Since mass can be computed by M = d ⋅ V M = 0.737 g mL ⋅ 51.48L ⋅ 1000 mL L M = 37,941g or M = 37.94kg One note: 1 US gallon = 3.7854 liters (as a Civil (Environmental) Engineer I use …

= 0.186 ⋅ mol ⋅ L−1 So what's unusual about this problem? It expresses volume in units of mL NOT L. But the prefix m stands for milli i.e. 10−3, just as c stands for centi i.e. 10−2, and k …

Divide thru by 10^3. 1*mL = 1xx10^-3*L, the m specifies "milli"=10^-3. Thus 87.98*mL = 87.98*cancel (mL)xx10^-3*L*cancel (mL^-1) = ??L

Molarity = g solute g/mol solute × 1000 mL/L mL solution 0.770 M = x 119 g/mol × 1000 mL/L 285 mL x = 26.1 grams

Dimensions of L, MT^(-2)L^2A^(-2) Dimensions of R, ML^2T^(-3)A^(-2) Firstly consider resistance. It's defining equation is, Ohm's law, V = IR implies R = V/I Now V has units of …

If 500 mL of HCl gas at 300 K and 100 kPa dissolved in pure water requires 12.50 mL of the NaOH solution to neutralize in a titration experiment, what is the concentration of the NaOH …

Well, to start you need a balanced equation to show molar equivalence....I gets approx. 0.2*mol*L^-1 with respect to sulfuric acid.... H_2SO_4 (aq) + 2KOH (aq) rarrK_2SO_4 (aq) + …

What is pH of the solution if a. 41 ⋅ mL of 0.310 ⋅ mol ⋅ L−1 N aOH (aq) is added to if 31 ⋅ mL of 0.310 ⋅ mol ⋅ L −1 H C l(aq)? And b. if a 31 ⋅ mL volume of 0.310 ⋅ mol ⋅ L−1 of H C l(aq) …

How many grams of solute are present in 705 mL of 0.900 M K B r?

22.5 ml of 0.102N KMnO4 is required to oxidized 10ml. mohr's salt solution,determination the amount of Fe++ of that solution??