Answer:
Refer to your periodic table. Lewis dot structures are based off the number of valence electrons an atom has.
Looking at the compounds, we can see that Gallium has three valence electrons in its outer shell and oxygen has six. Oxygen and Gallium are going to share electrons with one another, making a V shape in their diagram.
One Oxygen would make a double bond with a Gallium, leaving one valence electron to another oxygen. That oxygen takes that Final electron. It now has 7 in its outer shell. The remaining Gallium and Oxygen do the same double bond as the one before, leaving the 7 valence electron oxygen with one more electron.
First, we have to get:
1- The heat required to increase T of ice from -50 to 0 °C:
according to q formula:
q1 = m*C*ΔT
when m is the mass of ice = mol * molar mass
= 1 mol * 18 mol/g
= 18 g
and C is the specific heat capacity of ice = 2.09 J/g-K
and ΔT change in temperature = 0- (-50) = 50°C
by substitution:
∴q1 = 18 g * 2.09 J/g-K *50°C
= 1881 J = 1.881 KJ
2- the heat required to melt this mass of ice is :
q2 = n*ΔHfus
when n is the number of moles of ice = 1 mol
and ΔHfus = 6.01 KJ/mol
by substitution:
q2 = 1 mol * 6.01 KJ/mol
= 6.01 KJ
3- the heat required to increase the water temperature from 0°C to 60 °C is:
q3 = m*C*ΔT
when m is the mass of water = 18 g
C is the specific heat capacity of water = 4.18 J/g-K
ΔT is the change of Temperature of water = 60°C - 0°C = 60°C
by substitution:
∴q3 = 18 g * 4.18 J/g-K * 60°C
= 4514 J = 4.514 KJ
∴the total change of enthalpy = q1+q2+q3
= 1.881 KJ +6.01 KJ + 4.514 KJ
= 12.405 KJ
The solid form of a substance is usually more dense than its
liquid and gaseous forms. Similarly the liquid form is usually more dense than
the gaseous form. Ice floating in water is an exception that breaks the general
density rule. So option “A” is the correct option in regards to the given
question. In case of ice formation, actually the density of water decreases by
about 9%. This is the main reason behind ice floating in water. Pure water has
the maximum density at 4 degree centigrade.
From 5 L to moles, just divide 5 by 22.4. I got 0.22 moles of H2.
From 5 moles to liters, just multiply 5 by 22.4. I got 112 L of H2.
