Answer:
Lose two electrons.
Explanation:
Barium is present in group 2.
It is alkaline earth metal.
Its atomic number is 56.
Its electronic configuration is Ba₅₆ = [Xe] 6s².
In order to attain the noble gas electronic configuration it must loses its two valance electrons.
When barium loses it two electron its electronic configuration will equal to the Xenon.
The atomic number of xenon is 54 so barium must loses two electrons to becomes equal to the xenon.
Answer:
3.91 moles of Neon
Explanation:
According to Avogadro's Law, same volume of any gas at standard temperature (273.15 K or O °C) and pressure (1 atm) will occupy same volume. And one mole of any Ideal gas occupies 22.4 dm³ (1 dm³ = 1 L).
Data Given:
n = moles = <u>???</u>
V = Volume = 87.6 L
Solution:
As 22.4 L volume is occupied by one mole of gas then the 16.8 L of this gas will contain....
= ( 1 mole × 87.6 L) ÷ 22.4 L
= 3.91 moles
<h3>2nd Method:</h3>
Assuming that the gas is acting ideally, hence, applying ideal gas equation.
P V = n R T ∴ R = 0.08205 L⋅atm⋅K⁻¹⋅mol⁻¹
Solving for n,
n = P V / R T
Putting values,
n = (1 atm × 87.6 L)/(0.08205 L⋅atm⋅K⁻¹⋅mol⁻¹ × 273.15K)
n = 3.91 moles
Result:
87.6 L of Neon gas will contain 3.91 moles at standard temperature and pressure.
Answer:
The ΔH is 5.5 kJ/mol and the reaction is endothermic.
Explanation:
To calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction, you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient ( number of molecules of each compound participating in the reaction) and finally subtract them:
Combustion enthalpy = ΔH = ∑H products - ∑Hreactants
In this case:
ΔH = 15.7 kJ/mol - 10.2 kJ/mol= 5.5 kJ/mol
An endothermic reaction is one whose enthalpy value is positive, that is, the system absorbs heat from the environment (ΔH> 0).
<u><em>The ΔH is 5.5 kJ/mol and the reaction is endothermic.</em></u>
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.
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