Answer:
Mole fraction = 0,0166
Explanation:
Mole fraction is defined as mole of a compound per total moles of the mixture. In the solution, the solute is fructose and the solvent is water. That means you need to find moles of fructose and moles of water.
The molecular mass of fructose is 180,16g/mol and mass of water is 18,02 g/mol. Using these values:
91,7g fructose × (1mol / 180,16g) = <em>0,509 moles of fructose</em>
545g water × (1mol / 18,02g) = <em>30,24 moles of water</em>
Thus, mole fraction of fructose is:
<em>Mole fraction = 0,0166</em>
I hope it helps!
Answer:
Explanation:
Given that:
Pressure = 791 mmHg
Temperature = 20.0°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (20 + 273.15) K = 293.15 K
T = 293.15 K
Volume = 100 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 62.3637 L.mmHg/K.mol
Applying the equation as:
791 mmHg × 1.14 L = n × 62.3637 L.mmHg/K.mol × 293.15 K
⇒n of 
produced = 0.0493 moles
According to the reaction:-
1 mole of carbon dioxide is produced 1 mole of calcium carbonate reacts
0.0493 mole of carbon dioxide is produced 0.0493 mole of calcium carbonate reacts
Moles of calcium carbonate reacted = 0.0493 moles
Molar mass of 
= 100.0869 g/mol
The formula for the calculation of moles is shown below:
Thus,
Impure sample mass = 5.28 g
Percent mass is percentage by the mass of the compound present in the sample.
Answer:
Soil series as established by the National Cooperative Soil Survey of the United States Department of Agriculture (USDA) Natural Resources Conservation Service are a level of classification in the USDA Soil Taxonomy classification system hierarchy. The actual object of classification is the so-called soil individual, or pedon.[1] Soil series consist of pedons that are grouped together because of their similar pedogenesis, soil chemistry, and physical properties. More specifically, each series consists of pedons having soil horizons that are similar in soil color, soil texture, soil structure, soil pH, consistence, mineral and chemical composition, and arrangement in the soil profile.[2] These result in soils which perform similarly for land use purposes.
Explanation:
hope its correct
<span>1. Fill a beaker or graduated cylinder with enough water to completely immerse the sphere in. 2. Record the baseline initial measurement. 3. Drop the sphere in. 4 <span>Record final measurement.</span></span>
Answer:
0.35 atm
Explanation:
It seems the question is incomplete. But an internet search shows me these values for the question:
" At a certain temperature the vapor pressure of pure thiophene (C₄H₄S) is measured to be 0.60 atm. Suppose a solution is prepared by mixing 137. g of thiophene and 111. g of heptane (C₇H₁₆). Calculate the partial pressure of thiophene vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal."
Keep in mind that if the values in your question are different, your answer will be different too. <em>However the methodology will remain the same.</em>
First we <u>calculate the moles of thiophene and heptane</u>, using their molar mass:
- 137 g thiophene ÷ 84.14 g/mol = 1.63 moles thiophene
- 111 g heptane ÷ 100 g/mol = 1.11 moles heptane
Total number of moles = 1.63 + 1.11 = 2.74 moles
The<u> mole fraction of thiophene</u> is:
Finally, the <u>partial pressure of thiophene vapor is</u>:
Partial pressure = Mole Fraction * Vapor pressure of Pure Thiophene
- Partial Pressure = 0.59 * 0.60 atm
