Answer:
80.8 g
Explanation:
First, let's write a balanced equation of this reaction
MgO + 2HNO₃ → Mg(NO₃)₂ + H₂O
Now let's convert grams to moles
We gotta find the weight of MgO
24 + 16 = 40 g/mol
12/40 = 0.3 moles of MgO
We can use this to find out how much Magnesium Nitrate will be formed
0.3 x 1 MgO / 1 Mg(NO₃)₂ = 0.3 moles of Magnesium Nitrate formed
Convert moles to grams
Find the weight of Mg(NO₃)₂ but don't forget that 2 subscript acts as a multiplier of whatever is inside that parenthesis.
24 + 14 x 2 + 16 x 3 x 2 = 148 g/mol
148 x 0.3 = 80.8 g
Fine particles, ground level ozone, sulfur dioxide, nitrogen dioxide, lead
<u>Given:</u>
Change in internal energy = ΔU = -5084.1 kJ
Change in enthalpy = ΔH = -5074.3 kJ
<u>To determine:</u>
The work done, W
<u>Explanation:</u>
Based on the first law of thermodynamics,
ΔH = ΔU + PΔV
the work done by a gas is given as:
W = -PΔV
Therefore:
ΔH = ΔU - W
W = ΔU-ΔH = -5084.1 -(-5074.3) = -9.8 kJ
Ans: Work done is -9.8 kJ
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
