According to the law of conservation of mass, what is the same on both sides of a balanced chemical equation?
A. the volume of the substances
B. the subscripts
C. the total mass of atoms
D. the coefficients
Answer:
A balanced equation demonstrates the conservation of mass by having the same number of each type of atom on both sides of the arrow.
Explanation:
Every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. ... Use coefficients of products and reactants to balance the number of atoms of an element on both sides of a chemical equation.
Consider the balanced equation for the combustion of methane.
CH
4
+
2O
2
→
CO
2
+
2H
2
O
All balanced chemical equations must have the same number of each type of atom on both sides of the arrow.
In this equation, we have 1
C
atom, 4
H
atoms, and 4
O
atoms on each side of the arrow.
The number of atoms does not change, so the total mass of all the atoms is the same before and after the reaction. Mass is conserved.
Here is a video that discusses the importance of balancing a chemical equation.
Since the sign is positive, the entropy increased by 88.48 J/K.
Examine the phases of the species present to determine whether a physical or chemical process will cause an increase or decrease in entropy. Keep in mind "Silly Little Goats" to aid you in telling.
[1 Sf K+1 + 1 Sf Br-1 (aq)] ([1Sf(KBr (s))])
[1(102.5) + 1(82.42)] - [1(96.44)] = 88.48 J/K
If the entropy has grown, we say that Delta S is positive, and if it has dropped, we say that Delta S is negative. Due to its ionic nature, KBr is soluble in water and causes the 'K(+)' ions to hydrate.
Learn more about Entropy here-
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i. The dissolution of PbSO₄ in water entails its ionizing into its constituent ions:
---
ii. Given the dissolution of some substance
,
the Ksp, or the solubility product constant, of the preceding equation takes the general form
![K_{sp} = [B]^y [C]^z](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BB%5D%5Ey%20%5BC%5D%5Ez)
.
The concentrations of pure solids (like substance A) and liquids are excluded from the equilibrium expression.
So, given our dissociation equation in question i., our Ksp expression would be written as:
![K_{sp} = \mathrm{[Pb^{2+}] [SO_4^{2-}]}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5Cmathrm%7B%5BPb%5E%7B2%2B%7D%5D%20%5BSO_4%5E%7B2-%7D%5D%7D)
.
---
iii. Presumably, what we're being asked for here is the <em>molar </em>solubility of PbSO4 (at the standard 25 °C, as Ksp is temperature dependent). We have all the information needed to calculate the molar solubility. Since the Ksp tells us the ratio of equilibrium concentrations of PbSO4 in solution, we can consider either [Pb2+] or [SO4^2-] as equivalent to our molar solubility (since the concentration of either ion is the extent to which solid PbSO4 will dissociate or dissolve in water).
We know that Ksp = [Pb2+][SO4^2-], and we are given the value of the Ksp of for PbSO4 as 1.3 × 10⁻⁸. Since the molar ratio between the two ions are the same, we can use an equivalent variable to represent both:
So, the molar solubility of PbSO4 is 1.1 × 10⁻⁴ mol/L. The answer is given to two significant figures since the Ksp is given to two significant figures.
