Answer:
Q = -18118.5KJ
W = -18118.5KJ
∆U = 0
∆H = 0
∆S = -60.80KJ/KgK
Explanation:
W = RTln(P1/P2)
P1 = 1bar = 100KN/m^2, P2 = 1500bar = 1500×100 = 150000KN/m^2, T = 23°C = 23 + 273K = 298K
W = 8.314×298ln(100/150000) = 8.314×298×-7.313 = -18118.5KJ ( work is negative because the isothermal process involves compression)
∆U = Cv(T2 - T1)
For an isothermal process, temperature is constant, so T2 = T1
∆U = Cv(T1 - T1) = Cv × 0 = 0
Q = ∆U + W = 0 + (-18118.5) = 0 - 18118.5 = -18118.5KJ
∆H = Cp(T2 - T1)
T2 = T1
∆H = Cp(T1 - T1) = Cp × 0 = 0
∆S = Q/T
Mass of water = 1kg
Heat transferred (Q) per kilogram of water = -18118.5KJ/Kg
∆S = (-18118.5KJ/Kg)/298K = -60.80KJ/KgK
Answer:
The concentration is [-1 + sqrt(1+0.11t)]/0.1542 M
Explanation:
Let the concentration of CH3CHO after selected reaction times be y
Rate = Ky^2 = change in concentration of CH3CHO/time
K = 0.0771 M^-1 s^-1
Change in concentration of CH3CHO = 0.358 - y
0.0771y^2 = 0.358-y/t
0.0771ty^2 = 0.358 - y
0.0771ty^2 + y - 0.358 = 0
The value of y must be positive and is obtained in terms of t using the quadratic formula
y = [-1 + sqrt(1^2 -4(0.0771t)(-0.358)]/2(0.0771) = [-1 + sqrt(1+0.11t)]/0.1542 M
The product that is used on the natural nail prior to application to assist in adhesion and serves to chemically bond the enhancement product to the natural nail is known as nail primer.
<h3>What is a nail primer?</h3>
A nail primer is a chemical agent used in esthetic centers before applying a colored polish to the nails and serves as an adhesive product.
The nail primers are also very useful for improving the cleaning efficiency of the product before its application.
Nail care products include different types of chemical formulations such as, for example, creams that reinvigorate the cuticle.
In conclusion, the chemical formulation employed on the natural nail that is capable of enhancing and also assisting adhesion is called the nail primer.
Learn more about nail esthetic products here:
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an atom will gain or lose electrons to obtain 8 electrons in it's outer shell. In other words, elements want to reach an electron configuration of the nearest noble gas.
The concentration of the sodium chloride would be 0.082 M
<h3>Stoichiometric calculations</h3>
From the equation of the reaction, the ratio of AgCl produced to NaCl required is 1:1.
Mole of 46.6 g AgCl produced = 46.6/143.32 = 0.325 moles
Equivalent mole of NaCl = 0.325 moles.
Molarity of 0.325 moles, 3.95 L NaCl = mole/volume = 0.325/3.95 = 0.082 M
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