Answer:
38.152 g NaCl would be produced.
Explanation:
<u>M</u><u>a</u><u>r</u><u>k</u><u> </u><u>me</u><u> </u><u>as</u><u> </u><u>Brainliest</u><u> </u>
Considering ideal gas behavior, the volume of 1 mol of gas at STP is 22.4 L; then the volume occupied by 1.9 moles is 1.9mol*22.4L/mol
The answer is 43 L if I am correct.
Answer:
3
Explanation:
Lt= Loe^(-kt)
Data:
Lo = 10 mg/mL
Assume k = 0.23/da
1. Calculate L5
L5 = 10e^(-5×0.23) = 10e^-1.15 = 10 × 0.317 = 3.17 mg/mL
2. Calculate the dilution factor
You expect to find L5 to be about 3
You want L5 to be about 1.
You should use a dilute your sample by a factor of 3.
P = 3
Answer:
Look them up.
Explanation:
sulphate is SO4 (2-)
Sufide is S(-1)
Sulfite - look it up - there are several more right there you may need.