Answer:
the theoretical mass of cellulose in this mixture is 0.25 grams
Explanation:
Given the data in the question;
mass of the mixture = 5.0 gram
mass percentage of cellulose = 5%
mass percentage of caffeine = 47.5%
mass percentage of benzoic acid = 47.5%
so mass of caffeine in the mixture will be;
⇒ ( mass percentage of cellulose ) × ( mass of the mixture )
= 5% × 5.0 gram
= ( 5 / 100 ) × 5.0 grams
= 0.05 × 5.0 grams
= 0.25 grams
Therefore, the theoretical mass of cellulose in this mixture is 0.25 grams
Which of the following is true about oxidation-reduction reactions?
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One atom is oxidized and one is reduced
Both atoms are oxidized and reduced
The total number of electrons changes
One atom can be oxidized without one being reduced
Answer:
the last one probably
Explanation:
A certain first-order reaction is 27.5 percent complete in 8.90 min at 25°C. What is its rate constant?
Answer:
[tex]k= 0.145min^{-1}[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out necessary for us remember that the first-order kinetics is given by:
[tex]ln(A/A_0)=-kt[/tex]
Whereas the 27.5% complete means A/Ao=0.275, and thus, we solve for the rate constant as follows:
[tex]k=\frac{ln(A/A_0)}{-t}[/tex]
Then, we plug in the variables to obtain:
[tex]k=\frac{ln(0.275)}{-8.90min}\\\\k= 0.145min^{-1}[/tex]
Regards!
red litmus paper was used to test toothpaste and it turns blue.Explain what this tells about the toothpaste
5. A sample of benzene (C6H6), weighing 7.05 g underwent combustion in a bomb calorimeter by the following reaction:
2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (l)
The heat given off was absorbed by 500 g of water and caused the temperature of the water and the calorimeter to rise from 25.00 to 53.13 oC. The heat capacity of water = 4.18 J/g/oC and the heat capacity of the calorimeter = 10.5 kJ/oC. (1) what is the ΔH of the reaction? Using the definitions at the beginning of the module describe (2) the calorimeter + contents, (3) the type of process.
Answer:
A sample of benzene (C6H6), weighing 7.05 g underwent combustion in a bomb calorimeter by the following reaction:
2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (l)
The heat given off was absorbed by 500 g of water and caused the temperature of the water and the calorimeter to rise from 25.00 to 53.13 oC. The heat capacity of water = 4.18 J/g/oC and the heat capacity of the calorimeter = 10.5 kJ/oC. (1) what is the ΔH of the reaction?
Explanation:
The heat energy released by the reaction = heat absorbed by calorimeter + heat absorbed by water
Heat absorbed by water = mass of water x specific heat capacity of water x change in temperature
Heat absorbed by water = 500 g x 4.18 J/g. oC x (53.13-25.00)oC
= 58791.7 J
Heat absorbed by calorimeter = heat capacity of calorimeter x change in temperature
Heat absorbed by calorimeter = 10.5 x 10^3 J /oC x (53.13-25.00)oC
=295365 J
Total heat energy absorbed = 58791.7 J + 295365 J = 354156.7 J
Number of moles of benzene given is:
number of moles = goven mass of benzene /its molar mass
=7.05 g / 78.0 g/mol
=0.0903mol
Hence, the heat released by the reaction is:
= 354156.7 J / 0.0903 mol
= 3922.00 kJ/mol
Answer:
The heat released during the combustion of 7.05g of benzene is 3922.00kJ/mol.
A solution of the primary standard potassium hydrogen phthalate (KHP), KHC8H4O4 , was prepared by dissolving 0.4877 g of KHP in about 50 mL of water. Titration of the KHP solution with a KOH solution of unknown concentration required 28.49 mL to reach a phenolphthalein end point. What is the concentration of the KOH solution?
Answer:
0.08382 M
Explanation:
Step 1: Write the balanced neutralization equation
KHC₈H₄O₄ + KOH ⇒ K₂C₈H₄O₄+ H₂O
Step 2: Calculate the moles corresponding to 0.4877 g of KHC₈H₄O₄
The molar mass of KHC₈H₄O₄ is 204.22 g/mol.
0.4877 g × 1 mol/204.22 g = 2.388 × 10⁻³ mol
Step 3: Calculate the moles of KOH that react with 2.388 × 10⁻³ moles of KHC₈H₄O₄
The molar ratio of KHC₈H₄O₄ to KOH is 1:1. The moles of KOH that react are 1/1 × 2.388 × 10⁻³ mol = 2.388 × 10⁻³ mol.
Step 4: Calculate the molar concentration of KOH
2.388 × 10⁻³ moles of KOH are in 28.49 mL of solution.
2.388 × 10⁻³ mol / 0.02849 L = 0.08382 M
If a sample is said to contain a 98.0% enantiomeric excess of the compound (+)-camphor, what would be the observed specific rotation if the pure (+)-camphor was found to have a rotation of 44.1?
Answer:
the observed specific rotation is 43.218
Explanation:
Given the data in the question;
percentage of enantiomeric excess = 98.0%
observed specific rotation = ? { represented by x }
specific rotation of pure compound = 44.1
Now, we know that;
% of enantiomeric excess = ( observed specific rotation / specific rotation of pure compound ) × 100%
so we substitute
98.0 % = ( x / 44.1 ) × 100%
0.98 = x / 44.1
x = 0.98 × 44.1
x = 43.218
Therefore, the observed specific rotation is 43.218
Course Home P Acceptable units x + courseld=16709491&OpenVellumHMAC=f5c9929f4e4da0b5529475e262c91d79=10001 1 Review art A alculate the heat change in calories for condensation of 11.0 g of steam at 100°C. xpress your answer as a positive value using three significant figures and inc 2 MIKIN M HA Value CS
Answer:
The heat change in calories for condensation of 11.0 g of steam at 100°C is 5930 calories
Explanation:
Latent heat of condensation is the heat released when one mole of steam or water vapor condenses to form liquid droplets. The heat of condensation of water at 100° C is about 2,260 kJ/kg, which is equal to 40.68 kJ/mol. Since condensation of steam and vaporization of water occur at the same temperature and require the same amount of energy to occur, the heat of condensation is exactly equal to the heat vaporization, but has the opposite sign. In the vaporization, heat energy is absorbed by the substance, whereas in condensation heat energy is released by the substance.
The specific latent heat of vaporization of steam at 100° C = 40.68 kJ/mol
Number of moles of moles of water in 11.0 g of steam = mass/ molar mass
Molar mass of water = 18.0 g/mol
Number of moles of steam = 11.0 g / 18.0 g/mol = 0.61 moles
Heat released = 40.68 K/mol × 0.61 moles = 24.815 kJ
Converting to kcal by dividing 24.815 kJ by 4.184 = 5.93 kcal or 5930 calories
Therefore, the heat change in calories for condensation of 11.0 g of steam at 100°C is 5930 calories