A cotton gin is a machine that quickly and easily separates cotton fibers from their seeds, allowing for much greater productivity than manual cotton separation.[2] The fibers are then processed into various cotton goods such as linens, while any undamaged cotton is used largely for textiles including clothing. Seeds may be used to grow more cotton or to produce cottonseed oil.
Although simple handheld roller gins had been used in India and other countries since at earliest 500 AD,[3] the first modern mechanical cotton gin was created by American inventor Eli Whitney in 1793 and patented in 1794. However, the Indian worm-gear roller gin, invented some time around the sixteenth century,[4] has, according to Lakwete, remained virtually unchanged up to the present time. Whitney's gin used a combination of a wire screen and small wire hooks to pull the cotton through, while brushes continuously removed the loose cotton lint to prevent jams. It revolutionized the cotton industry in the United States, but also led to the growth of slavery in the American South as the demand for cotton workers rapidly increased. The invention has thus been identified as an inadvertent contributing factor to the outbreak of the American Civil War.[5] Modern automated cotton gins use multiple powered cleaning cylinders and saws, and offer far higher productivity than their hand-powered forebears.<span>[6]</span>
%yield = 91.8
<h3>Further explanation</h3>
Given
20 g NaCl
45 g AgCl
Required
%yield
Solution
Reaction
NaCl + AgNO₃ ⇒ AgCl + NaNO₃
mol NaCl :
= mass : MW
= 20 g : 58,44 g/mol
= 0.342
mol AgCl from equation :
= 1/1 x mol NaCl
= 1/1 x 0.342
= 0.342
Mass AgCl(theoretical) :
= mol x MW
= 0.342 x 143,32 g/mol
= 49.02 g
%yield = (actual/theoretical) x 100%
%yield = (45/49.02) x 100%
%yield = 91.8
Answer:
(a) Acid
(b) Base
(c) Acid
(d) Base
Explanation:
According to the Arrhenius acid-base theory:
- An acid is a substance that releases H⁺ in aqueous solution.
- A base is a substance that releases OH⁻ in aqueous solution.
(a) H₂SO₄ is an acid according to the following equation:
H₂SO₄(aq) ⇒ 2 H⁺(aq) + SO₄²⁻(aq)
(b) Sr(OH)₂ is a base according to the following equation:
Sr(OH)₂(aq) ⇄ Sr²⁺(aq) + 2 OH⁻(aq)
(c) HBr is an acid according to the following equation:
HBr(aq) ⇒ H⁺(aq) + Br⁻(aq)
(d) NaOH is a base according to the following equation:
NaOH(aq) ⇒ Na⁺(aq) + OH⁻(aq)
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 enthalpy of formation of 
is -396 kJ/mol
Explanation:
Calculating the enthalpy of formation of
The chemical equation for the combustion of propane follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![-198=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta -297)+(1\times (0))]\\\\\Delta H^o_f_{(SO_3(g))}=-396kJ/mol](https://tex.z-dn.net/?f=-198%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20-297%29%2B%281%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%3D-396kJ%2Fmol)
The enthalpy of formation of is -396 kJ/mol
