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”)

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

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

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!

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

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]

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

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.                                              

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

Answer:

the last one probably

Explanation: