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timurjin [86]
3 years ago
10

When an aqueous solution of AgNO3 is electrolyzed, a gas is observed to form at the anode. The gas is

Chemistry
1 answer:
iris [78.8K]3 years ago
6 0

Answer:

Oxygen

Explanation:

First, list out all the ions in the aqueous solution:

Ag+, NO3-

H+, OH-

In the anode, the substance lose electrons to undergo oxidation.

From the 4 ions, only OH- can lose electrons to form water and oxygen,

4OH- --->  O2 + 2H2O + 4e-

While others tend to gain electrons to form new substances instead (they undergo reduction).

Oxygen is the gas produced.

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The molar mass of an unknown gas was measured by an effusion experiment. It was found that it took 60 s for the gas to effuse, w
Marizza181 [45]

Answer:

The molar mass of the gas is 44 g/mol

Explanation:

It is possible to solve this problem using Graham's law that says: Rates of effusion are inversely dependent on the square of the mass of each gas. That is:

\frac{r_1}{r_2} =\frac{\sqrt{M_2} }{\sqrt{M_1} }

If rate of effusion of nitrogen is Xdistance / 48s and for the unknown gas is X distance / 60s and mass of nitrogen gas is 28g/mol (N₂):

\frac{X/48s}{X/60s} =\frac{\sqrt{M_2} }{\sqrt{28g/mol} }

6,61 = √M₂

44g/mol = M₂

<em>The molar mass of the gas is 44 g/mol</em>

<em></em>

I hope it helps!

4 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 <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

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

the diagram of the emission spectrum has been added.

<em>From the given</em><em> chart;</em>

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;

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

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;

  • <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>

<em />

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 <em>orange line</em> is 3.26 x 10⁻¹⁹ J.

Learn more here:brainly.com/question/15962928

6 0
2 years ago
What systems are working during the process of respiration?
Natali5045456 [20]

respiratory and lymphatic

3 0
3 years ago
It is possiblefor an object's weight to change while its mass remains constan?Explain.
Georgia [21]

Answer:

Yes. Weight is the product of mass times gravitational acceleration. So all you have to do is vary the gravitational field and you vary weight.

Explanation:

7 0
3 years ago
Attempt 2
Assoli18 [71]
Lets let our mass equal 3 on alletals and solve using d=m/v equation

Aluminum
V=3/2.70=1.11
Silver
V=3/10.5=.286
Rhenium
V=3/20.8=.144
Nickel
V=3/8.90=.337
This gives us the following list from largest to smallest Aluminum, Nickel, Silver, and Rhenium
4 0
3 years ago
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