Selenium, an element used in the manufacture of solar energy devices, forms an oxide that contains only one atom of selenium (SeOx) and is 37.8% oxygen by mass. What is the molecular formula of the oxide? (Hint: find “x”)

Answers

Answer 1

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]


Related Questions

What is the concentration of HNO3 if 5.00×10−2 mol are present in 905 mL of the solution?

Answers

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]

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

Answers

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

A certain first-order reaction is 27.5 percent complete in 8.90 min at 25°C. What is its rate constant?

Answers

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!

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.

Answers

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.                                              

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​

Answers

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

Help please! this is timed

Answers

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

Answers

It shows that the toothpaste contains chlorine. Because chlorine is present when it turns the litmus paper blue

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?

Answers

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

Which of the following is true about oxidation-reduction reactions?
=============================================================
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

Answers

Answer:

the last one probably

Explanation:

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?

Answers

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

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