A (0,1 M)
1.Weigh 0,8766 g ultrapure NaCl dried at 110 0C for 30 min.
2. Transfer NaCl in a clean 150 mL volumetric flask using a funnel. 3. Wash the funnel with 120 mL demineralized water.
4. Put the flask in a thermostat and maintain 30 min at 20 0C.
5. Add demineralized water up to the mark.
6. Stir vigorously and transfer in a clean bottle with stopper.
7. Add a label with necessary information.B (0,2 M)
1.Weigh 1, 7532 g ultrapure NaCl dried at 110 0C for 30 min.
2. Transfer NaCl in a clean 150 mL volumetric flask using a funnel. 3. Wash the funnel with 120 mL demineralized water.
4. Put the flask in a thermostat and maintain 30 min at 20 0C.
5. Add demineralized water up to the mark.
6. Stir vigorously and transfer in a clean bottle with stopper.
7. Add a label with necessary information.
A water solution containing 50 mM tris(hydroxymethyl)aminomethane and 150 mM sodium chloride has a pH of 7,6.
Also 150 mM of sodium.
Tyrode solution typically consists of sodium chloride, potassium chloride, calcium chloride, magnesium chloride, sodium bicarbonate, glucose, and HEPES buffer. The concentrations can vary, but a common composition is around 8 g/L of sodium chloride, 0.2 g/L of potassium chloride, 0.2 g/L of calcium chloride, 0.1 g/L of magnesium chloride, 1.0 g/L of glucose, and 5.0 g/L of HEPES.
The most common cation in the interstitial fluid is sodium at 150 mM. Next is calcium at 8.4 mM and potassium at 5 mM.
The needed mass is 0,0584 g.
The typical concentration of sodium is lower than potassium intracellularly. Sodium concentration is around 10-15 mM, while potassium concentration is around 140-150 mM inside the cell. This concentration gradient is maintained through the action of the sodium-potassium pump.
150 mm, there is 10 mm in 1cm, therefore 150 mm would be equal 15 cm.
The formula for this conversion is mEq = mg/atomic weight * valence. The atomic weight of sodium chloride is 23mg/mM.
Calculate the percent by weight of sodium (Na) and chlorine (Cl) in sodium chloride (NaCl) * Calculate the molecular mass (MM):MM = 22.99 + 35.45 = 58.44 * Calculate the total mass of Na present:1 Na is present in the formula, mass = 22.99 * Calculate the percent by weight of Na in NaCl:%Na = (mass Na ÷ MM) x 100 = (22.99 ÷ 58.44) x 100 = 39.34% * Calculate the total mass of Cl present:1 Cl is present in the formula, mass = 35.45 * Calculate the percent by weight of Cl in NaCl:%Cl = (mass Cl ÷ MM) x 100 = (35.45 ÷ 58.44) x 100 = 60.66% The answers above are probably correct if %Na + %Cl = 100, that is,39.34 + 60.66 = 100.
1 in = 25.4 mm. (150 mm)/(25.4 mm/in) = 5.9055 inches.
150 mm = 15 cm
25.4 mm = 1 in → 150 mm = 150 ÷ 25.4 in ≈ 5.91 in