<span>Take the inversion of density: 1mL/13.6 g and multiply it by the conversion factor 453.6 g/ 1 lb and the given 5.00 lb. Units for mass (grams) and units for weight (lbs) cancel leaving only units of volume. I believe it should be 167 mL or 0.167 L</span>
Rf value is the ratio of the distance traveled by the solute to that of the solvent front on the paper used in chromatographic separation.
From the image it is clear the distance traveled by solvent front = 7.3 cm
Distance traveled by the component -1 of the mixture = 1.4 cm
Distance traveled by the component -2 of the mixture = 3.0 cm
Distance traveled by the component -3 of the mixture = 4.5 cm
Distance traveled by the component -4 of the mixture = 6.5 cm
Rf value of component-1 = 
Rf value of component-2 = 
Rf value of component-3 = 
Rf value of component-4 = 
b) Samples can be separated from a mixture using chromatography as the relative affinities for the compounds towards the paper (stationary phase) and the solvent(mobile phase) are different. Each component spends different amounts of time on the stationary phase depending on it chemical nature. So, the components in a mixture can be separated based on their polarities and relative degrees of adsorption on the stationary phase.
Answer:
0.21 g
Explanation:
The equation of the reaction is;
NaCl(aq) + AgNO3(aq) -----> NaNO3(aq) + AgCl(s)
Number of moles of NaCl= 0.0860 g /58.5 g/mol = 0.00147 moles
Number of moles of AgNO3 = 30/1000 L × 0.050 M = 0.0015 moles
Since the reaction is 1:1, NaCl is the limiting reactant.
1 mole of NaCl yields 1 mole of AgCl
0.00147 moles of NaCl yields 0.00147 moles of AgCl
Mass of precipitate formed = 0.00147 moles of AgCl × 143.32 g/mol
= 0.21 g
Answer:
Answers are in the explanation
Explanation:
Based on the reaction:
CF₄ + 2Br₂ → CBr₄ + 2F₂
The mole ratio of CF₄ is:
CF₄:Br₂ = 1:2
CF₄:CBr₄ = 1:1
CF₄:F₂ = 1:2
<em>Moles F2:</em>
Molar mass CF₄: 88.0g/mol
57.0g * (1mol / 88.0g) = 0.6477 moles CF₄ * (2mol F₂ / 1mol CBr₄) =
<h3>1.30 moles F₂</h3><h3 />
<em>Mass Br2:</em>
Molar mass CBr₄: 331.63g/mol
250.0g * (1mol / 331.63g) = 0.7539 moles CBr₄ * (2mol Br₂ / 1mol CF₄) =
1.51 moles Br₂ * (159.808g / mol) =
<h3>241g Br2</h3><h3 /><h3 />
<em>Moles F2:</em>
4.8 moles CF₄ * (2mol F₂ / 1mol CF₄) =
<h3>9.6 moles F₂</h3><h3 />
<em />
Answer: Read Explanation
Explanation: Index fossils are useful because they tell the relative ages of the rock layers in which they occur. Geologists use particular types of organisms, such as trilobites, as index fossils.