The sample must be sufficiently soluble (fig. 2) to yield an NMR spectrum. For 1H and 1H observed NMR, it is recommended to dissolve between 2 and 10 mg in between 0.6 and 1 mL of solvent so that the sample depth is at least 4.5 cm in the tube (fig. 3).
<h3>
Answer:</h3>
19.3 g/cm³
<h3>
Explanation:</h3>
Density of a substance refers to the mass of the substance per unit volume.
Therefore, Density = Mass ÷ Volume
In this case, we are given;
Mass of the gold bar = 193.0 g
Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm
We are required to get the density of the gold bar
Step 1: Volume of the gold bar
Volume is given by, Length × width × height
Volume = 0.50 cm × 10.0 cm × 2.0 cm
= 10 cm³
Step 2: Density of the gold bar
Density = Mass ÷ volume
Density of the gold bar = 193.0 g ÷ 10 cm³
= 19.3 g/cm³
Thus, the density of the gold bar is 19.3 g/cm³
Answer:
a. The apparatus required to purify gypsum sample are: Bunsen burner, beaker, Filter Funnel, stirring rod, the filter paper.
b. Gypsum is a sulfate mineral that is made up of calcium sulfate dihydrate. Step-by-step instruction to purify gypsum sample is as follows:
1. Add water to the gypsum sample in a beaker.
2. Use the stirring rod to mix the mixture well.
3. Filter off the excess solid from the mixture using the filter paper and filter funnel.
4. Put the filtered mixture over the bunsen burner and evaporate the excess water from the mixture.
5. Allow the hot liquid to cool down and filter it again through the filter paper to get the pure gypsum.
Answer:
Because Oxygen shares 2 electrons with mutual bond interaction forming covalent bond . thus it is diatomic due to K shell 2 electrons mutual sharing .
Explanation:
Answer:
<h2>4.55 L</h2>
Explanation:
The new volume can be found by using the formula for Boyle's law which is
Since we're finding the new volume
We have
We have the final answer as
<h3>4.55 L</h3>
Hope this helps you
