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The element bromine is not a reddish-brown liquid. Liquid is the substance bromine.
M=DV
M=3.103 g/mL * 19.8 mL = 61.44 g
Answer:
for the given reaction is -238.7 kJ
Explanation:
The given reaction can be written as summation of three elementary steps such as:
![\Delta H_{1}= -393.5 kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7B1%7D%3D%20-393.5%20kJ)
![\Delta H_{2}= (2\times -285.8)kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7B2%7D%3D%20%282%5Ctimes%20-285.8%29kJ)
![\Delta H_{3}= 726.4 kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7B3%7D%3D%20726.4%20kJ)
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![C(graphite)+2H_{2}(g)+\frac{1}{2}O_{2}(g)\rightarrow CH_{3}OH(l)](https://tex.z-dn.net/?f=C%28graphite%29%2B2H_%7B2%7D%28g%29%2B%5Cfrac%7B1%7D%7B2%7DO_%7B2%7D%28g%29%5Crightarrow%20CH_%7B3%7DOH%28l%29)
![\Delta H=\Delta H_{1}+\Delta H_{2}+\Delta H_{3}=-238.7 kJ](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5CDelta%20H_%7B1%7D%2B%5CDelta%20H_%7B2%7D%2B%5CDelta%20H_%7B3%7D%3D-238.7%20kJ)
Answer:
17.3124 grams
Explanation:
Given;
Amount of heat to be produced = 175 kJ
Molar mass of the carbon monoxide = 12 + 16 = 28 grams
Now,
The standard molar enthalpy of carbon monoxide = 283 kJ/mol
Thus,
To produce 175 kJ heat, number of moles of CO required will be
= Amount heat to be produced / standard molar enthalpy of CO
or
= 175 / 283
= 0.6183
Also,
number of moles = Mass / Molar mass
therefore,
0.6183 = Mass / 28
or
Mass of the CO required = 0.6183 × 28 = 17.3124 grams
Answer is: empirical formula is Fe₂O₃.
m(Fe) = 7,50 g.
m(iron oxide) = 10,71 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 7,50 g ÷ 55,85 g/mol = 0,134 mol.
m(O) = m(iron oxide) - m(Fe).
m(O) = 10,71 g - 7,50 g = 3,21 g.
n(O) = 3,21 g ÷ 16 g/mol = 0,20 mol.
n(Fe) : n(O) = 0,134 mol : 0,2 mol = 2 : 3.