2017年7月8日 · There are 1000 mL (milliliters) in 1 L (liter). So, to convert from mL to L, divide by 1000. I hope this helps a lot! :)

2018年5月13日 · Q=mL Where, mis the mass and L is the latent heat of fusion or vaporization. ... mis the mass and L is the ...

2018年6月5日 · Divide by 1000. This is a case of the metric system. To convert from "milli-" to the standard unit, such as meters or liters, we divide by 1000. One milliliter is equal to a thousandth of a liter, and one liter is equal to one thousand milliliters. In numbers, that is: 1 \ "mL"=0.001 \ "L"=10^-3 \ "L" 1 \ "L"=1000 \ "mL"=10^3 \ "mL"

2014年10月18日 · NUMBER OF MOLES of HCl ; Molarity of solution x Volume of Solution of moles of HCl = (0.40 mol/L ) x 350 mL = (0.40 mol/L ) x 0.350 L = 0.14 mol mass of HCl that makes 0.14 mol of HCl = of moles x molar mass of HCl mass of HCl = 0.14 mol x 36.5 g/ mol mass of HCl = 5.11g As per Stoichiometry , 1g of HCl reacts with 1.015 g of Ca(OH)_2, 5.11g …

#7.24xx10^2*mL-=7.24xx10^2*cancel(mL)xx10^-3*L*cancel(mL^-1)=0.724*L# Such operations are known as #"dimensional analysis"#, and we did a little example in the problem; we wanted an answer with units of concentration and we got one. It is certainly not straightforward and intuitive, but it does work (eventually!). At your level you simply have ...

2016年12月13日 · The quantum numbers n, l, and m_l determine the probability pattern of an orbital. > The wave function of an orbital (its probability pattern) is determined by three quantum numbers. The principal quantum number n The principal quantum number determines the size of an orbital and its probability pattern. n can take the values 1, 2, 3, etc. The bigger the principal …

2014年9月15日 · You must dilute 401 mL with enough water to make 1.00 L of solution. Step 1. Calculate the molarity of the stock solution.

2017年4月28日 · Then that means you have "0.98 L H"_2"SO"_4 in "1 L" solution, for instance. Let's assume we are at "298.15 K". Then, assuming the density of 98% "v/v" "H"_2"SO"_4 is approximately the same as pure "H"_2"SO"_4, we have: "0.98 L H"_2"SO"_4 xx "1.84 g"/"mL" xx "1000 mL"/"L" ~~ "1803.2 g" and the mols of it would be: "1803.2 g H"_2"SO"_4 xx "1 …

How you would make 100.0 ml of a 1.00 mol/L buffer solution with a pH of 10.80 to be made using only sodium carbonate, sodium hydrogen carbonate and water? How much sodium carbonate and sodium hydrogen carbonate would you use?

2018年4月26日 · We need a volume of approx. 100*mL of the given HI(aq) solution. We assess the reaction... BaCO_3(s) + 2HI(aq) rarr BaI_2(aq) + H_2O(l) + CO_2(g)uarr And thus we need TWO EQUIV HI per equiv BaCO_3.