Answer : The energy removed must be, -67.7 kJ
Solution :
The process involved in this problem are :
The expression used will be:
![\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bp%2Cg%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bm%5Ctimes%20%5CDelta%20H_%7Bvap%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= heat released by the reaction = ?
m = mass of benzene = 125 g
= specific heat of gaseous benzene = 
= specific heat of liquid benzene = 
= enthalpy change for vaporization = 
Molar mass of benzene = 78.11 g/mole
Now put all the given values in the above expression, we get:
![\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B125g%5Ctimes%201.06J%2Fg.K%5Ctimes%20%28353.0-%28425.0%29%29K%5D%2B125g%5Ctimes%20-434.0J%2Fg%2B%5B125g%5Ctimes%201.73J%2Fg.K%5Ctimes%20%28335.0-353.0%29K%5D)
Therefore, the energy removed must be, -67.7 kJ
Answer:
Hydrogen chloride (HCl), a compound of the elements hydrogen and chlorine, a gas at room temperature and pressure. A solution of the gas in water is called hydrochloric acid.
Explanation:
hoped that helped
<span>The </span>elements are arranged<span> in order of increasing atomic number. Vertical columns(called groups) contain </span>elements with similar properties. Horizontal rows called periods elements with<span> the same number of atomic orbitals(That's why Hydrogen and Helium are separated from the rest of the table).
Hope this helps:)</span>
Answer:
Explanation:
This is a limiting reactant problem.
Mg(s)
+
2HCl(aq)
→
MgCl
2
(
aq
)
+ H
2
(
g
)
Determine Moles of Magnesium
Divide the given mass of magnesium by its molar mass (atomic weight on periodic table in g/mol).
4.86
g Mg
×
1
mol Mg
24.3050
g Mg
=
0.200 mol Mg
Determine Moles of 2M Hydrochloric Acid
Convert
100 cm
3
to
100 mL
and then to
0.1 L
.
1 dm
3
=
1 L
Convert
2.00 mol/dm
3
to
2.00 mol/L
Multiply
0.1
L
times
2.00 mol/L
.
100
cm
3
×
1
mL
1
cm
3
×
1
L
1000
mL
=
0.1 L HCl
2.00 mol/dm
3
=
2.00 mol/L
0.1
L
×
2.00
mol
1
L
=
0.200 mol HCl
Multiply the moles of each reactant times the appropriate mole ratio from the balanced equation. Then multiply times the molar mass of hydrogen gas,
2.01588 g/mol
0.200
mol Mg
×
1
mol H
2
1
mol Mg
×
2.01588
g H
2
1
mol H
2
=
0.403 g H
2
0.200
mol HCl
×
1
mol H
2
2
mol HCl
×
2.01588
g H
2
1
mol H
2
=
0.202 g H
2
The limiting reactant is
HCl
, which will produce
0.202 g H
2
under the stated conditions.
pls mark as brainliest ans
A.
Elements in the same group have similar properties.
B.
The similarity in their properties arises from the fact that they have an equal number of valence shell electrons.
C.
Fluorine, Chlorine, Bromine
