Entropy increases in reaction 1 due to increase in the number of particles while entropy decreases in reaction 2 due to decrease in the number of particles.
Entropy refers to the degree of disorderliness in a system. The higher the entropy, the more disorderly the system is and the lower the entropy, the less disorderly the system is. We must note that entropy increases with increase in the number of particles.
In the reaction, 2 KClO3 (s)⇌2 KCl (s)+3 O2 (g), entropy increases because one of the products is a gas and the number of particles increases from left to right. In the second reaction, CoCl2 (s)+6 H2O (g)⇌CoCl2⋅6H2O, entropy decreases because the number of particles decreases.
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Answer:
Ionic
Explanation:
Lithium is an alkali metal and form an ionic bond by donating an electron.
Fluorine is a halogen and forms ionic bonds by accepting an electron.
The electronegativity difference between lithium (0.98) and fluorine (3.98) is 3 Pauling units. Usually, when elements have a difference of 1.7 (or 2.0) Pauling units the bond is classified as ionic.
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In this problem, we need to use the ideal gas law. The following is the formula used in ideal gas law: PV = nRT, where n refers to the moles and R is the gas constant.
Given
P = 10130.0 kPa
V = 50 L
T = 300 degree celcius + 273.15 = 573.15 K
R = 8.314 L. kPa/K.mol
Solution
To get the moles which represent the "n" in the formula, we need to rearrange the equation.
PV = nRT PV
---- ------ ---> n = --------
RT RT RT
10130.0 kPa x 50 L
n= ---------------------------------------------
8.314 L. kPa/K.mol x 573.15 K
506,500
= ----------------------------
4,765.17 mol K
=106.29 mol Ar
So the moles of argon gas is 106.29 moles
Answer: a metal and a nonmental and a mental and a polyatomic anion :)
Explanation: i prayed ~.~
Answer:
Here's what I get
Explanation:
3. Molar concentration by formula.
(i) Comparison of molar concentrations
The formula gives a calculated value of 0.5302 mol·L⁻¹.
Dimensional analysis gives a calculated value of 0.1767 mol·L⁻¹.
The first value is three times the second.
It is wrong because the formula assumes that the acid supplies just enough moles of H⁺ to neutralize the OH⁻ from the NaOH.
Instead, I mol of H₃PO₄ provides 3 mol of H⁺, so your calculated concentration is three times the true value.
(ii) When is the formula acceptable?
The formula is acceptable only when the molar ratio of acid to base is 1:1.
Examples are
HCl + NaOH ⟶ NaCl + H₂O
H₂SO₄ + Ca(OH)₂ ⟶ CaSO₄ + 2H₂O
H₃PO₄ + Al(OH)₃ ⟶ AlPO₄ + 3H₂O
