Answer:
+125.4 KJmol-1
Explanation:
∆H C4H10(g) = -2877.6kJ/mol
∆H C(s)=-393.5kJ/mol
∆H H2(g) = -285.8
∆H reaction= ∆Hproducts - ∆H reactants
∆H reaction= (-2877.6kJ/mol) - [4(-393.5kJ/mol) +5(-285.8)]
∆H reaction= +125.4 KJmol-1
Na₂CrO₄ + PbCl₂ → PbCrO₄ + 2 NaCl
<u>Explanation:</u>
In a double displacement reaction, the reactants which are involved in the reaction exchanging their ions thereby produces 2 new compounds. Here sodium chromate and lead chloride are undergoing double displacement reaction, the ions exchanges their position there by forming sodium chloride and lead chromate. So the double displacement reaction is given as,
Na₂CrO₄ + PbCl₂ → PbCrO₄ + 2 NaCl
Ooooh boy alright. So, this may or may not be a limited reactant problem so we need to first find out of it is.
First, how many moles of each substance are there
the molar mass of BCl3 is <span>117.17 grams so 37.5 g / 117.17 is ~ .32 mol.
The molar mass of H2O is 18.02 so 60 / 18.02 is ~ 3.33 mol.
Now, for every 1 mole of BCl3, there are 3 moles of HCl created. Therefore, BCl3 can create ~ .96 moles.
For every 3 moles of H2O, there are 3 moles of HCl created. Therefore, HCl can create ~3.33 moles.
But, there is not enough BCl3 to support that 3.33 moles, only enough for .96 moles, therefore BCl3 is the limiting reactant. Now, to answer the question, simply multiply .96 moles by the molar mass of HCl.
.96 x 36.46 = ~35 g</span>
Look to be honest, I don't know how to work out the problem, but my teacher, and my says it takes 8 minutes for the Sun's light to reach
hope my answer works :)
