
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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What is the result of this expression:
3.62 x (22.3 –20.3) /0.6655=
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- 12. A buret is filled with deionized (DI) water and the stopcock is used to adjust the meniscus of the water in the buret to an initial value of 1.06 mL. The stopcock is opened and the DI water is dispensed in a flask. The stopcock is closed and the final buret reading is 15.72 mL. The volume of DI water (mL) in the flask is 14.66 35.34 16.8 14.7arrow_forwardDimensional analysis simply refers to the inclusion of units in an equation. After setting up the solution map, dimensional analysis can be used to set up the conversion factors that lead to the desired values vithin this map. The correct setup of an equation can be verified by checking if the result will only have the desired units after unit cancellation. Ba3 (PO4)2 (aq) + 3Na2SO4 (aq) –→ 3BASO4 (s) + 2NA3PO4(aq) For the described reaction, you have a 0.200 L solution of 0.660 M Ba3(PO4)2(aq). Using the solution map from Part A as a guide, complete the following dimensional analysis that allows you to calculate he moles of BaSO4 (s) that can be formed by placing the values of each conversion factor according to whether they should appear in the numerator or denominator. Drag the appropriate values to their respective targets. • View Available Hint(s) Reset Help 0.396 mol BaSO<(s) 1 mol Ba3 (PO4)2(aq) 0.101 mol BaS04(s) 0.044 mol BaS04(s) 3 mol BaSO4(s) 0.660 mol 0.909 mol BaSO4(s)…arrow_forward(65.431 + 12.09) X 1.03solve using appropriate number of significant figuresarrow_forward
- Given the equation 2N₂(g) + O₂(g) ⇒ 2N₂O(g) Kc = 1.2x10-¹2, what is Kc for the equation 4N2O(g) = 4N₂(g) + 20₂(g)? 6.9x1023 2.4x1012 1.2x10-12 Not enough information is provided to calculate Kc'.arrow_forward18/8 stainless steel is made up of 18% (m/m) chromium and 8% (m/m) nickel in iron. What mass of iron would be in 235.4g of stainless steel?arrow_forwardGiven the following data for Mass of test tube and stearic acid = 14.17 gMass of test tube = 11.40 gFreezing point of strearic acid = 69.59o CMass of weighing paper + naphthalene =1.230 gMass of weighing paper = 0.920 gFreezing point solution = 64.00o CKf = 4.5o C/m Determine the following1. mass of stearic acid in g (2 decimal places); _____2. mass of naphthalene in g (2 decimal places); _____3. freezing point depression (2 decimal places); _____4. molality of solution (3 significant figures); _____5. moles of naphthalene (3 significant figures); _____6. molar mass of naphthalene, experimentally (3 significant figures); _____7. % error if theoretical molar mass of naphthalene is 128.17 g/ mole, USE ABSOLUTE VALUE (3 significant figure); _____arrow_forward
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