If I have 0.010 M CaCl2, what is the molar concentrations of CaCl2, Ca2+, and Cl- in an aqueous solution?
CaCl2 would be 0.010M because that is stated in the problem. Ca2+ would also be 0.010M because there is only one mol of it. Cl- would be 0.020M because there are two mols of it so you multiply 0.010M x 2 to get the answer.
If solution A has a volume of 25 mL, what is the volume of solution B?
I did this problem just to see if I could remember how to do it... so I don't know for sure if this is the right answer or not... but from what i remember it would be: .24M CaCl2 * 1 mol * 25mL 1mol .49M CaCl2 = 12.25mL like I said, I could be totally wrong here... but yeah
Two solutions, "A" and "B", are labeled "0.23 M CaCl2" and "0.40 M CaCl2"....Cont...Thanks!!!?
Molarity is the ratio of the moles of solute to the volume of solution in liters. M = mol / L solve for moles mol(a) = M(a) x L(a) mol(b) = M(b) x L(b) since the moles are equal, then M(a) x L(a) = M(b) x L(b) solve for L(b) L(b) = M(a) x L(a) / M(b) L(b) = 0.23M x 25mL / 0.40M = 14mL (to two significant digits)
Grade 11 chemistry help!?
(A) Barium sulfate, BaSO4, is made by the following reaction. Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaNO3(aq) An experiment was begun with 75.00 g Ba(NO3)2 and an excess of Na2SO4. After collecting and drying the product, 64.45 g of BaSO4 was obtained. Calculate the theoretical yield and the percentage yield of BaSO4. (B) Two solutions, "A" and "B", are labeled "0.19 M CaCl2" and "0.34 M CaCl2", respectively. Both solutions contain the same number of moles of CaCl2. If solution B has a volume of 25 mL, what is the volume of solution A? (C) How many milliliters of 0.411 M Na3PO4 solution are needed to react completely with 19.7 mL of 0.335 M CaCl2 solution? How many grams of Ca3(PO4)2 will be formed? The reaction is 3CaCl2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaCl(aq)
Chemistry Help Please?
9.51 A dilute solution of boric acid, H₃BO₃ is often used as an eyewash. How would you prepare 500.00 mL of 0.50% of (w/v) boric acid solution? 9.53 Describe how you would prepare 1.0 L of a 7.5% (w/v) KBr solution. 9.55 The concentration of glucose in blood is approximately 90 mg/100 mL. What is the weight/volume percent concentration of glucose? What is the molarity of glucose? 9.57 Which of the following solutions is more concentrated? (a) 0.50 M KCl or 5.0% (w/v) KCl (b) 2.5% (w/v) NaHSO₄ or 0.025 M NaHSO₄ 9.59 Assuming a blood glucose concentration of 90 mg/100 mL and a blood volume of 5.0 L, how many grams of glucose are present in the blood of an average adult? 9.61 The maximum concentration set by the U.S. Public Health Service for arsenic in drinking water is 0.050 mg/kg. What is this concentration in ppm and in the ppb? 9.63 How many moles of solute are in the following solutions? (a) 200 mL of 0.30 M acetic acid, CH₃CO₂H (b) 1.50 L of 0.25 M NaOH (c) 750 mL of 2.5 M nitric acid, HNO₃
Chemistry Help?
I need help please. End of the semester is next week and the professor is hitting us with 3 chapters of homework due next week and I am having trouble. I will appreciate any help. Thanks in advance. 9.83 Which solution has the higher boiling point, 0.500 M glucose or 0.300 M KCl? Explain. 9.85 What is the boiling point of the solution produced by adding 350 g of ethylene glycol (molar mass = 62.1 g/mol) to 1.5 kg of water? For each mole of nonvolatile solute, the boiling point of 1 kg of water is raised 0.51˚C. 9.87 What dos it mean when we say that a 0.15 M NaCl solution isotonic with blood, whereas distilled water is hypotonic? 9.89 Which of the following solutions will give rise to a greater osmotic pressure at equilibrium: 5.00 g of NaCl in 350.0 mL water or 35.0 g of glucose in 400.0 mL water? For NaCl, MW = 58.5 amu; for glucose, MW = 180 amu. 9.91 An isotonic solution must be approximately 0.03 osmol. How much KCl is needed to prepare 320 mL of an isotonic solution?
CHEMISTRY! Determine the # of grams of NaOH needed to make 100.0mL of a 1.00M NaOH solution?
Alighty, this'ns gonna involve some Molarity calculatin'. Woo! Now, Molarity = moles of solute / Volume of solute (IN LITERS!) You best remeber that IN LITERS part. So right of the bat, your techers tryin to trick ya by given you a volume in mL! But Ha! That wont fool you because you know you need to convert that into Liters first. so... 100.0mL ( 1 L / 1000 mL) = 0.1000L Cool! So, we're given Molartiy... but we need grams... well lets find moles first because we can do that with the information we now have. moles NaOH = 1.00M * 0.1000L = 0.100 moles NaOH Now to figure out how many grams that would be, we use the molar mass on the periodic table. NaOH = 22.990 + 15.999 + 1.0079 = 39.997 g NaOH So... moles to grams... 0.100 moles NaOH ( 39.997g NaOH / 1 mole NaOH ) = 4.00 g NaOH So, voila! Answer to #1 is... You need 4.00 g NaOH. **** and since I type so damn slow... I got beat to the punch by someone! I'll review their answers for quality!!!! ******** Yeah, Rob, you win. But my answer has the correct # of sig figs! What!? ************Arnab too. I see the trick though, you find one you can do and just type a letter in to get your answer up first, then you type it all the way out... I'm a newb, I had no idea!