<u>Answer:</u>
<em>1) ∆H is positive 
Endothermic
</em>
<em>2) 
Endothermic </em>
<em>3) Energy is absorbed Endothermic
</em>
<em>4) 
Exothermic
</em>
<em>5) ∆H is negtive Exothermic
</em>
<em></em>
<u>Explanation:</u>
∆H is called as enthalpy change
It is also called as Heat of reaction
Energy is required for the bond to break a bond.
Energy is released when a bond is formed.
that is
We see in this equation, bonds between hydrogen and chlorine molecules gets broken and on the right side bond is formed in HCl.
If energy of products greater than energy of reactants then the reaction enthalpy change is endothermic .
If energy of products lesser than energy of reactants then the reaction enthalpy change is exothermic .
For example
(positive hence endothermic)
(negative hence exothermic)
Answer:
The initial temperature is 499 K
Explanation:
Step 1: Data given
initial volume = 12 cm3 = 12 mL
Final volume = 7 cm3 = 7mL
The final temperature = 18 °C = 291 K
Step 2: Calculate the initial temperature
V1/T1 = V2/T2
⇒with V1 = the initial volume = 0.012 L
⇒with T1 = the initial volume = ?
⇒with V2 = the final volume 0.007 L
⇒with T2 = The final temperature = 291 K
0.012 / T1 = 0.007 / 291
0.012/T1 = 2.4055*10^-5
T1 = 0.012/2.4055*10^-5
T1 = 499 K
The initial temperature is 499 K
Answer:
Empirical formula: BH3
Molecular Formula: B2H6
Explanation:
To solve the exercise, we need to know how many boron atoms and how many hydrogen atoms the compound has. We know that of the total weight of the compound, 78.14% correspond to boron and 21.86% to hydrogen. As the weight of the compound is between 27 g and 28 g, using the above percentages we can solve that the compound has between 21.1 g and 21.8 g of boron, and between 5.9 g and 6.1 g of hydrogen:
100% _____ 27 g
78.14% _____ x = 78.14% * 27g / 100% = 21.1 g boron
100% ______27 g
21.86% ______ x = 21.86% * 27g / 100% = 5.9 g hydrogen
100% _____ 28 g
78.14% _____ x = 78.14% * 28g / 100% = 21.8 g boron
100% _____ 28g
21.86% _____ x = 21.86% * 28g / 100% = 6.1 g hydrogen
So, if the atomic weight of boron is 10.8 g, there must be two boron atoms in the compound that sum 21.6 g. The weight of hydrogen is 1 g, so the compound must have six hydrogen atoms.
The molecular formula represents the real amount of atoms that form a compound. Therefore, the molecular formula of the compound is B2H6.
The empirical formula is the minimum expression that represents the proportion of atoms in a compound. For example, ethane has 2 carbon atoms and 6 hydrogen atoms, so its molecular formula is C2H6, however, its empirical formula is CH3. Therefore, the empirical formula of the boron compound is BH3.
Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄
Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.
For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:
Kp = 
where:
P(N₂O₄) and P(NO₂) are the partial pressure of each gas.
Calculating constant:
Kp = 
Kp = 0.0104
After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.
P(N₂O₄) + P(NO₂) = 200
P(N₂O₄) = 200 - P(NO₂)
Kp =
0.0104 = ![\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B200%20-%20P%28NO_%7B2%7D%29%20%20%7D%7B%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D%7D)
0.0104![[P(NO_{2} )]^{2}](https://tex.z-dn.net/?f=%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D)
+ 
- 200 = 0
Resolving the second degree equation:
= 
= 98.7
Find partial pressure of N₂O₄:
P(N₂O₄) = 200 - P(NO₂)
P(N₂O₄) = 200 - 98.7
P(N₂O₄) = 101.3
The partial pressures are = 98.7 MPa and P(N₂O₄) = 101.3 MPa
Explanation:
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