Need some help with these problems

b. a 0.5 M solution containing 250 g of manganese (II) chloride (MnCl2)
c. a 0.4 M solution containing 290 g of aluminum nitrate (Al(NO3)3)

complete question:

Calculate the volume of each solution, in liters.

b. a 0.5 M solution containing 250 g of manganese (II) chloride (MnCl₂)

c. a 0.4 M solution containing 290 g of aluminum nitrate (Al(NO₃)₃)

b. L ≈ 4.00 litres

c. L ≈ 3.4 litres

Explanation:

Molarity is the number of moles of solute per litre of solution. Together a solute and a solvent makes a solution. The formula for molarity can be represented below

M = number of moles of the solute(mol)/L

where

M = molarity

mol = number of moles of the solute

L = litre of the solution.

Therefore ,

The volume of each solution can be computed below

b. a 0.5 M solution containing 250 g of manganese (II) chloride (MnCl₂)

M = number of moles of the solute(mol)/L

number of moles of MnCl₂ = mass/molar mass

molar mass of MnCl₂ = 55 + 71 = 126 g/mol

number of moles of MnCl₂ = 250/126

number of moles of MnCl₂ = 1.9841 moles

0.5 = 1.9841/L

cross multiply

0.5L = 1.9841

L = 1.9841/0.5

L = 3.97

L ≈ 4.00 litres

c. a 0.4 M solution containing 290 g of aluminum nitrate Al(NO₃)₃

M = number of moles of the solute(mol)/L

number of moles = mass/molar mass

molar mass of Al(NO₃)₃ = 27 + 14 × 3 + 48 × 3 = 27 + 42 + 144 = 213

number of moles = 290/213 = 1.3615 moles

M = number of moles of the solute(mol)/L

0.4 =  1.3615/L

cross multiply

0.4L = 1.3615

divide both sides by 0.4

L = 1.3615/0.4

L = 3.4 litres

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.