Need Help calculating concentration.?
Since pH = 5.10 and pH = - log[H+], then- [H+] = 7.94x10^-6 moles/liter HNO3 is a strong (100% dissociated) acid HNO3 solution = 7.94x10^-6 moles/liter
Calculate the concentration of..?
Hi, Density of water = 1 g/mL So every mL contains 1 g of water. Now we need to find out g KNO3/100 g of H2O, I am using K for KNO3 and H for H2O 1) 8.00 (Vol water, mL) => 10 g K/ 8 g H => 10 * 100 g K/ (8 * 100) g H [multiplied and divided by 100] = 10/8*100 g K/ 100 g H = 125 g K/100 g H 2) 11.00 (Vol water, mL) => 10 g K / 11 g H = 10 * 100 g K/ (11 * 100) g H = 10/11*100 g K / 100 g H = 90.91 g K/ 100 g H 3) 14.00 (Vol water, mL) => 10 g K / 14 g H = 10 * 100 g K/ (14 * 100) g H = 10/14*100 g K / 100 g H = 71.43 g K/ 100 g H 4) 19.00 (Vol water, mL) => 10 g K / 194 g H = 10 * 100 g K/ (19 * 100) g H = 10/19*100 g K / 100 g H = 52.63 g K/ 100 g H 5) 27.00 (Vol water, mL) => 10 g K / 27 g H = 10 * 100 g K/ (27 * 100) g H = 10/27*100 g K / 100 g H = 37.04 g K/ 100 g H 6) 39.00 (Vol water, mL) => 10 g K / 39 g H = 10 * 100 g K/ (39 * 100) g H = 10/39*100 g K / 100 g H = 25.64 g K/ 100 g H Hope this is helpful. Keep smiling. Bye.
Chemistry help? calculating concentration?
Balanced equation: KI(aq)+ AgCl(s) → AgI(s) + KCl(aq) This reaction occurs because AgI is even less soluble than AgCl Ksp AgI = 8.52×10-17 Ksp = [Ag+]*[I-] but [Ag+] = [I-] Ksp = [I-]² [I-]² = 8.52*10^-17 [I-] = √8.52*10^-17 [I-] = 9.23*10^-9M
How do you calculate DNA concentration?
I will assume you mean DNA quantity(DNA mass per volume of diluted sample ) as a result of a sample being diluted or pre-processed before use in a Spectrophotometer or some other Quantitation methods.Either way, the following article can help (Unless you want the algorithmic reasoning leading to the formula):“A sample of dsDNA was diluted 50X. The diluted sample gave a reading of 0.65 on a spectrophotometer at OD260. To determine the concentration of DNA in the original sample, perform the following calculation:dsDNA concentration = 50 μg/mL × OD260 × dilution factordsDNA concentration = 50 μg/mL × 0.65 × 50dsDNA concentration = 1.63 mg/mL”Read more here:Quantitation of DNA and RNAor here where we had an ongoing discussion at ResearchGate:How i could determine DNA concentration ?
I need help calculating the concentration in this problem!!?
The molarity of KCl does not play any part in the concentration of S2SO3, assuming there is no chemical reaction, so we will ignore that bit of info. Use the formula M1V1 = M2V2. M1 = 0.0050 M V1 = 10.00 mL M2 = ? V2 = 50.0 mL Plug the numbers in and solve for M2, and you should get 0.001 M. Molarity is one way of measuring concentration, so there is your answer. Hope this helps, and good luck!
Help please..... Calculating Hydroxide ion (OH-) concentration.?
What's Kb? Kb = [HA+][OH-]/[A-]. What's the relationship? We know Kb. We know the molar concentration, 0.05M. We know the molar relationship is one to one. So, to solve, Kb = [x][x]/[CH3NH2] Define [CH3NH2] as [0.05 - x]. Resolve as Kb = x^2/0.05. (we assumed x was negligible when subtracted from the initial concentration of the base)
Calculating concentration of buffer components?
Start with the initial concentrations. To calculate these, you need moles, then divide by total volume = 0.225L + 0.435L = 0.660L [lactic acid]=0.85M x 0.225L / 0.660 L = 0.29M [lactate]=0.68M x 0.435L / 0.660 L = 0.45 M To a good approximation, these will be the actual concentrations of the buffer components, since the Ka of lactic acid is << that either [lactic acid] or [lactate].
How do you calculate Molar Concentration??? PLEASE HELP AHH?
Moles CaCl2 = 22.3 g / 110.984 g/mol = 0.201 moles Cl- = 2 x 0.201 = 0.402 [Cl-] = 0.402 / 0.325 L = 1.24 M Moles Cl- = 1.24 x 0.6 L = 0.744 mass = 0.744 x 35.453 g/mol = 26.4 g
CHEMISTRY PLEASE HELP: Calculate the concentration of OH− (aq) in a solution in which?
Calculations with [H+] and [OH-]...... Background: Kw = [H+] x [OH-] = 1.00x10^-14 ....... at about 25C (a) [H+] x [OH-] = 1.00x10^-14 [OH-] = 1.00x10^-14 / [H+] [OH-] = 1.00x10^-14 / 2x10^-6 [OH-] = 5x10^-9 M (b) [H+] = [OH-] [OH-] = (1.00x10^-14)^(½) [OH-] = 1.00x10^-7 (c) [H+] = 200 [OH-] [OH-] = [H+] / 200 [OH-] = Kw / [OH-] / 200 [OH-]² = Kw / 200 [OH-]² = 1.00x10^-14 / 200 = 5x10^-17 [OH-] = (5x10^-17)^(½) [OH-] = 7.07x10^-9