Which of the following compounds is the MOST polar?

How many sig figs are in 100.00?

d1i1m1o1n [39]

Answer:

5

Explanation:

6 0

3 years ago

An electrochemical cell is powered by the half reactions shown below.

andrezito [222]

Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation.

To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign.

Pb(s) --> Pb2+ +2e- E0 = +0.13 V
Ag+ + e- ---> Ag E0 = +0.80 V

Adding up the E0's would yield an overall electric cell potential of +0.93 V.

7 0

3 years ago

Mercury’s atomic emission spectrum is shown below. Estimate the wavelength of the orange line. What is its frequency? What is th

lions [1.4K]

The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this orange line is 3.26 x 10⁻¹⁹ J.

"Your question is not complete, it seems to be missing the diagram of the emission spectrum"

the diagram of the emission spectrum has been added.

From the given chart;

The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m

The frequency of this emission is calculated as follows;

c = fλ

where;

c is the speed of light = 3 x 10⁸ m/s
f is the frequency of the wave
λ is the wavelength

$f = \frac{c}{\lambda } \\\\f = \frac{3\times 10^8}{610 \times 10^{-9}} \\\\f = 4.92 \times 10^{14} \ Hz$

The energy of the emitted photon corresponding to the orange line is calculated as follows;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ Js

E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)

E = 3.26 x 10⁻¹⁹ J.

Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this orange line is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

6 0

2 years ago

Consider the reaction: CH4 + 2O2 = CO2 + 2H2O

crimeas [40]

Answer:

The answer to your question is:

a) 80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction CH4 + 2O2 ⇒ CO2 + 2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

MW CH4 = 16 g

MW O2 = 32 g

16 g of CH4 ---------------- 2(32) g of O2

20 g -------------- x

x = (20 x 64) / 16 = 80 g of O2

b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .

CH4 + 2O2 ⇒ CO2 + 2H2O

15 g 22 g

16 g of CH4 ---------------- 64 g of O2

15 g of CH4 --------------- x

x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

CH4 + 2O2 ⇒ CO2 + 2H2O

64 g of O2 ------------------ 44 g of CO2

22 g of O2 ------------------ x

x = (22 x 44)/ 64 = 15. 13 g of CO2

c. For the reaction CH4 + 2O2 = CO2 + 2H2O, if you have 10.31 g. of CH4 and an unknown amount of oxygen, and form 20.00 g. of CO2, i. Identify if there is a limiting reactant ______________ ii. Calculate the number of grams of the limiting reactant present if there is one. ______________

CH4 + 2O2 ⇒ CO2 + 2H2O

10.31 g 20 g

We can identify the limiting reactant if we know the quantity of the reactants, if we only know the quantity of one it is not posible to which is the limiting reactant.

4 0

3 years ago

nalin [4]

Osmosis and diffusion are related processes that display similarities. Both osmosis and diffusion equalize the concentration of two solutions. Both diffusion and osmosis are passive transport processes, which means they do not require any input of extra energy to occur. In both diffusion and osmosis, particles move from an area of higher concentration to one of lower concentration. Osmosis and facilitated diffusion both account for movement of molecules from a region of high concentration to a region of low concentration.

6 0

3 years ago

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