Response:
Tube 2: 8.26 * 10^-3; Tube 4: 6.83 * 10^-5
Explanation:
For the MIC test's serial dilutions, each tube should contain an equal volume of nutrient broth: 5.0 mL, while the agent's volume per dilution must also match: 0.5 mL.
The serial dilution process followed was:
- Tube 1: 0.5/5.5
- Tube 2: 0.5 mL from tube 1 was diluted with 5.0 mL of broth, resulting in a dilution of tube 2 as (1:11) * (1:11) = (0.5/5.5) * (0.5/5.5) = 1:121 = 8.26 * 10^-3
- Tube 3: similar calculations yield 1:1331 = 7.51 * 10^-4
- Tube 4: yields 1:14641 = 6.83 * 10^-5.
Answer:
Given that camphor is a sublime material while sand is not, a gradual heating of the mixture allows for the separation of camphor from sand. The camphor vapors can then be collected and allowed to cool. This process will result in the formation of solid camphor crystals.
Explanation:
Convert HCl and H2O to moles.
36.0 g of HCl = 0.987 moles HCl
98.0 g of H2O = 5.44 moles H2O
Based on the stoichiometric ratio for HCl,
there are 0.987 moles of H and 0.987 moles of Cl.
For H₂O, according to the stoichiometric ratio, you have 10.88 moles of H and 5.44 moles of O.
Combining them:
11.867 moles H
0.987 moles Cl
5.44 moles O
Revert the moles back to grams, then divide by the total mass and multiply by 100 for the percentage by mass.
11.867 moles H = 11.96 g H
0.987 moles Cl = 34.99 g Cl
5.44 moles O = 87.03 g O
11.96/(36.0+98.0)(100) = 8.93% for H
34.99/(36.0+98.0)(100) = 26.11% for Cl
87.03/(36.0+98.0)(100) = 64.96% for O.
Explanation:
Filtration serves as a method of separation where solid particles that are suspended in a liquid are isolated by passing the mixture through filter paper's pores. This process ensures that the solid particles accumulate on the filter paper and the liquid flows out through the filter paper's pores.
The ordered sequence of the steps provided is:
- Measure and fold the filter paper.
- Insert the filter paper into the funnel, then position the funnel above the Erlenmeyer flask.
- Let the solid/liquid mixture pass through the filter.
- Rinse the filter paper that holds the mixture with water.
- Measure the weight of the dry filter paper along with the copper.
For KNO₃, the mass is 346g. The molar mass can be computed as (39.098) + (14) + (15.99*3), which results in 101.068 gmol⁻¹. The volume of the solution is given as 750ml, equivalent to 0.75dm³. The formula for molarity is (mass of solute/molar mass of solute)*(1/volume of solution in dm³). Accordingly, molarity = (346/101.068)*(1/0.75), yielding 4.56 mol dm⁻³.
