The molarity of a solution is the number of moles of a substance in one liter of that substance.
The molar mass of ammonium sulfate (NH4)2SO4 is 132.14 grams/mole
Calculate the moles of ammonium sulfate:
(4.50 grams)/(132.14 grams/mole) = 0.0341 moles of ammonium sulfate
convert mL to Liters 250. mL becomes 0.250 liters
Take the number of moles over the number of liters
0.0341 moles / 0.250 liters = 0.136 molar or 0.136M = molarity of the solution
Answer:
The molecular geometry of SO3 is trigonal planar.
Explanation:
Look at the Lewis
NaH(s)+ H2O (l)=>NaOH(aq)+H2(g)
You want to calculate the mass of NaH, I assume. Otherwise, the question isn't clear. It simply says calculate the mass(??)
So, calculate the moles of H2 gas that satisfy the conditions of 982 ml at 28ºC and 765 torr. But you must subtract the vapor pressure of water at 28º to get the actual pressure of the H2 gas. So, the actual conditions are 982 ml (0.982 L) and 301 K and 765-28 = 737 torr.
PV = nRT
n = PV/RT = (737 torr)(0.982 L)/(62.4 L-torr/Kmol)(301 K)
n = 0.0385 moles H2
moles NaH needed = 0.0385 moles H2 x 1 mole NaH/mole H2 = 0.0385 moles NaH required
mass of NaH needed = 0.0385 moles x 24 g/mole = 0.925 g NaH
Brainliest Please :)
The answer is A, between 0 and 7.
In a pH scale from 0 to 14, we can groups these numbers into acidic, neutral, and alkaline. 7 is the neutral pH value, therefore, 0-7 is always acidic, and 7-14 is alkaline.
The smaller the number is, the more acidic the solution will be. This applies same in alkalis, the larger the pH value is, the more alkaline the solution is.
We can measure the pH of solution with many methods, the easiest way include using a pH paper, more advanced and accurate methods includes using a pH meter.
It's lone a little distinction (103 degrees versus 104 degrees in water), and I trust the standard rationalization is that since F is more electronegative than H, the electrons in the O-F bond invest more energy far from the O (and near the F) than the electrons in the O-H bond. That moves the powerful focal point of the unpleasant constrain between the bonding sets far from the O, and thus far from each other. So the shock between the bonding sets is marginally less, while the repugnance between the solitary matches on the O is the same - the outcome is the edge between the bonds is somewhat less.
