1 molecule of glucose contains 6 atoms of C, 12 atoms of H , and 6 atoms of 0.1 mole of glucose contains 6 moles of C atoms , 12 moles of H atoms , and 6 moles of O atoms .
i. The dissolution of PbSO₄ in water entails its ionizing into its constituent ions:
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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)
.
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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.
5.367 ml of the concentrated acid must be added to obtain a total volume of 100 ml of the dilute solution.
Dilution is defined as the process in which the concentration of a sample is decreased by adding more solvent. The dilution formula is given below.
C₁V₁ = C₂V₂
where C₁ = initial concentration of sample = 3.00 m
V₁ = initial volume of sample
C₂ = final concentration after dilution = 0.161 m
V₂ = total final volume after dilution = 100 ml
Plug in the values to the formula and solve for the volume of the concentrated acid that must be added.
C₁V₁ = C₂V₂
3.00 m (V₁) = 0.161 m (100 ml)
V₁ = 5.367 ml
Learn more about dilution here: brainly.com/question/1615979
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Answer:0.477 g/ml
Explanation:
Density=(40.14-33.79)/13.3 ml
Density=6.35/13.3
Density=0.477 g/ml
Answer: Significant figures in a measurement are all measured digits, and one estimated digit
Significant figures communicate the level of precision in measurements Significant figures are an indicator of the certainty in measurements.
Explanation:
Significant figures : The figures in a number which express the value or the magnitude of a quantity to a specific degree of accuracy or precision is known as significant digits.
The significant figures of a measured quantity are defined as all the digits known with certainty and the first uncertain or estimated digit.
Rules for significant figures:
1. Digits from 1 to 9 are always significant and have infinite number of significant figures.
2. All non-zero numbers are always significant.
3. All zero’s between integers are always significant.
4. All zero’s preceding the first integers are never significant.
5. All zero’s after the decimal point are always significant.
