Answer:
in 1983, Lake Mead had a recorded water level of 1,225 feet above sea level. The closest it's been to that level in modern times was in 2000. Explanation:
Answer:
Kr is a Noble Gas. Na is an alkali metal. F is halogen.
Group 17 is halogens. Inert is Noble Gases. Odourless and colourless is Noble Gases. Alkali metals do not occur freely in nature. Alkali metals are malleable
Explanation:
Answer:
![\boxed {\boxed {\sf P_2 \approx 3.53 \ atm}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%20P_2%20%5Capprox%203.53%20%5C%20atm%7D%7D)
Explanation:
In this problem, the temperature stays constant. The volume and pressure change, so we use Boyle's Law. This states that the pressure of a gas is inversely proportional to the volume. The formula is:
![P_1V_1=P_2V_2](https://tex.z-dn.net/?f=P_1V_1%3DP_2V_2)
Now we can substitute any known values into the formula.
Originally, the gas has a volume of 25.0 liters and a pressure of 2.05 atmospheres.
![25.0 \ L * 2.05 \ atm = P_2V_2](https://tex.z-dn.net/?f=25.0%20%5C%20L%20%2A%202.05%20%5C%20atm%20%3D%20P_2V_2)
The volume is decreased to 14.5 liters, but the pressure is unknown.
![25.0 \ L * 2.05 \ atm = P_2 * 14.5 \ L](https://tex.z-dn.net/?f=25.0%20%5C%20L%20%2A%202.05%20%5C%20atm%20%3D%20P_2%20%2A%2014.5%20%5C%20L)
Since we are solving for the new pressure, or P₂, we must isolate the variable. It is being multiplied by 14.5 liters and the inverse of multiplication is division. Divide both sides by 14.5 L .
![\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}=\frac{P_2 *14.5 \ L}{14.5 \ L}](https://tex.z-dn.net/?f=%5Cfrac%20%7B25.0%20%5C%20L%20%2A%202.05%20%5C%20atm%20%7D%7B14.5%20%5C%20L%7D%3D%5Cfrac%7BP_2%20%2A14.5%20%5C%20L%7D%7B14.5%20%5C%20L%7D)
![\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}= P_2](https://tex.z-dn.net/?f=%5Cfrac%20%7B25.0%20%5C%20L%20%2A%202.05%20%5C%20atm%20%7D%7B14.5%20%5C%20L%7D%3D%20P_2)
The units of liters cancel.
![\frac {25.0 * 2.05 \ atm }{14.5 }=P_2](https://tex.z-dn.net/?f=%5Cfrac%20%7B25.0%20%20%2A%202.05%20%5C%20atm%20%7D%7B14.5%20%7D%3DP_2)
![\frac {50.25\ atm }{14.5 }=P_2](https://tex.z-dn.net/?f=%5Cfrac%20%7B50.25%5C%20%20atm%20%7D%7B14.5%20%7D%3DP_2)
![3.53448276 \ atm = P_2](https://tex.z-dn.net/?f=3.53448276%20%5C%20atm%20%3D%20P_2)
The original values of volume and pressure have 3 significant figures, so our answer must have the same.
For the number we found, that is the hundredth place.
The 4 in the thousandth place (in bold above) tells us to leave the 3 in the hundredth place.
![3.53 \ atm \approx P_2](https://tex.z-dn.net/?f=3.53%20%5C%20atm%20%5Capprox%20P_2)
The new pressure is approximately <u>3.53 atmospheres.</u>
C. Formation of a new substance