Answer: Marcia made a more concentrated salt solution.
Explanation:
Marcia used 1 kg of salt, which the exercise tells us is 2.2 lb.
She dissolved it in 3L of water, so we have to find out how much that is in pints. We do that using cross multiplication.
If 1L equals 2.11 pints, then
xpints / 3L = 2.11pints / 1L, so
xpints = (3L . 2.11pints) / 1L = 6.33 pints
Now we know she dissolved 2.2lb in 6.33 pints of water. We use cross multiplication again to find out how much salt she would have in 10 pints, so we can know how concentrated it is compared to Bobby's solution, which is 10 pints.
xlb / 10pints = 2.2lb / 6.33pints
xlb = (10pints . 2.2lb) / 6.33pints = 3.47lb
So Marcia has a concentration of 3.47lb per 10 pints, whereas Bobby only has 3lb per 10 pints.
Answer:
I believe that the answer is c
Explanation:
The snails will split hope this helps! <3
Unfortunately the data provided doesn't include the DENSITY of the ammonium chloride solution and molarity is defined as moles per volume. So without the density, the calculation of the molarity is impossible. But fortunately, there are tables available that do provide the required density and for a 20% solution by weight, the density of the solution is 1.057 g/ml.
So 1 liter of solution will mass 1057 grams and the mass of ammonium chloride will be 0.2 * 1057 g = 211.4 g. The number of moles will then be 211.4 g / 53.5 g/mol = 3.951401869 mol. Rounding to 3 significant digits gives a molarity of 3.95.
Now assuming that your teacher wants you to assume that the solution masses 1.00 g/ml, then the mass of ammonium chloride will only be 200g, and that is only (200/53.5) = 3.74 moles.
So in conclusion, the expected answer is 3.74 M, although the correct answer using missing information is 3.95 M.
Answer:
mL
Explanation:
Atmospheres (atn), Torr, and mm of Hg are all units of pressure but mL is a unit of volume, not pressure.
pH of buffer can be calculated as:
pH=pKa+log[salt]/[Acid]
As ka = 4.58 x 10-4
Concentration of [Salt] that is NO2(-1)=0.380M
Concentration of [Acid] that is HNO2=0.500M
So, pH= -log(4.58*10^-4)+log((0.380)/0.500))
=3.21
So pH of solution will be 3.21
