Answer: The overall width of a crystal measures 1.65 mm.
Explanation:
Horizontal distance separating the two molecules is 16.5 nm.
Width of the 
molecules:
The overall width of a crystal measured in millimeters=
The overall width of a crystal is 1.65 mm.
<span>Using PV=nRT, which represents a universal constant for any state, we have:
P1V1/n1T1=R
and
P2V2/n2T2=R;
This implies that:
P1V1/n1T1=P2V2/n2T2
Thus we can express it as
V1/n1=V2/n2.
Rearranging yields:
V2=V1 x (n2/n1) = 750 mL x ((0.65+0.35)/(0.65)) = 1200 mL = 1.2 L... with 2 significant figures</span>
Answer:
Indeed, the chemist is capable of identifying the compound present in the sample.
Explanation:
In one mole of K₂O, potassium has a mass of 2 × 39.1 g = 78.2 g, while the total mass of K₂O is 94.2 g. The mass ratio of K compared to K₂O is calculated as 78.2 g / 94.2 g = 0.830.
For 1 mole of K₂O₂, potassium's mass remains the same at 78.2 g, but the total mass of K₂O₂ is 110.2 g. The mass ratio of K to K₂O₂ then equates to 78.2 g / 110.2 g = 0.710.
When the chemist measures the mass of K in relation to the overall sample, the mass ratio can be computed.
- If the mass ratio is 0.830, then it indicates a pure K₂O compound.
- If the mass ratio is 0.710, it indicates a pure K₂O₂ compound.
- If the mass ratio falls outside of 0.830 or 0.710, the sample is assessed to be a mixture.
Answer:
Explanation:
Considering the reaction: 2X + 3Y = 3Z, combining 2.00 moles of X with 2.00 moles of Y results in the production of 1.75 moles of Z.
2 mol 2 mol 1.75 mol
2X + 3Y = 3Z
2 mol is required with 3 mol to yield 3 mol.
3 mol Z / 3 mol Y = 1 to 1
should yield 2 mol Z
1.75 / 2 = 87.5 % production yield
The classification is
disaccharide.
<span>Lactose qualifies as a disaccharide since it's made up of two monosaccharides, which are simple sugars (glucose and galactose). Disaccharides constitute a subgroup of carbohydrates, and besides lactose, other common examples include sucrose and maltose.</span>
