Answer:
2a. 3mL.
2b. 10mL.
3. 38°C.
Explanation:
2a. We read the volume under the meniscus i.e under the curve.
The volume of the solution is 3mL.
2b. The volume of the pipette 10mL and the solution is at the calibration mark. Therefore, the volume of the solution is 10mL.
3. From the diagram given, we can see that 1 line represents 1 unit.
Now a careful look at the picture shows that the temperature is 38°C.
Answer:
I believe that it is B. 1
Explanation: I could be wrong but I believe its 1
<span>Answer:
.01 moles of D to .005 moles of L ~ so, .01+.005 = .015 total; using this total value, divide the portions of D and L.
so .01/.015 to .005/.015 ~ 67% D to 33% L.
And thus, the enantiomer excess will be 34%.</span>
"O2 was produced at a faster rate.(C)"
<span>0.0797 g
Looking at the formula, 1 mole of KIO3 and 5 moles of KI will react and produce moles of iodine molecules or 6 moles of iodine atoms. So first, determine the number of moles of KIO3 and KI provided
moles KIO3 = 0.0121 * 0.097 = 0.0011737 mol
moles KI = 0.0308 * 0.017 = 0.0005236 mol
The limiting reactant is KI at 0.0005236 mol so divide by 5 and multiply by 6 to get the number of moles of iodine atoms.
0.0005236 / 5 * 6 = 0.00062832 mol
Lookup the atomic weight of iodine which is 126.90447
And multiply that by the number of moles of iodine produced
126.90447 g/mol * 0.00062832 mol = 0.079736617 g
Rounding to 4 decimal places gives 0.0797 g</span>
