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Answer:
The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.
Explanation:
The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:
n = 1.375x10⁻⁵ mol
The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):
1.375x10⁻⁵ mol _________ 19.9 units
x _________ 25.2 units
x = 1.741x10⁻⁵mol
Finally, we can calculate the Cu²⁺ concentration :
C = 1.741x10⁻⁵mol / 0.025 L
C = 6.964x10⁻⁴ M
Answer:
The empirical formula is SF6 (option E)
Explanation:
Step 1: Data given
Mass of sulfur = 3.21 grams
Mass of fluorine = 11.4 grams
Molar mass sulfur = 32.065 g/mol
Molar mass fluorine = 19.00 g/mol
Step 2: Calculate moles
Moles = mass /molar mass
Moles sulfur = 3.21 grams / 32.065 g/mol
Moles sulfur = 0.100 moles
Moles fluorine = 11.4 grams / 19.00 g/mol
Moles fluorine = 0.600 moles
Step 3: Calculate mol ratio
We divide by the smallest amount of moles
S: 0.100 / 0.100 = 1
F : 0.600 / 0.100 = 6
The empirical formula is SF6 (option E)
The purpose of the uninoculated control tubes used in this test is that two uninoculated control tubes are needed to show the results of the medium in both aerobic and anaerobic environments. It is used to show it is sterile and also as a color comparison, used also to show that the medium remains green under both conditions.
