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r-ruslan [8.4K]
3 years ago
10

[Please see the attatched photo]​

Chemistry
2 answers:
Klio2033 [76]3 years ago
6 0

Answer:

C

Explanation:

Trust me

mixer [17]3 years ago
5 0

Answer:

Its C

Explanation:

Got 100% on test edge 2020

You might be interested in
Calculate the energy required to ionize a hydrogen atom to an excited state where the electron is initially in the n = 5 energy
SVETLANKA909090 [29]

. The energy of shells in a hydrogen atom is calculated by the formula E = -Eo/n^2 where n is any integer, and Eo = 2.179X10^-18 J. So, the energy of a ground state electron in hydrogen is:

E = -2.179X10^-18 J / 1^2 = -2.179X10^-21 kJ

Consequently, to ionize this electron would require the input of 2.179X10^-21 kJ


2. The wavelength of a photon with this energy would be:

Energy = hc/wavelength

wavelength = hc/energy

wavelength = 6.626X10^-34 Js (2.998X10^8 m/s) / 2.179X10^-18 J = 9.116X10^-8 m

Converting to nanometers gives: 91.16 nm


3. Repeat the calculation in 1, but using n=5.


4. Repeat the calculation in 2 using the energy calculated in 3.

7 0
3 years ago
Read 2 more answers
To what volume should 25ml of 15m nitric acid be diluted to prepare a 3m solution
Maurinko [17]

We can use the dilution formula to find the volume of the diluted solution to be prepared

c1v1 = c2v2

Where c1 is concentration and v1 is volume of the concentrated solution

And c2 is concentration and v2 is volume of the diluted solution to be prepared

Substituting the values in the equation

15 M x 25 mL = 3 M x v2

v2 = 125 mL

The 25 mL concentrated solution should be diluted with distilled water upto 125 mL to make a 3 M solution

6 0
4 years ago
Isotopes of an element contain the same number of protons but different numbers of _________________.
sp2606 [1]
Neutrons is your answer

7 0
3 years ago
In a certain electrolysis experiment, 1.24 g of Ag were deposited in one cell (containing an aqueous AgNO3 solution), while 0.65
eduard

Answer:

The correct answer is 169.56 g/mol.

Explanation:

Based on the given information, the mass of Ag deposited is 1.24 g, and the mass of unknown metal X deposited in another cell is 0.650 g. The number of moles of electrons can be determined as,

= 1.24 g Ag * 1mol Ag/107.87 g/mol Ag * 1 mol electron/1 mol Ag ( the molecular mass of Ag is 107.87 g/mol)

= 0.0115 mole of electron

The half cell reaction for the metal X is,  

X^3+ (aq) + 3e- = X (s)

From the reaction, it came out that 3 faraday will reduce one mole of X^3+.  

The molar mass of X will be,  

= 0.650 g/0.0115 *3 mol electron/1 mol

= 56.52 * 3

= 169.56 g/mol

7 0
3 years ago
A sample of 10.6 g of KNO3 was dissolved in 251.0 g of water at 25 oC in a calorimeter. The final temperature of the solution wa
finlep [7]

Answer:

36.55kJ/mol

Explanation:

The heat of solution is the change in heat when the KNO3 dissolves in water:

KNO3(aq) → K+(aq) + NO3-(aq)

As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.

To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:

<em>Moles KNO3 -Molar mass: 101.1032g/mol-</em>

10.6g * (1mol/101.1032g) = 0.1048 moles KNO3

<em>Change in heat:</em>

q = m*S*ΔT

<em>Where q is heat in J,</em>

<em>m is the mass of the solution: 10.6g + 251.0g = 261.6g</em>

S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-

And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C

q = 261.6g*4.184J/g°C*3.5°C

q = 3830.87J

<em>Molar heat of solution:</em>

3830.87J/0.1048 moles KNO3 =

36554J/mol =

<h3>36.55kJ/mol</h3>

<em />

6 0
3 years ago
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