Answer:
1) Length - Meter
2) Mass - Pound
3) Time - Minute
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Branched chain alkanes
The alkanes don't contain a functional group and so the branches are numbered from the end that gives the lowest set of position numbers for the branches.
Use the above rules to see how the names of the alkanes below are built up.
The structure of 2-methylbutane is a butane molecule (C4H10) but with a methyl group (CH3) replacing a hydrogen on the second carbon atom in the chain. The structure of 3-methylpentane could be drawn as butane with an ethyl group (C2H5) replacing a hydrogen on the second carbon. Note that this is not 2-ethylbutane. The structure of 2,2-dimethylbutane is butane with two methyl groups replacing the two hydrogens on the second carbon.
The most common pH indicator used in Simmons Citrate Agar is Bromthymol Blue (BTB)
Simmons Citrate Agar is a selective and differential medium used for the detection and differentiation of Enterobacteriaceae (gram-negative bacteria).
The medium contains sodium citrate as the sole carbon source, which is used to differentiate organisms based on their ability to utilize citrate as a sole carbon source.
The medium also contains pH indicators that change color based on the pH of the medium. The most common pH indicator used in Simmons Citrate Agar is Bromthymol Blue (BTB).
BTB is a pH indicator that turns yellow in acidic conditions and blue in basic conditions. As the bacteria metabolize the citrate in the medium, they produce acids, which cause the medium to become acidic.
This change in pH is detected by the BTB, which changes color from blue to yellow. The yellow coloration of the medium is an indication that the organism is utilizing citrate as a sole carbon source.
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According to boiling point elevation equation:
Δ T = i Kb m
when ΔT (change in boiling point) = 7.10 C°
and i (van't Hoff factor)= 1
and Kb = 0.520
so, by substitution:
7.10 = 1*0.520 *m
m = 7.1 / 0.52 = 13.65 m
Answer:
1. mol/L
2. 0.120 M
Explanation:
1. Molarity is equal to the moles of solute divided by the liters of solution. The units of molarity are mol/L.
2.
Step 1: Given data
- Mass of sodium chloride (solute): 5.25 g
- Volume of solution (V): 750.0 mL = 0.7500 L
Step 2: Calculate the moles of solute (n)
The molar mass of NaCl is 58.44 g/mol.
n = 5.25 g × 1 mol/58.44 g = 0.0898 mol
Step 3: Determine the molarity of the solution
We will use the definition of molarity
M = n/V
M = 0.0898 mol / 0.7500 L = 0.120 M
