To determine the equivalent weight of NaCl, we can follow these steps:1. Understand the Concept of Equivalent Weight: The equivalent weight of a substance is defined as the molecular weight divided by its valency. The valenc
From equation (1) and (2) 1 mole of AgNO3 is equivalent to 1 mole of NaCl and 1 mole of AgNO3 is equivalent to 1 mole of NH4SCN. Therefore 1 mole of AgNO3 is equivalent to 76.12 g of NH4SCN. Then 0.1 mole of AgNO3 is equivalent to 7.612 g of NH4SCN. ...
(Note: the molecular weight of NaCl is 58.44 g/mol.) This time, the solute has a valence of 2, as NaCl separates into Na+ and Cl-. The equation to get mEq is therefore [(30 mg)(2)]/(58.44 mg/mmol) = 1.027 mEq. Since there are 400 mL = 0.4 L, the concentration in mEq/L...
我也觉得你写错了,应该是equivalent weight中文解释为:当量当量:当量指与特定或俗成的数值相当的量;化学专业用语,用作物质相互作用时的质量比值的称谓。化学专业用语:在任何化学反应中,物质的质量比等于它们的当量比。泛 当量指化学方面的当量术。诸如 当量、克当量、当量浓度、酸碱盐当量、电化当量 等。根据定组成...
In the laboratory, NaCl, tap water and distilled water are used for measuring conductivity and ESDD. In addition, different sizes of glass plates are used as an insulating medium. A conductivity-measuring instrument (Cond 300i) is used to measure the conductivity of the salt-solution. Based ...
Just before the study end, GFR was estimated in conscious mice using the transcutaneous decay of retro-orbitally injected FITC-sinistrin (10 mg/100 g body weight dissolved in 0.9% NaCl), as previously described58. Plasma cystatin C was also measured by ELISA at the study end (BioVendor...
For analysis of the clay fraction, approximately 0.5 g samples were shaken with distilled water for 15 min, mixed with 10 ml of 1 M NaCl, and then repeatedly dispersed and centrifuged to recover the <2 μm fraction. This suspension was treated with acetic acid to remove carbonate minerals....
Weigh 0.5711 g Na2HPO4 (Mr 141.96 g/mol), 0.24 g KH2PO4 (Mr 136.09 g/mol), 8 g NaCl (Mr 58.44 g/mol), and 0.2 g KCl (Mr 74.55 g/mol) and dissolve in ~900 mL distilled water. Adjust to pH 7.4 with 1 M HCl and make up to 1000 mL with distilled water. • Oxidi...
approximately 0.15 g was quickly weighed into a 100 ml titration cup. The sample of known dry weight was then allowed to soak in the titration cup for 15 minutes in 5 ml of deionized water and 5 ml of ethanol. To the soaked sample, 55 ml of 2.0N NaCl solution were then added. A ...
To the soaked sample, 55 ml of 2.0N NaCl solution were then added. A back titration method using a TIM900 Titration Manager (Radiometer Analytical S. A., Lyon, France) was then started beginning with the addition of 5 ml of 0.05N NaOH solution. The entire blend was then stirred for ...