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luda_lava [24]
2 years ago
6

During the process of electrolysis, liquid water (H2O) chemically changes into hydrogen gas (H2) and oxygen gas (O2). Based on t

he model below, how many
H2 MOLECULES would be produced?
Chemistry
1 answer:
NNADVOKAT [17]2 years ago
5 0

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wpazsebu

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How many grams of water are produced when 4.50 L of
MA_775_DIABLO [31]

The answer for the following problem has been mentioned below.

  • <em><u>Therefore the mass of the water is 5.802 grams.</u></em>

Explanation:

Given:

volume of oxygen (V) = 4.50 L

Temperature (T) = 425 K

pressure of oxygen (P) = 2.50 atm

Gram molecular mass of oxygen (M) = 16.0 grams

To calculate:

mass of water (m)

We know;

According to the ideal gas equation;

     P × V = n × R × T

As we know;

no of moles = \frac{m}{M}

m represents the mass of oxygen (m)

M represents the Gram molecular mass (M)

According to above mentioned equation;

           P × V = n × R × T

P represents the pressure of the oxygen

V represents the volume of the oxygen

n represents the no of moles of the oxygen

R represents the universal gas constant

where,

the value of R is 0.0821 L atm/K moles

Substituting the values in the above equation;

                  2.50 × 4.50 = \frac{m}{16.0} × 0.0821 × 425

                   11.25 =  \frac{m}{16.0} × 34.8925

                  180 = m × 34.8925

                  m = \frac{180}{34.8925}

                  m = 5.158 grams

Therefore the mass of the of oxygen is 5.158 grams

Now;

As we know;

           \frac{m_{1} }{M_{1} } = \frac{m_{2} }{M_{2} }

where;

m_{1} represents the mass of the oxygen

M_{1} represents the gram molecular mass of the oxygen

m_{2} represents the mass of the water

M_{2} represents the gram molecular mass of water

    From the above given formula,

      \frac{5.158}{16.0} = \frac{m_{2} }{18}

where;

Gram molecular weight of water = 18.0 u

    m_{2} = 5.802 grams

<em><u>Therefore the mass of the water is 5.802 grams.</u></em>

5 0
3 years ago
How many photons are produced in a laser pulse of 0.862 J at 691 nm?
Inga [223]
To calculate how many photons are in a certain amount of energy (joules) we need to know how much energy is in one photon.
Start by using two equations:
Energy of a photon = Frequency * Planck's constant (6.626 * 10^(-34) J-s)
Speed of light (constant 3 * 10^8 m/s) = Frequency * Wavelength
Which means:
frequency = Speed of Light / Wavelength
So energy of a photon = (Speed of light * Planck's constant)/(Wavelength)
You may have seen this equation as E = hc/<span>λ</span>
We have a wavelength of 691 nm or 691 * 10^-9 meters
So we can plug in all of our knowns:
E = (6.626 * 10^(-34) J-s) * (3.00 * 10^8 m/s) / (691 * 10^-9 m) = 
2.88 * 10^(-19) joules per photon
Now we have joules per photon, and the total number of joules (0.862 joules)
,so divide joules by joules per photon, and we have the number of photons:
0.862 J/ (2.88 * 10^(-19) J/photon) = 3.00 * 10^18 photons.

4 0
3 years ago
Which type of energy uses the photoelectric effect?
ikadub [295]

Answer:

My lovely people the answer is SOLAR

Explanation:

i just know

8 0
2 years ago
I need help please j
Sidana [21]
Reactants Hydrogen: 5
Products Hydrogen: 5

Reactants Carbon: 3
Products Carbon: 3

Reactants Oxygen: 4
Products Oxygen: 5
3 0
3 years ago
Which of the following elements is the least reactive metal? Sodium, Rubidium, Chlorine, and Magnesium.​
LuckyWell [14K]
First off chlorine is not a metal so you can ignore that one.

Sodium and Rubidium are in group 1 of the periodic table and Magnesium is in group 2.

Group one metals are more reactive than group two because it is harder for the group two metals to lose their 2 valence (outer most) electrons.

As you go down group 1 there is an increase in the reactivity this is because as you go down there is an increase in the atomic radius which leads to more shielding. This weakens the electrostatic forces of attraction making it easier to lose the outermost electrons, therefore they are more reactive.
6 0
2 years ago
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