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Hold on here, I need to do some heavy calculation ... okay, done. One tablespoon of table salt contains 1.0 tablespoons of sodium chloride. Since the atomic weight of sodium chloride is 58.5, 1 meq of salt is 58.5 mg. Since one tablespoon = 14.8 ml and salt weights 1.02 g/ml, one tablespoon of salt weighs 15.2 g or 15,200 mg. So one tablespoon of salt = 15,200 / 58.5 = 260 mEQ.
First:Comparing the electropotential values of both redox couples (being equal, see below 1. and 2.)AND considering that Cl2 gas might escape the mixture giving toxic fumes,it does NOT look a good titration method.2Cl- --> Cl2,g + 2e- (delta)Eo= -1.36VCr2O72- + 14H+ + 6e- --> 2Cr3+ + 7H2O (delta)Eo= +1.36VSecond:There is no exact answer possible because the titration end point can never be established, you'll never know when the last Cl2 is escaped, and there is no sharp (sudden) change in colour, and the reaction at the endpoint is very slow.Third: The answer is 60.9 mLThe equivalent point (theorethical end point) can be calculated as follows:mEq reductor (HCl) = mEq oxidator (K2Cr2O7)0.325 (mEq/mL HCl) * VHCl (mL) = 0.526 (mEq/mL K2Cr2O7) * 37.6 (mL K2Cr2O7) So VHCl (mL) = 0.526 (mEq/mL K2Cr2O7) * 37.6 (mL K2Cr2O7) / 0.325 (mEq/mL HCl) = 60.854 = 60.9 mL
5 L x 1000 mL/1L = 5000 mL
To convert ml to kl, multiply ml x 0.000001 (1.0E-06). 1000 ml = 0.001 kl To convert kl to ml, divide kl by 0.000001 (1.0E-06). 0.001 kl = 1000 ml To convert ml to kl, multiply ml x 0.000001 (1.0E-06). 1000 ml = 0.001 kl To convert kl to ml, divide kl by 0.000001 (1.0E-06). 0.001 kl = 1000 ml
an equilibrium between dissolved KCl and solid KCl
2,5 mL solution = 0,0025 L
Hold on here, I need to do some heavy calculation ... okay, done. One tablespoon of table salt contains 1.0 tablespoons of sodium chloride. Since the atomic weight of sodium chloride is 58.5, 1 meq of salt is 58.5 mg. Since one tablespoon = 14.8 ml and salt weights 1.02 g/ml, one tablespoon of salt weighs 15.2 g or 15,200 mg. So one tablespoon of salt = 15,200 / 58.5 = 260 mEQ.
The answer is 0,86 meq.
If a teaspoon has approx. 5 mL - approx. 35 mEq.
no they are not the same thing. mEq is a milliequivlant. cc and ml are the same thing.
it depends on what you are measuring. Meq is a measure of charge. If you are measuring H+ which has one positive charge, 1 mmol = 1 meq for Ca2+, 1 mmol = 2 meq so the conversion will depend on the charge of the ion or molecule
Lactated Ringer's is a sterile, nonpyrogenic solution containing isotonic concentrations of electrolytes in water for injection. Lactated Ringer's is administered by intravenous infusion for parenteral replacement of extracellular losses of fluid and electrolytes. Each 100 mL of Lactated Ringer's Injection, USP contains sodium chloride 600 mg, sodium lactate, anhydrous 310 mg, potassium chloride 30 mg and calcium chloride, dihydrate 20 mg. May contain hydrochloric acid and/or sodium hydroxide for pH adjustment. A liter provides 9 calories (from lactate), sodium (Na+), 130 mEq, potassium (K+) 4 mEq, calcium (Ca-++) 3 mEq, chloride (Cl-) 109 mEq and lactate [CH3CH(OH) COO-] 28 mEq. The electrolyte content is isotonic (273 mOsmol/liter, calc.) in relation to the extracellular fluid (approx. 280 mOsmol/liter). The pH of the solution is 6.6 (6.0-7.5). The solution contains no bacteriostat, antimicrobial agent or added buffer (except for pH adjustment) and each is intended only for use as a single-dose injection. When smaller doses are required the unused portion should be discarded.ctated Ringer's is a sterile, nonpyrogenic solution containing isotonic concentrations of electrolytes in water for injection. Lactated Ringer's is administered by intravenous infusion for parenteral replacement of extracellular losses of fluid and electrolytes.
It is like 0.0024g of KCl per 100g of acetonitrile at 25oC. Link: http://potassiumchloride.in/
First:Comparing the electropotential values of both redox couples (being equal, see below 1. and 2.)AND considering that Cl2 gas might escape the mixture giving toxic fumes,it does NOT look a good titration method.2Cl- --> Cl2,g + 2e- (delta)Eo= -1.36VCr2O72- + 14H+ + 6e- --> 2Cr3+ + 7H2O (delta)Eo= +1.36VSecond:There is no exact answer possible because the titration end point can never be established, you'll never know when the last Cl2 is escaped, and there is no sharp (sudden) change in colour, and the reaction at the endpoint is very slow.Third: The answer is 60.9 mLThe equivalent point (theorethical end point) can be calculated as follows:mEq reductor (HCl) = mEq oxidator (K2Cr2O7)0.325 (mEq/mL HCl) * VHCl (mL) = 0.526 (mEq/mL K2Cr2O7) * 37.6 (mL K2Cr2O7) So VHCl (mL) = 0.526 (mEq/mL K2Cr2O7) * 37.6 (mL K2Cr2O7) / 0.325 (mEq/mL HCl) = 60.854 = 60.9 mL
dont ask the internet to do your homework it wont go well
1 mL solution contain 1 mg potassium permanganate.
You can not convert mg (weight) to volume (ml).