45 m. If each student needs 750 mm of tubing, the teacher should order 45 m of tubing.
a) Find the <em>length in millimetres</em>
Length = 60 students x (750 mm tubing/1 student) = 45 000 mm tubing
b) Convert <em>millimetres to metres
</em>
Length = 45 000 mm tubing x (1 m tubing/1000 mm tubing) = 45 m tubing
Answer:
1. 7 , 2. 0 crackers, 6 choc pieces and 1 marshmallow
Explanatio
Find the number of moles
C = n / V
C(Concentration) = 0.30 moles / L
V ( Volume) = 2 L
n = ??
n = C * V
n = 0.30 mol / L * 2 L
n = 0.60 mol
Find the molar mass
2Na = 23 * 2 = 46 grams
1S = 32 * 1 = 32 grams
O4 = 16 * 4 = 64 grams
Total = 142 grams / mol
Find the mass
n = given mass / molar mass
n = 0.06 mol
molar Mass = 142 grams / mol
given mass = ???
given mass = molar mass * mols
given mass = 142 * 0.6
given mass = 85.2 grams.
85.2 are in a 2 L solution that has a concentration of 0.6 mol/L
Answer: It will be produced 276,3 mg of product
Explanation: The reaction of anthracene (C14H10) and maleic anhydride (C4H2O3) produce a compound named 9,10-dihydroanthracene-9,10-α,β-succinic anhydride (C18H12O3), as described below:
C14H10 + C4H2O3 → C18H12O3
The reaction is already balanced, which means to produce 1 mol of C18H12O3 is necessary 1 mol of anthracene and 1 mol of maleic anhydride.
1 mol of C14H10 equals 178,23 g. As it is used 180 mg of that reagent, we have 0,001 mol of anthracene. With it, the reaction produces 0,001 mol of C18H12O3.
As 1 mol of C18H12O3 equals 276,3 g, the mass produced is 276,3 mg.
