Explanation:
SOLID
Sodium hydroxide exists in the solid phase at room temperature. You would find it in the lab as hemispherical white solid pellets. The phase of a substance depends on temperature and pressure. As you heat a solid, it will melt and change to the liquid phase.
<em>mC₃H₈: 44 g/mol</em>
<em>mCO₂: 44 g/mol</em>
---------------------
C₃H₈ + 5O₂ ----> 3CO₂ + 4H₂O
44g (44·3)g
44g C₃H₈ ------ 132g CO₂
15g C₃H₈ ------ X
X = (15×132)/44
<u>X = 45g CO₂
</u>
_____
:)
Recall; pH + pOH = 14
In this case [OH-] =0.100 m
therefore;
pOH = -LOG[OH-]
= - Log (0.100)
= 1.00
Therefore; the pOH is 1.00
And since, pH +pOH = 14
Then pH = 14-pOH
= 14 -1
= 13
Thus the pH is 13.00
Answer: both are solid at room temperture
Explanation:
Answer:
ΔHr = -103,4 kcal/mol
Explanation:
<u>Using:</u>
<u>AH° (kcal/mol)
</u>
<u>Metano (CH)
</u>
<u>-17,9
</u>
<u>Cloro (CI)
</u>
<u>tetraclorometano (CCI)
</u>
<u>- 33,3
</u>
<u>Acido cloridrico (HCI)
</u>
<u>-22</u>
It is possible to obtain the ΔH of a reaction from ΔH's of formation for each compound, thus:
ΔHr = (ΔH products - ΔH reactants)
For the reaction:
CH₄(g) + Cl₂(g) → CCl₄(g) + HCl(g)
The balanced reaction is:
CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g)
The ΔH's of formation for these compounds are:
ΔH CH₄(g): -17,9 kcal/mol
ΔH Cl₂(g): 0 kcal/mol
ΔH CCl₄(g): -33,3 kcal/mol
ΔH HCl(g): -22 kcal/mol
The ΔHr is:
-33,3 kcal/mol × 1 mol + -22 kcal/mol× 4 mol - (-17,9 kcal/mol × 1 mol + 0kcal/mol × 4mol)
<em>ΔHr = -103,4 kcal/mol</em>
<em></em>
I hope it helps!
