(4 mol H2O) x (112 kJ / 3 mol H2O) = 149 kJ
<span>(14.5 g HCl) / (36.4611 g HCl/mol) x (112 kJ / 3 mol HCl) = 14.9 kJ </span>
<span>(475 kJ) / (181 kJ / 2 mol HgO) x (216.5894 g HgO/mol) = 1137 g HgO </span>
<span>(179 kJ) / (181 kJ / 1 mol O2) x (31.99886 g O2/mol) = 31.6 g O2 </span>
<span>(145 kJ) / (112 kJ / 3 mol Cl2) x (70.9064 g Cl2/mol) = 275 g Cl2 </span>
<span>(14.5 g S2Cl2) / (135.0360 g S2Cl2/mol) x (112 kJ / 1 mol S2Cl2) = 12.0 kJ </span>
<span>CaCO3 + 2 NH3 → CaCN2 + 3 H2O; ∆H = –90.0 kJ </span>
<span>(798 kJ) / (90.0 kJ / 2 mol HN3) x (17.03056 g NH3/mol) = 302 g NH3 </span>
<span>(19.7 g H2O) / (18.01532 g H2O/mol) x (90.0 kJ / 3 mol H2O) = 32.8 kJ</span>
Answer:
The heat produced by this reaction is 2887J
Explanation:
The reaction of HCl with NaOH occurs as follows:
HCl + NaOH → H2O + NaCl + ΔH
<em>Where ΔH is released heat due the reaction</em>
<em />
The equation of a coffee cup calorimeter is:
Q = C*m*ΔT
<em>Where Q is heat produced, </em>
<em>C is specific heat of the solution (4.18J/g°C), </em>
<em>m is mass of the solution (50mL + 50mL = 100mL = 100g -Because density is 1g/mL) </em>
<em>and ΔT is change in temperature (28.9°C - 22.0°C = 6.9°C)</em>
<em />
Replacing:
Q = 4.184J/g°C*100g*6.9°C
Q = 2887J
The heat produced by this reaction is 2887J
<em />
Answer:
4.76
Explanation:
In this case, we have to start with the <u>buffer system</u>:
We have an acid () and a base (). Therefore we can write the <u>henderson-hasselbach reaction</u>:
If we want to calculate the pH, we have to <u>calculate the pKa</u>:
According to the problem, we have the <u>same concentration</u> for the acid and the base 0.1M. Therefore:
If we divide:
If we do the Log of 1:
So:
With this in mind, the pH is 4.76.
I hope it helps!
Answer:
yes
Explanation:
it is important for electrolysis
Answer:
FeCl₃ is present in excess.
Explanation:
Given data:
Number of moles of FeCl₃ = 6 mol
Number of moles of oxygen = 3 mol
Excess reactant = ?
Solution:
Chemical equation:
4FeCl₃ + 3O₂ → 2Fe₂O₃ + 6Cl₂
Now we will compare the moles of reactant with product.
FeCl₃ : Fe₂O₃
4 : 2
6 : 2/4×6= 3
FeCl₃ : Cl₂
4 : 6
6 : 6/4×6= 9
O₂ : Fe₂O₃
3 : 2
O₂ : Cl₂
3 : 6
Less number of moles of product are formed by the oxygen thus it will act as limiting reactant while FeCl₃ is present in excess.