Answer:
Mg
Explanation:
The standard reduction potentials are
<u>E°/V
</u>
Au³⁺(aq ) + 3e⁻ ⟶ Au(s); 1.42
Hg²⁺(aq) + 2e⁻ ⟶ Hg(l); 0.85
Ag⁺(aq) + e⁻ ⟶ Ag(s); 0.80
Cu²⁺(aq) + 2e⁻ ⟶ Cu(s); 0.34
Mg2+(aq) + 2e- ⟶ Mg(s); -2.38
The more negative the standard reduction potential, the stronger the metal is as a reducing agent.
Mg is the only metal with a standard reduction potential lower than that of Cu, so
Only Mg will react spontaneously with Cu²⁺.
Kepler's
third law shows the relationship between the orbital period of an object and
the distance between the object and the object it orbits.
The
simplified version of this law is: P^2 = a^3
Where,
P =
period of the orbit in years = 0.62 years
a =
average distance from the object to the object it orbits in AU. The
astronomical unit AU is a unit of length which is roughly equivalent to the
distance from Earth to the Sun.
Therefore
calculating for a:
0.62
^ 2 = a ^ 3
a =
0.62 ^ (2/3)
a =
0.727 AU = 0.72 AU
Therefore we can interpret this as: The distance from Venus to the Sun is about 72% of the distance from Earth to
Sun.
<span>Answer:
B. 0.72 AU</span>
Answer:
kJ/mol
Explanation: <u>Enthalpy</u> <u>Change</u> is the amount of energy in a reaction - absorption or release - at a constant pressure. So, <u>Standard</u> <u>Enthalpy</u> <u>of</u> <u>Formation</u> is how much energy is necessary to form a substance.
The standard enthalpy of formation of HCl is calculated as:

→ 
Standard Enthalpy of formation for the other compounds are:
Calcium Hydroxide:
-1002.82 kJ/mol
Calcium chloride:
-795.8 kJ/mol
Water:
-285.83 kJ/mol
Enthalpy is given per mol, which means we have to multiply by the mols in the balanced equation.
Calculating:
![-17.2=[-795.8+2(285.85)]-[-1002.82+2\Delta H]](https://tex.z-dn.net/?f=-17.2%3D%5B-795.8%2B2%28285.85%29%5D-%5B-1002.82%2B2%5CDelta%20H%5D)



So, the standard enthalpy of formation of HCl is -173.72 kJ/mol
The volume of oxygen at STP required would be 252.0 mL.
<h3>Stoichiometic problem</h3>
The equation for the complete combustion of C2H2 is as below:

The mole ratio of C2H2 to O2 is 2:5.
1 mole of a gas at STP is 22.4 L.
At STP, 100.50 mL of C2H2 will be:
100.50 x 1/22400 = 0.0045 mole
Equivalent mole of O2 according to the balanced equation = 5/2 x 0.0045 = 0.01125 moles
0.01125 moles of O2 at STP = 0.01125 x 22400 = 252.0 mL
Thus, 252.0 mL of O2 gas will be required at STP.
More on stoichiometric problems can be found here: brainly.com/question/14465605
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