Answer:
3.1atm
Explanation:
Given parameters:
Volume of gas = 2L
Number of moles = 0.25mol
Temperature = 25°C = 25 + 273 = 298K
Unknown:
Pressure of the gas = ?
Solution:
To solve this problem, we use the ideal gas equation.
This is given as;
PV = nRT
P is the pressure
V is the volume
n is the number of moles
R is the gas constant = 0.082atmdm³mol⁻¹K⁻¹
T is the temperature
P = 
Now insert the parameters and solve;
P = 
= 3.1atm
B: The total thermal energy is greater in a large body of water than one much smaller
Explanation:
A large lake filled filled with cool water will have more thermal energy than smaller pond filled with warmer water because the total thermal energy is greater in a large body of water than one that is much smaller.
Thermal energy is a form of kinetic energy usually due to transfer of heat energy.
Amount of heat energy is dependent on the differences in temperature, mass and specific heat capacity of a body.
Both lake water will have the same specific heat capacity. Since larger body of water has more mass, it will possess more thermal energy.
learn more:
Specific heat capacity brainly.com/question/7210400
Thermal energy brainly.com/question/914750
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Answer:
36.55kJ/mol
Explanation:
The heat of solution is the change in heat when the KNO3 dissolves in water:
KNO3(aq) → K+(aq) + NO3-(aq)
As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.
To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:
<em>Moles KNO3 -Molar mass: 101.1032g/mol-</em>
10.6g * (1mol/101.1032g) = 0.1048 moles KNO3
<em>Change in heat:</em>
q = m*S*ΔT
<em>Where q is heat in J,</em>
<em>m is the mass of the solution: 10.6g + 251.0g = 261.6g</em>
S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-
And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C
q = 261.6g*4.184J/g°C*3.5°C
q = 3830.87J
<em>Molar heat of solution:</em>
3830.87J/0.1048 moles KNO3 =
36554J/mol =
<h3>36.55kJ/mol</h3>
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Answer:
There are typically three ways that it is accomplished: use of erythropoietin (EPO) or synthetic oxygen carriers and blood transfusions. While transfusions of large volumes of blood or use of EPO can be detected, microdosing EPO or transfusing smaller volumes of packed red blood cells is much harder to detect.
