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Mrrafil [7]
3 years ago
11

How does atomic hydrogen torch function for cutting and welding purposes​

Chemistry
1 answer:
lbvjy [14]3 years ago
5 0

Explanation:

atomic hydrogen torch utilizes an electric arc whereby two closely - but not touching electrodes - result in the release of powerful electric spark as the current tries to flow through the gap. The gap is filled with hydrogen gas in an atomic hydrogen torch rather than air. The electric arch is split the hydrogen gas molecules into hydrogen atoms (some in plasma form). When the hydrogen atoms land on cooler objects like the metal being welded or cut, they region back to H₂ molecules releasing enormous amounts of heat on the surface. Surface temperatures can reach   4000 °C. The use of hydrogen gas protects the metal being welded from oxidation. Oxidation may compromise the quality of the weld.

Learn More:

For more on other welding torches check out;

brainly.com/question/13335545

brainly.com/question/13056701

#LearnWithBrainly

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Some SO2 and O2 are mixed together in a flask at 1100 K in such a way ,that at the instant of mixing, their partial pressures ar
kakasveta [241]

Answer:

The answer is "\bold{0.525\ \ atm^{-1}}"

Explanation:

Given equation:

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\Delta x = 2-(2+1)\\\\

     = 2-(2+1)\\\\= 2-(3)\\\\= -1

\left\begin{array}{cccc}I\ (atm)&1&0.5&0\\C\ (atm)&2x&-x&2x\\E\ (atm)     &1-2x&0.5-x&2x\end{array}\right

calculating the total pressure on equilibrium=  (1-2x)+(0.5-x)+2x \ atm\\\\

                                                                         = 1-2x+0.5-x+2x\\\\= 1+0.5-x\\\\=1.5-x\\

\therefore \\\\\to 1.50-x= 1.35 \\\\\to 1.50-1.35= x\\\\\to 0.15= x\\\\ \to  x= 0.15\\\\

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= (1-2 \times 0.15)

= 1-0.30 \\\\ =0.70 \ atm

calculating the pressure in  O_2:

= (0.5- 0.15)\\\\= 0.35 \ atm \\

calculating the pressure in  So_3:

= (2 \times 0.15)\\\\= (.30) \ atm \\\\

Calculating the Kp at 1100 K:

= \frac{(Pressure(So_3))^2}{(Pressure(So_2))^2 \times Pressure(O_2)}\\\\= \frac{0.30^2}{0.70^2 \times 0.35}\\\\= \frac{0.30 \times 0.30 }{0.70\times 0.70 \times 0.35}\\\\= \frac{0.09 }{0.49\times 0.35} \\\\= \frac{0.09 }{0.1715} \\\\=  0.5247 \ \  or \ \  0.525 \ \ atm^{-1}  \\\\

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Answer:

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