Here's an example of a bar graph.
Boiling point<span> is the </span>temperature<span> at which the vapor pressure of the liquid equals the surrounding pressure.
Above boiling point point, liquid get converted into vapour.
Now, boiling point of water is 100 oC at room pressure. Room pressure is equal to 760 torr. Thus, at 100 oC, vapour pressure of water becomes equal to 760 torr.
Now, if external pressure is increased to 880 torr, more heat is to be supplied so that vapour pressure of water equals 880 torr.
So, at 880 torr, boiling point of water will be more than 100 oC. In present case, most like the boiling point of water is equal to 105 oC.
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Considering that CCL3F gas behave like an ideal gas then we can use the Ideal Gas Law
<span>PV = nRT, however is an approximation and not the only way to resolve this problem with the given data..So,at the end of the solution I am posting some sources for further understanding and a expanded point of view. </span>
<span>Data: P= 856torr, T = 300K, V= 1.1L, R = 62.36 L Torr / KMol </span>
<span>Solving and substituting in the Gas equation for n = PV / RT = (856)(1.1L) /( 62.36)(300) = 0.05 Mol. This RESULT is of any gas. To tie it up to our gas we need to look for its molecular weight:MW of CCL3F = 137.7 gm/mol. </span>
<span>Then : 0.05x 137.5 = 6.88gm of vapor </span>
<span>If we sustract the vapor weight from the TOTAL weight of liquid we have: 11.5gm - 6.88gm = 4.62 gm of liquid.d</span>
Answer:
29.575%
Explanation:
Data provided:
Calories taken in daily diet = 2000
Recommended amount of fat = 65 grams
Average number of calories for fat = 9.1 calories / g
Thus,
Number of calories in the diet with average number of calories for fat
= Recommended amount of fat × Average number of calories for fat
= 65 × 9.1
= 591.5 calories
Therefore,
the percentage of calories in his diet supplied = ( 591.5 / 2000 ) × 100
= 29.575%
