Answer: The molecular formula of the compound will be [tex]SeO_3[/tex]
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Let the mass of the compound be 100 g
Given values:
% of O = 37.8%
% of Se = [100 - 37.8] = 62.2%
Mass of O = 37.8 g
Mass of Se = 62.2 g
We know:
Molar mass of Se = 79 g/mol
Molar mass of O = 16 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of Se}=\frac{62.2g}{79g/mol}=0.787 mol[/tex]
[tex]\text{Moles of O}=\frac{37.8g}{16g/mol}=2.36 mol[/tex]
Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.787 moles
[tex]\text{Mole fraction of Se}=\frac{0.787 }{0.787 }=1[/tex]
[tex]\text{Mole fraction of O}=\frac{2.36}{0.787 }=2.99\approx 3[/tex]
Taking the mole ratio as their subscripts.
The ratio of Se : O = 1 : 3
Hence, the molecular formula of the compound will be [tex]SeO_3[/tex]
For the reaction N2(g) + 2H2(g) → N2H4(l), if the percent yield for this reaction is 100.0%, what is the actual mass of hydrazine (N2H4) produced when 59.20 g of nitrogen reacts with 6.750 g of hydrogen?
a. Molar mass of N2 = 28.01 g/mol
b. Molar mass of H2 = 2.016 g/mol
c. Molar mass of N2H4 = 32.05 g/mol
Answer:
53.6 g of N₂H₄
Explanation:
The begining is in the reaction:
N₂(g) + 2H₂(g) → N₂H₄(l)
We determine the moles of each reactant:
59.20 g / 28.01 g/mol = 2.11 moles of nitrogen
6.750 g / 2.016 g/mol = 3.35 moles of H₂
1 mol of N₂ react to 2 moles of H₂
Our 2.11 moles of N₂ may react to (2.11 . 2) /1 = 4.22 moles of H₂, but we only have 3.35 moles. The hydrogen is the limiting reactant.
2 moles of H₂ produce at 100 % yield, 1 mol of hydrazine
Then, 3.35 moles, may produce (3.35 . 1)/2 = 1.67 moles of N₂H₄
Let's convert the moles to mass:
1.67 mol . 32.05 g/mol = 53.6 g
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
What is the concentration of HNO3 if 5.00×10−2 mol are present in 905 mL of the solution?
Answer:
0.05525 M or 55.25 mM
Explanation:
Concentration = moles/volume
*Note that volume is expressed in L so you will need to convert mL > L here
[tex]C= \frac{n}{V}\\C= \frac{0.05}{0.905}\\C=0.05525[/tex]
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
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
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:
Help please! this is timed
Answer:
C, P, P, C, P
Explanation:
is it still the same thing but the physical property change or did the thing change too? that's what it's asking
red litmus paper was used to test toothpaste and it turns blue.Explain what this tells about the toothpaste
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!