic Table B. Avogadro's Number C. Dictionary 4. What is the mass of 1 mole of Magnesium? A. 6.02 x 10 23 atoms B. 24.3 g C. 48.6 g D. 22.4 L 5. In the equation, 2 H2 + O2 ---> 2 H2O, what substances are the reactants? A. H2 B. O2 C. H2O D. A and b only E. All of the above 6. Given the following equation: Cu + 2AgNO3 ---> Cu(NO3)2 + 2 Ag How many moles of Cu are needed to react with 3.50 moles of AgNO3? A. 2 moles Cu B. 1 mole Cu C. 1.75 moles Cu D. 7 moles Cu 7. Given the following equation: 2 NaClO3 --->2 NaCl + 3 O2 12.00 moles of NaClO3 will produce how many moles of O2? A. 3 moles O2 B. 18 moles O2 C. 36 moles O2 D. 2 moles O2 8. How many moles of hydrogen gas (H2) is produced if 20.0 mol of Zn are added to hydrochloric acid (HCl) according to the equation; Zn + 2HCl ---> ZnCl2 + H2 A. 20 moles H2 B. 10 moles H2 C. 2 moles H2 D. 1 mole H2 9. 1. What mass in grams of sodium hydroxide (NaOH) is produced if 2.0 moles of sodium (Na) reacts with water according to the chemical equation: 2Na(s) + 2H2O ---> 2NaOH(aq) + H2(g)? A. 22.98 g Na B. 45.96 g Na C. 39.97 g Na D. 79.9 g Na Thanks to whoever helps. Answered by Sheilah Trunzo 1 year ago.
1) b - number of moles of each - can be used for mole to mole ratio 2) a - see above explanation 3) a - periodic table lists elemental atomic masses 4) b - 24.3050g 5) d - reactants on left side of equation 6) c - use the ratios from the balanced equation; 3.50 mol AgNO3 x (1 mol Cu / 2 mol AgNO3) = 1.75 mol Cu 7) b - use the ratios from the balanced equation; 12 mol NaClO3 x (3 mol O2 / 2 mol NaClO3) = 18 mol O2 8) d - this question is actually a trick!! - you will still only be able to produce 1 mol of H2 at the end of this reaction because even though you are adding 20 moles of Zinc (Zn), you still only have enough moles of Hydrochloric Acid (HCl) to react with 1 mole of Zinc. Only if you add more HCl with the Zn, then you will be able to produce more moles of H2. 9) d - first you would need to convert your given moles of Na to moles of NaOH. once you are in moles of NAOH then you need to find how many grams (mass) are in one mole of NaOH so you can use it as a conversion factor to find your final answer. 2.0 mol Na x (2 mol NaOH / 2 mol Na) x (39.9971g NaOH / 1 mol NaOH) = 79.9942g NaOH Answered by Winfred Reily 1 year ago.
1.a 2.a 3.a 4.b 5.d 6.c 7.c 8.c 9.b While I'm here, in your last question regarding Q. 2, volcano lava. Where do you get the idea that solidifying lava is not freezing......Anything Solidifying at whatever temperature, means Freezing......NOT necessarily about ICE....B is correct. Answered by Kristine Glebocki 1 year ago.
the answer to the first question must be 'c' by using the undeniable fact that's addition polymerisation and a double bond enables effortless addition. look at of your textual content e book. As to question 4, all combustion launch CO2 and H2O. relax of the first answer is high-quality. Answered by Sharyn Naz 1 year ago.
1.a 2.a 3.a 4.b 5.d 6.c 7.b 8.a 9.d Answered by Audrea Cooks 1 year ago.
1B 2A 3A 4B 5D 6C 7B 8A 9D Answered by Scottie Greengo 1 year ago.
Grade 11 Chemistry question...empirical formula...?
typo: (CuO) not (CuO0!
Asked by Kristel Daughetee 1 year ago.
Question: A 9.54g sample of a compound containing only copper and oxygen was converted into copper by a reduction process involving hydrogen gas. The mass of the copper that remained after the reaction was 7.62g. Determine: a) the empirical formula for the compound b) if the relative formula mass of the substance is 79.55, what is its true formula and its proper name? Note: I don't know how to show all of the necessary working. Could you please show working before the answer and then at the bottom explain the steps. I live in Australia. My answer for a) is CUO and for b) is copper (II) oxide. Are these corret? I don't know how to show the working for my teacher. Thanks. Answered by Virgilio Veazey 1 year ago.
I was able to produce the same answer as yours. Here's my solution. GIVEN: mass of Cu = 7.62 g mass of Sample = 9.54 g FIND: (a) Empirical Formula: 7.62 g Cu * (1 mol Cu / 63.546 g Cu) = 0.12 mol Cu mass of O = 9.54 g sample - 7.62 g Cu, thus mass of O = 1.92 g 1.92 g O * (1 mol O / 16.00 g O) = 0.12 mol O Therefore, the ratio of Cu to O (Cu:O) is 1:1. .:. The empirical formula of the compound is CuO. Copper must have a charge of (2+). (b) If the if the relative formula mass of the substance is 79.55, what is its true formula and its proper name? From the data above, mass of Cu = 7.62 g mass of O = 1.92 g mass of sample in letter (a) = 9.54 g We can compute for the % composition of each element in the compound. % Cu = 7.62 g Cu / 9.54 g sample = 79.87% Cu % O = 1.92 g O / 9.54 g sample = 20.13% O We can use the two percentages in our solution. mol element = (% composition of element)(mass of sample)(molar mass) 0.7987 Cu * 79.55 g sample * (1 mol Cu / 63.546 g Cu) = 1 mol Cu 0.2013 O * 79.55 g sample * (1 mol O / 16.00 g O) = 1 mol O Therefore, the true formula of the compound is CuO, and its proper name would be Copper(II) Oxide or if you want to use the classical method of naming it, it's Cupric Oxide. I hope that helped! =) Answered by Anderson Kerlin 1 year ago.
Alright, let me see if I remember how to do this. Well, you know you have 9.54 g of something and 7.62 g of that is copper... so you know the compound is about 79.9% copper (7.62/9.54). Therefore, if you had 100 g of the stuff, 79.9 g would be copper. That means you'd have 20.1 g oxygen as well. OK, the trick to finding an empirical formula is you have to figure out what the MOLE ratio of copper to oxygen is. That means you have to figure out how many moles of copper you have if you've got 79.9 g. (I reverted to having 100 g of the compound because it's just easier to work with. You don't actually have to do that, but the ratios work out better.) So convert 79.9 g copper to moles using the fact that there are 63.546 g copper in a mole (this you get from the periodic table): 79.9 g Cu x (1 mol Cu)/(63.546 g Cu) = 1.26 mol OK, so if you have 20.1 g OXYGEN, how many moles is THAT? Well, there are about 16 g oxygen in a mole... 20.1 g O x (1 mol O)/(16 g O) = 1.26 mol Hey, look at that! Same number of moles! Well that means Cu and O actually exist in the same molar amounts in the formula - there's just so many more GRAMS of Cu because it's a heavier metal. That means your EMPIRICAL formula is, as you correctly stated, CuO. The true formula could be Cu2O2, or Cu3O3, or whatever had them in EQUAL molar amounts. OK, for part b, let's just suppose for a second that the true formula were CuO. Then what would the mass of a mole of that stuff be? Well, the mass of a mole of Cu is, as I said, 63.546 g, and it's about 16 g for a mole of oxygen. So add 'em up and you get 79.546, which is pretty dang close to what they said the formula mass is... so that means likely you've got CuO on your hands, as you said, and not Cu2O2 or something else (which would be much heavier per mole). And yes, according to the rules of nomenclature that'd be copper(II) oxide. Hope I helped! Answered by Ferne Fremont 1 year ago.
To find the empirical formula, first you find the number of moles of each element present and then find the molar ratio betwwen them. mass of Cu=7.62g mass of O= 9.54 - 7.62=1.92g no. of mole(Cu)= 7.62/63.5 = 0.12 no. of mole (O) = 1.92/16.0 = 0.12 Now divide each of the moles with the smallest no. of moles. In this case both of them are equal. molar ratio (Cu) = 0.12/0.12 = 1 molar rario (O) = 0.12/0.12 = 1 So there is 1 copper and 1 oxygen in the formula. therefore the empirical formula is CuO. For the true formula, find the empirical formula mass and divide the real mass by it. Empirical formula mass (CuO) = 63.5+16.0 = 79.5g n = Empirical formula mass/molecular formula mass = 79.5/79.55 = 1 Now multiply this 'n' number to the empirical formula. As the number in this case is 1, the formula remains the same, that is CuO. And the proper name is Copper(II) oxide because in this formula oxygen has a valence of 2 and so does copper. Answered by Evonne Hardung 1 year ago.
Is turquoise 8.43% Cu by weight?
Turquoise is CuAl6(PO4)4(OH)8 + 4H2O I took Cu's amu put it over the total amu and multiplied by 100 Did I get this right? If so, how about the sig figs? If not, please let me know the answer and how to do it so I can learn Thanks in advance! :)
Asked by Marquitta Poinson 1 year ago.
I did not go to the trouble of doing all the calculations. I used a molar mass / % composition programme. If you include the 4 molecules of water in the formula ( as you should) then the % Cu = 7.81% If you do not include the 4 molecules of water, then the % Cu is 8.57%, Your method is in principle correct, But possibly made some mistake in the molar mass of the elements: Or you have possibly rounded the atomic mass values too drastically. My programme that did the above calculations used the following data: 1) including the 4 molecules H2O Molar mass = 813.4418g/mol For each individual element: CuCopper63.546317.8120 % AlAluminum26.98153868619.9018 % HHydrogen1.007947161.9826 % OOxygen15.999432855.0727 % PPhosphorus30.9737622415.2310 % Reading each line : element symbol : Element name : Element atomic mass : number of atoms of element in 1 molecule : Mass % of element 2) Not including the 4 molecules of H2O Molar mass = 741.3805 g / mol For each individual element CuCopper63.546318.5713 % AlAluminum26.98153868621.8362 % OOxygen15.999432451.7934 % HHydrogen1.00794781.0876 % PPhosphorus30.9737622416.7114 % As far as sig figs go, you will see that the lowest number of figures used in the calculation is 6 - that is in the atomic mass of Cu. Therefore your quoted answer must not contain more than 6 sig figs. But I am sure that there is no restriction in using less than 6 sig figs, correctly rounded. In case 1 the answer can safely be quoted to 4 sig figs 7.81% and in case 2 as 8.57%. Answered by Jill Scheider 1 year ago.
I haven't checked your numbers, but your principle is correct. Significant figues, well 2 is normally sufficient, but if the question says different then always answer the question Answered by Johnathon Olivar 1 year ago.