If, in the reaction between magnesium and HCl, we use 3 M HCl rather than 6 M HCl, would the rate of reaction?
444111
Rate of reaction depends upon the concentration of reactants. Since, 6M HCl has a greater conc. than 3M HCl, rate of reaction decreases when Mg reacts with 3M HCl
1014284
2.70\n\nregard what ppl saying below, I can make this a little bit more clear.\n\nFirst of all, since the concentration of HCl is 0.00200M and HCl is a strong acid, so we can safely assume 100% dissociation. So, we have 0.02M hydronium ions [one to one ratio]; even if we consider the dissociation of water, it will not make any difference. the dissocation of water is depressed by the presence of "high" concentration of hydronium ions. \n\nyou can even calculate how much water actually dissociates into hydroxide and hydronium ions using the disociation constant of water, 10^-14. But I can guarantee you this will not make ANY difference in pH, say less than 1% for sure.\n\ndissociation of water need to be put into consideration ONLY IF the concentration of hydronium ions is VERY low, say 10^-6 -10^-7, or pH around 7, then it will make a difference. \n\nSo, pH = -log[H3O+]= -log[0.002]= 2.70
431100
further explanation - when certain compounds are mixed, for instance an amine (primary or secondary) and an acid chloride, they will  form an amide and an equivalent of HCl  which will form an equilibrium with the starting materials.   Using pyridine (or another tertiary amine such as triethylamine or diisopropylethylamine) will trap the HCl and drive the reaction forward.
397908
The different acids have a different pH so turn a differetn colour.\nThe concentrations are different. The HCl is more concentrated as it will fully dissociate.
787375
1. AgCl(s) + NaNO3\n\n2. AgOH(s) + NaNO3\n\n3. They would not react as both are nitrates already
923888
CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)
834501
First let's find how many moles of each one we have\n\nmole= M*V, so for \nFeCl3 =0.234*25*10^-3= 5.85*10^-3\nNaOH =0.453*42.5*10^-3 =19.25*10^-3\n\nFeCl3 + 3NaOH -> Fe(OH)3 + 3NaCl\n\nso\n\n1mole FeCl3 reacts with 3mole NaOH\n5.85*10^-3 mole react with x\n\nThus x =3*5.85*10^-3=17.55*10^-3 mole NaOH but we have 19.25*10^-3 mole NaOH, therefore NaOH is in excess\nTherefore the amount of Fe(OH)3 will be calculated according to the limiting reageant FeCl3.\n\nso from the reaction\n\n1 mole FeCl3 gives 1 mole Fe(OH)3\nor better yet, since the molecular weight of Fe(OH)3 is 106.85 \n1 mole FeCl3 gives 106.85 g Fe(OH)3\n5.85*10^-3 mole give x g\nx= 106.85*5.85*10^-3= 0.625 g\n\nWe calculated that 17.55*10^-3 mole NaOH reacted.\nSo we have remaining (19.25-17.55) *10^-3= 1.7*10^-3 mole NaOH\n\nThese are in a total volume of 25.0+42.5=67.5 ml= 67.5*10^-3L.\n\nSo the concentration of the excess NaOH is\nC=mole/V= 1.7*10^-3/67.5*10^-3= 0.025 M
823593
When magnesium reacts with oxygen gas: \n 2 Mg  +  O2   ---> 2 (MgO)  the ratio of Mg to O is 1:1\n(1 mole of Mg reacted produces 40.3 g of MgO)\n\nWhen it reacts with nitrogen gas\n 3 Mg +   N2   --->  Mg3N2  the ratio of  Mg to N is 1: 0.67\n(1 mole of Mg reacted gives 33.6 g of Mg3N2)\n\nSo if some Mg3N2 gets left behind in what you think is pure MgO, then the mass will be lower than it should be. Your calculated mass of oxygen reacted will be lower than it should be, too.\n\nWhen you convert to moles, you will have fewer moles of oxygen \nthan you should have,  i.e., the ratio of Mg to O will be too high.
904934
2HNO3 + Mg(OH)2 -->Mg(NO3)2 +2H20\n\n2H+(aq) + 2NO3-(aq) + Mg+(aq) + 2(OH)-(aq) -->Mg+(aq) + 2(NO3)-(aq) +2H20(l)\n\n2+(aq) + 2(OH)-(aq) --> 2H20(l)
805472
The first step in working with limiting reagents problem is to balance the chemical equation.\n\n2 Al (s) + 6 HCl (aq) ====> 2 AlCl3 (aq) + 3 H2 (g)\n\nNow calculate the number of moles of each reactant\nMoles of Al = 14.6g / M.W. of Al\nMoles of Al = 14.6g / 27g/mol = 0.5407 moles\nMoles of HCl = 14.6g / M.W. of HCl\nMoles of HCl = 14.6g / 36.45g/mol = 0.4005 moles\n\n1.        Look at the coefficients of the reactants in the balanced equation. For every 2 moles of Al  6 moles of HCl is required or for every 1 mole of Al 3 moles of HCl is required\nIf you used up 0.5407 moles of Al then 3 * 0.5407 (1.6621) moles of HCl will be needed. Since only 0.4005 moles of HCl were used instead of 1.6621 moles, therefore it is the limiting reagent.\n\n2.\nMass of AlCl3 = Moles of HCl * coefficient of AlCl3 / coefficent of HCl * Molecular weight of AlCl3\n\n3.\nMass of H2 = Moles of HCl * coefficient of H2 / coefficent of HCl * Molecular weight of H2
794643
1. highter since given weight of magnesium looks to be a larger percent of finished product.\n2.It would depend on molecular weight of other reactants.  If they were heavier than Oxygen the Mg would seem to be a lower percentage.  If only other possibility for Rx was with nitrogen and ratio of recatants was the same as with Oxygen(unlikely) then the resulting compound would be lighter than MgO making the Mg percent higher.\n3.  Mg oxydizes in air at room temp.  This white film was MgO already oxydized which would interfere with your results, raising the calculated percentage of Mg.
136146
The 1st answer is wrong. The reaction is\n\nNa2CO3(aq) + SrCl2 (aq)--> 2NaCl (aq) +SrCO3 (s)
386784
Well first of all you have to specify if the H's you want to account for are either comming from HCl alone or both HCl and some H's comming from the dissociation of water.\n\nI suppose that only the number of moles of H's comming from HCl is really worth while working the question for, since the H's of the dissociated water does not change the pH of the 0.10M solution.\n\nTherefore:\n\n0.10moles are in 1000mL\n? moles are in 50mL\n\n50 x 0.1 / 1000 = 0.005moles\n\nTherefore we can say that 0.005moles of HCl in 50mL of water results in a concentration of 0.10M
501821
(a) because Mg has valency +2 and chlorine -1
30238
No, HCL has nothing to do with the number of moles of CaCO3 that you used.  So you asume you used enough HCl so that all of your CaCO3 is dissolved.  All you need to do is devide 0.5215 g  with the molar weight of CaCO3 and you'll get your number of moles.\n\nHope this helps,\nUltimeaciax
64647
2x (G2 + M2 ---> 2GM)      DH=-280.0*2= -560.0 kJ\n2x (L + 2M2 ---> LM4)      DH=-350.0*2= -700.0 kJ\n   (L2G4 ---> 2L + 2 G2)   DH=122.0*(-1)=-122.0 kJ (change sign when flip over)\n\nAdd them together, we'll have:\n2G2 + 2L + 6M2 + L2G4 ---> 4GM + 2LM4 + 2L + 2G2\nthen cancel 2G2 and 2L on both side, we end up with:\nL2G4 + 6M2 ---> 2LM4 + 4GM\nthe enthalpy of this reaction is, therefore the sum of each enthalpies calculated above.\nDH=(-560.0) + (-700.0) + (-122.0)= -1382.0 kJ (exothermic process)\nhope that'll help!
810145
depends what the magnesium is for. If this is for a chemical reaction, it's because copper is less reactive than magnesium (it's lower on the activity series list).