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>
The true statement is B. With identical masses for both metals, the final temperature of the two will be more aligned with 498 K rather than 298 K, as iron's specific heat capacity is significantly greater than that of gold's.
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:
The correct options include choice 2, 3, and 6.
Explanation:
Density is identified as the mass of a substance per unit volume occupied by that substance.
The density remains constant for a given substance, regardless of variations in mass and volume hence it is considered an intensive property.
2. 20.2 g of silver in 21.6 mL of water and 12.0 g of silver also in 21.6 mL of water.
3. 15.2 g of copper in 21.6 mL of water and 50.0 g of copper in 23.4 mL of water.
6. 11.2 g of gold in 21.6 mL of water and 14.9 g of gold in 23.4 mL of water.
The same metals in both instances will yield consistent densities due to the fixed density of the metal.
