- Scan Question
- + Post
- Get Coins
a 25.7027g sample of well water was placed in an aluminum dish as evaporated to dryness. The residue weighed 0.0035g, what is the concentration of total solids in the well water in ppm?
mass of solute (residue) = 0.0035 g
mass of solution = 25.7027 g
concentration of total solids in the well water = mass of solute * 1000000 / mass of solution
= 0.0035 * 1000000 / 25.7027
= 136.2 ppm (answer)
a 25.7027g sample of well water was placed in an aluminum dish as evaporated to dryness....
Problem #2 A 100-mL water sample is collected from the activated sludge process of municipal wastewater treatment. The sample is placed in a drying dish (weight-0.5000 g before the sample is added), and then placed in an oven at 104°C until all the moisture is evaporated. The weight of the dried dish is recorded as 0.5625 g. A similar 100-mL sample is filtered and the 100-mL liquid sample that passes through the filter is collected and placed in another drying...
If 276 mL of a 2.30 M aqueous solution of Ca(NO3) is evaporated to dryness, what is the mass of the residue?
If 242 mL of a 2.50 M aqueous solution of Ca(NO3)2 is evaporated to dryness, what is the mass of the residue?
PLEASE SHOW THE SOLUTION ORDERLY THANK YOU. PART A. Total Dissolved Solids (TDS) A 25.0 mL aliquot of a well-shaken and filtered sample of river water is pipetted into an evaporating dish. The sample was heated to dryness. The following data were obtained: • Mass of evaporating dish = 25.034 g • Mass of dried sample plus evaporating dish = 25.636 g PART B. Total Solids (TS) and Total Suspended Solids (TSS) A 25.0 mL aliquot of a well-shaken sample...
5. If 230 mL of a 2.40 M aqueous solution of Ca(NO3)2 is evaporated to dryness, what is the mass of the residue ? Enter the correct mumerical value with the appropriate number of significant figures. Express scientific notation like 4.29E-15 Grade This Exercise 1 41 11/
In a dehydration experiment a student uses MgSO4*nH20. The student placed an evaporating dish which contained 3.8457 g of MgSO4*nH20 in an oven at 110 degrees Celcius.At the temperature, some but not all the water in this hydrate was lost. When a constant mass was achieved, the salt weighed 2.4402 g.The dish was placed back in the oven at 160 degrees C and the residue became a constant mass of 1.8779 g. Stronger heating produced no further loss in mass...
A 12.0 mL sample of NaCl solution has a mass of 12.65g after the solution is evaporated to dryness, the dry residue has a mass of 4.25g. Calculate the concentration of the solution in m/m% m/v% and M
Solve this please Problem-1) The solids test results for 150 mL wastewater sample are: Item Weight (g) Clean dish 42.61 Dry filter 1.55 Dish and total dry solid after evaporation 42.66 Dish and dry filtered solids 44.19 Dish and residue solids after burning on 600°C 42.64 Dish and residue after burning filtered solids on 44.18 600°C Find the Total, Total Dissolves, Total Suspended, Total Volatile, Total Fixed, Volatile Suspended, Fixed Suspended, Volatile Dissolved and Fixed Dissolved solids in mg/L,
Instructor Name: Student Name: DATA (EXP #2 ): Sink wek In keng Origin of your water sample: Part 1: Total Dissolved Solids Water sample size (mL): 5mL Mass after further heating Mass initial 43-3269 A3 3459 Mass empty evaporating dish (g) 43-3369 43-34 Mass of dish +water residue (g) Mass Difference (g) 0:0lg Part 2: Hardness Titration Molarity of EDTA (from carboy): Trial 2 Trial 3 Trial 1 0.00 me Initial Volume (mL) 000mL 7ML 그오at 8.1mc Final Volume (mL)...
A 25 gram sphere of aluminum is heated to 95.5°C and placed in a well-insulated container containing 100.0 mL water at 25.5°C. What is the temperature of the water after thermal equilibrium is reached?