Answer:
What about hot air balloons? They work by similar principles. If you heat up a gas it expands. In the case of a hot air balloon, when the gas inside the balloon expands the extra gas is pushed out the bottom of the balloon, meaning that there are fewer atoms inside the balloon, meaning that the air in the balloon is lighter than the air outside the balloon.
The amount of lifting power is controlled by how hot the air is. If you heat the air inside the balloon 100 degrees F hotter than the outside air temperature, then the air inside the balloon will be about 25 percent lighter than the air outside the balloon. So a cubic foot of air weighs about 35 grams at 32 degrees F. A cubic foot of hot air at 132 degrees F will weigh 25 percent less, or about 26.5 grams. The difference is 8.5 grams or so. So a hot air balloon has to be much bigger to support the same weight, but it will float because hotter air is lighter than cooler air.
Explanation:
The amount of lifting power
Answer:
The answer is 5.10
Explanation:
<h3><u>Given</u>;</h3>
<h3>
<u>To </u><u>Find</u>;</h3>
We know that
pH + pOH = 7
pOH = 7 – pH
pOH = 7 – 1.90
pOH = 5.10
Thus, The pOH of the solution is 5.10
The complete balanced chemical reaction is:
2 AgNO3 + Na2S --> 2 NaNO3 + Ag2S
First let us calculate the number of moles of AgNO3.
moles AgNO3 = 0.315 M * 0.035 L
moles AgNO3 = 0.011025 mol
From the reaction, 1 mole of Na2S is needed for every 2
moles of AgNO3 hence:
moles Na2S required = 0.011025 mol AgNO3 * (1 mol Na2S / 2
mol AgNO3)
moles Na2S required = 5.5125 x 10^-3 mol
Therefore volume required is:
volume Na2S = 5.5125 x 10^-3 mol / 0.260 M
<span>volume Na2S = 0.0212 L = 21.2 mL</span>
<h3>
Answer:</h3>
0.35 M
<h3>
Explanation:</h3>
<u>We are given;</u>
- Initial volume as 35.0 mL or 0.035 L
- Initial molarity as 12.0 M
- Final volume is 1.20 L
We are required to determine the final molarity of the solution;
- Dilution involves adding solvent to a solution to make it more dilute which reduces the concentration and increases the solvent while maintaining solute constant.
- Using dilution formula we can determine the final molarity.
M1V1 = M2V2
M2 = M1V1 ÷ V2
= (12.0 M × 0.035 L) ÷ 1.2 L
= 0.35 M
Thus, the final concentration of the solution is 0.35 M
