WHEN 2g of NaOH were dissolved in 53.g H2O in a calorimeter at 24.0*C the tem of the solution went up to 33.7*
372088
exothermic is the release of heat to the surroundings. endothermic is the absorption of heat from the surroundings.  Your temperature went up...think about it.  Part 2) specific heat of water is 4.18 j/gK, you have 55g substance. and temperature change was 9.7K...do the math.\nPart 3) well you have qH2O...it is the negative of deltaH!!\nPart 4) you have half as much NaOH...Half as much deltaH\nPart 5) divide delta H from part 3 by the number of moles NaOH in 2g NaOH
170755
Specific heat 0.13 J/gK
963248
The heat given off by the burning reaction (delta E) is equal to the heat gained by the calorimeter.  (m * change in temp)\n\nYou can calculate the heat released by burning.\n.553 g with a molecular wt. of 88g/mole \n(.553 grams) / (88 (grams / mole)) = 0.00628409091 moles\n(975 (kJ / mole)) * 0.00628409091 moles = 6 126.98864 joules\n\nYou can calculate the heat absorbed by the water (3000 g * .356)\n3 000 * .356 = 1 068 joules\n\nyou can calculate the difference, it must have heated up the apparatus.\n6 127 - 1 068 = 5 059 joules \n\n5059 joules = x grams * .356\n5059 joules / .356 = 14210.67 grams of water\n\nThe metal apparatus was had a heat capacity equivalent to 14210.67 grams of water.
237697
O.k., so you are given the final temperature (-0.430 C). Since the freezing point of water is usually O degrees C, the temperature decrease is 0.43 degrees C. Like in the other problem, let's set up some dimensional analysis.\n\n(0.43 deg.C decr)*(1 mole compound / 1.86 deg.C decr) = X\n\nHere X would equal 0.23 moles compound\n\nNext, you want to find the grams of solute in solution.\n\n(8.02g solute/0.861 kg solvent) = Y\n\nY here is equal to 9.31 g compound/kg solution\n\nNow that you have the mass in grams and the number of moles of the solute, you can divide Y by X, which gives the units for molecular mass (g/ mol.)\n\nThe answer should be around 40.48 g/mol. The solute cannot be determined from this information because no additional information is given concerning what is most likely a compound (not a pure element).
952215
[H3O+][OH-] = 1.0 x 10^-14 M always at 25C\nso if [H3O+]= 5.0 x 10^-10M\n[OH-]= 2.0x10^-5
358982
NaOH is completely dissociated when it is dissolved in water.  A .0025M KOH solution therefore produces a [OH-] equal to .0025M.  The ion product of [H30+] and [OH-] is a constant, 1.0 x 10^-14 M^2.  By substitution, the [H30+] can be determined.  The pH can then be calculated.\n\n[H30+] = 1.0 x 10^-14 M^2/[OH-] = 1.0 x 10^-14 M^2/2.5 x 10^-3 \n\naaaand all that equals .4 x 10^-11... and you put that in correct scientific notation... 4 x 10^-12 M for the concentration of [H30+].\n\nWe then plug that in...   \n\npH= -log[H30+] = -log (4 x 10^-12) = 11.4\n\nThe pH of a .0025M solution of KOH is 11.4, which is greater than 7, making it a basic solution.\n\nStrong base at that.\n\nP.S.- If it is supposed to have the correct number of significant figures/digits, the answer would then be 11.
237268
.001565mol / .075L = .021M NaCl
773826
the answer that i got is 3.10 M. \n\n3.25 g NaOH/40.00 g NaOH = .0813 mol NaOH\n\n26.2 mL / 1000mL = .0262 L\n\nM= .0813 mol / .0262 L = 3.10 M
621533
NaOH ----------------> Na+   +   OH-\nFind the number of moles of NaOH\nMoles of NaOH = 0.10M * 0.150L = 0.015 moles\nSince the coefficients in fornt of NaOH, Na+ and OH- are all 1, therefore if 0.015 moles of NaOH disssociates, 0.015 moles of Na+ and 0.015 moles of OH- are produced.\n\nSimilarly we will work with HCl\nHCl --------------------> H+      +      Cl-\nMoles of HCL = 0.30M * 0.05L = 0.015 moles\nThis means the 0.015 moles of H+ will be produced\n\nSince H+ and OH- have equal amiunts, they will completely neutralize each other leaving the solution at neutral pH. Since the neutral pH is 7, therefore, the pH of this solution will be 7.
787311
A. H2O. Molecular water is 2H2O
886093
2NaOH + H2C2O4 ===> Na2C2O4 + 2H2O\n\nC2O4= is oxalate ion.
781203
2H3PO4(aq) + 3Mn(OH)2(s)  --> Mn3(PO4)2(s) + 6H2O(l)
1097630
Na2CO3 + 2HCl-> NaCl + H2O + CO2
136369
reverse reaction one(and change the sign of delta H), add the two reactions together, cancel out on both sides. then add the delta H for both reactions, should get equals 54.08kJ
474531
CH3COOH + NAOH =.CH3COONA +H2O
779566
17.13 mL = 0.01713 L\n\n0.01713 L x 0.3150 M = 0.005396 mol\n\nNaOH is a strong base, HCl is a strong acid.\nEach NaOH releases one hydroxide anion.\nEach HCl releases one proton.\n\nTherefore it takes 0.005396 mol of H+ to neutralize 0.005396 mol of -OH.\n\n0.005396 mol / 0.025 L = 0.2158 M\n\nThe concentration of the NaOH solution is 0.2158 M.
47432
3AgC2H3O2 + Na3PO4 -----> Ag3PO4 + 3NaC2H3O2