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victus00 [196]
3 years ago
8

The combustion of ethyne, shown below unbalance, produces heat which can be used to weld metals:

Chemistry
1 answer:
Andreyy893 years ago
3 0

Answer:

3.69 g

Explanation:

Given that:

The mass m = 325 g

The change in temperature ΔT = ( 1540 - 165)° C

= 1375 ° C

Heat capacity c_p = 0.490 J/g°C

The amount of heat required:

q = mcΔT

q =  325 × 0.490 × 1375

q = 218968.75 J

q = 218.97 kJ

The equation for the reaction is expressed as:

C_2H_{2(g)} + 5O_{2(g)} \to 2CO_{2(g)} + H_2O_{(g)}   \ \   \ \ \  \Delta H^o_{reaction} = -1544 \ kJ

Then,

1 mole of the ethyne is equal to 26 g of ethyne required for 1544 kJ heat.

Thus, for 218.97 kJ, the amount of ethyne gas required will be:

= \dfrac{26 \ g}{1544 \ kJ} \times 218.97 \ kJ

= 3.69 g

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Answer: 0.150 m Na_2SO_4(aq) will have highest boiling point.

Explanation:

Formula used for Elevation in boiling point :

\Delta T_b=i\times k_b\times m

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A) 0.100 m NiBr_2

i = 3 as NiBr_2\rightarrrow Ni^{2+}+2Br^-

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B) 0.250 m CH_3OH

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The solution having the highest concentration of ions will have the highest boiling point and thus 0.150 m Na_2SO_4(aq) will have highest boiling point.

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