Take an oxo acid, sulfuric acid, nitric acid, phosphoric acid, take away the of water, and you form the so-called $"acid anhydride"$, literally the acid sans water.
Add the elements of water, you get back the acid. Clearly, sulfur trioxide is the acid anhydride of sulfuric acid, $H_2SO_4$, which would be the strongest acid available, so strong that dissociation to sulfate ion would be almost quantitative .
This is a limiting reactant problem. We know that we will need a balanced equation and moles of each reactant. 1. Gather all the information in one place with...
The first thing to check here is how much hydrogen chloride, $"HCl"$, can you dissolve in water at room temperature. This will help you make sure that all the mass...
The trick here is to realize that protein $"X"$ has a molar mass of $"15,000 g mol"^(-1)$. First of all, I should mention that the problem is actually giving you...
For starters, convert the mass of potassium bromide to moles by using the molar mass of the compound. $15.6 color(red)(cancel(color(black)("g"))) * "1 mole KBr"/(119.002color(red)(cancel(color(black)("g")))) = "0.1311 moles KBr"$...
We can solve this problem using some calculations: $"molarity" = "mol solute"/"L soln"$ We should convert the given mass of $"(NH"_4")"_2"SO"_4$ to moles using its molar mass (calculated to...
$"Molarity"$ $=$ $"Moles of solute"/"Volume of solution"$, and so we work out the quotient..... $((117*g)/(58.44*g*mol^-1))/(1*L)=???*mol*L^-1$. The initial dilution was there to distract you. We only need the final volume, and...
$C$ $=$ $n/V$, where $n=(66.6*g)/(110.98*g*mol^-1)$ $=$ $0.600$ $mol$. The final volume of the solution is $550xx10^-3*L$ So $"Concentration with respect to "CaCl_2=(0.600*mol)/(550xx10^-3*L)$ $=$ $??$ What is the concentration with respect...
Use Raoults Law $ p_{i}=p_{i}^{\star }x_{i},$ $" Calculate number of moles in each of the substance"$ $"Mol of H20 = 250 g H2O * 1 mol/18.02 g H2O = 13.9...
$"A."$ The silver salt is quite soluble, and proceeds in the forward direction. Thus $DeltaG^@$ is negative. $"B."$ Given the first answer, $DeltaS^@$ is also positive. Most (all!) spontaneous reactions...