If 122 mL of allM glucose solution is diluted to 550.0 mL, what is the molarity of the diluted solution? Express your answer using two significant figures.
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A: Given: M1 = 1.2 M V1 = 124 mL. V2 = 550.0 mL
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- A solution of HCl was titrated against sodium carbonate. What is the average normality of acid in the given data? (MW Na₂CO3 = 106) T2 0.3479 0.3562 0.3042 Weight (g) Initial V (ml) 0.80 1.60 35.20 36.70 39.80 Final V (mL) Vol HCl used (mL) 35.10 39.40 N of HCl (eq/L) 0.1954 0.1870 Average N of HCl (eq/L)asia Fuqua Leam X e Home-Liferay www-awu.aleks.com = CHEMICAL REACTIONS Calculating molarity using solute mass mol/L. X F 12PM SPAN 101 Instructions Explanation Check X X A chemist prepares a solution of potassium permanganate (KMnO) by measuring out 1.5 g of potassium permanganate into a 50. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's potassium permanganate solution. Be sure your answer has the correct number of significant digits. S VHL Central Code Activation X O M + 0/5 ©2023 McGraw Hill LLC. All Rights Reserved. Terms of Use Terms of Use | Privacy Center Ala: Acce Jan 1Density of solution:Trial 1: 1.2 g/mLTrial 2: 1.2 g/mLTrial 3: 1.2 g/mL Average density = 1.2 g/mL What is the relative average deviaion, %?
- VIII. A primary standard of HCI is sodium carbonate, Na2CO3. A measured mass of solid Na2CO3, mNa2CO3' in grams, is dissolved in water and then titrated with HCI solution using methyl 2 NaCl + CO2 + H2O. orange indicator. The reaction is: Na2CO3 + 2 HCI- Complete the following formulas using the context described above: MHCI × VHCI in mL at the methyl orange endpoint = mNa2CO3 × ( )= mol of Na2CO3 mol of Na2CO3 X (— ( - - )= mol of HCl30.8 mg NaCl in 127 5 mL of solution Express the molarity in moles per liter to three significant figures • View Available Hint(s) M(NaCl) = Submit Provide Feedback F6 F7 F8 F9 F10 F11 24 AAktiv Chemistry → C < + X b Success Confirmation of QX M Your Chemistry answer is rex b Answered: How many mL o x b Answered: A student combi x app.101edu.co Time's Up! A 0.660 M solution of KCI needs to be prepared through dilution. A 2.00 M stock solution will be added to a 0.250 L volumetric flask and then water will be added to the 0.250 L mark. Determine the volume (in mL) of the 2.00 M stock solution of KCI needed to produce this solution. 1 4 7 +/- New Tab 2 5 8 mL 3 6 9 G 0 x + Submit X C x 100
- Table 10.2 Samples Placed in Water (data) Sample Dialysis tubing containing sugar solution ass Before Soakin lon 5.083 Prune Mass After Soaki 5.480 12.652 11.278 Table 10.3 (report) hange in Sample as urroundin irectiono onc. of Solution (inside sample Sample lativ en Dialysis tubing + water Dialysis tubing + sugar solution Prune Sugar solution Pure water Sugar solution Pure water PruneRewrite as a logarithmic equation. 10 = 2 Dlog O Olog0 No D=0 solutionThe solubility of BaSO, in water at 25 °C is measured to be 0.0023 Use this information to calculate K, L for BaSO4: sp Round your answer to 2 significant digits.
- Part 2 Data Mass of Urea (g): Buret Initial Volume (mL): water Part 2 Calculation: 4.05819 24.90 A. Find the Keq for dissolving urea (saturated solution) Mass of urea used 4.05819 Moles urea = Final Temperature of saturated solution ("C): Buret Final Volume (ml): fikcal - In 1.Write the chemical equation for urea dissolving: CityN₂0 (s) + H₂0 (1) → (H₂N₂O) (aq) Volume in Liters = 2.Calculate the molarity of the urea, [Urea]eam for the saturated solution of urea in water. You must first find the moles of urea (molar mass 60.05 g/mol). Then you must use the volume given in the data (above) to find the molarity of the urea in the saturated solution. Show calculations here. 3. Calculate AG soln Volume of water used: 3.88 ML Molarity of saturated solution (NOTE: this molarity = Keq ) = To do this, use: AG soln = - RT In Keq AH-TAS soln 4. Calculate AS soln Use: AG First, re-arrange the equation given above to solve for AS soln 6 27.0 W AS soln AGOsoln Then plug in the values you got for AG°…In the standardization of HCl using pure anhydrous sodium carbonate as primarystandard for methyl orange as indicator, 1.0 mL HCl was found to be equivalent to 0.05gof sodium carbonate (MW =106). The normality of HCl is: (Use whole numbers for the MM and 2 decimal places for the Final Answer)afrter estimating, i got 98.38 but apparentky its .37. i would likr to confirm which is right 2.3 g of a substance with a relative molecular mass of 668 are dissolved in water to make a total volume of 35 mL. What is the millimolar concentration? State your answer to 2 decimal places. 12. X 98.38. 98.37 millimolar -10 out of 1