Density = 0.186 g/cm^3
Step 1. Calculate the <em>volume</em> of the block
V = lwh = 9.8 cm × 5.6 cm × 4.2 cm = 230 cm^3
Step 2. Calculate the <em>density
</em>
Density = mass/volume = 42.8 g/230 cm^3 = 0.186 g/cm^3
(<em>There is no metal with this low a density</em>.)
Being landlandlord dies not matter in chemistry in anyway..
So the answer is it doesn't matter in any way
Answer:
The answer to your question is ΔHrxn = 0 kJ
Explanation:
Process
1.- Multiply Equation 1 by 2
2C(coal) + 2H₂O ⇒ 2CO (g) + 2H₂ ΔH rxn = 259.4 kJ
2.- Sum equation 2
CO(g) + H₂O ⇒ CO₂ (g) + H₂ (g) ΔHrxn = -41 kJ
Result
2C + 3H₂O + CO ⇒ 2CO + 3H₂ + CO₂ ΔHrxn = 218.4 kJ
Simplification
2C + 3H₂O ⇒ CO + 3H₂ + CO₂
3.- Sum equation 3
CO(g) + 3H₂ (g) ⇒ CH₄ (g) + H₂O (g) ΔHrxn = -218.4 kJ
Result
2C + 3H₂O + 3H₂ + CO ⇒ CO + 3H₂ + CO₂ + CH₄ + H₂O
Simplification
2C + 2H₂O ⇒ CO₂ + CH₄ ΔHrxn = 0 kJ
Answer:
Final temperature of solution is 27.48
Explanation:
Total volume of mixture = (60.0+60.0) mL = 120.0 mL
We know, density = (mass)/(volume)
So mass of mixture = 
Amount of heat released per mol of
= 
Where, m represents mass , C represents specific heat,
represents change in temperature and n is number of moles
As this reaction is an exothermic reaction therefore temperature of mixture will be higher than it's initial temperature.
Let's say final temperature of mixture is T 
So, 
Here
and 
Moles of H_{3}PO_{4} are added =
= 0.012 moles
So, ![(173.2\times 10^{3})J=\frac{[(135.6g)\times (3.78J.g^{-1}.^{0}\textrm{C}^{-1})\times (T-23.43)^{0}\textrm{C}]}{0.012}](https://tex.z-dn.net/?f=%28173.2%5Ctimes%2010%5E%7B3%7D%29J%3D%5Cfrac%7B%5B%28135.6g%29%5Ctimes%20%283.78J.g%5E%7B-1%7D.%5E%7B0%7D%5Ctextrm%7BC%7D%5E%7B-1%7D%29%5Ctimes%20%28T-23.43%29%5E%7B0%7D%5Ctextrm%7BC%7D%5D%7D%7B0.012%7D)
or, T = 27.48
So, final temperature of solution is 27.48