Physical changes occur when the properties of a substance are retained and/or the materials can be recovered after the change. Chemical changes involve the formation of a new substance. Formation of a gas, solid, light, or heat are possible evidence of chemical change.
Answer:
The equation for the reaction of one sodium bicarbonate ( NaHCO3 ) molecule with one citric acid (C6H8O7) molecule is the following:
Sodium Bicarbonate + Citric Acid ⇒ Water + Carbon Dioxide + Sodium Citrate
NaHCO3 + C6H8O7 ⇒ 3 CO2 + 3 H2O + Na3C6H5O7
Explanation:
The reaction is in balance, that is, the whole H2CO3 is not finished, but a little bit of this acid is left in the solution. Therefore, when sodium bicarbonate is added to the solution with citric acid, sodium citrate salt (C6H5O7Na3) and carbonic acid (H2CO3) are formed, which is rapidly broken down into water (H2O) and carbonic oxide (CO2).
C6H8O7 + NaHCO3 ⇒ C6H5O7Na3 + 3 H2CO3
C6H5O7Na3 + 3 H2CO3 ⇔ C6H5O7Na3 + 3 H2O + 3 CO2
Answer is: the boiling point of the resulting solution of sucrose is 100.42°C.
m(H₂<span>O) = 15.2 g ÷ 1000 g/kg = 0.0152 kg.
</span>m(C₁₂H₂₂O₁₁<span>) = 4.27 g.
n</span>(C₁₂H₂₂O₁₁) = m(C₁₂H₂₂O₁₁) ÷ M(C₁₂H₂₂O₁₁).
n(C₁₂H₂₂O₁₁) = 4.27 g ÷ 342.3 g/mol.
n(C₁₂H₂₂O₁₁) = 0.0125 mol.
b(solution) = n(C₁₂H₂₂O₁₁) ÷ m(H₂O).
b(solution) = 0.0125 mol ÷ 0.0152 kg.
b(solution) = 0.82 m.
ΔT = b(solution) · Kb(H₂O).
ΔT = 0.82 m · 0.512°C/m.
ΔT = 0.42°C.
Tb = 100°C + 0.42°C = 100.42°C.
Answer:
Explanation:
1. Concentration of SO₄²⁻
SrSO₄(s) ⇌ Sr²⁺(aq) +SO₄²⁻(aq); Ksp = 3.44 × 10⁻⁷
0.0150 x
![K_{sp} =\text{[Sr$^{2+}$][SO$_{4}^{2-}$]} = 0.0150x = 3.44 \times 10^{-7}\\x = \dfrac{3.44 \times 10^{-7}}{0.0150} = \mathbf{2.293 \times 10^{-5}} \textbf{ mol/L}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BSr%24%5E%7B2%2B%7D%24%5D%5BSO%24_%7B4%7D%5E%7B2-%7D%24%5D%7D%20%3D%200.0150x%20%3D%203.44%20%5Ctimes%2010%5E%7B-7%7D%5C%5Cx%20%3D%20%5Cdfrac%7B3.44%20%5Ctimes%2010%5E%7B-7%7D%7D%7B0.0150%7D%20%3D%20%5Cmathbf%7B2.293%20%5Ctimes%2010%5E%7B-5%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D)
2. Concentration of Pb²⁺
PbSO₄(s) ⇌ Pb²⁺(aq) + SO₄²⁻(aq); Ksp = 2.53 × 10⁻⁸
x 2.293 × 10⁻⁵
![K_{sp} =\text{[Pb$^{2+}$][SO$_{4}^{2-}$]} = x \times 2.293 \times 10^{-5} = 2.53 \times 10^{-8}\\\\x = \dfrac{2.53 \times 10^{-8}}{2.293 \times 10^{-5}} = \mathbf{1.10 \times 10^{-3}} \textbf{ mol/L}\\\\\text{The concentration of Pb$^{2+}$ is $\large \boxed{\mathbf{1.10 \times 10^{-3}}\textbf{ mol/L}}$}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%5Ctext%7B%5BPb%24%5E%7B2%2B%7D%24%5D%5BSO%24_%7B4%7D%5E%7B2-%7D%24%5D%7D%20%3D%20x%20%5Ctimes%202.293%20%5Ctimes%2010%5E%7B-5%7D%20%3D%202.53%20%5Ctimes%2010%5E%7B-8%7D%5C%5C%5C%5Cx%20%3D%20%5Cdfrac%7B2.53%20%5Ctimes%2010%5E%7B-8%7D%7D%7B2.293%20%5Ctimes%2010%5E%7B-5%7D%7D%20%3D%20%5Cmathbf%7B1.10%20%5Ctimes%2010%5E%7B-3%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D%5C%5C%5C%5C%5Ctext%7BThe%20concentration%20of%20Pb%24%5E%7B2%2B%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B1.10%20%5Ctimes%2010%5E%7B-3%7D%7D%5Ctextbf%7B%20mol%2FL%7D%7D%24%7D)
