Answer:
Explanation:
M(s) → M (g ) + 20.1 kJ --- ( 1 )
X₂ ( g ) → 2X (g ) + 327.3 kJ ---- ( 2 )
M( s) + 2 X₂(g) → M X₄ (g ) - 98.7 kJ ----- ( 3 )
( 3 ) - 2 x ( 2 ) - ( 1 )
M( s) + 2 X₂(g) - 2 X₂ ( g ) - M(s) → M X₄ (g ) - 98.7 kJ - 2 [ 2X (g ) + 327.3 kJ ] - M (g ) - 20.1 kJ
0 = M X₄ (g ) - 4 X (g ) - M (g ) - 773.4 kJ
4 X (g ) + M (g ) = M X₄ (g ) - 773.4kJ
heat of formation of M X₄ (g ) is - 773.4 kJ
Bond energy of one M - X bond = 773.4 / 4 = 193.4 kJ / mole
Answer:−329.4 calorie
Explanation:Heat released by metal = q=m_mc_m(T_2-T1)q=m
m
c
m
(T
2
−T1)
m_m=50\ g\\c_m=0.108\ cal\ g^{-1}\ \degree C^{-1}\\q=50\times0.108(17-78)=-329.4\ caloriem
m
=50 g
c
m
=0.108 cal g
−1
°C
−1
q=50×0.108(17−78)=−329.4 calorie
Here minus sign indicate that heat is released
Answer:
3.918 mol Al
Explanation:
To convert between moles and grams, you have to use the molar mass of the substance. The molar mass of aluminum is 26.98 g/mol. You use this as the unit converter.
Round the number to the lowest number of significant figures; 3.918 mol Al
Answer:
Explanation:
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