Hello!
1.00 L of a gas at STP is compressed to 473 mL. What is the new pressure of gas?
- <u><em>We have the following data:</em></u>
Vo (initial volume) = 1.00 L
V (final volume) = 473 mL → 0.473 L
Po (initial pressure) = 1 atm (pressure exerted by the atmosphere - in STP)
P (final pressure) = ? (in atm)
- <u><em>We have an isothermal transformation, that is, its temperature remains constant, if the volume of the gas in the container decreases, so its pressure increases. Applying the data to the equation Boyle-Mariotte, we have:</em></u>






<u><em>Answer: </em></u>
<u><em>The new pressure of the gas is 2.11 atm </em></u>
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C; The Valence electrons spend more time around the atom of F
Answer:

Explanation:
We are asked to find the volume of a solution given the moles of solute and molarity.
Molarity is a measure of concentration in moles per liter. It is calculated using the following formula:

We know there are 0.14 moles of potassium chloride (KCl), which is the solute. The molarity of the solution is 1.8 molar or 1.8 moles of potassium chloride per liter.
- moles of solute = 0.14 mol KCl
- molarity= 1.8 mol KCl/ L
- liters of solution=x
Substitute these values/variables into the formula.

We are solving for x, so we must isolate the variable. First, cross multiply. Multiply the first numerator and second denominator, then the first denominator and second numerator.



Now x is being multiplied by 1.8 moles of potassium chloride per liter. The inverse operation of multiplication is division, so we divide both sides by 1.8 mol KCl/L.


The units of moles of potassium chloride cancel.


The original measurements of moles and molarity have 2 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 7 in the ten-thousandth place tells us to round the 7 up to a 8.

There are approximately <u>0.078 liters of solution.</u>
Cruising at 35,000 feet in an airliner, straight toward the east,
at 500 miles per hour.
Since your speed is not changing and you're moving in a straight line,
your acceleration is zero.
Answer:
Explanation:
Examples of pure substances include tin, sulfur, diamond, water, pure sugar (sucrose), table salt (sodium chloride) and baking soda (sodium bicarbonate). Crystals, in general, are pure substances. Tin, sulfur, and diamond are examples of pure substances that are chemical elements.