Prepare a graph with the mass as abscissa and the volume as the ordinate. Plot the tabulated data of "density of water". Examine the curves and make a generalization regarding the relationship between mass and volume.
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- Hello can this problem be simplified In administering a mixture of NO2 and O2 by flow meters the target total flow is 14 L/min. What flow , in L/min should the NO2 have to administer 31 % NO2 ? (answer to 1/10 X.X and include units) Answer unitsExpress 20,000ppb as a percentage strength A. 2% B. 0.02% C. 0.00002% D. 0.002%Please answer super fast and answer all questions and show calculations For the image attached For 1. a Mass of metal: Trial 1 is 35.0228 g Trial 2 is 35.0915 g Trial 3 is 34.0821 g Mass of water: Trial 1 is 20.0177 g Trial 2 is 20.0250 g Trial 3 is 20.0168 g For delta t of water: Trial 1 is 15.5 C Trial 2 is 15.7 C Trial 3 is 15.1 C For delta t of metal Trial 1 is 80.1 C Trial 2 is 80.2 C Trial 3 is 79.5 C For B my calculated Specific heat is: Trial 1 is 0.462 Trial 2 is 0.467 Trial 3 is 0.466
- One hundred grams (100 g) of water is mixed to 150 g of alcohol (density = 790 kg/m3). Calculate the SG of the total mixture. Please show all complete solution with units to its finals answer thank youWhat's 3 products available on the market where its specific gravity must be measured with picture and strength of them?Need help with question #24 Results Trial 1 : Initial Volume = 0 0.5143g=KHP 14.3 = Buret Measurment Trial 2: Initial Volume= 0 0.5615g = KHP 14.1 = Buret Measurement
- 1. Calculate the experimental density of a salt solution and the percent error (same as relative error percent) using some or all the data given below. solubility of NaCl salt in water: 0.357 g/mLmass of empty graduated cylinder: 25.19g mass of graduated cylinder + salt solution: 30.47g total volume of salt solution: 4.98 mLtrue density of salt solution: 1.07 g/mLWhat is the density of the vinegar sample used? TRIAL 1 TRIAL 2 TRIAL 3 Wt. of vinegar sample 8.2000 g 8.1000 g 8.2500 g Vol. of vinegar sample 10.00 mL 10.00 mL 10.00 mLPlease answer fast it’s very important and urgent I say very urgent so please answer super super fast please For the image attached For 1. a Mass of metal: Trial 1 is 35.0228 g Trial 2 is 35.0915 g Trial 3 is 34.0821 g Mass of water: Trial 1 is 20.0177 g Trial 2 is 20.0250 g Trial 3 is 20.0168 g For delta t of water: Trial 1 is 15.5 C Trial 2 is 15.7 C Trial 3 is 15.1 C For delta t of metal Trial 1 is 80.1 C Trial 2 is 80.2 C Trial 3 is 79.5 C For B my calculated Specific heat is: Trial 1 is 0.462 Trial 2 is 0.467 Trial 3 is 0.466
- Ethyl alcohol (C2H5OH) may be prepared by the fermentation of glucose (C6H12O6) as indicated by the equation: yeastC6H12O6 ----> C2H5OH + CO2 74.12 mL of ethyl alcohol (specific gravity = 0.790) was collected by this fermentation pro- cess. What mass of glucose was used? SET-UP: Answer:You wait an empty glass filter on the balance and record a way of 12.4361 g. After a 1.00 L sample of river water has been filtered and the filter has been dried for one week at 105°C, the dried filter and TSS weigh 13.682 g. What is the weight of the TSS? Make sure to give the answer in correct significant figures in with the correct units. Thank you!Data of milk: (first trial) Mass of milk: 104.4579g Concentration of NaOH (M): 0.09639 Volume of NaOH solution used: 5.5 mL Data of milk: (2nd trial) Mass of milk: 103.8405g Concentration of NaOH (M): 0.09639 Volume of NaOH solution used: 5.3 mL